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https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/back-matter/appendix-a-useful-information-constants-units-formulae/ Thu, 29 Jun 2017 22:14:41 +0000 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/back-matter/appendix-a-useful-information-constants-units-formulae/ This appendix is broken into several tables.

• Table 3, Important Constants
• Table 4, Submicroscopic Masses
• Table 5, Solar System Data
• Table 6, Metric Prefixes for Powers of Ten and Their Symbols
• Table 7, The Greek Alphabet
• Table 8, SI units
• Table 9, Selected British Units
• Table 10, Other Units
• Table 11, Useful Formulae

Table 3. Important Constants¹

Symbol	Meaning	Best Value	Approximate Value
$latex c $	Speed of light in vacuum	$latex 2.99792458 \times 10^8 \text{m/s} $	$latex 3.00 \times 10^8 \text{m/s}$
$latex G $	Gravitational constant	$latex 6.67408(31) \times 10^{-11} \;\text{N} \cdot \text{m}^2 / \text{kg}^2 $	$latex 6.67 \times 10^{-11} \;\text{N} \cdot \text{m}^2 / \text{kg}^2 $
$latex N_A$	Avogadro’s number	$latex 6.02214129(27) \times 10^{23}$	$latex 6.02 \times 10^{23} $
$latex k $	Boltzmann’s constant	$latex 1.3806488(13) \times 10^{-23} \;\text{J} / \text{K}$	$latex 1.38 \times 10^{-23} \;\text{J} / \text{K}$
$latex R $	Gas constant	$latex 8.3144621(75) \;\text{J} / \text{mol} \cdot \text{K}$	$latex 8.31 \;\text{J} / \text{mol} \cdot \text{K} = 1.99 \text{cal} / \text{mol} \cdot \text{K} = 0.0821 \text{atm} \cdot \text{L} / \text{mol} \cdot \text{K}$
$latex \sigma $	Stefan-Boltzmann constant	$latex 5.670373(21) \times 10^{-8} \;\text{W} / \text{m}^2 \cdot \text{K}$	$latex 5.67 \times 10^{-8} \;\text{W} / \text{m}^2 \cdot \text{K}$
$latex k $	Coulomb force constant	$latex 8.987551788 \dots \times 10^9 \;\text{N} \cdot \text{m}^2 / \text{C}^2$	$latex 8.99 \times 10^9 \;\text{N} \cdot \text{m}^2 / \text{C}^2$
$latex q_e$	Charge on electron	$latex -1.602176565(35) \times 10^{-19} \;\text{C}$	$latex -1.60 \times 10^{-19} \;\text{C}$
$latex \varepsilon _0$	Permittivity of free space	$latex 8.854187817 \dots \times 10^{-12} \;\text{C}^2 / \text{N} \cdot \text{m}^2 $	$latex 8.85 \dots \times 10^{-12} \;\text{C}^2 / \text{N} \text{m}^2 $
$latex \mu _0 $	Permeability of free space	$latex 4 \pi \times 10^{-7} \;\text{T} \cdot \;\text{m}/ \text{A}$	$latex 1.26 \times 10^{-6} \;\text{T} \cdot \text{m} / \text{A}$
$latex h $	Planck’s constant	$latex 6.62606957(29) \times 10^{-34} \;\text{J} \cdot \text{s}$	$latex 6.63 \times 10^{-34} \;\text{J} \cdot \text{s}$

Symbol	Meaning	Best Value	Approximate Value
$latex m_e$	Electron mass	$latex 9.10938291(40) \times 10^{-31} \text{kg} $	$latex 9.11 \times 10^{-31} \text{kg} $
$latex m_p $	Proton mass	$latex 1.672621777(74) \times 10^{-27} \text{kg} $	$latex 1.6726 \times 10^{-27} \text{kg} $
$latex m_n $	Neutron mass	$latex 1.674927351(74) \times 10^{-27} \text{kg} $	$latex 1.6749 \times 10^{-27} \text{kg} $
$latex \text{u} $	Atomic mass unit	$latex 1.660538921(73) \times 10^{-27} \text{kg} $	$latex 1.6605 \times 10^{-27} \text{kg} $
Table 4. Submicroscopic Masses2

Sun	mass	$latex 1.99 \times 10^{30} \text{kg} $
	average radius	$latex 6.96 \times 10^8 \text{m}$
	Earth-sun distance (average)	$latex 1.496 \times 10^{11} \text{m} $
Earth	mass	$latex 5.9736 \times 10^{24} \text{kg} $
	average radius	$latex 6.376 \times 10^6 \text{m}$
	orbital period	$latex 3.16 \times 10^7 \text{s} $
Moon	mass	$latex 7.35 \times 10^{22} \text{kg} $
	average radius	$latex 1.74 \times 10^6 \text{m} $
	orbital period (average)	$latex 2.36 \times 10^6 \text{s} $
	Earth-moon distance (average)	$latex 3.84 \times 10^8 \text{m} $
Table 5. Solar System Data

Prefix	Symbol	Value	Prefix	Symbol	Value
tera	T	$latex 10^{12}$	deci	d	$latex 10^{-1}$
giga	G	$latex 10^{9}$	centi	c	$latex 10^{-2}$
mega	M	$latex 10^{6}$	milli	m	$latex 10^{-3}$
kilo	k	$latex 10^{3}$	micro	$latex \mu $	$latex 10^{-6}$
hecto	h	$latex 10^{2}$	nano	n	$latex 10^{-9}$
deka	da	$latex 10^{1}$	pico	p	$latex 10^{-12}$
—	—	$latex 10^{0} (= 1)$	femto	f	$latex 10^{-15}$
Table 6. Metric Prefixes for Powers of Ten and Their Symbols

Alpha	A	$latex \alpha $	Eta	Η	$latex \eta $	Nu	Ν	$latex \nu $	Tau	Τ	$latex \tau $
Beta	Β	$latex \beta $	Theta	Θ	$latex \theta $	Xi	Ξ	$latex \xi $	Upsilon	Υ	$latex \upsilon $
Gamma	Γ	$latex \gamma $	Iota	Ι	$latex \iota $	Omicron	Ο	$latex \o $	Phi	Φ	$latex \phi $
Delta	Δ	$latex \delta $	Kappa	Κ	$latex \kappa $	Pi	Π	$latex \pi $	Chi	Χ	$latex \chi $
Epsilon	Ε	$latex \epsilon $	Lambda	Λ	$latex \lambda $	Rho	Ρ	$latex \rho $	Psi	Ψ	$latex \psi $
Zeta	Ζ	$latex \zeta $	Mu	Μ	$latex \mu $	Sigma	Σ	$latex \sigma $	Omega	Ω	$latex \omega $
Table 7. The Greek Alphabet

	Entity	Abbreviation	Name
Fundamental units	Length	$latex \text{m} $	meter
	Mass	$latex \text{kg} $	kilogram
	Time	$latex \text{s} $	second
	Current	$latex \text{A} $	ampere
Supplementary unit	Angle	$latex \text{rad}$	radian
Derived units	Force	$latex \text{N} = \text{kg} \cdot \text{m} / \text{s}^2 $	newton
	Energy	$latex \text{J} = \text{kg} \cdot \text{m}^2 / \text{s}^2 $	joule
	Power	$latex \text{W} = \text{J} / \text{s}$	watt
	Pressure	$latex \text{Pa} = \text{N} / \text{m}^2 $	pascal
	Frequency	$latex \text{Hz} = 1/ \text{s} $	hertz
	Electronic potential	$latex \text{V} = \text{J} / \text{C}$	volt
	Capacitance	$latex \text{F} = \text{C} / \text{V}$	farad
	Charge	$latex \text{C} = \text{s} \cdot \text{A}$	coulomb
	Resistance	$latex \Omega = \text{V} / \text{A}$	ohm
	Magnetic field	$latex \text{T} = \text{N} / (\text{A} \cdot \text{m}) $	tesla
	Nuclear decay rate	$latex \text{Bq} = 1 / \text{s}$	becquerel
Table 8. SI Units

Length	$latex 1 \;\text{inch} \; (\text{in.}) = 2.54 \;\text{cm} \; (\text{exactly}) $
	$latex 1 \;\text{foot} \; (\text{ft}) = 0.3048 \;\text{m} $
	$latex 1 \;\text{mile} \; (\text{mi}) = 1.609 \;\text{km} $
Force	$latex 1 \;\text{pound} \; (\text{lb}) = 4.448 \;\text{N} $
Energy	$latex 1 \;\text{British thermal unit} \; (\text{Btu}) = 1.055 \times 10^3 \text{J}$
Power	$latex 1 \;\text{horsepower} \; (\text{hp}) = 746 \;\text{W}$
Pressure	$latex 1 \;\text{lb} / \text{in}^2 = 6.895 \times 10^3 \;\text{Pa}$
Table 9. Selected British Units

Length	$latex 1 \;\text{light year} \; (\text{ly}) = 9.46 \times 10^{15} \; \text{m}$
	$latex 1 \; \text{astronomical unit} \;(\text{au}) = 1.50 \times 10^{11} \; \text{m}$
	$latex 1 \;\text{nautical mile} = 1.852 \;\text{km}$
	$latex 1 \;\text{angstrom} ( \AA ) = 10^{-10} \text{m}$
Area	$latex 1 \;\text{acre} \;(\text{ac}) = 4.05 \times 10^3 \text{m}^2$
	$latex 1 \;\text{square foot} \; (\text{ft}^2) = 9.29 \times 10^{-2} \;\text{m}^2 $
	$latex 1 \;\text{barn} \;(b) = 10^{-28} \;\text{m}^2 $
Volume	$latex 1 \;\text{liter} \;(L) = 10^{-3} \; \text{m}^3 $
	$latex 1 \;\text{U.S. gallon} \;(\text{gal}) = 3.785 \times 10^{-3} \;\text{m}^3 $
Mass	$latex 1 \;\text{solar mass} = 1.99 \times 10^{30} \; \text{kg}$
	$latex 1 \;\text{metric ton} = 10^3 \;\text{kg}$
	$latex 1 \;\text{atomic mass unit} \;(u)= 1.6605 \times 10^{-27} \;\text{kg}$
Time	$latex 1 \;\text{year} \;(\text{y}) = 3.16 \times 10^7 \;\text{s}$
	$latex 1 \;\text{day} \;(\text{d}) = 86,400 \;\text{s}$
Speed	$latex 1 \;\text{mile per hour} \;(\text{mph}) = 1.609 \;\text{km} / \text{h} $
	$latex 1 \;\text{nautical mile per hour} \;(\text{naut}) = 1.852 \;\text{km} / \text{h}$
Angle	$latex 1 \;\text{degree} \;(^{\circ}) = 1.745 \times 10^{-2} \;\text{rad} $
	$latex 1 \;\text{minute of arc} \; ($' $latex ) = 1/60 \;\text{degree} $
	$latex 1 \;\text{second of arc} \; (") = 1/60 \;\text{minute of arc} $
	$latex 1 \;\text{grad} = 1.571 \times 10^{-2} \;\text{rad} $
Energy	$latex 1 \;\text{kiloton TNT} \;(\text{kT}) = 4.2 \times 10^{12} \;\text{J} $
	$latex 1 \; \text{kilowatt hour} \;(\text{kW} \cdot h) = 3.60 \times 10^6 \;\text{J}$
	$latex 1 \;\text{food calorie} \;(\text{kcal}) = 4186 \;\text{J} $
	$latex 1 \;\text{calorie} \;(\text{cal}) = 4.186 \;\text{J}$
	$latex 1 \;\text{electron volt} \;(\text{eV}) = 1.60 \times 10^{-19} \;\text{J}$
Pressure	$latex 1 \;\text{atmosphere} \;(\text{atm}) = 1.013 \times 10^5 \;\text{Pa}$
	$latex 1 \;\text{millimeter of mercury} \;(\text{mm Hg}) = 133.3 \;\text{Pa}$
	$latex 1 \;\text{torricelli} \;(\text{torr}) = 1 \;\text{mm Hg} = 133.3 \;\text{Pa}$
Nuclear decay rate	$latex 1 \;\text{curie} \;(\text{Ci}) = 3.70 \times 10^{10} \; \text{Bq}$
Table 10. Other Units

$latex \text{Circumference of a circle with radius} \; \text{r} \; \text{or diameter} \;\text{d}$	$latex C = 2 \pi r = \pi d $
$latex \text{Area of a circle with radius} \; r \;\text{or diameter} d $	$latex A = \pi r^2 = \pi d^{2}/4 $
$latex \text{Area of a sphere with radius} \; r $	$latex A = 4 \pi r^2 $
$latex \text{Volume of a sphere with radius} \; r $	$latex V = (4/3)( \pi r^3) $
Table 11. Useful Formulae

Units of Length
meter (m)	= 39.37 inches (in.) = 1.094 yards (yd)	angstrom (Å)	= 10–8 cm (exact, definition) = 10–10 m (exact, definition)
centimeter (cm)	= 0.01 m (exact, definition)	yard (yd)	= 0.9144 m
millimeter (mm)	= 0.001 m (exact, definition)	inch (in.)	= 2.54 cm (exact, definition)
kilometer (km)	= 1000 m (exact, definition)	mile (US)	= 1.60934 km

Units of Volume
liter (L)	= 0.001 m3 (exact, definition) = 1000 cm3 (exact, definition) = 1.057 (US) quarts	liquid quart (US)	= 32 (US) liquid ounces (exact, definition) = 0.25 (US) gallon (exact, definition) = 0.9463 L
milliliter (mL)	= 0.001 L (exact, definition) = 1 cm3 (exact, definition)	dry quart	= 1.1012 L
microliter (μL)(μL)	= 10–6 L (exact, definition) = 10–3 cm3 (exact, definition)	cubic foot (US)	= 28.316 L

Units of Mass
gram (g)	= 0.001 kg (exact, definition)	ounce (oz) (avoirdupois)	= 28.35 g
milligram (mg)	= 0.001 g (exact, definition)	pound (lb) (avoirdupois)	= 0.4535924 kg
kilogram (kg)	= 1000 g (exact, definition) = 2.205 lb	ton (short)	=2000 lb (exact, definition) = 907.185 kg
ton (metric)	=1000 kg (exact, definition) = 2204.62 lb	ton (long)	= 2240 lb (exact, definition) = 1.016 metric ton

Units of Energy
4.184 joule (J)	= 1 thermochemical calorie (cal)
1 thermochemical calorie (cal)	= 4.184 × 107  erg
erg	= 10–7 J (exact, definition)
electron-volt (eV)	= 1.60218 × 10−19 J = 23.061 kcal mol−1
liter∙atmosphere	= 24.217 cal = 101.325 J (exact, definition)
nutritional calorie (Cal)	= 1000 cal (exact, definition) = 4184 J
British thermal unit (BTU)	= 1054.804 J  BTU is the amount of energy needed to heat one pound of water by one degree Fahrenheit. Therefore, the exact relationship of BTU to joules and other energy units depends on the temperature at which BTU is measured. 59 °F (15 °C) is the most widely used reference temperature for BTU definition in the United States. At this temperature, the conversion factor is the one provided in this table.

Footnotes

1. 1 Stated values are according to the National Institute of Standards and Technology Reference on Constants, Units, and Uncertainty, www.physics.nist.gov/cuu (accessed May 18, 2012). Values in parentheses are the uncertainties in the last digits. Numbers without uncertainties are exact as defined.
2. 2 Stated values are according to the National Institute of Standards and Technology Reference on Constants, Units, and Uncertainty, www.physics.nist.gov/cuu (accessed May 18, 2012). Values in parentheses are the uncertainties in the last digits. Numbers without uncertainties are exact as defined.

]]> 728 0 2 0 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/cnx_chem_00_bb_dependence_img/ Thu, 29 Jun 2017 22:14:42 +0000 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/wp-content/uploads/sites/215/2017/06/CNX_Chem_00_BB_Dependence_img.jpg 729 0 0 0 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/cnx_chem_00_bb_function_img/ Thu, 29 Jun 2017 22:14:42 +0000 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/wp-content/uploads/sites/215/2017/06/CNX_Chem_00_BB_Function_img.jpg 730 0 0 0 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/back-matter/appendix-b-useful-mathematics-originally-from-open-stax-chemistry/ Thu, 29 Jun 2017 22:14:42 +0000 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/back-matter/appendix-b-useful-mathematics-originally-from-open-stax-chemistry/

Exponential Arithmetic

Exponential notation is used to express very large and very small numbers as a product of two numbers. The first number of the product, the digit term, is usually a number not less than 1 and not greater than 10. The second number of the product, the exponential term, is written as 10 with an exponent. Some examples of exponential notation are:

$latex \begin{array}{r @{{}={}} l} 1000 & 1\;\times\;10^3 \\[0.75em] 100 & 1\;\times\;10^2 \\[0.75em] 10 & 1\;\times\;10^1 \\[0.75em] 1 & 1\;\times\;10^0 \\[0.75em] 0.1 & 1\;\times\;10^{-1} \\[0.75em] 0.001 & 1\;\times\;10^{-3} \\[0.75em] 2386 & 2.386\;\times\;1000 = 2.386\;\times\;10^3 \\[0.75em] 0.123 & 1.23\;\times\;0.1 = 1.23\;\times\;10^{-1} \end{array}$

The power (exponent) of 10 is equal to the number of places the decimal is shifted to give the digit number. The exponential method is particularly useful notation for every large and very small numbers. For example, 1,230,000,000 = 1.23 × 10⁹, and 0.00000000036 = 3.6 × 10⁻¹⁰.

Addition of Exponentials

Convert all numbers to the same power of 10, add the digit terms of the numbers, and if appropriate, convert the digit term back to a number between 1 and 10 by adjusting the exponential term.

Example 1

Adding Exponentials Add 5.00 × 10⁻⁵ and 3.00 × 10⁻³.

Solution

$latex 3.00\;\times\;10^{-3} = 300\;\times\;10^{-5}$ $latex (5.00\;\times\;10^{-5})\;+\;(300\;\times\;10^{-5}) = 305\;\times\;10^{-5} = 3.05\;\times\;10^{-3} $

Subtraction of Exponentials

Convert all numbers to the same power of 10, take the difference of the digit terms, and if appropriate, convert the digit term back to a number between 1 and 10 by adjusting the exponential term.

Example 2

Subtracting Exponentials Subtract 4.0 × 10⁻⁷ from 5.0 × 10⁻⁶.

Solution

$latex 4.0\;\times\;10^{-7} = 0.40\;\times\;10^{-6}$ $latex (5.0\;\times\;10^{-6})\;-\;(0.40\;\times\;10^{-6}) = 4.6\;\times\;10^{-6}$

Multiplication of Exponentials

Multiply the digit terms in the usual way and add the exponents of the exponential terms.

Example 3

Multiplying Exponentials Multiply 4.2 × 10⁻⁸ by 2.0 × 10³.

Solution

$latex (4.2\;\times\;10^{-8})\;\times\;(2.0\;\times\;10^3) = (4.2\;\times\;2.0)\;\times\;10^{(-8)+(+3)} = 8.4\;\times\;10^{-5}$

Division of Exponentials

Divide the digit term of the numerator by the digit term of the denominator and subtract the exponents of the exponential terms.

Example 4

Dividing Exponentials Divide 3.6 × 10⁵ by 6.0 × 10⁻⁴.

Solution

$latex \frac{3.6\;\times\;10^{-5}}{6.0\;\times\;10^{-4}} = (\frac{3.6}{6.0})\;\times\;10^{(-5)-(-4)} = 0.60\;\times\;10^{-1} = 6.0\;\times\;10^{-2}$

Squaring of Exponentials

Square the digit term in the usual way and multiply the exponent of the exponential term by 2.

Example 5

Squaring Exponentials Square the number 4.0 × 10⁻⁶.

Solution

$latex (4.0\;\times\;10^{-6})^2 = 4\;\times\;4\;\times\;10^{2\;\times\;(-6)} = 16\;\times\;10^{-12} = 1.6\;\times\;10^{-11}$

Cubing of Exponentials

Cube the digit term in the usual way and multiply the exponent of the exponential term by 3.

Example 6

Cubing Exponentials Cube the number 2 × 10⁴.

Solution

$latex (2\;\times\;10^4)^3 = 2\;\times\;2\;\times\;2\;\times\;10^{3\;\times\;4} = 8\;\times\; 10^{12}$

Taking Square Roots of Exponentials

If necessary, decrease or increase the exponential term so that the power of 10 is evenly divisible by 2. Extract the square root of the digit term and divide the exponential term by 2.

Example 7

Finding the Square Root of Exponentials Find the square root of 1.6 × 10⁻⁷.

Solution

$latex 1.6\;\times\;10^{-7} = 16\;\times\;10^{-8}$ $latex \sqrt{16\;\times\;10^{-8}} = \sqrt{16}\;\times\;\sqrt{10^{-8}} = \sqrt{16}\;\times\;\sqrt{10}^{-\frac{8}{2}} = 4.0\;\times\;10^{-4}$

Significant Figures

A beekeeper reports that he has 525,341 bees. The last three figures of the number are obviously inaccurate, for during the time the keeper was counting the bees, some of them died and others hatched; this makes it quite difficult to determine the exact number of bees. It would have been more accurate if the beekeeper had reported the number 525,000. In other words, the last three figures are not significant, except to set the position of the decimal point. Their exact values have no meaning useful in this situation. In reporting any information as numbers, use only as many significant figures as the accuracy of the measurement warrants.

The importance of significant figures lies in their application to fundamental computation. In addition and subtraction, the sum or difference should contain as many digits to the right of the decimal as that in the least certain of the numbers used in the computation (indicated by underscoring in the following example).

Example 8

Addition and Subtraction with Significant Figures Add 4.383 g and 0.0023 g.

Solution

$latex \begin{array}{l} 4.38\underline{3}\;\text{g} \\ 0.002\underline{3}\;\text{g} \\ \hline 4.38\underline{5}\;\text{g} \end{array}$

In multiplication and division, the product or quotient should contain no more digits than that in the factor containing the least number of significant figures.

Example 9

Multiplication and Division with Significant Figures Multiply 0.6238 by 6.6.

Solution

$latex 0.623\underline{8}\;\times\;6.\underline{6} = 4.\underline{1}$

When rounding numbers, increase the retained digit by 1 if it is followed by a number larger than 5 (“round up”). Do not change the retained digit if the digits that follow are less than 5 (“round down”). If the retained digit is followed by 5, round up if the retained digit is odd, or round down if it is even (after rounding, the retained digit will thus always be even).

The Use of Logarithms and Exponential Numbers

The common logarithm of a number (log) is the power to which 10 must be raised to equal that number. For example, the common logarithm of 100 is 2, because 10 must be raised to the second power to equal 100. Additional examples follow.

Number	Number Expressed Exponentially	Common Logarithm
1000	103	3
10	101	1
1	100	0
0.1	10−1	−1
0.001	10−3	−3
Table 1. Logarithms and Exponential Numbers

What is the common logarithm of 60? Because 60 lies between 10 and 100, which have logarithms of 1 and 2, respectively, the logarithm of 60 is 1.7782; that is,

$latex 60 = 10^{1.7782}$

The common logarithm of a number less than 1 has a negative value. The logarithm of 0.03918 is −1.4069, or

$latex 0.03918 = 10^{-1.4069} = \frac{1}{10^{1.4069}}$

To obtain the common logarithm of a number, use the log button on your calculator. To calculate a number from its logarithm, take the inverse log of the logarithm, or calculate 10^x (where x is the logarithm of the number).

The natural logarithm of a number (ln) is the power to which e must be raised to equal the number; e is the constant 2.7182818. For example, the natural logarithm of 10 is 2.303; that is,

$latex 10 = e^{2.303} = 2.7182818^{2.303}$

To obtain the natural logarithm of a number, use the ln button on your calculator. To calculate a number from its natural logarithm, enter the natural logarithm and take the inverse ln of the natural logarithm, or calculate e^x (where x is the natural logarithm of the number).

Logarithms are exponents; thus, operations involving logarithms follow the same rules as operations involving exponents.

1. The logarithm of a product of two numbers is the sum of the logarithms of the two numbers.
   $latex \text{log}\;xy = \text{log}\;x\;+\;\text{log}\;y\text{,\;and\;ln}\;xy = \text{ln}\;x\;+\;\text{ln}\;y$
2. The logarithm of the number resulting from the division of two numbers is the difference between the logarithms of the two numbers.
   $latex \text{log}\;\frac{x}{y} = \text{log}\;x\;-\;\text{log}\;y\text{,\;and\;ln}\;\frac{x}{y} = \text{ln}\;x\;-\;\text{ln}\;y$
3. The logarithm of a number raised to an exponent is the product of the exponent and the logarithm of the number.
   $latex \text{log}\;x^n = n\text{log}\;x\;\text{and\;ln}\;x^n = n\text{ln}\;x$

The Solution of Quadratic Equations

Mathematical functions of this form are known as second-order polynomials or, more commonly, quadratic functions.

$latex ax^2\;+\;bx\;+\;c = 0$

The solution or roots for any quadratic equation can be calculated using the following formula:

$latex x = \frac{-b\;{\pm}\;\sqrt{b^2\;-\;4ac}}{2a}$

Example 10

Solving Quadratic Equations Solve the quadratic equation 3x² + 13x − 10 = 0.

Solution Substituting the values a = 3, b = 13, c = −10 in the formula, we obtain

$latex x = \frac{-13\;{\pm}\;\sqrt{(13)^2\;-\;4\;\times\;3\;\times\;(-10)}}{2\;\times\;3}$

$latex x = \frac{-13\;{\pm}\;\sqrt{169\;+\;120}}{6} = \frac{-13\;{\pm}\;\sqrt{289}}{6} = \frac{-13\;{\pm}\;17}{6}$

The two roots are therefore

$latex x = \frac{-13\;+\;17}{6} = \frac{2}{3}\;\text{and}\;x = \frac{-13\;-\;17}{6} = -5$

Quadratic equations constructed on physical data always have real roots, and of these real roots, often only those having positive values are of any significance.

Two-Dimensional (x-y) Graphing

The relationship between any two properties of a system can be represented graphically by a two-dimensional data plot. Such a graph has two axes: a horizontal one corresponding to the independent variable, or the variable whose value is being controlled (x), and a vertical axis corresponding to the dependent variable, or the variable whose value is being observed or measured (y).

When the value of y is changing as a function of x (that is, different values of x correspond to different values of y), a graph of this change can be plotted or sketched. The graph can be produced by using specific values for (x,y) data pairs.

Example 11

Graphing the Dependence of y on x

x	y
1	5
2	10
3	7
4	14
Table 2.

This table contains the following points: (1,5), (2,10), (3,7), and (4,14). Each of these points can be plotted on a graph and connected to produce a graphical representation of the dependence of y on x.

If the function that describes the dependence of y on x is known, it may be used to compute x,y data pairs that may subsequently be plotted.

Example 12

Plotting Data Pairs If we know that y = x² + 2, we can produce a table of a few (x,y) values and then plot the line based on the data shown here.

x	y = x2 + 2
1	3
2	6
3	11
4	18
Table 3.

]]> 731 0 3 0 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/999992a1a87b3d301b2407777ee5f950-1/ Thu, 29 Jun 2017 22:14:42 +0000 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/wp-content/uploads/sites/215/2017/06/999992a1a87b3d301b2407777ee5f950-1.jpg 732 0 0 0 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/ae1d9db8c6cef79a4db550933c8c02cd-1/ Thu, 29 Jun 2017 22:14:46 +0000 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/wp-content/uploads/sites/215/2017/06/ae1d9db8c6cef79a4db550933c8c02cd-1.jpg 733 0 0 0 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/back-matter/appendix-c-periodic-table-of-the-elements-from-open-stax-chemistry-1st-canadian-edition/ Thu, 29 Jun 2017 22:14:50 +0000 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/back-matter/appendix-c-periodic-table-of-the-elements-from-open-stax-chemistry-1st-canadian-edition/ Appendix: Periodic Table of the Elements In this chapter, we present some data on the chemical elements. The periodic table, lists all the known chemical elements, arranged by atomic number (that is, the number of protons in the nucleus). The periodic table is arguably the best tool in all of science; no other branch of science can summarize its fundamental constituents in such a concise and useful way. Many of the physical and chemical properties of the elements are either known or understood based on their positions on the periodic table. Periodic tables are available with a variety of chemical and physical properties listed in each element’s box. What follows here is a more complex version of the periodic table than what was presented in . The Internet is a great place to find periodic tables that contain additional information. One item on most periodic tables is the atomic mass of each element. For many applications, only one or two decimal places are necessary for the atomic mass. However, some applications (especially nuclear chemistry; seerequire more decimal places. The atomic masses in represent the number of decimal places recognized by the International Union of Pure and Applied Chemistry, the worldwide body that develops standards for chemistry. The atomic masses of some elements are known very precisely, to a large number of decimal places. The atomic masses of other elements, especially radioactive elements, are not known as precisely. Some elements, such as lithium, can have varying atomic masses depending on how their isotopes are isolated. The web offers many interactive periodic table resources. For example, see http://www.ptable.com. Table 17.1 The Basics of the Elements of the Periodic Table

Name	Atomic Symbol	Atomic Number	Atomic Mass	Footnotes
actinium*	Ac	89
aluminum	Al	13	26.9815386(8)
americium*	Am	95
antimony	Sb	51	121.760(1)	g
argon	Ar	18	39.948(1)	g, r
arsenic	As	33	74.92160(2)
astatine*	At	85
barium	Ba	56	137.327(7)
berkelium*	Bk	97
beryllium	Be	4	9.012182(3)
bismuth	Bi	83	208.98040(1)
bohrium*	Bh	107
boron	B	5	10.811(7)	g, m, r
bromine	Br	35	79.904(1)
cadmium	Cd	48	112.411(8)	g
caesium (cesium)	Cs	55	132.9054519(2)
calcium	Ca	20	40.078(4)	g
californium*	Cf	98
carbon	C	6	12.0107(8)	g, r
cerium	Ce	58	140.116(1)	g
chlorine	Cl	17	35.453(2)	g, m, r
chromium	Cr	24	51.9961(6)
cobalt	Co	27	58.933195(5)
copernicium*	Cn	112
copper	Cu	29	63.546(3)	r
curium*	Cm	96
darmstadtium*	Ds	110
dubnium*	Db	105
dysprosium	Dy	66	162.500(1)	g
einsteinium*	Es	99
erbium	Er	68	167.259(3)	g
europium	Eu	63	151.964(1)	g
fermium*	Fm	100
fluorine	F	9	18.9984032(5)
francium*	Fr	87
gadolinium	Gd	64	157.25(3)	g
gallium	Ga	31	69.723(1)
germanium	Ge	32	72.64(1)
gold	Au	79	196.966569(4)
hafnium	Hf	72	178.49(2)
hassium*	Hs	108
helium	He	2	4.002602(2)	g, r
holmium	Ho	67	164.93032(2)
hydrogen	H	1	1.00794(7)	g, m, r
indium	In	49	114.818(3)
iodine	I	53	126.90447(3)
iridium	Ir	77	192.217(3)
iron	Fe	26	55.845(2)
krypton	Kr	36	83.798(2)	g, m
lanthanum	La	57	138.90547(7)	g
lawrencium*	Lr	103
lead	Pb	82	207.2(1)	g, r
lithium	Li	3	[6.941(2)]†	g, m, r
lutetium	Lu	71	174.967(1)	g
magnesium	Mg	12	24.3050(6)
manganese	Mn	25	54.938045(5)
meitnerium*	Mt	109
mendelevium*	Md	101
mercury	Hg	80	200.59(2)
molybdenum	Mo	42	95.94(2)	g
neodymium	Nd	60	144.242(3)	g
neon	Ne	10	20.1797(6)	g, m
neptunium*	Np	93
nickel	Ni	28	58.6934(2)
niobium	Nb	41	92.90638(2)
nitrogen	N	7	14.0067(2)	g, r
nobelium*	No	102
osmium	Os	76	190.23(3)	g
oxygen	O	8	15.9994(3)	g, r
palladium	Pd	46	106.42(1)	g
phosphorus	P	15	30.973762(2)
platinum	Pt	78	195.084(9)
plutonium*	Pu	94
polonium*	Po	84
potassium	K	19	39.0983(1)
praseodymium	Pr	59	140.90765(2)
promethium*	Pm	61
protactinium*	Pa	91	231.03588(2)
radium*	Ra	88
radon*	Rn	86
roentgenium*	Rg	111
rhenium	Re	75	186.207(1)
rhodium	Rh	45	102.90550(2)
rubidium	Rb	37	85.4678(3)	g
ruthenium	Ru	44	101.07(2)	g
rutherfordium*	Rf	104
samarium	Sm	62	150.36(2)	g
scandium	Sc	21	44.955912(6)
seaborgium*	Sg	106
selenium	Se	34	78.96(3)	r
silicon	Si	14	28.0855(3)	r
silver	Ag	47	107.8682(2)	g
sodium	Na	11	22.98976928(2)
strontium	Sr	38	87.62(1)	g, r
sulfur	S	16	32.065(5)	g, r
tantalum	Ta	73	180.94788(2)
technetium*	Tc	43
tellurium	Te	52	127.60(3)	g
terbium	Tb	65	158.92535(2)
thallium	Tl	81	204.3833(2)
thorium*	Th	90	232.03806(2)	g
thulium	Tm	69	168.93421(2)
tin	Sn	50	118.710(7)	g
titanium	Ti	22	47.867(1)
tungsten	W	74	183.84(1)
ununhexium*	Uuh	116
ununoctium*	Uuo	118
ununpentium*	Uup	115
ununquadium*	Uuq	114
ununtrium*	Uut	113
uranium*	U	92	238.02891(3)	g, m
vanadium	V	23	50.9415(1)
xenon	Xe	54	131.293(6)	g, m
ytterbium	Yb	70	173.04(3)	g
yttrium	Y	39	88.90585(2)
zinc	Zn	30	65.409(4)
zirconium	Zr	40	91.224(2)	g
*Element has no stable nuclides. However, three such elements (Th, Pa, and U) have a characteristic terrestrial isotopic composition, and for these an atomic mass is tabulated.
†Commercially available Li materials have atomic weights that range between 6.939 and 6.996; if a more accurate value is required, it must be determined for the specific material.
g Geological specimens are known in which the element has an isotopic composition outside the limits for normal material. The difference between the atomic mass of the element in such specimens and that given in the table may exceed the stated uncertainty.
m Modified isotopic compositions may be found in commercially available material because it has been subjected to an undisclosed or inadvertent isotopic fractionation. Substantial deviations in the atomic mass of the element from that given in the table can occur.
r Range in isotopic composition of normal terrestrial material prevents a more precise Ar(E) being given; the tabulated Ar(E) value and uncertainty should be applicable to normal material.

Source: Adapted from Pure and Applied Chemistry 78, no. 11 (2005): 2051–66. © IUPAC (International Union of Pure and Applied Chemistry).]]> 734 0 4 0 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/back-matter/appendix-d-atomic-masses/ Thu, 29 Jun 2017 22:14:50 +0000 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/back-matter/appendix-d-atomic-masses/ Atomic Number, Z Name Atomic Mass Number, A Symbol Atomic Mass (u) Percent Abundance or Decay Mode Half-life, t_1/2 0 neutron 1 $latex \textit{\textbf{n}}$ 1.008 665 $latex \beta ^{-}$ 10.37 min 1 Hydrogen 1 $latex ^{1} \textbf{H}$ 1.007 825 99.985% Deuterium 2 $latex ^{2}\textbf{H or D}$ 2.014 102 0.015% Tritium 3 $latex ^{3}\textbf{H or T}$ 3.016 050 $latex \beta ^{-}$ 12.33 y 2 Helium 3 $latex ^{3} \textbf{He}$ 3.016 030 $latex 1.38 \times 10^{-4} \% $ 4 $latex ^{4} \textbf{He}$ 4.002 603 $latex \approx 100 \% $ 3 Lithium 6 $latex ^6 \textbf{Li}$ 6.015 121 7.5% 7 $latex ^7 \textbf{Li}$ 7.016 003 92.5% 4 Beryllium 7 $latex ^7 \textbf{Be}$ 7.016 928 EC 53.29 d 9 $latex ^9 \textbf{Be}$ 9.012 182 100% 5 Boron 10 $latex ^{10} \textbf{B}$ 10.012 937 19.9% 11 $latex ^{11} \textbf{B}$ 11.009 305 80.1% 6 Carbon 11 $latex ^{11} \textbf{C}$ 11.011 432 EC, $latex \beta ^{+}$ 12 $latex ^{12} \textbf{C}$ 12.000 000 98.90% 13 $latex ^{13} \textbf{C}$ 13.003 355 1.10% 14 $latex ^{14} \textbf{C}$ 14.003 241 $latex \beta ^{-}$ 5730 y 7 Nitrogen 13 $latex ^{13} \textbf{N}$ 13.005 738 $latex \beta ^{+}$ 9.96 min 14 $latex ^{14} \textbf{N}$ 14.003 074 99.63% 15 $latex ^{15} \textbf{N}$ 15.000 108 0.37% 8 Oxygen 15 $latex ^{15} \textbf{O}$ 15.003 065 EC, $latex \beta ^{+}$ 122 s 16 $latex ^{16} \textbf{O}$ 15.994 915 99.76% 18 $latex ^{18} \textbf{O}$ 17.999 160 0.200% 9 Fluorine 18 $latex ^{18} \textbf{F}$ 18.000 937 EC, $latex \beta ^{+}$ 1.83 h 19 $latex ^{19} \textbf{F}$ 18.998 403 100% 10 Neon 20 $latex ^{20} \textbf{Ne}$ 19.992 435 90.51% 22 $latex ^{22} \textbf{Ne}$ 21.991 383 9.22% 11 Sodium 22 $latex ^{22} \textbf{Na}$ 21.994 434 $latex \beta ^{+}$ 2.602 y 23 $latex ^{23} \textbf{Na}$ 22.989 767 100% 24 $latex ^{24} \textbf{Na}$ 23.990 961 $latex \beta ^{-}$ 14.96 h 12 Magnesium 24 $latex ^{24} \textbf{Mg}$ 23.985 042 78.99% 13 Aluminum 27 $latex ^{27} \textbf{Al}$ 26.981 539 100% 14 Silicon 28 $latex ^{28} \textbf{Si}$ 27.976 927 92.23% 2.62h 31 $latex ^{31} \textbf{Si}$ 30.975 362 $latex \beta ^{-}$ 15 Phosphorus 31 $latex ^{31} \textbf{P}$ 30.973 762 100% 32 $latex ^{32} \textbf{P}$ 31.973 907 $latex \beta ^{-}$ 14.28 d 16 Sulfur 32 $latex ^{32} \textbf{S}$ 31.972 070 95.02% 35 $latex ^{35} \textbf{S}$ 34.969 031 $latex \beta ^{-}$ 87.4 d 17 Chlorine 35 $latex ^{35} \textbf{Cl}$ 34.968 852 75.77% 37 $latex ^{37} \textbf{Cl}$ 36.965 903 24.23% 18 Argon 40 $latex ^{40} \textbf{Ar}$ 39.962 384 99.60% 19 Potassium 39 $latex ^{39} \textbf{K}$ 38.963 707 93.26% 40 $latex ^{40} \textbf{K}$ 39.963 999 0.0117%, EC, $latex \beta ^{-}$ $latex 1.28 \times 10^9 \text{y}$ 20 Calcium 40 $latex ^{40} \textbf{Ca}$ 39.962 591 96.94% 21 Scandium 45 $latex ^{45} \textbf{Sc}$ 44.955 910 100% 22 Titanium 48 $latex ^{48} \textbf{Ti}$ 47.947 947 73.8% 23 Vanadium 51 $latex ^{51} \textbf{V}$ 50.943 962 99.75% 24 Chromium 52 $latex ^{52} \textbf{Cr}$ 51.940 509 83.79% 25 Manganese 55 $latex ^{55} \textbf{Mn}$ 54.938 047 100% 26 Iron 56 $latex ^{56} \textbf{Fe}$ 55.934 939 91.72% 27 Cobalt 59 $latex ^{59} \textbf{Co}$ 58.933 198 100% 60 $latex ^{60} \textbf{Co}$ 59.933 819 $latex \beta ^{-}$ 5.271 y 28 Nickel 58 $latex ^{58} \textbf{Ni}$ 57.935 346 68.27% 60 $latex ^{60} \textbf{Ni}$ 59.930 788 26.10% 29 Copper 63 $latex ^{63} \textbf{Cu}$ 62.939 598 69.17% 65 $latex ^{65} \textbf{Cu}$ 64.927 793 30.83% 30 Zinc 64 $latex ^{64} \textbf{Zn}$ 63.929 145 48.6% 66 $latex ^{66} \textbf{Zn}$ 65.926 034 27.9% 31 Gallium 69 $latex ^{69} \textbf{Ga}$ 68.925 580 60.1% 32 Germanium 72 $latex ^{72} \textbf{Ge}$ 71.922 079 27.4% 74 $latex ^{74} \textbf{Ge}$ 73.921 177 36.5% 33 Arsenic 75 $latex ^{75} \textbf{As}$ 74.921 594 100% 34 Selenium 80 $latex ^{80} \textbf{Se}$ 79.916 520 49.7% 35 Bromine 79 $latex ^{79} \textbf{Br}$ 78.918 336 50.69% 36 Krypton 84 $latex ^{84} \textbf{Kr}$ 83.911 507 57.0% 37 Rubidium 85 $latex ^{85} \textbf{Rb}$ 84.911 794 72.17% 38 Strontium 86 $latex ^{86} \textbf{Sr}$ 85.909 267 9.86% 88 $latex ^{88} \textbf{Sr}$ 87.905 619 82.58% 90 $latex ^{90} \textbf{Sr}$ 89.907 738 $latex \beta ^{-}$ 28.8 y 39 Yttrium 89 $latex ^{89} \textbf{Y}$ 88.905 849 100% 90 $latex ^{90} \textbf{Y}$ 89.907 152 $latex \beta ^{-}$­ 64.1 h 40 Zirconium 90 $latex ^{90} \textbf{Zr}$ 89.904 703 51.45% 41 Niobium 93 $latex ^{93} \textbf{Nb}$ 92.906 377 100% 42 Molybdenum 98 $latex ^{98} \textbf{Mo}$ 97.905 406 24.13% 43 Technetium 98 $latex ^{98} \textbf{Tc} $ 97.907 215 $latex \beta ^{-}$­ $latex 4.2 \times 10^6 \text{y} $ 44 Ruthenium 102 $latex ^{102} \textbf{Ru}$ 101.904 348 31.6% 45 Rhodium 103 $latex ^{103} \textbf{Rh}$ 102.905 500 100% 46 Palladium 106 $latex ^{106} \textbf{Pd}$ 105.903 478 27.33% 47 Silver 107 $latex ^{107} \textbf{Ag}$ 106.905 092 51.84% 109 $latex ^{109} \textbf{Ag}$ 108.904 757 48.16% 48 Cadmium 114 $latex ^{114} \textbf{Cd}$ 113.903 357 28.73% 49 Indium 115 $latex ^{115} \textbf{In}$ 114.903 880 95.7%, $latex \beta ^{-}$­ $latex 4.4 \times 10^{14} \text{y}$ 50 Tin 120 $latex ^{120} \textbf{Sn}$ 119.902 200 32.59% 51 Antimony 121 $latex ^{121} \textbf{Sb}$ 120.903 821 57.3% 52 Tellurium 130 $latex ^{130} \textbf{Te}$ 129.906 229 33.8%, $latex \beta ^{-}$­ $latex 2.5 \times 10^{21} \text{y}$ 53 Iodine 127 $latex ^{127} \textbf{I}$ 126.904 473 100% 131 $latex ^{131} \textbf{I}$ 130.906 114 $latex \beta ^{-}$­ 8.040 d 54 Xenon 132 $latex ^{132} \textbf{Xe}$ 131.904 144 26.9% 136 $latex ^{136} \textbf{Xe}$ 135.907 214 8.9% 55 Cesium 133 $latex ^{133} \textbf{Cs}$ 132.905 429 100% 134 $latex ^{134} \textbf{Cs}$ 133.906 696 EC, $latex \beta ^{-}$­ 2.06 y 56 Barium 137 $latex ^{137} \textbf{Ba}$ 136.905 812 11.23% 138 $latex ^{138} \textbf{Ba}$ 137.905 232 71.70% 57 Lanthanum 139 $latex ^{139} \textbf{La}$ 138.906 346 99.91% 58 Cerium 140 $latex ^{140} \textbf{Ce}$ 139.905 433 88.48% 59 Praseodymium 141 $latex ^{141} \textbf{Pr}$ 140.907 647 100% 60 Neodymium 142 $latex ^{142} \textbf{Nd}$ 141.907 719 27.13% 61 Promethium 145 $latex ^{145} \textbf{Pm}$ 144.912 743 EC, $latex \alpha$ 17.7 y 62 Samarium 152 $latex ^{152} \textbf{Sm}$ 151.919 729 26.7% 63 Europium 153 $latex ^{153} \textbf{Eu}$ 152.921 225 52.2% 64 Gadolinium 158 $latex ^{158} \textbf{Gd}$ 157.924 099 24.84% 65 Terbium 159 $latex ^{159} \textbf{Tb}$ 158.925 342 100% 66 Dysprosium 164 $latex ^{164} \textbf{Dy}$ 163.929 171 28.2% 67 Holmium 165 $latex ^{165} \textbf{Ho}$ 164.930 319 100% 68 Erbium 166 $latex ^{166} \textbf{Er}$ 165.930 290 33.6% 69 Thulium 169 $latex ^{169} \textbf{Tm}$ 168.934 212 100% 70 Ytterbium 174 $latex ^{174} \textbf{Yb}$ 173.938 859 31.8% 71 Lutecium 175 $latex ^{175} \textbf{Lu}$ 174.940 770 97.41% 72 Hafnium 180 $latex ^{180} \textbf{Hf}$ 179.946 545 35.10% 73 Tantalum 181 $latex ^{181} \textbf{Ta}$ 180.947 992 99.98% 74 Tungsten 184 $latex ^{184} \textbf{W}$ 183.950 928 30.67% 75 Rhenium 187 $latex ^{187} \textbf{Re}$ 186.955 744 62.6%, $latex \beta ^{-}$ $latex 4.6 \times 10^{10} \text{y}$ 76 Osmium 191 $latex ^{191} \textbf{Os}$ 190.960 920 $latex \beta ^{-}$­ 15.4 d 192 $latex ^{192} \textbf{Os}$ 191.961 467 41.0% 77 Iridium 191 $latex ^{191} \textbf{Ir}$ 190.960 584 37.3% 193 $latex ^{193} \textbf{Ir}$ 192.962 917 62.7% 78 Platinum 195 $latex ^{195} \textbf{Pt}$ 194.964 766 33.8% 79 Gold 197 $latex ^{197} \textbf{Au}$ 196.966 543 100% 198 $latex ^{198} \textbf{Au}$ 197.968 217 $latex \beta ^{-}$­ 2.696 d 80 Mercury 199 $latex ^{199} \textbf{Hg}$ 198.968 253 16.87% 202 $latex ^{202} \textbf{Hg}$ 201.970 617 29.86% 81 Thallium 205 $latex ^{205} \textbf{Tl}$ 204.974 401 70.48% 82 Lead 206 $latex ^{206} \textbf{Pb}$ 205.974 440 24.1% 207 $latex ^{207} \textbf{Pb}$ 206.975 872 22.1% 208 $latex ^{208} \textbf{Pb}$ 207.976 627 52.4% 210 $latex ^{210} \textbf{Pb}$ 209.984 163 $latex \alpha $, $latex \beta ^{-}$­ 22.3 y 211 $latex ^{211} \textbf{Pb}$ 210.988 735 $latex \beta ^{-}$ 36.1 min 212 $latex ^{212} \textbf{Pb}$ 211.991 871 $latex \beta ^{-}$ 10.64 h 83 Bismuth 209 $latex ^{209} \textbf{Bi}$ 208.980 374 100% 211 $latex ^{211} \textbf{Bi}$ 210.987 255 $latex \alpha $, $latex \beta ^{-}$ 2.14 min 84 Polonium 210 $latex ^{210} \textbf{Po}$ 209.982 848 $latex \alpha $ 138.38 d 85 Astatine 218 $latex ^{218} \textbf{At}$ 218.008 684 $latex \alpha $, $latex \beta ^{-}$ 1.6 s 86 Radon 222 $latex ^{222} \textbf{Rn}$ 222.017 570 $latex \alpha $ 3.82 d 87 Francium 223 $latex ^{223} \textbf{Fr}$ 223.019 733 $latex \alpha $, $latex \beta ^{-}$ 21.8 min 88 Radium 226 $latex ^{226} \textbf{Ra}$ 226.025 402 $latex \alpha $ $latex 1.60 \times 10^3 \text{y}$ 89 Actinium 227 $latex ^{227} \textbf{Ac}$ 227.027 750 $latex \alpha $, $latex \beta ^{-}$ 21.8 y 90 Thorium 228 $latex ^{228} \textbf{Th}$ 228.028 715 $latex \alpha$ 1.91 y 232 $latex ^{232} \textbf{Th}$ 232.038 054 100%, $latex \alpha $ $latex 1.41 \times 10^{10} \text{y}$ 91 Protactinium 231 $latex ^{231} \textbf{Pa}$ 231.035 880 $latex \alpha $ $latex 3.28 \times 10^4 \text{y}$ 92 Uranium 233 $latex ^{233} \textbf{U}$ 233.039 628 $latex \alpha $ $latex 1.59 \times 10^3 \text{y}$ 235 $latex ^{235} \textbf{U}$ 235.043 924 0.720%, $latex \alpha $ $latex 7.04 \times 10^8 \text{y}$ 236 $latex ^{236} \textbf{U}$ 236.045 562 $latex \alpha $ $latex 2.34 \times 10^7 \text{y}$ 238 $latex ^{238} \textbf{U}$ 238.050 784 99.2745%, $latex \alpha $ $latex 4.47 \times 10^9 \text{y}$ 239 $latex ^{239} \textbf{U}$ 239.054 289 $latex \beta ^{-}$ 23.5 min 93 Neptunium 239 $latex ^{239} \textbf{Np}$ 239.052 933 $latex \beta ^{-}$ 2.355 d 94 Plutonium 239 $latex ^{239} \textbf{Pu}$ 239.052 157 $latex \alpha $ $latex 2.41 \times 10^4 \text{y} $ 95 Americium 243 $latex ^{243} \textbf{Am}$ 243.061 375 $latex \alpha $, fission $latex 7.37 \times 10^3 \text{y}$ 96 Curium 245 $latex ^{245} \textbf{Cm}$ 245.065 483 $latex \alpha $ $latex 8.50 \times 10^3 \text{y}$ 97 Berkelium 247 $latex ^{247} \textbf{Bk} $ 247.070 300 $latex \alpha $ $latex 1.38 \times 10^3 \text{y}$ 98 Californium 249 $latex ^{249} \textbf{Cf} $ 249.074 844 $latex \alpha $ 351 y 99 Einsteinium 254 $latex ^{254} \textbf{Es}$ 254.088 019 $latex \alpha $, $latex \beta ^{-}$­ 276 d 100 Fermium 253 $latex ^{253} \textbf{Fm}$ 253.085 173 EC, $latex \alpha $ 3.00 d 101 Mendelevium 255 $latex ^{255} \textbf{Md}$ 255.091 081 EC, $latex \alpha $ 27 min 102 Nobelium 255 $latex ^{255} \textbf{No} $ 255.093 260 EC, $latex \alpha $ 3.1 min 103 Lawrencium 257 $latex ^{257} \textbf{Lr}$ 257.099 480 EC, $latex \alpha $ 0.646 s 104 Rutherfordium 261 $latex ^{261} \textbf{Rf} $ 261.108 690 $latex \alpha $ 1.08 min 105 Dubnium 262 $latex ^{262} \textbf{Db}$ 262.113 760 $latex \alpha $, fission 34 s 106 Seaborgium 263 $latex ^{263} \textbf{Sg}$ 263.11 86 $latex \alpha $, fission 0.8 s 107 Bohrium 262 $latex ^{262} \textbf{Bh}$ 262.123 1 $latex \alpha $ 0.102 s 108 Hassium 264 $latex ^{264} \textbf{Hs}$ 264.128 5 $latex \alpha $ 0.08 ms 109 Meitnerium 266 $latex ^{266} \textbf{Mt}$ 266.137 8 $latex \alpha $ 3.4 ms Table 1: Atomic Masses ]]> 735 0 5 0 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/back-matter/appendix-e-half-lives-for-several-radioactive-isotopes-originally-from-openstax-chemistry/ Thu, 29 Jun 2017 22:14:50 +0000 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/back-matter/appendix-e-half-lives-for-several-radioactive-isotopes-originally-from-openstax-chemistry/ Half-Lives for Several Radioactive Isotopes Isotope Half-Life[footnote]y = years, d = days, h = hours, m = minutes, s = seconds[/footnote] Type of Emission[footnote]E.C. = electron capture, S.F. = Spontaneous fission[/footnote] Isotope Half-Life[footnote]y = years, d = days, h = hours, m = minutes, s = seconds[/footnote] Type of Emission[footnote]E.C. = electron capture, S.F. = Spontaneous fission[/footnote] $latex _6^{14}\text{C}$ 5730 y $latex ({\beta}-)$ $latex _{83}^{210}\text{Bi}$ 5.01 d $latex ({\beta}-)$ $latex _7^{13}\text{N}$ 9.97 m $latex ({\beta}+)$ $latex _{83}^{212}\text{Bi}$ 60.55 m $latex ({\alpha}\;\text{or}\;{\beta}-)$ $latex _9^{15}\text{F}$ 4.1 × 10⁻²² s $latex (p)$ $latex _{84}^{210}\text{Po}$ 138.4 d $latex ({\alpha})$ $latex _{11}^{24}\text{Na}$ 15.00 h $latex ({\beta}-)$ $latex _{84}^{212}\text{Po}$ 3 × 10⁻⁷ s $latex ({\alpha})$ ]]> 736 0 6 0 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/back-matter/appendix-f-glossary-of-key-symbols-and-notation/ Thu, 29 Jun 2017 22:14:50 +0000 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/back-matter/appendix-f-glossary-of-key-symbols-and-notation/ In this glossary, key symbols and notation are briefly defined.

Symbol	Definition
$latex \overline{\text{any symbol}}$	average (indicated by a bar over a symbol—e.g., $latex \bar{v}$ is average velocity)
$latex ^{\circ} \text{C}$	Celsius degree
$latex ^{\circ} \text{F}$	Fahrenheit degree
$latex // $	parallel
$latex \bot $	perpendicular
$latex \propto $	proportional to
$latex \pm $	plus or minus
$latex _0 $	zero as a subscript denotes an initial value
$latex \alpha $	alpha rays
$latex \alpha $	angular acceleration
$latex \alpha $	temperature coefficient(s) of resistivity
$latex \beta $	beta rays
$latex \beta $	sound level
$latex \beta $	volume coefficient of expansion
$latex \beta ^{-} $	electron emitted in nuclear beta decay
$latex \beta ^{+} $	positron decay
$latex \gamma $	gamma rays
$latex \gamma $	surface tension
$latex \gamma = 1/ \sqrt{1 - v^2 / c^2} $	a constant used in relativity
$latex \Delta $	change in whatever quantity follows
$latex \delta $	uncertainty in whatever quantity follows
$latex \mathit\Delta E $	change in energy between the initial and final orbits of an electron in an atom
$latex \mathit\Delta E $	uncertainty in energy
$latex \mathit\Delta m $	difference in mass between initial and final products
$latex \mathit\Delta N $	number of decays that occur
$latex \mathit\Delta p $	change in momentum
$latex \mathit\Delta p $	uncertainty in momentum
$latex \mathit\Delta \text{PE}_{\text{g}} $	change in gravitational potential energy
$latex \mathit\Delta \theta $	rotation angle
$latex \mathit\Delta s $	distance traveled along a circular path
$latex \mathit\Delta t $	uncertainty in time
$latex \mathit\Delta t_0 $	proper time as measured by an observer at rest relative to the process
$latex \mathit\Delta V $	potential difference
$latex \mathit\Delta x $	uncertainty in position
$latex \epsilon _0 $	permittivity of free space
$latex \eta $	viscosity
$latex \theta $	angle between the force vector and the displacement vector
$latex \theta $	angle between two lines
$latex \theta $	contact angle
$latex \theta $	direction of the resultant
$latex \theta _b $	Brewster's angle
$latex \theta _c $	critical angle
$latex \kappa $	dielectric constant
$latex \lambda $	decay constant of a nuclide
$latex \lambda $	wavelength
$latex \lambda _n $	wavelength in a medium
$latex \mu _0 $	permeability of free space
$latex \mu _k $	coefficient of kinetic friction
$latex \mu _s $	coefficient of static friction
$latex v_e $	electron neutrino
$latex \pi ^+ $	positive pion
$latex \pi ^- $	negative pion
$latex \pi ^0 $	neutral pion
$latex \rho $	density
$latex \rho _{\text{c}} $	critical density, the density needed to just halt universal expansion
$latex \rho _{\text{fl}} $	fluid density
$latex \overline{\rho} _{\text{obj}} $	average density of an object
$latex \rho / \rho _{\text{w}} $	specific gravity
$latex \tau $	characteristic time constant for a resistance and inductance ($latex RL $) or resistance and capacitance ($latex RC $) circuit
$latex \tau $	characteristic time for a resistor and capacitor ($latex RC $) circuit
$latex \tau $	torque
$latex \Upsilon $	upsilon meson
$latex \Phi $	magnetic flux
$latex \phi $	phase angle
$latex \Omega $	ohm (unit)
$latex \omega $	angular velocity
$latex \text{A}$	ampere (current unit)
$latex A $	area
$latex A $	cross-sectional area
$latex A $	total number of nucleons
$latex a $	acceleration
$latex a_{\text{B}} $	Bohr radius
$latex a_{\text{c}} $	centripetal acceleration
$latex a_{\text{t}} $	tangential acceleration
$latex \text{AC}$	alternating current
$latex \text{AM}$	amplitude modulation
$latex \text{atm}$	atmosphere
$latex B $	baryon number
$latex B $	blue quark color
$latex \overline{B} $	antiblue (yellow) antiquark color
$latex b $	quark flavor bottom or beauty
$latex B $	bulk modulus
$latex B $	magnetic field strength
$latex B_{\text{int}} $	electron’s intrinsic magnetic field
$latex B_{\text{orb}} $	orbital magnetic field
$latex \text{BE} $	binding energy of a nucleus—it is the energy required to completely disassemble it into separate protons and neutrons
$latex \text{BE/A} $	binding energy per nucleon
$latex \text{Bq} $	becquerel—one decay per second
$latex C $	capacitance (amount of charge stored per volt)
$latex C $	coulomb (a fundamental SI unit of charge)
$latex C_{\text{p}} $	total capacitance in parallel
$latex C_{\text{s}} $	total capacitance in series
$latex \text{CG} $	center of gravity
$latex \text{CM} $	center of mass
$latex c $	quark flavor charm
$latex c $	specific heat
$latex c $	speed of light
$latex \text{Cal} $	kilocalorie
$latex \text{cal} $	calorie
$latex \textit{\text{COP}}_{\text{hp}} $	heat pump’s coefficient of performance
$latex \textit{\text{COP}}_{\text{ref}} $	coefficient of performance for refrigerators and air conditioners
$latex \text{cos} \theta $	cosine
$latex \text{cot} \theta $	cotangent
$latex \text{csc} \theta $	cosecant
$latex D $	diffusion constant
$latex d $	displacement
$latex d $	quark flavor down
$latex \text{dB} $	decibel
$latex d_i $	distance of an image from the center of a lens
$latex d_o $	distance of an object from the center of a lens
$latex \text{DC} $	direct current
$latex E $	electric field strength
$latex \epsilon $	emf (voltage) or Hall electromotive force
$latex \text{emf} $	electromotive force
$latex E $	energy of a single photon
$latex E $	nuclear reaction energy
$latex E $	relativistic total energy
$latex E $	total energy
$latex E_0 $	ground state energy for hydrogen
$latex E_0 $	rest energy
$latex \text{EC} $	electron capture
$latex E_{\text{cap}} $	energy stored in a capacitor
$latex \textit{\text{Eff}} $	efficiency—the useful work output divided by the energy input
$latex \textit{\text{Eff}}_{\textit{\text{C}}} $	Carnot efficiency
$latex E_{\text{in}} $	energy consumed (food digested in humans)
$latex E_{\text{ind}} $	energy stored in an inductor
$latex E_{\text{out}} $	energy output
$latex e $	emissivity of an object
$latex e^+ $	antielectron or positron
$latex \text{eV} $	electron volt
$latex \text{F} $	farad (unit of capacitance, a coulomb per volt)
$latex \text{F} $	focal point of a lens
$latex \textbf{\text{F}} $	force
$latex F $	magnitude of a force
$latex F $	restoring force
$latex F_{\text{B}} $	buoyant force
$latex F_{\text{c}} $	centripetal force
$latex F_{\text{i}} $	force input
$latex \textbf{F}_{\text{net}} $	net force
$latex F_{\text{o}} $	force output
$latex \text{FM} $	frequency modulation
$latex f $	focal length
$latex f $	frequency
$latex f_0 $	resonant frequency of a resistance, inductance, and capacitance ($latex RLC $) series circuit
$latex f_0 $	threshold frequency for a particular material (photoelectric effect)
$latex f_1 $	fundamental
$latex f_2 $	first overtone
$latex f_3 $	second overtone
$latex f_{\text{B}} $	beat frequency
$latex f_{\text{k}} $	magnitude of kinetic friction
$latex f_{\text{s}} $	magnitude of static friction
$latex G $	gravitational constant
$latex G $	green quark color
$latex \overline{G} $	antigreen (magenta) antiquark color
$latex g $	acceleration due to gravity
$latex g $	gluons (carrier particles for strong nuclear force)
$latex h $	change in vertical position
$latex h $	height above some reference point
$latex h $	maximum height of a projectile
$latex h $	Planck's constant
$latex hf $	photon energy
$latex h_i $	height of the image
$latex h_o $	height of the object
$latex I $	electric current
$latex I $	intensity
$latex I $	intensity of a transmitted wave
$latex I $	moment of inertia (also called rotational inertia)
$latex I_0 $	intensity of a polarized wave before passing through a filter
$latex I_{\text{ave}} $	average intensity for a continuous sinusoidal electromagnetic wave
$latex I_{\text{rms}} $	average current
$latex J $	joule
$latex J / \psi $	Joules/psi meson
$latex \text{K} $	kelvin
$latex k $	Boltzmann constant
$latex k $	force constant of a spring
$latex K_{\alpha} $	x rays created when an electron falls into an $latex n = 1 $ shell vacancy from the $latex n = 3 $ shell
$latex K_{\beta} $	x rays created when an electron falls into an $latex n = 2 $ shell vacancy from the $latex n = 3 $ shell
$latex \text{kcal} $	kilocalorie
$latex \text{KE} $	translational kinetic energy
$latex \text{KE} + \text{PE} $	mechanical energy
$latex \text{KE}_e $	kinetic energy of an ejected electron
$latex \text{KE}_{\text{rel}} $	relativistic kinetic energy
$latex \text{KE}_{\text{rot}} $	rotational kinetic energy
$latex \overline{\text{KE}} $	thermal energy
$latex \text{kg} $	kilogram (a fundamental SI unit of mass)
$latex L $	angular momentum
$latex \text{L} $	liter
$latex L $	magnitude of angular momentum
$latex L $	self-inductance
$latex \ell $	angular momentum quantum number
$latex L_{\alpha} $	x rays created when an electron falls into an $latex n = 2 $ shell from the $latex n = 3 $ shell
$latex L_e $	electron total family number
$latex L_{\mu} $	muon family total number
$latex L_{\tau} $	tau family total number
$latex L_{\text{f}} $	heat of fusion
$latex L_{\text{f}}$ and $latex L_{\text{v}} $	latent heat coefficients
$latex L_{\text{orb}} $	orbital angular momentum
$latex L_{\text{s}} $	heat of sublimation
$latex L_{\text{v}} $	heat of vaporization
$latex L_z $	z - component of the angular momentum
$latex M $	angular magnification
$latex M $	mutual inductance
$latex \text{m} $	indicates metastable state
$latex m $	magnification
$latex m $	mass
$latex m $	mass of an object as measured by a person at rest relative to the object
$latex \text{m} $	meter (a fundamental SI unit of length)
$latex m $	order of interference
$latex m $	overall magnification (product of the individual magnifications)
$latex m(^AX) $	atomic mass of a nuclide
$latex \text{MA} $	mechanical advantage
$latex m_{\text{e}} $	magnification of the eyepiece
$latex m_e $	mass of the electron
$latex m_{\ell} $	angular momentum projection quantum number
$latex m_n $	mass of a neutron
$latex m_{\text{o}} $	magnification of the objective lens
$latex \text{mol} $	mole
$latex m_p $	mass of a proton
$latex m_{\text{s}} $	spin projection quantum number
$latex N $	magnitude of the normal force
$latex \text{N} $	newton
$latex \textbf{\text{N}}$	normal force
$latex N $	number of neutrons
$latex n $	index of refraction
$latex n $	number of free charges per unit volume
$latex N_A $	Avogadro's number
$latex N_{\text{r}} $	Reynolds number
$latex \text{N} \cdot \text{m} $	newton-meter (work-energy unit)
$latex \text{N} \cdot \text{m} $	newtons times meters (SI unit of torque)
$latex \text{OE} $	other energy
$latex P $	power
$latex P $	power of a lens
$latex P $	pressure
$latex \textbf{\text{p}} $	momentum
$latex p $	momentum magnitude
$latex p $	relativistic momentum
$latex \textbf{\text{p}}_{\text{tot}} $	total momentum
$latex \textbf{\text{p}}^{\prime}_{\text{tot}} $	total momentum some time later
$latex p_{\text{abs}} $	absolute pressure
$latex p_{\text{atm}} $	atmospheric pressure
$latex p_{\text{atm}} $	standard atmospheric pressure
$latex \text{PE} $	potential energy
$latex \text{PE}_{el} $	elastic potential energy
$latex \text{PE}_{\text{elec}} $	electric potential energy
$latex \text{PE}_s $	potential energy of a spring
$latex P_g $	gauge pressure
$latex P_{in} $	power consumption or input
$latex P_{out} $	useful power output going into useful work or a desired, form of energy
$latex Q $	latent heat
$latex Q $	net heat transferred into a system
$latex Q $	flow rate—volume per unit time flowing past a point
$latex +Q $	positive charge
$latex -Q $	negative charge
$latex q $	electron charge
$latex q_p $	charge of a proton
$latex q $	test charge
$latex \text{QF} $	quality factor
$latex R $	activity, the rate of decay
$latex R $	radius of curvature of a spherical mirror
$latex R $	red quark color
$latex \overline{R} $	antired (cyan) quark color
$latex R $	resistance
$latex \text{R} $	resultant or total displacement
$latex R $	Rydberg constant
$latex R $	universal gas constant
$latex r $	distance from pivot point to the point where a force is applied
$latex r $	internal resistance
$latex r_{\bot} $	perpendicular lever arm
$latex r $	radius of a nucleus
$latex r $	radius of curvature
$latex r $	resistivity
$latex \text{r or rad} $	radiation dose unit
$latex \text{rem} $	roentgen equivalent man
$latex \text{rad} $	radian
$latex \text{RBE} $	relative biological effectiveness
$latex RC $	resistor and capacitor circuit
$latex \text{rms} $	root mean square
$latex r_n $	radius of the nth H-atom orbit
$latex R_p $	total resistance of a parallel connection
$latex R_s $	total resistance of a series connection
$latex R_s $	Schwarzschild radius
$latex S $	entropy
$latex \textbf{\text{S}} $	intrinsic spin (intrinsic angular momentum)
$latex S $	magnitude of the intrinsic (internal) spin angular momentum
$latex S $	shear modulus
$latex S $	strangeness quantum number
$latex s $	quark flavor strange
$latex \text{s} $	second (fundamental SI unit of time)
$latex s $	spin quantum number
$latex \textbf{\text{s}} $	total displacement
$latex \text{sec} \theta $	secant
$latex \text{sin} \theta $	sine
$latex s_z $	z-component of spin angular momentum
$latex T $	period—time to complete one oscillation
$latex T $	temperature
$latex T_c $	critical temperature—temperature below which a material becomes a superconductor
$latex T $	tension
$latex \text{T} $	tesla (magnetic field strength B)
$latex t $	quark flavor top or truth
$latex t $	time
$latex t_{1/2} $	half-life—the time in which half of the original nuclei decay
$latex \text{tan} \theta $	tangent
$latex U $	internal energy
$latex u $	quark flavor up
$latex \text{u} $	unified atomic mass unit
$latex \textbf{\text{u}} $	velocity of an object relative to an observer
$latex \textbf{\text{u}}^{\prime} $	velocity relative to another observer
$latex V $	electric potential
$latex V $	terminal voltage
$latex \text{V} $	volt (unit)
$latex V $	volume
$latex \textbf{\text{v}} $	relative velocity between two observers
$latex v $	speed of light in a material
$latex \textbf{\text{v}} $	velocity
$latex \overline{\textbf{\text{v}}} $	average fluid velocity
$latex V_B - V_A $	change in potential
$latex \textbf{\text{v}}_d $	drift velocity
$latex V_p $	transformer input voltage
$latex V_{\text{rms}} $	rms voltage
$latex V_s $	transformer output voltage
$latex \textbf{\text{v}}_{\text{tot}} $	total velocity
$latex v_{\text{w}} $	propagation speed of sound or other wave
$latex \textbf{\text{v}}_{\text{w}} $	wave velocity
$latex W $	work
$latex W $	net work done by a system
$latex \text{W} $	watt
$latex w $	weight
$latex w_{\text{fl}} $	weight of the fluid displaced by an object
$latex W_c $	total work done by all conservative forces
$latex W_{nc} $	total work done by all nonconservative forces
$latex W_{out} $	useful work output
$latex X $	amplitude
$latex \text{X} $	symbol for an element
$latex \begin{matrix} Z \\ A \end{matrix} K_N $	notation for a particular nuclide
$latex x $	deformation or displacement from equilibrium
$latex x $	displacement of a spring from its undeformed position
$latex x $	horizontal axis
$latex X_C $	capacitive reactance
$latex X_L $	inductive reactance
$latex x_{\text{rms}} $	root mean square diffusion distance
$latex y $	vertical axis
$latex Y $	elastic modulus or Young's modulus
$latex Z $	atomic number (number of protons in a nucleus)
$latex Z $	impedance

]]> 737 0 7 0 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/?attachment_id=739 Thu, 29 Jun 2017 22:24:08 +0000 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/?attachment_id=739 739 17 0 0 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/chapter/chapter-1/ Thu, 29 Jun 2017 18:16:33 +0000 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/chapter/chapter-1/ 5 23 13 0 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/chapter/1-0-introduction/ Thu, 29 Jun 2017 22:12:56 +0000 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/chapter/1-0-introduction/ [caption id="" align="aligncenter" width="1000"] Galaxies are as immense as atoms are small. Yet the same laws of physics describe both, and all the rest of nature—an indication of the underlying unity in the universe. The laws of physics are surprisingly few in number, implying an underlying simplicity to nature’s apparent complexity. (credit: NASA, JPL-Caltech, P. Barmby, Harvard-Smithsonian Center for Astrophysics)[/caption]

What is your first reaction when you hear the word “physics”? Did you imagine working through difficult equations or memorizing formulas that seem to have no real use in life outside the physics classroom? Many people come to the subject of physics with a bit of fear. But as you begin your exploration of this broad-ranging subject, you may soon come to realize that physics plays a much larger role in your life than you first thought, no matter your life goals or career choice.

For example, take a look at the image above. This image is of the Andromeda Galaxy, which contains billions of individual stars, huge clouds of gas, and dust. Two smaller galaxies are also visible as bright blue spots in the background. At a staggering 2.5 million light years from the Earth, this galaxy is the nearest one to our own galaxy (which is called the Milky Way). The stars and planets that make up Andromeda might seem to be the furthest thing from most people’s regular, everyday lives. But Andromeda is a great starting point to think about the forces that hold together the universe. The forces that cause Andromeda to act as it does are the same forces we contend with here on Earth, whether we are planning to send a rocket into space or simply raise the walls for a new home. The same gravity that causes the stars of Andromeda to rotate and revolve also causes water to flow over hydroelectric dams here on Earth. Tonight, take a moment to look up at the stars. The forces out there are the same as the ones here on Earth. Through a study of physics, you may gain a greater understanding of the interconnectedness of everything we can see and know in this universe.

Think now about all of the technological devices that you use on a regular basis. Computers, smart phones, GPS systems, MP3 players, and satellite radio might come to mind. Next, think about the most exciting modern technologies that you have heard about in the news, such as trains that levitate above tracks, “invisibility cloaks” that bend light around them, and microscopic robots that fight cancer cells in our bodies. All of these groundbreaking advancements, commonplace or unbelievable, rely on the principles of physics. Aside from playing a significant role in technology, professionals such as engineers, pilots, physicians, physical therapists, electricians, and computer programmers apply physics concepts in their daily work. For example, a pilot must understand how wind forces affect a flight path and a physical therapist must understand how the muscles in the body experience forces as they move and bend. As you will learn in this text, physics principles are propelling new, exciting technologies, and these principles are applied in a wide range of careers.

In this text, you will begin to explore the history of the formal study of physics, beginning with natural philosophy and the ancient Greeks, and leading up through a review of Sir Isaac Newton and the laws of physics that bear his name. You will also be introduced to the standards scientists use when they study physical quantities and the interrelated system of measurements most of the scientific community uses to communicate in a single mathematical language. Finally, you will study the limits of our ability to be accurate and precise, and the reasons scientists go to painstaking lengths to be as clear as possible regarding their own limitations.

]]> 25 23 1 0 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/chapter/1-1-physics-an-introduction/ Thu, 29 Jun 2017 22:12:57 +0000 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/chapter/1-1-physics-an-introduction/

Summary

• Explain the difference between a principle and a law.
• Explain the difference between a model and theory.

[caption id="" align="aligncenter" width="325"]Figure 1. The flight formations of migratory birds such as Canada geese are governed by the laws of physics. (credit: David Merrett).[/caption]

The physical universe is enormously complex in its detail. Every day, each of us observes a great variety of objects and phenomena. Over the centuries, the curiosity of the human race has led us collectively to explore and catalog a tremendous wealth of information. From the flight of birds to the colors of flowers, from lightning to gravity, from quarks to clusters of galaxies, from the flow of time to the mystery of the creation of the universe, we have asked questions and assembled huge arrays of facts. In the face of all these details, we have discovered that a surprisingly small and unified set of physical laws can explain what we observe. As humans, we make generalizations and seek order. We have found that nature is remarkably cooperative—it exhibits the underlying order and simplicity we so value.

It is the underlying order of nature that makes science in general, and physics in particular, so enjoyable to study. For example, what do a bag of chips and a car battery have in common? Both contain energy that can be converted to other forms. The law of conservation of energy (which says that energy can change form but is never lost) ties together such topics as food calories, batteries, heat, light, and watch springs. Understanding this law makes it easier to learn about the various forms energy takes and how they relate to one another. Apparently unrelated topics are connected through broadly applicable physical laws, permitting an understanding beyond just the memorization of lists of facts.

The unifying aspect of physical laws and the basic simplicity of nature form the underlying themes of this text. In learning to apply these laws, you will, of course, study the most important topics in physics. More importantly, you will gain analytical abilities that will enable you to apply these laws far beyond the scope of what can be included in a single book. These analytical skills will help you to excel academically, and they will also help you to think critically in any professional career you choose to pursue. This module discusses the realm of physics (to define what physics is), some applications of physics (to illustrate its relevance to other disciplines), and more precisely what constitutes a physical law (to illuminate the importance of experimentation to theory).

Science and the Realm of Physics

Science consists of the theories and laws that are the general truths of nature as well as the body of knowledge they encompass. Scientists are continually trying to expand this body of knowledge and to perfect the expression of the laws that describe it. Physics is concerned with describing the interactions of energy, matter, space, and time, and it is especially interested in what fundamental mechanisms underlie every phenomenon. The concern for describing the basic phenomena in nature essentially defines the realm of physics.

Physics aims to describe the function of everything around us, from the movement of tiny charged particles to the motion of people, cars, and spaceships. In fact, almost everything around you can be described quite accurately by the laws of physics. Consider a smart phone (Figure 2). Physics describes how electricity interacts with the various circuits inside the device. This knowledge helps engineers select the appropriate materials and circuit layout when building the smart phone. Next, consider a GPS system. Physics describes the relationship between the speed of an object, the distance over which it travels, and the time it takes to travel that distance. When you use a GPS device in a vehicle, it utilizes these physics equations to determine the travel time from one location to another.

[caption id="" align="aligncenter" width="244"]Figure 2. The Apple “iPhone” is a common smart phone with a GPS function. Physics describes the way that electricity flows through the circuits of this device. Engineers use their knowledge of physics to construct an iPhone with features that consumers will enjoy. One specific feature of an iPhone is the GPS function. GPS uses physics equations to determine the driving time between two locations on a map. (credit: @gletham GIS, Social, Mobile Tech Images).[/caption]

Applications of Physics

You need not be a scientist to use physics. On the contrary, knowledge of physics is useful in everyday situations as well as in nonscientific professions. It can help you understand how microwave ovens work, why metals should not be put into them, and why they might affect pacemakers. (See Figure 3 and Figure 4.) Physics allows you to understand the hazards of radiation and rationally evaluate these hazards more easily. Physics also explains the reason why a black car radiator helps remove heat in a car engine, and it explains why a white roof helps keep the inside of a house cool. Similarly, the operation of a car’s ignition system as well as the transmission of electrical signals through our body’s nervous system are much easier to understand when you think about them in terms of basic physics.

Physics is the foundation of many important disciplines and contributes directly to others. Chemistry, for example—since it deals with the interactions of atoms and molecules—is rooted in atomic and molecular physics. Most branches of engineering are applied physics. In architecture, physics is at the heart of structural stability, and is involved in the acoustics, heating, lighting, and cooling of buildings. Parts of geology rely heavily on physics, such as radioactive dating of rocks, earthquake analysis, and heat transfer in the Earth. Some disciplines, such as biophysics and geophysics, are hybrids of physics and other disciplines.

Physics has many applications in the biological sciences. On the microscopic level, it helps describe the properties of cell walls and cell membranes (Figure 5 and Figure 6). On the macroscopic level, it can explain the heat, work, and power associated with the human body. Physics is involved in medical diagnostics, such as x-rays, magnetic resonance imaging (MRI), and ultrasonic blood flow measurements. Medical therapy sometimes directly involves physics; for example, cancer radiotherapy uses ionizing radiation. Physics can also explain sensory phenomena, such as how musical instruments make sound, how the eye detects color, and how lasers can transmit information.

It is not necessary to formally study all applications of physics. What is most useful is knowledge of the basic laws of physics and a skill in the analytical methods for applying them. The study of physics also can improve your problem-solving skills. Furthermore, physics has retained the most basic aspects of science, so it is used by all of the sciences, and the study of physics makes other sciences easier to understand.

Models, Theories, and Laws; The Role of Experimentation

The laws of nature are concise descriptions of the universe around us; they are human statements of the underlying laws or rules that all natural processes follow. Such laws are intrinsic to the universe; humans did not create them and so cannot change them. We can only discover and understand them. Their discovery is a very human endeavour, with all the elements of mystery, imagination, struggle, triumph, and disappointment inherent in any creative effort. (See Figure 7 and Figure 8.) The cornerstone of discovering natural laws is observation; science must describe the universe as it is, not as we may imagine it to be.

[caption id="" align="aligncenter" width="290"]Figure 8. Marie Curie (1867–1934) sacrificed monetary assets to help finance her early research and damaged her physical well-being with radiation exposure. She is the only person to win Nobel prizes in both physics and chemistry. One of her daughters also won a Nobel Prize. (credit: Wikimedia Commons).[/caption]

We all are curious to some extent. We look around, make generalizations, and try to understand what we see—for example, we look up and wonder whether one type of cloud signals an oncoming storm. As we become serious about exploring nature, we become more organized and formal in collecting and analyzing data. We attempt greater precision, perform controlled experiments (if we can), and write down ideas about how the data may be organized and unified. We then formulate models, theories, and laws based on the data we have collected and analyzed to generalize and communicate the results of these experiments.

A model is a representation of something that is often too difficult (or impossible) to display directly. While a model is justified with experimental proof, it is only accurate under limited situations. An example is the planetary model of the atom in which electrons are pictured as orbiting the nucleus, analogous to the way planets orbit the Sun. (See Figure 9.) We cannot observe electron orbits directly, but the mental image helps explain the observations we can make, such as the emission of light from hot gases (atomic spectra). Physicists use models for a variety of purposes. For example, models can help physicists analyze a scenario and perform a calculation, or they can be used to represent a situation in the form of a computer simulation. A theory is an explanation for patterns in nature that is supported by scientific evidence and verified multiple times by various groups of researchers. Some theories include models to help visualize phenomena, whereas others do not. Newton’s theory of gravity, for example, does not require a model or mental image, because we can observe the objects directly with our own senses. The kinetic theory of gases, on the other hand, is a model in which a gas is viewed as being composed of atoms and molecules. Atoms and molecules are too small to be observed directly with our senses—thus, we picture them mentally to understand what our instruments tell us about the behavior of gases.

A law uses concise language to describe a generalized pattern in nature that is supported by scientific evidence and repeated experiments. Often, a law can be expressed in the form of a single mathematical equation. Laws and theories are similar in that they are both scientific statements that result from a tested hypothesis and are supported by scientific evidence. However, the designation law is reserved for a concise and very general statement that describes phenomena in nature, such as the law that energy is conserved during any process, or Newton’s second law of motion, which relates force, mass, and acceleration by the simple equation $$\vec{\text{F}} = m\vec{a}$$. A theory, in contrast, is a less concise statement of observed phenomena. For example, the Theory of Evolution and the Theory of Relativity cannot be expressed concisely enough to be considered a law. The biggest difference between a law and a theory is that a theory is much more complex and dynamic. A law describes a single action, whereas a theory explains an entire group of related phenomena. And, whereas a law is a postulate that forms the foundation of the scientific method, a theory is the end result of that process.

Less broadly applicable statements are usually called principles (such as Pascal’s principle, which is applicable only in fluids), but the distinction between laws and principles often is not carefully made.

MODELS, THEORIES, AND LAWS

Models, theories, and laws are used to help scientists analyze the data they have already collected. However, often after a model, theory, or law has been developed, it points scientists toward new discoveries they would not otherwise have made.

The models, theories, and laws we devise sometimes imply the existence of objects or phenomena as yet unobserved. These predictions are remarkable triumphs and tributes to the power of science. It is the underlying order in the universe that enables scientists to make such spectacular predictions. However, if experiment does not verify our predictions, then the theory or law is wrong, no matter how elegant or convenient it is. Laws can never be known with absolute certainty because it is impossible to perform every imaginable experiment in order to confirm a law in every possible scenario. Physicists operate under the assumption that all scientific laws and theories are valid until a counterexample is observed. If a good-quality, verifiable experiment contradicts a well-established law, then the law must be modified or overthrown completely.

The study of science in general and physics in particular is an adventure much like the exploration of uncharted ocean. Discoveries are made; models, theories, and laws are formulated; and the beauty of the physical universe is made more sublime for the insights gained.

THE SCIENTIFIC METHOD

As scientists inquire and gather information about the world, they follow a process called the scientific method. This process typically begins with an observation and question that the scientist will research. Next, the scientist typically performs some research about the topic and then devises a hypothesis. Then, the scientist will test the hypothesis by performing an experiment. Finally, the scientist analyzes the results of the experiment and draws a conclusion. Note that the scientific method can be applied to many situations that are not limited to science, and this method can be modified to suit the situation.

Consider an example. Let us say that you try to turn on your car, but it will not start. You undoubtedly wonder: Why will the car not start? You can follow a scientific method to answer this question. First off, you may perform some research to determine a variety of reasons why the car will not start. Next, you will state a hypothesis. For example, you may believe that the car is not starting because it has no engine oil. To test this, you open the hood of the car and examine the oil level. You observe that the oil is at an acceptable level, and you thus conclude that the oil level is not contributing to your car issue. To troubleshoot the issue further, you may devise a new hypothesis to test and then repeat the process again.

The Evolution of Natural Philosophy into Modern Physics

Physics was not always a separate and distinct discipline. It remains connected to other sciences to this day. The word physics comes from Greek, meaning nature. The study of nature came to be called “natural philosophy.” From ancient times through the Renaissance, natural philosophy encompassed many fields, including astronomy, biology, chemistry, physics, mathematics, and medicine. Over the last few centuries, the growth of knowledge has resulted in ever-increasing specialization and branching of natural philosophy into separate fields, with physics retaining the most basic facets. (See Figure 10, Figure 11, and Figure 12.) Physics as it developed from the Renaissance to the end of the 19th century is called classical physics. It was transformed into modern physics by revolutionary discoveries made starting at the beginning of the 20th century.

[caption id="" align="aligncenter" width="299"]Figure 10. Over the centuries, natural philosophy has evolved into more specialized disciplines, as illustrated by the contributions of some of the greatest minds in history. The Greek philosopher Aristotle (384–322 B.C.) wrote on a broad range of topics including physics, animals, the soul, politics, and poetry. (credit: Jastrow (2006)/Ludovisi Collection).[/caption]

[caption id="" align="aligncenter" width="297"]Figure 11. Galileo Galilei (1564–1642) laid the foundation of modern experimentation and made contributions in mathematics, physics, and astronomy. (credit: Domenico Tintoretto).[/caption]

[caption id="" align="aligncenter" width="337"]Figure 12. Niels Bohr (1885–1962) made fundamental contributions to the development of quantum mechanics, one part of modern physics. (credit: United States Library of Congress Prints and Photographs Division).[/caption]

Classical physics is not an exact description of the universe, but it is an excellent approximation under the following conditions: Matter must be moving at speeds less than about 1% of the speed of light, the objects dealt with must be large enough to be seen with a microscope, and only weak gravitational fields, such as the field generated by the Earth, can be involved. Because humans live under such circumstances, classical physics seems intuitively reasonable, while many aspects of modern physics seem bizarre. This is why models are so useful in modern physics—they let us conceptualize phenomena we do not ordinarily experience. We can relate to models in human terms and visualize what happens when objects move at high speeds or imagine what objects too small to observe with our senses might be like. For example, we can understand an atom’s properties because we can picture it in our minds, although we have never seen an atom with our eyes. New tools, of course, allow us to better picture phenomena we cannot see. In fact, new instrumentation has allowed us in recent years to actually “picture” the atom.

LIMITS ON THE LAWS OF CLASSICAL PHYSICS

For the laws of classical physics to apply, the following criteria must be met: Matter must be moving at speeds less than about 1% of the speed of light, the objects dealt with must be large enough to be seen with a microscope, and only weak gravitational fields (such as the field generated by the Earth) can be involved.

[caption id="" align="aligncenter" width="398"]Figure 13. Using a scanning tunneling microscope (STM), scientists can see the individual atoms that compose this sheet of gold. (credit: Erwinrossen).[/caption]

Some of the most spectacular advances in science have been made in modern physics. Many of the laws of classical physics have been modified or rejected, and revolutionary changes in technology, society, and our view of the universe have resulted. Like science fiction, modern physics is filled with fascinating objects beyond our normal experiences, but it has the advantage over science fiction of being very real. Why, then, is the majority of this text devoted to topics of classical physics? There are two main reasons: Classical physics gives an extremely accurate description of the universe under a wide range of everyday circumstances, and knowledge of classical physics is necessary to understand modern physics.

Modern physics itself consists of the two revolutionary theories, relativity and quantum mechanics. These theories deal with the very fast and the very small, respectively. Relativity must be used whenever an object is traveling at greater than about 1% of the speed of light or experiences a strong gravitational field such as that near the Sun. Quantum mechanics must be used for objects smaller than can be seen with a microscope. The combination of these two theories is relativistic quantum mechanics, and it describes the behavior of small objects traveling at high speeds or experiencing a strong gravitational field. Relativistic quantum mechanics is the best universally applicable theory we have. Because of its mathematical complexity, it is used only when necessary, and the other theories are used whenever they will produce sufficiently accurate results. We will find, however, that we can do a great deal of modern physics with the algebra and trigonometry used in this text.

Check Your Understanding

1: A friend tells you he has learned about a new law of nature. What can you know about the information even before your friend describes the law? How would the information be different if your friend told you he had learned about a scientific theory rather than a law?

PHET EXPLORATIONS: EQUATION GRAPHER

Learn about graphing polynomials. The shape of the curve changes as the constants are adjusted. View the curves for the individual terms (e.g. y=bx) to see how they add to generate the polynomial curve.

[caption id="" align="aligncenter" width="450"] Figure 14. Equation Grapher.[/caption]

Summary

• Science seeks to discover and describe the underlying order and simplicity in nature.
• Physics is the most basic of the sciences, concerning itself with energy, matter, space and time, and their interactions.
• Scientific laws and theories express the general truths of nature and the body of knowledge they encompass. These laws of nature are rules that all natural processes appear to follow.

Conceptual Questions

1: Models are particularly useful in relativity and quantum mechanics, where conditions are outside those normally encountered by humans. What is a model?

2: How does a model differ from a theory?

3: If two different theories describe experimental observations equally well, can one be said to be more valid than the other (assuming both use accepted rules of logic)?

4: What determines the validity of a theory?

5: Certain criteria must be satisfied if a measurement or observation is to be believed. Will the criteria necessarily be as strict for an expected result as for an unexpected result?

6: Can the validity of a model be limited, or must it be universally valid? How does this compare to the required validity of a theory or a law?

7: Classical physics is a good approximation to modern physics under certain circumstances. What are they?

8: When is it necessary to use relativistic quantum mechanics?

9: Can classical physics be used to accurately describe a satellite moving at a speed of 7500 m/s? Explain why or why not.

 Glossary

classical physics

physics that was developed from the Renaissance to the end of the 19th century

physics

the science concerned with describing the interactions of energy, matter, space, and time; it is especially interested in what fundamental mechanisms underlie every phenomenon

model

representation of something that is often to difficult (or impossible) to display directly

theory

an explanation for patterns in nature that is supported by scientific evidence and verified multiple times by various groups of researchers

law

a description, using concise language or a mathematical formula, a generalized pattern in nature that is supported by scientific evidence and repeated examples

scientific method

a method that typically begins with an observation and question that the scientist will research; next, the scientist typically performs some research about the topic and then devises a hypothesis; then, the scientist will test the hypothesis by performing an experiment; finally, the scientist analyzes the results of the experiment and draws a conclusion

modern physics

the study of relativity, quantum mechanics, or both

relativity

the study of objects moving at speeds greater than about 1% of the speed of light, or of objects being affected by a strong gravitational field

quantum mechanics

the study of objects smaller than can be seen with a microscope

Solutions

Check Your Understanding 1: Without knowing the details of the law, you can still infer that the information your friend has learned conforms to the requirements of all laws of nature: it will be a concise description of the universe around us; a statement of the underlying rules that all natural processes follow. If the information had been a theory, you would be able to infer that the information will be a large-scale, broadly applicable generalization.

]]> 40 23 2 0 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/chapter/1-2-physical-quantities-and-units/ Thu, 29 Jun 2017 22:12:58 +0000 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/chapter/1-2-physical-quantities-and-units/

Vectors in Two Dimensions

A vector is a quantity that has magnitude and direction. Displacement, velocity, acceleration, and force, for example, are all vectors. In one-dimensional, or straight-line, motion, the direction of a vector can be given simply by a plus or minus sign. In two dimensions (2-d), however, we specify the direction of a vector relative to some reference frame (i.e., coordinate system), using an arrow having length proportional to the vector’s magnitude and pointing in the direction of the vector.

Figure 2 shows such a graphical representation of a vector, using as an example the total displacement for the person walking in a city considered in Chapter 3.1 Kinematics in Two Dimensions: An Introduction. We shall use the notation that a symbol with an arrow over it, such as $$\vec{\text{D}}$$, stands for a vector. Its magnitude is represented by the symbol in italics, D, and its direction by θ.

VECTORS IN THIS TEXT

In this text, we will represent a vector with an arrow over a symbol. For example, we will represent the quantity force with the vector $$\vec{\text{F}}$$, which has both magnitude and direction. The magnitude of the vector will be represented by a variable in italics, such as F, and the direction of the variable will be given by an angle θ.

[caption id="" align="aligncenter" width="325"]Figure 2. A person walks 9 blocks east and 5 blocks north. The displacement is 10.3 blocks at an angle 29.1^o north of east.[/caption]

[caption id="" align="aligncenter" width="333"]Figure 3. To describe the resultant vector for the person walking in a city considered in Figure 2 graphically, draw an arrow to represent the total displacement vector D. Using a protractor, draw a line at an angle θ relative to the east-west axis. The length D of the arrow is proportional to the vector’s magnitude and is measured along the line with a ruler. In this example, the magnitude D of the vector is 10.3 units, and the direction θ is 29.1^o north of east.[/caption]

Vector Addition: Head-to-Tail Method

The head-to-tail method is a graphical way to add vectors, described in Figure 4 below and in the steps following. The tail of the vector is the starting point of the vector, and the head (or tip) of a vector is the final, pointed end of the arrow.

Step 1. Draw an arrow to represent the first vector (9 blocks to the east) using a ruler and protractor. [caption id="" align="aligncenter" width="344"]Figure 5.[/caption] Step 2. Now draw an arrow to represent the second vector (5 blocks to the north). Place the tail of the second vector at the head of the first vector. [caption id="" align="aligncenter" width="343"]Figure 6.[/caption]

Step 3. If there are more than two vectors, continue this process for each vector to be added. Note that in our example, we have only two vectors, so we have finished placing arrows tip to tail.

Step 4. Draw an arrow from the tail of the first vector to the head of the last vector. This is the resultant, or the sum, of the other vectors. [caption id="" align="aligncenter" width="200"]Figure 7.[/caption]

Step 5. To get the magnitude of the resultant, measure its length with a ruler. (Note that in most calculations, we will use the Pythagorean theorem to determine this length.)

Step 6. To get the direction of the resultant, measure the angle it makes with the reference frame using a protractor. (Note that in most calculations, we will use trigonometric relationships to determine this angle.)

The graphical addition of vectors is limited in accuracy only by the precision with which the drawings can be made and the precision of the measuring tools. It is valid for any number of vectors.

Example 1: Adding Vectors Graphically Using the Head-to-Tail Method: A Women Takes a Walk

Use the graphical technique for adding vectors to find the total displacement of a person who walks the following three paths (displacements) on a flat field. First, she walks 25.0 m in a direction 49.0° north of east. Then, she walks 23.0 m heading 15.0° north of east. Finally, she turns and walks 32.0 m in a direction 68.0° south of east.

Strategy

Represent each displacement vector graphically with an arrow, labeling the first $$\vec{\text{A}}$$, the second $$\vec{\text{B}}$$, and the third $$\vec{\text{C}}$$, making the lengths proportional to the distance and the directions as specified relative to an east-west line. The head-to-tail method outlined above will give a way to determine the magnitude and direction of the resultant displacement, denoted $$\vec{\text{R}}$$.

Solution

(1) Draw the three displacement vectors. [caption id="" align="aligncenter" width="500"]Figure 8.[/caption] (2) Place the vectors head to tail retaining both their initial magnitude and direction. [caption id="" align="aligncenter" width="250"]Figure 9.[/caption] (3) Draw the resultant vector, $$\vec{\text{R}}$$. [caption id="" align="aligncenter" width="250"]Figure 10.[/caption] (4) Use a ruler to measure the magnitude of $$\vec{\text{R}}$$ and a protractor to measure the direction of $$\vec{\text{R}}$$. While the direction of the vector can be specified in many ways, the easiest way is to measure the angle between the vector and the nearest horizontal or vertical axis. Since the resultant vector is south of the eastward pointing axis, we flip the protractor upside down and measure the angle between the eastward axis and the vector. [caption id="" align="aligncenter" width="220"]Figure 11.[/caption]

In this case, the total displacement R is seen to have a magnitude of 50.0 m and to lie in a direction 7.0° south of east. By using its magnitude and direction, this vector can be expressed as R = 50.0 m and θ = 7.0° south of east.

Discussion

The head-to-tail graphical method of vector addition works for any number of vectors. It is also important to note that the resultant is independent of the order in which the vectors are added. Therefore, we could add the vectors in any order as illustrated in Figure 12 and we will still get the same solution. [caption id="" align="aligncenter" width="275"]Figure 12.[/caption]

Here, we see that when the same vectors are added in a different order, the result is the same. This characteristic is true in every case and is an important characteristic of vectors. Vector addition is commutative. Vectors can be added in any order.

[latex]\vec{\text{A}}+\vec{\text{B}}=\vec{\text{B}}+\vec{\text{A}}.[/latex]

(This is true for the addition of ordinary numbers as well—you get the same result whether you add 2+3 or 3+2, for example).

Vector Subtraction

Vector subtraction is a straightforward extension of vector addition. To define subtraction (say we want to subtract $$\vec{\text{B}}$$ from $$\vec{\text{A}}$$, written $$\vec{\text{A}}-\vec{\text{B}}$$, we must first define what we mean by subtraction. The negative of a vector $$\vec{\text{B}}$$ is defined to be $$-\vec{\text{B}}$$; that is, graphically the negative of any vector has the same magnitude but the opposite direction, as shown in Figure 13. In other words, $$\vec{\text{B}}$$ has the same length as $$-\vec{\text{B}}$$, but points in the opposite direction. Essentially, we just flip the vector so it points in the opposite direction.

The subtraction of vector $$\vec{\text{B}}$$ from vector $$\vec{\text{A}}$$ is then simply defined to be the addition of $$-\vec{\text{B}}$$ to $$\vec{\text{A}}$$. Note that vector subtraction is the addition of a negative vector. The order of subtraction does not affect the results.

$$\vec{\text{A}}-\vec{\text{B}}=\vec{\text{A}}+(-\vec{\text{B}})$$

This is analogous to the subtraction of scalars (where, for example, 5-2=5+(-2)). Again, the result is independent of the order in which the subtraction is made. When vectors are subtracted graphically, the techniques outlined above are used, as the following example illustrates.

Example 2: Subtracting Vectors Graphically: A Women Sailing a Boat

A woman sailing a boat at night is following directions to a dock. The instructions read to first sail 27.5 m in a direction 66.0° north of east from her current location, and then travel 30.0 m in a direction 112° north of east (or 22.0° west of north). If the woman makes a mistake and travels in the opposite direction for the second leg of the trip, where will she end up? Compare this location with the location of the dock. [caption id="" align="aligncenter" width="450"]Figure 14.[/caption]

Strategy

We can represent the first leg of the trip with a vector $$\vec{\text{A}}$$, and the second leg of the trip with a vector $$\vec{\text{B}}$$. The dock is located at a location $$\vec{\text{A}}+\vec{\text{B}}$$. If the woman mistakenly travels in the opposite direction for the second leg of the journey, she will travel a distance $$\vec{\text{B}}$$ (30.0 m) in the direction 180°-112°=68° south of east. We represent this as $$-\vec{\text{B}}$$, as shown below. The vector $$-\vec{\text{B}}$$ has the same magnitude as $$\vec{\text{B}}$$ but is in the opposite direction. Thus, she will end up at a location $$\vec{\text{A}}+(-\vec{\text{B}})$$, or $$\vec{\text{A}}-\vec{\text{B}}$$. [caption id="" align="aligncenter" width="200"]Figure 15.[/caption]

We will perform vector addition to compare the location of the dock, $$\vec{\text{A}}+\vec{\text{B}}$$, with the location at which the woman mistakenly arrives, $$\vec{\text{A}}+(-\vec{\text{B}})$$.

Solution

(1) To determine the location at which the woman arrives by accident, draw vectors $$\vec{\text{A}}$$ and $$-\vec{\text{B}}$$.

(2) Place the vectors head to tail.

(3) Draw the resultant vector $$\vec{\text{R}}$$.

(4) Use a ruler and protractor to measure the magnitude and direction of $$\vec{\text{R}}$$. [caption id="" align="aligncenter" width="300"]Figure 16. [/caption]

In this case, R=23.0 m and θ=7.5° south of east.

(5) To determine the location of the dock, we repeat this method to add vectors $$\vec{\text{A}}$$ and $$\vec{\text{B}}$$. We obtain the resultant vector $$\vec{\text{R}}^\prime$$: [caption id="figure17" align="aligncenter" width="250"]Figure 17.[/caption]

In this case R=52.9 m and θ=90.1° north of east.

We can see that the woman will end up a significant distance from the dock if she travels in the opposite direction for the second leg of the trip.

Discussion

Because subtraction of a vector is the same as addition of a vector with the opposite direction, the graphical method of subtracting vectors works the same as for addition.

Multiplication of Vectors and Scalars

If we decided to walk three times as far on the first leg of the trip considered in the preceding example, then we would walk 3 × 27.5 m, or 82.5 m, in a direction 66.0° north of east. This is an example of multiplying a vector by a positive scalar. Notice that the magnitude changes, but the direction stays the same.

If the scalar is negative, then multiplying a vector by it changes the vector’s magnitude and gives the new vector the opposite direction. For example, if you multiply by –2, the magnitude doubles but the direction changes. We can summarize these rules in the following way: When vector $$\vec{\text{A}}$$ is multiplied by a scalar c,

• the magnitude of the vector becomes the absolute value of c$$\vec{\text{A}}$$,
• if c is positive, the direction of the vector does not change,
• if c is negative, the direction is reversed.

In our case, c=3 and A=27.5 m. Vectors are multiplied by scalars in many situations. Note that division is the inverse of multiplication. For example, dividing by 2 is the same as multiplying by the value (1/2). The rules for multiplication of vectors by scalars are the same for division; simply treat the divisor as a scalar between 0 and 1.

Resolving a Vector into Components

In the examples above, we have been adding vectors to determine the resultant vector. In many cases, however, we will need to do the opposite. We will need to take a single vector and find what other vectors added together produce it. In most cases, this involves determining the perpendicular components of a single vector, for example the x- and y-components, or the north-south and east-west components.

For example, we may know that the total displacement of a person walking in a city is 10.3 blocks in a direction 29.0° north of east and want to find out how many blocks east and north had to be walked. This method is called finding the components (or parts) of the displacement in the east and north directions, and it is the inverse of the process followed to find the total displacement. It is one example of finding the components of a vector. There are many applications in physics where this is a useful thing to do. We will see this soon in Chapter 3.4 Projectile Motion, and much more when we cover forces in Chapter 4 Dynamics: Newton’s Laws of Motion. Most of these involve finding components along perpendicular axes (such as north and east), so that right triangles are involved. The analytical techniques presented in Chapter 3.3 Vector Addition and Subtraction: Analytical Methods are ideal for finding vector components.

PHET EXPLORATIONS: MAZE GAME

Learn about position, velocity, and acceleration in the "Arena of Pain". Use the green arrow to move the ball. Add more walls to the arena to make the game more difficult. Try to make a goal as fast as you can.

[caption id="" align="aligncenter" width="450"] Figure 18. Maze Game[/caption]

Summary

• The graphical method of adding vectors $$\vec{\text{A}}$$ and $$\vec{\text{B}}$$ involves drawing vectors on a graph and adding them using the head-to-tail method. The resultant vector $$\vec{\text{R}}$$ is defined such that $$\vec{\text{A}}+\vec{\text{B}}=\vec{\text{R}}$$. The magnitude and direction of $$\vec{\text{R}}$$ are then determined with a ruler and protractor, respectively.
• The graphical method of subtracting vector $$\vec{\text{B}}$$ from $$\vec{\text{A}}$$ involves adding the opposite of vector $$\vec{\text{B}}$$, which is defined as $$-\vec{\text{B}}$$. In this case, $$\vec{\text{A}}-\vec{\text{B}}=\vec{\text{A}}+(-\vec{\text{B}})=\vec{\text{R}}$$. Then, the head-to-tail method of addition is followed in the usual way to obtain the resultant vector $$\vec{\text{R}}$$.
• Addition of vectors is commutative such that $$\vec{\text{A}}+\vec{\text{B}}=\vec{\text{B}}+\vec{\text{A}}$$.
• The head-to-tail method of adding vectors involves drawing the first vector on a graph and then placing the tail of each subsequent vector at the head of the previous vector. The resultant vector is then drawn from the tail of the first vector to the head of the final vector.
• If a vector $$\vec{\text{A}}$$ is multiplied by a scalar quantity c, the magnitude of the product is given by c$$\vec{\text{A}}$$. If c is positive, the direction of the product points in the same direction as $$\vec{\text{A}}$$; if c is negative, the direction of the product points in the opposite direction as$$\vec{\text{A}}$$.

Conceptual Questions

1: Which of the following is a vector: a person’s height, the altitude on Mt. Everest, the age of the Earth, the boiling point of water, the cost of this book, the Earth’s population, the acceleration of gravity?

2: Give a specific example of a vector, stating its magnitude, units, and direction.

3: What do vectors and scalars have in common? How do they differ?

4: Two campers in a national park hike from their cabin to the same spot on a lake, each taking a different path, as illustrated below. The total distance traveled along Path 1 is 7.5 km, and that along Path 2 is 8.2 km. What is the final displacement of each camper?

[caption id="" align="aligncenter" width="275"]Figure 19.[/caption]

5: If an airplane pilot is told to fly 123 km in a straight line to get from San Francisco to Sacramento, explain why he could end up anywhere on the circle shown in Figure 20. What other information would he need to get to Sacramento?

[caption id="" align="aligncenter" width="374"]Figure 20.[/caption]

6: Suppose you take two steps A and B (that is, two nonzero displacements). Under what circumstances can you end up at your starting point? More generally, under what circumstances can two nonzero vectors add to give zero? Is the maximum distance you can end up from the starting point A+B the sum of the lengths of the two steps?

7: Explain why it is not possible to add a scalar to a vector.

8: If you take two steps of different sizes, can you end up at your starting point? More generally, can two vectors with different magnitudes ever add to zero? Can three or more?

Problems & Exercises

Use graphical methods to solve these problems. You may assume data taken from graphs is accurate to three digits.

1: Find the following for path A in Figure 21: (a) the total distance traveled, and (b) the magnitude and direction of the displacement from start to finish.

2: Find the following for path B in Figure 21: (a) the total distance traveled, and (b) the magnitude and direction of the displacement from start to finish.

3: Find the north and east components of the displacement for the hikers shown in Figure 19.

4: Suppose you walk 18.0 m straight west and then 25.0 m straight north. How far are you from your starting point, and what is the compass direction of a line connecting your starting point to your final position? (If you represent the two legs of the walk as vector displacements $$\vec{\text{A }}$$ and $$\vec{\text{B}}$$, as in Figure 22, then this problem asks you to find their sum $$\vec{\text{R}}=\vec{\text{A}}+\vec{\text{B}}$$.)

5: Suppose you first walk 12.0 m in a direction 20° west of north and then 20.0 m in a direction 40.0° south of west. How far are you from your starting point, and what is the compass direction of a line connecting your starting point to your final position? (If you represent the two legs of the walk as vector displacements $$\vec{\text{A}}$$ and $$\vec{\text{B}}$$, as in Figure 23, then this problem finds their sum $$\vec{\text{R}}=\vec{\text{A}}+\vec{\text{B}}$$.)

[caption id="" align="aligncenter" width="250"]Figure 23.[/caption]

6: Repeat the problem above, but reverse the order of the two legs of the walk; show that you get the same final result. That is, you first walk leg $$\vec{\text{B}}$$, which is 20.0 m in a direction exactly 40° south of west, and then leg $$\vec{\text{A}}$$, which is 12.0 m in a direction exactly 20° west of north. (This problem shows that $$\vec{\text{A}}+\vec{\text{B}}=\vec{\text{B}}+\vec{\text{A}}$$.)

7: (a) Repeat the problem two problems prior, but for the second leg you walk 20.0 m in a direction 40.0° north of east (which is equivalent to subtracting $$\vec{\text{B}}$$ from $$\vec{\text{A}}$$ —that is, to finding $$\vec{\text{R}}^\prime=\vec{\text{A}}-\vec{\text{B}}$$). (b) Repeat the problem two problems prior, but now you first walk 20.0 m in a direction 40.0° south of west and then 12.0 m in a direction 20.0° east of south (which is equivalent to subtracting $$\vec{\text{A}}$$ from $$\vec{\text{B}}$$—that is, to finding $$\vec{\text{R}}^{\prime\prime}=\vec{\text{B}}-\vec{\text{A}}=-\vec{\text{R}}^{\prime}$$). Show that this is the case.

8: Show that the order of addition of three vectors does not affect their sum. Show this property by choosing any three vectors $$\vec{\text{A}}$$, $$\vec{\text{B}}$$, and $$\vec{\text{C}}$$, all having different lengths and directions. Find the sum $$\vec{\text{A}}+\vec{\text{B}}+\vec{\text{C}}$$ then find their sum when added in a different order and show the result is the same. (There are five other orders in which $$\vec{\text{A}}$$, $$\vec{\text{B}}$$, and $$\vec{\text{C}}$$ can be added; choose only one.)

9: Show that the sum of the vectors discussed in Example 2 gives the result shown in Figure 17.

10: Find the magnitudes of velocities v_A and v_B in Figure 24

11: Find the components of v_tot along the x- and y-axes in Figure 24.

12: Find the components of v_tot along a set of perpendicular axes rotated 30° counterclockwise relative to those in Figure 24.

Glossary

component (of a 2-d vector)

a piece of a vector that points in either the vertical or the horizontal direction; every 2-d vector can be expressed as a sum of two vertical and horizontal vector components

commutative

refers to the interchangeability of order in a function; vector addition is commutative because the order in which vectors are added together does not affect the final sum

direction (of a vector)

the orientation of a vector in space

head (of a vector)

the end point of a vector; the location of the tip of the vector’s arrowhead; also referred to as the “tip”

head-to-tail method

a method of adding vectors in which the tail of each vector is placed at the head of the previous vector

magnitude (of a vector)

the length or size of a vector; magnitude is a scalar quantity

resultant

the sum of two or more vectors

resultant vector

the vector sum of two or more vectors

scalar

a quantity with magnitude but no direction

tail

the start point of a vector; opposite to the head or tip of the arrow

Solutions

Problems & Exercises

1: (a) [latex]\boldsymbol{480\textbf{ m}}[/latex] (b) [latex]\boldsymbol{379\textbf{ m}},\boldsymbol{18.4^o}[/latex] east of north

3: north component $$\boldsymbol{3.21\textbf{ km}}$$, east component $$\boldsymbol{3.83\textbf{ km}}$$ 5: [latex]\boldsymbol{19.5\textbf{ m}},\boldsymbol{4.65^o}[/latex] south of west

7: (a) [latex]\boldsymbol{26.6\textbf{ m}},\boldsymbol{65.1^o}[/latex] north of east (b) [latex]\boldsymbol{26.6\textbf{ m}},\boldsymbol{65.1^o}[/latex] south of west

9: [latex]\boldsymbol{52.9\textbf{ m}},\boldsymbol{90.1^o}[/latex] with respect to the x-axis.

11: x-component $$\boldsymbol{4.41\textbf{ m/s}}$$, y-component $$\boldsymbol{5.07\textbf{ m/s}}$$

]]> 228 194 3 0 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/chapter/3-3-vector-addition-and-subtraction-analytical-methods/ Thu, 29 Jun 2017 22:13:23 +0000 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/chapter/3-3-vector-addition-and-subtraction-analytical-methods/

Summary

• Understand the rules of vector addition and subtraction using analytical methods.
• Apply analytical methods to determine vertical and horizontal component vectors.
• Apply analytical methods to determine the magnitude and direction of a resultant vector.

Analytical methods of vector addition and subtraction employ geometry and simple trigonometry rather than the ruler and protractor of graphical methods. Part of the graphical technique is retained, because vectors are still represented by arrows for easy visualization. However, analytical methods are more concise, accurate, and precise than graphical methods, which are limited by the accuracy with which a drawing can be made. Analytical methods are limited only by the accuracy and precision with which physical quantities are known.

Resolving a Vector into Perpendicular Components

Analytical techniques and right triangles go hand-in-hand in physics because (among other things) motions along perpendicular directions are independent. We very often need to separate a vector into perpendicular components. For example, given a vector like $$\vec{\text{A}}$$ in Figure 1, we may wish to find which two perpendicular vectors, $$\text{A}_x$$ and $$\text{A}_y$$, add to produce it. Notice these perpendicular component vectors will not have vector arrows in this text.

$$\text{A}_x$$ and $$\text{A}_y$$ are defined to be the components of $$\vec{\text{A}}$$ along the x- and y-axes. The three vectors $$\vec{\text{A}}$$, $$\text{A}_x$$, and $$\text{A}_y$$ form a right triangle:

[latex]\boldsymbol{\textbf{A}_x +\textbf{A}_y =\vec{\textbf{A}}.}[/latex]

Note that this relationship between vector components and the resultant vector holds only for vector quantities (which include both magnitude and direction). The relationship does not apply for the magnitudes alone. For example, if $$\text{A}_x=3\text{ m}$$ east, $$\text{A}_y=4\text{ m}$$ north, and $$\vec{\text{A}}=5\text{ m}$$ north-east, then it is true that the vectors $$\text{A}_x+\text{A}_y=\vec{\text{A}}$$. However, it is not true that the sum of the magnitudes of the vectors is also equal. That is,

[latex]\boldsymbol{3\textbf{ m}+4\textbf{ m}\neq 5\textbf{ m}}[/latex]

Thus,

[latex]\boldsymbol{\textbf{A}_x+\textbf{A}_y\neq\vec{\textbf{A}}}[/latex]

If the vector $$\vec{\text{A}}$$ is known, then its magnitude A and its angle θ (its direction) are known. To find $$\text{A}_x$$ and $$\text{A}_y$$, its x- and y-components, we use the following relationships for a right triangle.

[latex]\boldsymbol{\textbf{A}_x=A \cos\:\theta}[/latex]

and

[latex]\boldsymbol{\textbf{A}_y=A \sin\:\theta}.[/latex]

Suppose, for example, that $$\vec{\text{A}}$$ is the vector representing the total displacement of the person walking in a city considered in Chapter 3.1 Kinematics in Two Dimensions: An Introduction and Chapter 3.2 Vector Addition and Subtraction: Graphical Methods.

Then A=10.3 blocks and θ=29.1°, so that

[latex]\boldsymbol{\textbf{A}_x=\textbf{A cos}\:\theta=(10.3\textbf{ blocks})(\textbf{cos}29.1^o)=9.0\textbf{ blocks}}[/latex]

[latex]\boldsymbol{\textbf{A}_y=\textbf{A sin}\:\theta=(10.3\textbf{ blocks})(\textbf{sin}29.1^o)=5.0\textbf{ blocks}.}[/latex]

Calculating a Resultant Vector

If the perpendicular components $$\text{A}_x$$ and $$\text{A}_y$$ of a vector $$\vec{\text{A}}$$ are known, then $$\vec{\text{A}}$$ can also be found analytically. To find the magnitude A and direction θ of a vector from its perpendicular components $$\text{A}_x$$ and $$\text{A}_y$$, we use the following relationships:

[latex]\boldsymbol{A=\sqrt{\textbf{A}_x\:^2\:+\:\textbf{A}_y\:^2}}[/latex]

[latex]\boldsymbol{\theta=\textbf{tan}^{-1}(\textbf{A}_y\:/\:\textbf{A}_x).}[/latex]

Note that the equation [latex]A=\sqrt{\text{A}_x\:^2\:+\:\text{A}_y\:^2}[/latex] is just the Pythagorean theorem relating the legs of a right triangle to the length of the hypotenuse. For example, if $$\text{A}_x$$ and $$\text{A}_y$$ are 9 and 5 blocks, respectively, then [latex]A=\sqrt{9^2\:+\:5^2}=10.3[/latex] blocks, again consistent with the example of the person walking in a city. Finally, the direction is $$\theta=\tan^{-1} (5/9)=29.1^0$$, as before.

DETERMINING VECTORS AND VECTOR COMPONENTS WITH ANALYTICAL METHODS

Equations $$\boldsymbol{\textbf{A}_x=A \cos \theta}$$ and $$\boldsymbol{\textbf{A}_y=A \sin \theta}$$ are used to find the perpendicular components of a vector—that is, to go from A and θ to $$\boldsymbol{\textbf{A}_x}$$ and $$\boldsymbol{\textbf{A}_y}$$. Equations [latex]\boldsymbol{A=\sqrt{{\textbf{A}_x}^2\:+\:{\textbf{A}_y}^2}}[/latex] and $$\boldsymbol{\theta = \tan^{-1} \left( \textbf{A}_y\: / \: \textbf{A}_x \right) }$$ are used to find a vector from its perpendicular components—that is, to go from $$\boldsymbol{\textbf{A}_x}$$ and $$\boldsymbol{\textbf{A}_y}$$ to A and θ. Both processes are crucial to analytical methods of vector addition and subtraction.

Adding Vectors Using Analytical Methods

To see how to add vectors using perpendicular components, consider Figure 5, in which the vectors $$\vec{\text{A}}$$ and $$\vec{\text{B}}$$ are added to produce the resultant $$\vec{\text{R}}$$.

If $$\vec{\text{A}}$$ and $$\vec{\text{B}}$$ represent two legs of a walk (two displacements), then $$\vec{\text{R}}$$ is the total displacement. The person taking the walk ends up at the tip of $$\vec{\text{R}}$$. There are many ways to arrive at the same point. In particular, the person could have walked first in the x-direction and then in the y-direction. Those paths are the x- and y-components of the resultant, $$\text{R}_x$$ and $$\text{R}_y$$. If we know $$\text{R}_x$$ and $$\text{R}_y$$, we can find R and θ using the equations [latex]\boldsymbol{A=\sqrt{{\textbf{A}_x}^2\:+\:{\textbf{A}_y}^2}}[/latex] and $$\boldsymbol{\theta=\tan^{-1} \left( \text{A}_y\: / \: \text{A}_x \right)}$$. When you use the analytical method of vector addition, you can determine the components or the magnitude and direction of a vector. Step 1. Identify the x- and y-axes that will be used in the problem. Then, find the components of each vector to be added along the chosen perpendicular axes. Use the equations $$\text{A}_x=A \cos \theta$$ and $$\text{A}_y=A \sin \theta$$ to find the components. In Figure 6, these components are $$\text{A}_x$$, $$\text{A}_y$$, $$\text{B}_x$$, and $$\text{B}_y$$. The angles that vectors $$\vec{\text{A}}$$ and $$\vec{\text{B}}$$ make with the x-axis are θ_A and θ_B, respectively.

Step 2. Find the components of the resultant along each axis by adding the components of the individual vectors along that axis. That is, as shown in Figure 7,

[latex]\boldsymbol{\textbf{R}_x=\textbf{A}_x+\textbf{B}_x}[/latex]

and

[latex]\boldsymbol{\textbf{R}_y=\textbf{A}_y+\textbf{B}_y.}[/latex]

Components along the same axis, say the x-axis, are vectors along the same line and, thus, can be added to one another like ordinary numbers. The same is true for components along the y-axis. (For example, a 9-block eastward walk could be taken in two legs, the first 3 blocks east and the second 6 blocks east, for a total of 9, because they are along the same direction.) So resolving vectors into components along common axes makes it easier to add them. Now that the components of $$\vec{\text{R}}$$ are known, its magnitude and direction can be found.

Step 3. To get the magnitude R of the resultant, use the Pythagorean theorem:

[latex]\boldsymbol{ \textbf{R}=\sqrt{ {\textbf{R}_x}^2\:+\:{\textbf{R}_y}^2 }. }[/latex]

Step 4. To get the direction of the resultant:

[latex]\boldsymbol{\theta=\textbf{tan}^{-1}(\textbf{R}_y\:/\:\textbf{R}_x).}[/latex]

The following example illustrates this technique for adding vectors using perpendicular components.

Example 1: Adding Vectors Using Analytical Methods

Add the vector $$\vec{\textbf{A}}$$ to the vector $$\vec{\textbf{B}}$$ shown in Figure 8, using perpendicular components along the x- and y-axes. The x- and y-axes are along the east–west and north–south directions, respectively. Vector $$\vec{\textbf{A}}$$ represents the first leg of a walk in which a person walks 53.0 m in a direction 20.0° north of east. Vector $$\vec{\textbf{B}}$$ represents the second leg, a displacement of 34.0 m in a direction 63.0° north of east.

Strategy

The components of $$\vec{\textbf{A}}$$ and $$\vec{\textbf{B}}$$ along the x- and y-axes represent walking due east and due north to get to the same ending point. Once found, they are combined to produce the resultant.

Solution

Following the method outlined above, we first find the components of $$\vec{\textbf{A}}$$ and $$\vec{\textbf{B}}$$ along the x- and y-axes. Note that A=53.0 m, θ_A=20.0°, B=34.0 m, and θ_B=63.0°. We find the x-components by using $$\boldsymbol{\textbf{A}_x=A \cos \theta}$$, which gives

[latex]\boldsymbol{\textbf{A}_x=A \cos\:\theta_A=(53.0\textbf{ m})(\textbf{cos} 20.0^o)}[/latex]

[latex]\boldsymbol{=(53.0\textbf{ m})(0.940)=49.8\textbf{ m}}\:\:\:\:\:[/latex]

and

[latex]\boldsymbol{\textbf{B}_x=B \cos\:\theta_B=(34.0\textbf{ m})(\textbf{cos} 63.0^o)}[/latex]

[latex]\boldsymbol{=(34.0\textbf{ m})(0.454)=15.4\textbf{ m}.}\:\:\:\:\:[/latex]

Similarly, the y-components are found using $$\textbf{A}_y=A \sin \theta$$:

[latex]\boldsymbol{\textbf{A}_y=A \sin\:\theta_A=(53.0\textbf{ m})(\textbf{sin} 20.0^o)}[/latex]

[latex]\boldsymbol{=(53.0\textbf{ m})(0.342)=18.1\textbf{ m}}\:\:\:\:\:[/latex]

and

[latex]\boldsymbol{\textbf{B}_y=B \sin\:\theta_B=(34.0\textbf{ m})(\textbf{sin} 63.0^o)}[/latex]

[latex]\boldsymbol{=(34.0\textbf{ m})(0.891)=30.3\textbf{ m}.}\:\:\:\:\:[/latex]

The x- and y-components of the resultant are thus

[latex]\boldsymbol{\textbf{R}_x=\textbf{A}_x\:+\:\textbf{B}_x=49.8\textbf{ m}\:+\:15.4\textbf{ m}=65.2\textbf{ m}}[/latex]

and

[latex]\boldsymbol{\textbf{R}_y=\textbf{A}_y\:+\:\textbf{B}_y=18.1\textbf{ m}\:+\:30.3\textbf{ m}=48.4\textbf{ m}.}[/latex]

Now we can find the magnitude of the resultant by using the Pythagorean theorem:

[latex]\boldsymbol{R=\sqrt{\textbf{R}_x^2+\textbf{R}_y^2}=\sqrt{(65.2)^2+(48.4)^2\textbf{ m}}}[/latex]

so that

[latex]\boldsymbol{R=81.2\textbf{ m}.}[/latex]

Finally, we find the direction of the resultant:

[latex]\boldsymbol{\theta=\textbf{tan}^{-1}(\textbf{R}_y\:/\:\textbf{R}_x)=+\textbf{tan}^{-1}(48.4/65.2).}[/latex]

Thus,

[latex]\boldsymbol{\theta=\textbf{tan}^{-1}(0.742)=36.6^o.}[/latex]

Discussion

This example illustrates the addition of vectors using perpendicular components. Vector subtraction using perpendicular components is very similar—it is just the addition of a negative vector.

Subtraction of vectors is accomplished by the addition of a negative vector. That is, $$\vec{\text{A}}-\vec{\text{B}}\equiv\vec{\text{A}}+(-\vec{\text{B}})$$. Thus, the method for the subtraction of vectors using perpendicular components is identical to that for addition. The components of $$-\vec{\text{B}}$$ are the negatives of the components of $$\vec{\text{B}}$$. The x- and y-components of the resultant $$\vec{\text{A}}-\vec{\text{B}}=\vec{\text{R}}$$ are thus

[latex]\boldsymbol{\textbf{R}_x=\textbf{A}_x\:+\:(-\textbf{B}_x)}[/latex]

and

[latex]\boldsymbol{\textbf{R}_y=\textbf{A}_y\:+\:(-\textbf{B}_y)}[/latex]

and the rest of the method outlined above is identical to that for addition. (See Figure 10.)

Analyzing vectors using perpendicular components is very useful in many areas of physics, because perpendicular quantities are often independent of one another. The next module, Chapter 3.4 Projectile Motion, is one of many in which using perpendicular components helps make the picture clear and simplifies the physics. [caption id="" align="aligncenter" width="300"]Figure 10. The subtraction of the two vectors shown in Figure 5. The components of -B are the negatives of the components of B. The method of subtraction is the same as that for addition.[/caption]

PHET EXPLORATIONS: VECTOR ADDITION

Learn how to add vectors. Drag vectors onto a graph, change their length and angle, and sum them together. The magnitude, angle, and components of each vector can be displayed in several formats.

[caption id="" align="aligncenter" width="450"] Figure 11. Vector Addition[/caption]

Summary

• The analytical method of vector addition and subtraction involves using the Pythagorean theorem and trigonometric identities to determine the magnitude and direction of a resultant vector.
• The steps to add vectors $$\vec{\textbf{A}}$$ and $$\vec{\textbf{B}}$$ using the analytical method are as follows:

  Step 1: Determine the coordinate system for the vectors. Then, determine the horizontal and vertical components of each vector using the equations

  [latex]\boldsymbol{\textbf{A}_x=A \cos\:\theta}[/latex]
  [latex]\boldsymbol{\textbf{B}_x=B \cos\:\theta}[/latex]

  and

  [latex]\boldsymbol{\textbf{A}_y=A \sin\:\theta}[/latex]
  [latex]\boldsymbol{\textbf{B}_y=B \sin\:\theta}.[/latex]

  Step 2: Add the horizontal and vertical components of each vector to determine the components $$\textbf{R}_x$$ and $$\textbf{R}_y$$ of the resultant vector, $$\vec{\textbf{R}}$$:

  [latex]\boldsymbol{\textbf{R}_x=\textbf{A}_x+\textbf{B}_x}[/latex]

  and

  [latex]\boldsymbol{\textbf{R}_y=\textbf{A}_y+\textbf{B}_y}.[/latex]

  Step 3: Use the Pythagorean theorem to determine the magnitude, R, of the resultant vector $$\vec{\textbf{R}}$$:

  [latex]\boldsymbol{R=\sqrt{{\textbf{R}_x}^2\:+\:{\textbf{R}_y}^2}.}[/latex]

  Step 4: Use a trigonometric identity to determine the direction, θ, of R:

  [latex]\boldsymbol{\theta=\textbf{tan}^{-1}(\textbf{R}_y\:/\:\textbf{R}_x).}[/latex]

Conceptual Questions

1: Suppose you add two vectors $$\vec{\text{A}}$$ and $$\vec{\text{B}}$$. What relative direction between them produces the resultant with the greatest magnitude? What is the maximum magnitude? What relative direction between them produces the resultant with the smallest magnitude? What is the minimum magnitude?

2: Give an example of a nonzero vector that has a component of zero.

3: Explain why a vector cannot have a component greater than its own magnitude.

4: If the vectors $$\vec{\text{A}}$$ and $$\vec{\text{B}}$$ are perpendicular, what is the component of $$\vec{\text{A}}$$ along the direction of $$\vec{\text{B}}$$? What is the component of $$\vec{\text{B}}$$ along the direction of $$\vec{\text{A}}$$?

Problems & Exercises

1: Find the following for path C in Figure 12: (a) the total distance traveled and (b) the magnitude and direction of the displacement from start to finish. In this part of the problem, explicitly show how you follow the steps of the analytical method of vector addition.

2: Find the following for path D in Figure 12: (a) the total distance traveled and (b) the magnitude and direction of the displacement from start to finish. In this part of the problem, explicitly show how you follow the steps of the analytical method of vector addition.

3: Find the north and east components of the displacement from San Francisco to Sacramento shown in Figure 13.

[caption id="" align="aligncenter" width="250"]Figure 13.[/caption]

4: Solve the following problem using analytical techniques: Suppose you walk 18.0 m straight west and then 25.0 m straight north. How far are you from your starting point, and what is the compass direction of a line connecting your starting point to your final position? (If you represent the two legs of the walk as vector displacements $$\vec{\text{A}}$$ and $$\vec{\text{B}}$$, as in Figure 14, then this problem asks you to find their sum $$\vec{\text{R}}=\vec{\text{A}}+\vec{\text{B}}$$.)

Note that you can also solve this graphically. Discuss why the analytical technique for solving this problem is potentially more accurate than the graphical technique.

5: Repeat Exercise 4 using analytical techniques, but reverse the order of the two legs of the walk and show that you get the same final result. (This problem shows that adding them in reverse order gives the same result—that is, $$\vec{\text{B}}+\vec{\text{A}} =\vec{\text{A}}+\vec{\text{B}}$$.) Discuss how taking another path to reach the same point might help to overcome an obstacle blocking you other path.

6: You drive 7.50 km in a straight line in a direction 15° east of north. (a) Find the distances you would have to drive straight east and then straight north to arrive at the same point. (This determination is equivalent to find the components of the displacement along the east and north directions.) (b) Show that you still arrive at the same point if the east and north legs are reversed in order.

7: Do Exercise 4 again using analytical techniques and change the second leg of the walk to 25.0 m straight south. (This is equivalent to subtracting $$\vec{\text{B}}$$ from $$\vec{\text{A}}$$ —that is, finding $$\vec{\text{R}}^\prime=\vec{\text{A}}-\vec{\text{B}}$$) (b) Repeat again, but now you first walk 25.0 m north and then 18.0 m east. (This is equivalent to subtract $$\vec{\text{A}}$$ from $$\vec{\text{B}}$$ —that is, to find $$\vec{\text{A}}=\vec{\text{B}}+\vec{\text{C}}$$. Is that consistent with your result?)

8: A new landowner has a triangular piece of flat land she wishes to fence. Starting at the west corner, she measures the first side to be 80.0 m long and the next to be 105 m. These sides are represented as displacement vectors $$\vec{\text{A}}$$ from $$\vec{\text{B}}$$ in Figure 15. She then correctly calculates the length and orientation of the third side $$\vec{\text{C}}$$. What is her result?

[caption id="" align="aligncenter" width="250"]Figure 15.[/caption]

9: You fly 32.0 km in a straight line in still air in the direction 35.0° south of west. (a) Find the distances you would have to fly straight south and then straight west to arrive at the same point. (This determination is equivalent to finding the components of the displacement along the south and west directions.) (b) Find the distances you would have to fly first in a direction 45.0° south of west and then in a direction 45.0° west of north. These are the components of the displacement along a different set of axes—one rotated 45°.

10: A farmer wants to fence off his four-sided plot of flat land. He measures the first three sides, shown as $$\vec{\text{A}}$$ , $$\vec{\text{B}}$$, and $$\vec{\text{C}}$$ in Figure 16, and then correctly calculates the length and orientation of the fourth side $$\vec{\text{D}}$$. What is his result?

[caption id="" align="aligncenter" width="300"]Figure 16.[/caption]

11: In an attempt to escape his island, Gilligan builds a raft and sets to sea. The wind shifts a great deal during the day, and he is blown along the following straight lines: 2.50 km 45.0° north of west; then 4.70 km 60.0° south of east; then 1.30 km 25.0° south of west; then 5.10 km straight east; then 1.70 km 5.00° east of north; then 7.20 km 55.0° south of west; and finally 2.80 km 10.0° north of east. What is his final position relative to the island?

12: Suppose a pilot flies 40.0 km in a direction 60° north of east and then flies 30.0 km in a direction 15° north of east as shown in Figure 17. Find her total distance R from the starting point and the direction θ of the straight-line path to the final position. Discuss qualitatively how this flight would be altered by a wind from the north and how the effect of the wind would depend on both wind speed and the speed of the plane relative to the air mass.

[caption id="" align="aligncenter" width="275"]Figure 17.[/caption]

Glossary

analytical method

the method of determining the magnitude and direction of a resultant vector using the Pythagorean theorem and trigonometric identities

Solutions

Problems & Exercises

1: (a) $$\boldsymbol{1.56\textbf{ km}}$$ (b) $$\boldsymbol{120\textbf{ m}}$$ east

3: North-component $$\boldsymbol{87.0\textbf{ km}}$$, east-component $$\boldsymbol{87.0\textbf{ km}}$$ 5: $$\boldsymbol{30.8\textbf{ m}}$$, $$\boldsymbol{35.8^0}$$ west of north

7: (a)[latex]\boldsymbol{30.8\textbf{ m}},\boldsymbol{54.2^o}[/latex] south of west (b)[latex]\boldsymbol{30.8\textbf{ m}},\boldsymbol{54.2^o}[/latex] north of east

9: (a) $$\boldsymbol{18.4\textbf{ km}}$$ south, then $$\boldsymbol{26.2\textbf{ km}}$$ west (b) $$\boldsymbol{31.5\textbf{ km}}$$ at [latex]\boldsymbol{45.0^o}[/latex] south of west, then $$\boldsymbol{5.56\textbf{ km}}$$ at[latex]\boldsymbol{45.0^o}[/latex] west of north 11: [latex]\boldsymbol{7.34\textbf{ km}},\boldsymbol{63.5^o}[/latex] south of east

]]> 246 194 4 0 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/chapter/3-4-projectile-motion/ Thu, 29 Jun 2017 22:13:25 +0000 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/chapter/3-4-projectile-motion/

Summary

• Identify and explain the properties of a projectile, such as acceleration due to gravity, range, maximum height, and trajectory.
• Determine the location and velocity of a projectile at different points in its trajectory.
• Apply the principle of independence of motion to solve projectile motion problems.

Projectile motion is the motion of an object thrown or projected into the air, subject to only the acceleration of gravity. The object is called a projectile, and its path is called its trajectory. The motion of falling objects, as covered in Chapter 2.6 Problem-Solving Basics for One-Dimensional Kinematics, is a simple one-dimensional type of projectile motion in which there is no horizontal movement. In this section, we consider two-dimensional projectile motion, such as that of a football or other object for which air resistance is negligible.

The most important fact to remember here is that motions along perpendicular axes are independent and thus can be analyzed separately. This fact was discussed in Chapter 3.1 Kinematics in Two Dimensions: An Introduction, where vertical and horizontal motions were seen to be independent. The key to analyzing two-dimensional projectile motion is to break it into two motions, one along the horizontal axis and the other along the vertical. (This choice of axes is the most sensible, because acceleration due to gravity is vertical—thus, there will be no acceleration along the horizontal axis when air resistance is negligible.) As is customary, we call the horizontal axis the x-axis and the vertical axis the y-axis. Figure 1 illustrates the notation for displacement, where [latex]\vec{\textbf{s}}[/latex] is defined to be the total displacement and x and y are its components along the horizontal and vertical axes, respectively. The magnitudes of these vectors are s, x, and y. (Note that in the last section we used the notation [latex]\vec{\textbf{A}}[/latex] to represent a vector with components A_x and A_y. If we continued this format, we would call displacement [latex]\vec{\textbf{s}}[/latex] with components s_x and s_y. However, to simplify the notation, we will simply represent the component vectors as x and y.)

Of course, to describe motion we must deal with velocity and acceleration, as well as with displacement. We must find their components along the x- and y-axes, too. We will assume all forces except gravity (such as air resistance and friction, for example) are negligible. The components of acceleration are then very simple: a_y=-g=-9.80 m/s². (Note that this definition assumes that the upwards direction is defined as the positive direction. If you arrange the coordinate system instead such that the downwards direction is positive, then acceleration due to gravity takes a positive value.) Because gravity is vertical, a_x=0. Both accelerations are constant, so the kinematic equations can be used.

REVIEW OF KINEMATIC EQUATIONS (CONSTANT α)

[latex]\boldsymbol{x=x_0+\bar{v}t}[/latex]

[latex]\boldsymbol{\bar{v}=}[/latex][latex size="2"]\boldsymbol{\frac{v_0+v}{2}}[/latex]

[latex]\boldsymbol{v=v_0+at}[/latex]

[latex]\boldsymbol{x=x_0+v_0t\:+}[/latex][latex size="2"]\boldsymbol{\frac{1}{2}}[/latex][latex]\boldsymbol{at^2}[/latex]

[latex]\boldsymbol{v_2=v_0^2+2a(x-x_0).}[/latex]

Given these assumptions, the following steps are then used to analyze projectile motion:

Step 1. Resolve or break the motion into horizontal and vertical components along the x- and y-axes. These axes are perpendicular, so A_x=A cos θ and A_y=A cos θ are used. The magnitude of the components of displacement $$\vec{\textbf{s}}$$ along these axes are x and y. The magnitudes of the components of the velocity $$\vec{\textbf{v}}$$ are v_x = v cos θ and v_y = v sin θ, where v is the magnitude of the velocity and θ is its direction, as shown in Figure 2. Initial values are denoted with a subscript 0, as usual.

Step 2. Treat the motion as two independent one-dimensional motions, one horizontal and the other vertical. The kinematic equations for horizontal and vertical motion take the following forms:

[latex]\boldsymbol{\textbf{Horizontal Motion}(a_x=0)}[/latex]

[latex]\boldsymbol{\textbf{x}=\textbf{x}_0+\textbf{v}_xt}[/latex]

[latex]\boldsymbol{\textbf{v}_x=\textbf{v}_{0x}=\textbf{v}_x=\textbf{velocity is a constant.}}[/latex]

[latex]\boldsymbol{\textbf{Vertical Motion(assuming positive is up} \; a_y=-g=-9.80\textbf{ m/s}^2)}[/latex]

[latex]\boldsymbol{\textbf{y}=\textbf{y}_0\:+}[/latex][latex size="2"]\boldsymbol{\frac{1}{2}}[/latex][latex]\boldsymbol{(\textbf{v}_{0y}+\textbf{v}_y)t}[/latex]

[latex]\boldsymbol{\textbf{v}_y=\textbf{v}_{0y}-gt}[/latex]

[latex]\boldsymbol{\textbf{y}=\textbf{y}_0+\textbf{v}_{0y}t\:-}[/latex][latex size="2"]\boldsymbol{\frac{1}{2}}[/latex][latex]\boldsymbol{gt^2}[/latex]

[latex]\boldsymbol{\textbf{v}_y^2=\textbf{v}_{0y}^2-2g(\textbf{y}-\textbf{y}_0).}[/latex]

Step 3. Solve for the unknowns in the two separate motions—one horizontal and one vertical. Note that the only common variable between the motions is time t. The problem solving procedures here are the same as for one-dimensional kinematics and are illustrated in the solved examples below.

Step 4. Recombine the two motions to find the total displacement [latex]\vec{\textbf{s}}[/latex] and velocity [latex]\vec{\textbf{v}}.[/latex] Because the x - and y -motions are perpendicular, we determine these vectors by using the techniques outlined in the Chapter 3.3 Vector Addition and Subtraction: Analytical Methods and employing [latex]\boldsymbol{A=\sqrt{\textbf{A}_x^2+\textbf{A}_y^2}}[/latex] and θ = tan^-1 ( A_y / A_x) in the following form, where θ is the direction of the displacement [latex]\vec{\textbf{s}}[/latex] and θ_v is the direction of the velocity $$\vec{\textbf{v}}$$:

Total displacement and velocity

[latex]\boldsymbol{s=\sqrt{\textbf{x}^2+\textbf{y}^2}}[/latex]

[latex]\boldsymbol{\theta=\textbf{tan}^{-1}(\textbf{y}\: / \:\textbf{x} )}[/latex]

[latex]\boldsymbol{v=\sqrt{\textbf{v}_x^2+\textbf{v}_y^2}}[/latex]

[latex]\boldsymbol{\theta_{v}=\textbf{tan}^{-1}(\textbf{v}_y\:/\:\textbf{v}_x).}[/latex]

[caption id="" align="aligncenter" width="937"]Figure 2. (a) We analyze two-dimensional projectile motion by breaking it into two independent one-dimensional motions along the vertical and horizontal axes. (b) The horizontal motion is simple, because a_x=0 and v_x is thus constant. (c) The velocity in the vertical direction begins to decrease as the object rises; at its highest point, the vertical velocity is zero. As the object falls towards the Earth again, the vertical velocity increases again in magnitude but points in the opposite direction to the initial vertical velocity. (d) The x - and y -motions are recombined to give the total velocity at any given point on the trajectory.[/caption]

Example 1: A Fireworks Projectile Explodes High and Away

During a fireworks display, a shell is shot into the air with an initial speed of 70.0 m/s at an angle of 75.0° above the horizontal, as illustrated in Figure 3. The fuse is timed to ignite the shell just as it reaches its highest point above the ground. (a) Calculate the height at which the shell explodes. (b) How much time passed between the launch of the shell and the explosion? (c) What is the horizontal displacement of the shell when it explodes?

Strategy

Because air resistance is negligible for the unexploded shell, the analysis method outlined above can be used. The motion can be broken into horizontal and vertical motions in which a_x=0 and a_y=-g. We can then define x₀ and y₀ to be zero and solve for the desired quantities.

Solution for (a)

By “height” we mean the altitude or vertical position y above the starting point. The highest point in any trajectory, called the apex, is reached when v_y = 0. Since we know the initial and final velocities as well as the initial position, we use the following equation to find y:

[latex]\boldsymbol{v_y^2=v_{0y}^2-2g(y-y_0).}[/latex]

Because y₀ and v_y are both zero, the equation simplifies to

[latex]\boldsymbol{0=v_{0y}^2-2gy.}[/latex]

Solving for y gives

[latex]\boldsymbol{y\:=}[/latex][latex size="2"]\boldsymbol{\frac{v_{0y}^2}{2g}.}[/latex]

Now we must find v_0y, the component of the initial velocity in the y-direction. It is given by v_0y = v₀ sin θ, where v₀ is the initial velocity of 70.0 m/s, and θ₀=75.0° is the initial angle. Thus,

[latex]\boldsymbol{v_{0y}=v_0\:\textbf{sin}\:\theta_0=(70.0\textbf{ m/s})(\textbf{sin }75^o)=67.6\textbf{ m/s}.}[/latex]

and y is

[latex]\boldsymbol{y\:=}[/latex][latex size="2"]\boldsymbol{\frac{(67.6\textbf{ m/s})^2}{2(9.80\textbf{ m/s}^2)}},[/latex]

so that

[latex]\boldsymbol{y=233\textbf{ m}.}[/latex]

Discussion for (a)

Note that because up is positive, the initial velocity is positive, as is the maximum height, but the acceleration due to gravity is negative. Note also that the maximum height depends only on the vertical component of the initial velocity, so that any projectile with a 67.6 m/s initial vertical component of velocity will reach a maximum height of 233 m (neglecting air resistance). The numbers in this example are reasonable for large fireworks displays, the shells of which do reach such heights before exploding. In practice, air resistance is not completely negligible, and so the initial velocity would have to be somewhat larger than that given to reach the same height.

Solution for (b)

As in many physics problems, there is more than one way to solve for the time to the highest point. In this case, the easiest method is to use [latex]\boldsymbol{y=y_0+\frac{1}{2}(v_{0y}+v_y)t}.[/latex] Because y₀ is zero, this equation reduces to simply

[latex]\boldsymbol{y\:=}[/latex][latex size="2"]\boldsymbol{\frac{1}{2}}[/latex][latex]\boldsymbol{(v_{0y}+v_y)t.}[/latex]

Note that the final vertical velocity, v_y, at the highest point is zero. Thus,

[latex]\boldsymbol{t=\frac{2y}{(v_{0y}+v_y)}=\frac{2(233\textbf{ m})}{(67.6\textbf{ m/s})}}[/latex]

[latex]\boldsymbol{=6.90\textbf{ s}.}\:\:\:\:\:\:\:\:\:[/latex]

Discussion for (b)

This time is also reasonable for large fireworks. When you are able to see the launch of fireworks, you will notice several seconds pass before the shell explodes. (Another way of finding the time is by using[latex]\boldsymbol{y=y_0+v_{0y}t-\frac{1}{2}gt^2},[/latex]and solving the quadratic equation for t.)

Solution for (c)

Because air resistance is negligible, a_x = 0 and the horizontal velocity is constant, as discussed above. The horizontal displacement is horizontal velocity multiplied by time as given by [latex]\boldsymbol{x=x_0+v_xt},[/latex] where x₀ is equal to zero:

[latex]\boldsymbol{x=v_xt,}[/latex]

where v_x is the x-component of the velocity, which is given by v_x= v₀ cos θ₀. Now,

[latex]\boldsymbol{v_x=v_0\textbf{cos}\theta_0=(70.0\textbf{ m/s})(\textbf{cos }75.0^o)=18.1\textbf{ m/s}.}[/latex]

The time t for both motions is the same, and so x is

[latex]\boldsymbol{x=(18.1\textbf{ m/s})(6.90\textbf{ s})=125\textbf{ m}.}[/latex]

Discussion for (c)

The horizontal motion is a constant velocity in the absence of air resistance. The horizontal displacement found here could be useful in keeping the fireworks fragments from falling on spectators. Once the shell explodes, air resistance has a major effect, and many fragments will land directly below.

In solving part (a) of the preceding example, the expression we found for y is valid for any projectile motion where air resistance is negligible. Call the maximum height y = h; then,

[latex]\boldsymbol{h\:=}[/latex][latex size="2"]\boldsymbol{\frac{v_{0y}^2}{2g}.}[/latex]

This equation defines the maximum height of a projectile and depends only on the vertical component of the initial velocity.

DEFINING A COORDINATE SYSTEM

It is important to set up a coordinate system when analyzing projectile motion. One part of defining the coordinate system is to define an origin for the x and y positions. Often, it is convenient to choose the initial position of the object as the origin such that x₀ = 0 and y₀ = 0. It is also important to define the positive and negative directions in the x and y directions. Typically, we define the positive vertical direction as upwards, and the positive horizontal direction is usually the direction of the object’s motion. When this is the case, the vertical acceleration, g, takes a negative value (since it is directed downwards towards the Earth). However, it is occasionally useful to define the coordinates differently. For example, if you are analyzing the motion of a ball thrown downwards from the top of a cliff, it may make sense to define the positive direction downwards since the motion of the ball is solely in the downwards direction. If this is the case, g takes a positive value.

Example 2: Calculating Projectile Motion: Hot Rock Projectile

Kilauea in Hawaii is the world’s most continuously active volcano. Very active volcanoes characteristically eject red-hot rocks and lava rather than smoke and ash. Suppose a large rock is ejected from the volcano with a speed of 25.0 m/s and at an angle 35.0° above the horizontal, as shown in Figure 4. The rock strikes the side of the volcano at an altitude 20.0 m lower than its starting point. (a) Calculate the time it takes the rock to follow this path. (b) What are the magnitude and direction of the rock’s velocity at impact?

Strategy

Again, resolving this two-dimensional motion into two independent one-dimensional motions will allow us to solve for the desired quantities. The time a projectile is in the air is governed by its vertical motion alone. We will solve for t first. While the rock is rising and falling vertically, the horizontal motion continues at a constant velocity. This example asks for the final velocity. Thus, the vertical and horizontal results will be recombined to obtain v and θ_v at the final time t determined in the first part of the example.

Solution for (a)

While the rock is in the air, it rises and then falls to a final position 20.0 m lower than its starting altitude. We can find the time for this by using

[latex]\boldsymbol{y=y_0+v_{0y}t\:-}[/latex][latex size="2"]\boldsymbol{\frac{1}{2}}[/latex][latex]\boldsymbol{gt^2.}[/latex]

If we take the initial position y₀ to be zero, then the final position is y = -20.0 m. Now the initial vertical velocity is the vertical component of the initial velocity, found from v_0y = v₀ sin θ₀= (25.0 m/s)(sin 35.0°) = 14.3 m/s. Substituting known values yields

[latex]\boldsymbol{-20.0\textbf{ m}=(14.3\textbf{ m/s})t-4.90\textbf{ m/s}^2t^2.}[/latex]

Rearranging terms gives a quadratic equation in t:

[latex]\boldsymbol{4.90\textbf{ m/s}^2t^2-14.3\textbf{ m/s}t-20.0\textbf{ m}=0.}[/latex]

This expression is a quadratic equation of the form at²+bt+c = 0, where the constants are a = 4.90, b = -14.3, and c=-20.0. Its solutions are given by the quadratic formula:

[latex]\boldsymbol{t\:=}[/latex][latex size="2"]\boldsymbol{\frac{-b\pm\sqrt{b^2-4ac}}{2a}}[/latex]

This equation yields two solutions: t = 3.96 and t = -1.03. (It is left as an exercise for the reader to verify these solutions.) The time is t = 3.96 s or -1.03 s. The negative value of time implies an event before the start of motion, and so we discard it. Thus,

[latex]\boldsymbol{t=3.96\textbf{ s}.}[/latex]

Discussion for (a)

The time for projectile motion is completely determined by the vertical motion. So any projectile that has an initial vertical velocity of 14.3 m/s and lands 20.0 m below its starting altitude will spend 3.96 s in the air.

Solution for (b)

From the information now in hand, we can find the final horizontal and vertical velocities v_x and v_y and combine them to find the total velocity v and the angle θ₀ it makes with the horizontal. Of course, v_x is constant so we can solve for it at any horizontal location. In this case, we chose the starting point since we know both the initial velocity and initial angle. Therefore:

[latex]\boldsymbol{v_x=v_0\:\textbf{cos}\:\theta_0=(25.0\textbf{ m/s})(\textbf{cos }35^o)=20.5\textbf{ m/s}.}[/latex]

The final vertical velocity is given by the following equation:

[latex]\boldsymbol{v_y=v_{0y}-gt,}[/latex]

where v_0y was found in part (a) to be 14.3 m/s. Thus,

[latex]\boldsymbol{v_y=14.3\textbf{ m/s}-(9.80\textbf{ m/s}^2)(3.96\textbf{ s})}[/latex]

so that

[latex]\boldsymbol{v_y=-24.5\textbf{ m/s}.}[/latex]

To find the magnitude of the final velocity v we combine its perpendicular components, using the following equation:

[latex]\boldsymbol{v=\sqrt{v_x^2+v_y^2}=\sqrt{(20.5\textbf{ m/s})^2+(-24.5\textbf{ m/s})^2},}[/latex]

which gives

[latex]\boldsymbol{v=31.9\textbf{ m/s}.}[/latex]

The direction θ_v is found from the equation:

[latex]\boldsymbol{\theta_v=\textbf{tan}^{-1}(v_y\:/\:v_x)}[/latex]

so that

[latex]\boldsymbol{\theta_v=\textbf{tan}^{-1}(-24.5/20.5)=\textbf{tan}^{-1}(-1.19).}[/latex]

Thus,

[latex]\boldsymbol{\theta_v=-50.1^o.}[/latex]

Discussion for (b)

The negative angle means that the velocity is 50.1° below the horizontal. This result is consistent with the fact that the final vertical velocity is negative and hence downward—as you would expect because the final altitude is 20.0 m lower than the initial altitude. (See Figure 4.)

One of the most important things illustrated by projectile motion is that vertical and horizontal motions are independent of each other. Galileo was the first person to fully comprehend this characteristic. He used it to predict the range of a projectile. On level ground, we define range to be the horizontal distance R traveled by a projectile. Galileo and many others were interested in the range of projectiles primarily for military purposes—such as aiming cannons. However, investigating the range of projectiles can shed light on other interesting phenomena, such as the orbits of satellites around the Earth. Let us consider projectile range further.

How does the initial velocity of a projectile affect its range? Obviously, the greater the initial speed v₀, the greater the range, as shown in Figure 5(a). The initial angle θ₀ also has a dramatic effect on the range, as illustrated in Figure 5(b). For a fixed initial speed, such as might be produced by a cannon, the maximum range is obtained with θ₀= 45°. This is true only for conditions neglecting air resistance. If air resistance is considered, the maximum angle is approximately 38°. Interestingly, for every initial angle except 45° there are two angles that give the same range—the sum of those angles is 90°. The range also depends on the value of the acceleration of gravity g. The lunar astronaut Alan Shepherd was able to drive a golf ball a great distance on the Moon because gravity is weaker there. The range R of a projectile on level ground for which air resistance is negligible is given by

[latex]\boldsymbol{R\:=}[/latex][latex size="2"]\boldsymbol{\frac{v_0^2\:\textbf{sin}\:2\theta_0}{g},}[/latex]

where v₀ is the initial speed and θ₀ is the initial angle relative to the horizontal. The proof of this equation is left as an end-of-chapter problem (hints are given), but it does fit the major features of projectile range as described.

When we speak of the range of a projectile on level ground, we assume that R is very small compared with the circumference of the Earth. If, however, the range is large, the Earth curves away below the projectile and acceleration of gravity changes direction along the path. The range is larger than predicted by the range equation given above because the projectile has farther to fall than it would on level ground. (See Figure 6.) If the initial speed is great enough, the projectile goes into orbit. This possibility was recognized centuries before it could be accomplished. When an object is in orbit, the Earth curves away from underneath the object at the same rate as it falls. The object thus falls continuously but never hits the surface. These and other aspects of orbital motion, such as the rotation of the Earth, will be covered analytically and in greater depth later in this text.

Once again we see that thinking about one topic, such as the range of a projectile, can lead us to others, such as the Earth orbits. In Chapter 3.5 Addition of Velocities, we will examine the addition of velocities, which is another important aspect of two-dimensional kinematics and will also yield insights beyond the immediate topic.

PHET EXPLORATIONS: PROJECTILE MOTION

Blast a Buick out of a cannon! Learn about projectile motion by firing various objects. Set the angle, initial speed, and mass. Add air resistance. Make a game out of this simulation by trying to hit a target.

[caption id="" align="aligncenter" width="450"] Figure 7. Projectile Motion[/caption]

Summary

• Projectile motion is the motion of an object through the air that is subject only to the acceleration of gravity.
• To solve projectile motion problems, perform the following steps:
  1. Determine a coordinate system. Then, resolve the position and/or velocity of the object in the horizontal and vertical components. The components of position [latex]\vec{\textbf{s}}[/latex] are given by the quantities x and y, and the components of the velocity [latex]\vec{\textbf{v}}[/latex] are given by v_x = v cos θ and v_y = v sin θ, where v is the magnitude of the velocity and θ is its direction.
  2. Analyze the motion of the projectile in the horizontal direction using the following equations:
     [latex]\boldsymbol{\textbf{Horizontal motion}(a_x=0)}[/latex]
     [latex]\boldsymbol{x=x_0+v_xt}[/latex]
     [latex]\boldsymbol{v_x=v_{0x}=\textbf{v}_x=\textbf{velocity is a constant.}}[/latex]
  3. Analyze the motion of the projectile in the vertical direction using the following equations:
     [latex]\boldsymbol{\textbf{Vertical motion (Assuming positive direction is up;}\:a_y=-g=-9.80\textbf{ m/s}^2)}[/latex]
     [latex]\boldsymbol{y=y_0\:+}[/latex][latex size="2"]\boldsymbol{\frac{1}{2}}[/latex][latex]\boldsymbol{(v_{0y}+v_y)t}[/latex]
     [latex]\boldsymbol{v_y=v_{0y}-gt}[/latex]
     [latex]\boldsymbol{y=y_0+v_{0y}t\:-}[/latex][latex size="2"]\boldsymbol{\frac{1}{2}}[/latex][latex]\:\boldsymbol{gt^2}[/latex]
     [latex]\boldsymbol{v_y^2=v_{0y}^2-2g(y-y_0).}[/latex]
  4. Recombine the horizontal and vertical components of location and/or velocity using the following equations:
     [latex]\boldsymbol{s=\sqrt{x^2+y^2}}[/latex]
     [latex]\boldsymbol{\theta=\textbf{tan}^{-1}(y/x)}[/latex]
     [latex]\boldsymbol{v=\sqrt{v_x^2+v_y^2}}[/latex]
     [latex]\boldsymbol{\theta_v=\textbf{tan}^{-1}(v_y\:/\:v_x).}[/latex]

• The maximum height h of a projectile launched with initial vertical velocity v_0y is given by
  [latex]\boldsymbol{h\:=}[/latex][latex size="2"]\boldsymbol{\frac{v_{0y}^2}{2g}.}[/latex]
• The maximum horizontal distance traveled by a projectile is called the range. The range R of a projectile on level ground launched at an angle θ₀ above the horizontal with initial speed v₀ is given by
  [latex]\boldsymbol{\textbf{R}\:=}[/latex][latex size="2"]\boldsymbol{\frac{v_0^2\:\textbf{sin}2\:\theta_0}{g}.}[/latex]

Conceptual Questions

1: Answer the following questions for projectile motion on level ground assuming negligible air resistance (the initial angle being neither 0° nor 90°): (a) Is the velocity ever zero? (b) When is the velocity a minimum? A maximum? (c) Can the velocity ever be the same as the initial velocity at a time other than at t = 0? (d) Can the speed ever be the same as the initial speed at a time other than at t = 0?

2: Answer the following questions for projectile motion on level ground assuming negligible air resistance (the initial angle being neither 0° nor 90°): (a) Is the acceleration ever zero? (b) Is the acceleration ever in the same direction as a component of velocity? (c) Is the acceleration ever opposite in direction to a component of velocity?

3: For a fixed initial speed, the range of a projectile is determined by the angle at which it is fired. For all but the maximum, there are two angles that give the same range. Considering factors that might affect the ability of an archer to hit a target, such as wind, explain why the smaller angle (closer to the horizontal) is preferable. When would it be necessary for the archer to use the larger angle? Why does the punter in a football game use the higher trajectory?

4: During a lecture demonstration, a professor places two coins on the edge of a table. She then flicks one of the coins horizontally off the table, simultaneously nudging the other over the edge. Describe the subsequent motion of the two coins, in particular discussing whether they hit the floor at the same time.

Problems & Exercises

1: A projectile is launched at ground level with an initial speed of 50.0 m/s at an angle of 30.0° above the horizontal. It strikes a target above the ground 3.00 seconds later. What are the x and y distances from where the projectile was launched to where it lands?

2: A ball is kicked with an initial velocity of 16 m/s in the horizontal direction and 12 m/s in the vertical direction. (a) At what speed does the ball hit the ground? (b) For how long does the ball remain in the air? (c)What maximum height is attained by the ball?

3: A ball is thrown horizontally from the top of a 60.0-m building and lands 100.0 m from the base of the building. Ignore air resistance. (a) How long is the ball in the air? (b) What must have been the initial horizontal component of the velocity? (c) What is the vertical component of the velocity just before the ball hits the ground? (d) What is the velocity (including both the horizontal and vertical components) of the ball just before it hits the ground?

4: (a) A daredevil is attempting to jump his motorcycle over a line of buses parked end to end by driving up 32° ramp at a speed of 40.0 m/s (144 km/h). How many buses can he clear if the top of the takeoff ramp is at the same height as the bus tops and the buses are 20.0 m long? (b) Discuss what your answer implies about the margin of error in this act—that is, consider how much greater the range is than the horizontal distance he must travel to miss the end of the last bus. (Neglect air resistance.)

5: An archer shoots an arrow at a 75.0 m distant target; the bull’s-eye of the target is at same height as the release height of the arrow. (a) At what angle must the arrow be released to hit the bull’s-eye if its initial speed is 35.0 m/s? In this part of the problem, explicitly show how you follow the steps involved in solving projectile motion problems. (b) There is a large tree halfway between the archer and the target with an overhanging horizontal branch 3.50 m above the release height of the arrow. Will the arrow go over or under the branch?

6: A rugby player passes the ball 7.00 m across the field, where it is caught at the same height as it left his hand. (a) At what angle was the ball thrown if its initial speed was 12.0 m/s, assuming that the smaller of the two possible angles was used? (b) What other angle gives the same range, and why would it not be used? (c) How long did this pass take?

7: Verify the ranges for the projectiles in Figure 5(a) for θ = 45° and the given initial velocities.

8: Verify the ranges shown for the projectiles in Figure 5(b) for an initial velocity of 50 m/s at the given initial angles.

9: The cannon on a battleship can fire a shell a maximum distance of 32.0 km. (a) Calculate the initial velocity of the shell. (b) What maximum height does it reach? (At its highest, the shell is above 60% of the atmosphere—but air resistance is not really negligible as assumed to make this problem easier.) (c) The ocean is not flat, because the Earth is curved. Assume that the radius of the Earth is 6.37 × 10³ km. How many meters lower will its surface be 32.0 km from the ship along a horizontal line parallel to the surface at the ship? Does your answer imply that error introduced by the assumption of a flat Earth in projectile motion is significant here?

10: An arrow is shot from a height of 1.5 m toward a cliff of height H. It is shot with a velocity of 30 m/s at an angle of 60° above the horizontal. It lands on the top edge of the cliff 4.0 s later. (a) What is the height of the cliff? (b) What is the maximum height reached by the arrow along its trajectory? (c) What is the arrow’s impact speed just before hitting the cliff?

11: In the standing broad jump, one squats and then pushes off with the legs to see how far one can jump. Suppose the extension of the legs from the crouch position is 0.600 m and the acceleration achieved from this position is 1.25 times the acceleration due to gravity, g. How far can they jump? State your assumptions. (Increased range can be achieved by swinging the arms in the direction of the jump.)

12: The world long jump record is 8.95 m (Mike Powell, USA, 1991). Treated as a projectile, what is the maximum range obtainable by a person if he has a take-off speed of 9.5 m/s? State your assumptions.

13: Serving at a speed of 170 km/h, a tennis player hits the ball at a height of 2.5 m and an angle θ below the horizontal. The service line is 11.9 m from the net, which is 0.91 m high. What is the angle θ such that the ball just crosses the net? Will the ball land in the service box, whose out line is 6.40 m from the net?

14: A football quarterback is moving straight backward at a speed of 2.00 m/s when he throws a pass to a player 18.0 m straight downfield. (a) If the ball is thrown at an angle of 25° relative to the ground and is caught at the same height as it is released, what is its initial speed relative to the ground? (b) How long does it take to get to the receiver? (c) What is its maximum height above its point of release?

15: Gun sights are adjusted to aim high to compensate for the effect of gravity, effectively making the gun accurate only for a specific range. (a) If a gun is sighted to hit targets that are at the same height as the gun and 100.0 m away, how low will the bullet hit if aimed directly at a target 150.0 m away? The muzzle velocity of the bullet is 275 m/s. (b) Discuss qualitatively how a larger muzzle velocity would affect this problem and what would be the effect of air resistance.

16: An eagle is flying horizontally at a speed of 3.00 m/s when the fish in her talons wiggles loose and falls into the lake 5.00 m below. Calculate the velocity of the fish relative to the water when it hits the water.

17: An owl is carrying a mouse to the chicks in its nest. Its position at that time is 4.00 m west and 12.0 m above the center of the 30.0 cm diameter nest. The owl is flying east at 3.50 m/s at an angle 30.0° below the horizontal when it accidentally drops the mouse. Is the owl lucky enough to have the mouse hit the nest? To answer this question, calculate the horizontal position of the mouse when it has fallen 12.0 m.

18: Suppose a soccer player kicks the ball from a distance 30 m toward the goal. Find the initial speed of the ball if it just passes over the goal, 2.4 m above the ground, given the initial direction to be 40° above the horizontal.

19: Can a goalkeeper at her/ his goal kick a soccer ball into the opponent’s goal without the ball touching the ground? The distance will be about 95 m. A goalkeeper can give the ball a speed of 30 m/s.

20: The free throw line in basketball is 4.57 m (15 ft) from the basket, which is 3.05 m (10 ft) above the floor. A player standing on the free throw line throws the ball with an initial speed of 7.15 m/s, releasing it at a height of 2.44 m (8 ft) above the floor. At what angle above the horizontal must the ball be thrown to exactly hit the basket? Note that most players will use a large initial angle rather than a flat shot because it allows for a larger margin of error. Explicitly show how you follow the steps involved in solving projectile motion problems.

21: In 2007, Michael Carter (U.S.) set a world record in the shot put with a throw of 24.77 m. What was the initial speed of the shot if he released it at a height of 2.10 m and threw it at an angle of 38.0° above the horizontal? (Although the maximum distance for a projectile on level ground is achieved at 45° when air resistance is neglected, the actual angle to achieve maximum range is smaller; thus, 38° will give a longer range than 45° in the shot put.)

22: A basketball player is running at 5.00 m/s directly toward the basket when he jumps into the air to dunk the ball. He maintains his horizontal velocity. (a) What vertical velocity does he need to rise 0.750 m above the floor? (b) How far from the basket (measured in the horizontal direction) must he start his jump to reach his maximum height at the same time as he reaches the basket?

23: A football player punts the ball at a 45.0° angle. Without an effect from the wind, the ball would travel 60.0 m horizontally. (a) What is the initial speed of the ball? (b) When the ball is near its maximum height it experiences a brief gust of wind that reduces its horizontal velocity by 1.50 m/s. What distance does the ball travel horizontally?

24: Prove that the trajectory of a projectile is parabolic, having the form y = ax + bx². To obtain this expression, solve the equation x = v_0xt for t and substitute it into the expression for y = v_0yt - (1/2)gt² (These equations describe the x and y positions of a projectile that starts at the origin.) You should obtain an equation of the form y = ax + bx² where a and b are constants.

25: Derive [latex]\text{R}=\frac{v_0^2\:\textbf{sin}\:2\theta_0}{g}[/latex] for the range of a projectile on level ground by finding the time t at which y becomes zero and substituting this value of t into the expression for x - x₀, noting that R = x - x₀.

26: Unreasonable Results (a) Find the maximum range of a super cannon that has a muzzle velocity of 4.0 km/s. (b) What is unreasonable about the range you found? (c) Is the premise unreasonable or is the available equation inapplicable? Explain your answer. (d) If such a muzzle velocity could be obtained, discuss the effects of air resistance, thinning air with altitude, and the curvature of the Earth on the range of the super cannon.

27: Construct Your Own Problem Consider a ball tossed over a fence. Construct a problem in which you calculate the ball’s needed initial velocity to just clear the fence. Among the things to determine are; the height of the fence, the distance to the fence from the point of release of the ball, and the height at which the ball is released. You should also consider whether it is possible to choose the initial speed for the ball and just calculate the angle at which it is thrown. Also examine the possibility of multiple solutions given the distances and heights you have chosen.

Glossary

air resistance

a frictional force that slows the motion of objects as they travel through the air; when solving basic physics problems, air resistance is assumed to be zero

kinematics

the study of motion without regard to mass or force

motion

displacement of an object as a function of time

projectile

an object that travels through the air and experiences only acceleration due to gravity

projectile motion

the motion of an object that is subject only to the acceleration of gravity

range

the maximum horizontal distance that a projectile travels

trajectory

the path of a projectile through the air

Solutions

Problems & Exercises 1: [latex]\boldsymbol{x=1.30\textbf{ m}\times10^2}\:\:\boldsymbol{y=30.9\textbf{ m}.}[/latex]

3: (a) $$\boldsymbol{3.50\textbf{ s}}$$ (b) $$\boldsymbol{28.6\textbf{ m/s}}$$ (c) $$\boldsymbol{34.3\textbf{ m/s}}$$ (d) $$\boldsymbol{44.7\textbf{ m/s}}$$, $$\boldsymbol{50.2^0}$$ below horizontal

5: (a)[latex]\boldsymbol{18.4^0}[/latex] (b) The arrow will go over the branch.

7: [latex]\boldsymbol{\textbf{R}=\frac{v_0^2}{\textbf{sin }2\theta_0g}}[/latex] [latex]\boldsymbol{\textbf{For }\theta=45^0},\:\boldsymbol{\textbf{R}=\frac{v_0^2}{g}}[/latex] [latex]\boldsymbol{\textbf{R}=91.8\textbf{ m}}[/latex] for [latex]\boldsymbol{v_0=30\textbf{ m/s}};\:\boldsymbol{\textbf{R}=163\textbf{ m}}[/latex] for [latex]\boldsymbol{v_0=40\textbf{ m/s}};\:\boldsymbol{\textbf{R}=255\textbf{ m}}[/latex] for [latex]\boldsymbol{v_0=50\textbf{ m/s}}.[/latex]

9: (a) $$\boldsymbol{560\textbf{ m/s}}$$ (b) [latex]\boldsymbol{8.00 \times 10^3\textbf{ m}}[/latex] (c) $$\boldsymbol{80.0\textbf{ m}}$$. This error is not significant because it is only 1% of the answer in part (b).

11: $$\boldsymbol{1.50\textbf{ m}}$$, assuming launch angle of[latex]\boldsymbol{45^0}[/latex]

13: [latex]\boldsymbol{\theta=6.1^0}[/latex] yes, the ball lands at 5.3 m from the net

15: (a) $$\boldsymbol{-0.486\textbf{ m}}$$ (b) The larger the muzzle velocity, the smaller the deviation in the vertical direction, because the time of flight would be smaller. Air resistance would have the effect of decreasing the time of flight, therefore increasing the vertical deviation.

17: $$\boldsymbol{4.23\textbf{ m}}$$. No, the owl is not lucky; he misses the nest. 19: No, the maximum range (neglecting air resistance) is about $$\boldsymbol{92\textbf{ m}}$$. 21: $$\boldsymbol{15.0\textbf{ m/s}}$$

23: (a) $$\boldsymbol{24.2\textbf{ m/s}}$$ (b) The ball travels a total of $$\boldsymbol{57.4\textbf{ m}}$$ with the brief gust of wind.

25: [latex]\boldsymbol{y-y_0=0=v_{0y}t-\frac{1}{2}gt^2=(v_0\:\textbf{sin}\:\theta)t-\frac{1}{2}gt^2},[/latex]

so that [latex]\boldsymbol{t=\frac{2(v_0\:\textbf{sin}\:\theta)}{g}}[/latex]

[latex]\boldsymbol{x-x_0=v_{0x}t=(v_0\:\textbf{cos}\:\theta)t=R},[/latex] and substituting for [latex]\boldsymbol{t}[/latex] gives:

[latex]\boldsymbol{R=v_0\:\textbf{cos}\:\theta(\frac{2v_0\:\textbf{sin}\:\theta}{g})=(\frac{2v_0^2\:\textbf{sin}\:\theta\textbf{ cos}\:\theta}{g})}[/latex]

since [latex]\boldsymbol{2\textbf{sin}\theta\textbf{cos}\theta=\textbf{sin}2\theta},[/latex] the range is: [latex]\boldsymbol{\textbf{R}=\frac{v_0^2\:\textbf{sin}\:2\theta}{g}}.[/latex]

]]> 254 194 5 0 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/chapter/3-5-addition-of-velocities/ Thu, 29 Jun 2017 22:13:26 +0000 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/chapter/3-5-addition-of-velocities/

Summary

• Apply principles of vector addition to determine relative velocity.
• Explain the significance of the observer in the measurement of velocity.

Relative Velocity

If a person rows a boat across a rapidly flowing river and tries to head directly for the other shore, the boat instead moves diagonally relative to the shore, as in Figure 1. The boat does not move in the direction in which it is pointed. The reason, of course, is that the river carries the boat downstream. Similarly, if a small airplane flies overhead in a strong crosswind, you can sometimes see that the plane is not moving in the direction in which it is pointed, as illustrated in Figure 2. The plane is moving straight ahead relative to the air, but the movement of the air mass relative to the ground carries it sideways.

[caption id="" align="aligncenter" width="350"]Figure 1. A boat trying to head straight across a river will actually move diagonally relative to the shore as shown. Its total velocity (solid arrow) relative to the shore is the sum of its velocity relative to the river plus the velocity of the river relative to the shore.[/caption]

[caption id="" align="aligncenter" width="300"]Figure 2. An airplane heading straight north is instead carried to the west and slowed down by wind. The plane does not move relative to the ground in the direction it points; rather, it moves in the direction of its total velocity (solid arrow).[/caption]

In each of these situations, an object has a velocity relative to a medium (such as a river) and that medium has a velocity relative to an observer on solid ground. The velocity of the object relative to the observer is the sum of these velocity vectors, as indicated in Figure 1 and Figure 2. These situations are only two of many in which it is useful to add velocities. In this module, we first re-examine how to add velocities and then consider certain aspects of what relative velocity means.

How do we add velocities? Velocity is a vector (it has both magnitude and direction); the rules of vector addition discussed in Chapter 3.2 Vector Addition and Subtraction: Graphical Methods and Chapter 3.3 Vector Addition and Subtraction: Analytical Methods apply to the addition of velocities, just as they do for any other vectors. In one-dimensional motion, the addition of velocities is simple—they add like ordinary numbers. For example, if a field hockey player is moving at 5 m/s straight toward the goal and drives the ball in the same direction with a velocity of 30 m/s relative to her body, then the velocity of the ball is 35 m/s relative to the stationary, profusely sweating goalkeeper standing in front of the goal.

In two-dimensional motion, either graphical or analytical techniques can be used to add velocities. We will concentrate on analytical techniques. The following equations give the relationships between the magnitude and direction of velocity (v and θ) and its components (v_x and v_y) along the x- and y-axes of an appropriately chosen coordinate system:

[latex]\boldsymbol{v_x=v\:\textbf{cos}\:\theta}[/latex]

[latex]\boldsymbol{v_y=v\:\textbf{sin}\:\theta}[/latex]

[latex]\boldsymbol{v=\sqrt{v_x^2+v_y^2}}[/latex]

[latex]\boldsymbol{\theta=\textbf{tan}^{-1}(v_y/v_x).}[/latex]

[caption id="" align="aligncenter" width="250"]Figure 3. The velocity, v, of an object traveling at an angle θ to the horizontal axis is the sum of component vectors v_x and v_y.[/caption]

These equations are valid for any vectors and are adapted specifically for velocity. The first two equations are used to find the components of a velocity when its magnitude and direction are known. The last two are used to find the magnitude and direction of velocity when its components are known.

TAKE-HOME EXPERIMENT: RELATIVE VELOCITY OF A BOAT

Fill a bathtub half-full of water. Take a toy boat or some other object that floats in water. Unplug the drain so water starts to drain. Try pushing the boat from one side of the tub to the other and perpendicular to the flow of water. Which way do you need to push the boat so that it ends up immediately opposite? Compare the directions of the flow of water, heading of the boat, and actual velocity of the boat.

Example 1: Adding Velocities: A Boat on a River

[caption id="" align="aligncenter" width="400"]Figure 4. A boat attempts to travel straight across a river at a speed 0.75 m/s. The current in the river, however, flows at a speed of 1.20 m/s to the right. What is the total displacement of the boat relative to the shore?[/caption]

Refer to Figure 4, which shows a boat trying to go straight across the river. Let us calculate the magnitude and direction of the boat’s velocity relative to an observer on the shore, v_tot. The velocity of the boat, v_boat, is 0.75 m/s in the y-direction relative to the river and the velocity of the river, v_river, is 1.20 m/s to the right.

Strategy

We start by choosing a coordinate system with its x-axis parallel to the velocity of the river, as shown in Figure 4. Because the boat is directed straight toward the other shore, its velocity relative to the water is parallel to the y-axis and perpendicular to the velocity of the river. Thus, we can add the two velocities by using the equations [latex]\boldsymbol{v_{tot}=\sqrt{v_x^2+v_y^2}}[/latex] and θ = tan^-1 (v_y / v_x) directly.

Solution

The magnitude of the total velocity is

[latex]\boldsymbol{v_{tot}=\sqrt{v_x^2+v_y^2}},[/latex]

where

[latex]\boldsymbol{v_x=v_{river}=1.20\textbf{ m/s}}[/latex]

and

[latex]\boldsymbol{v_y=v_{boat}=0.750\textbf{ m/s}}.[/latex]

Thus,

[latex]\boldsymbol{v_{tot}=\sqrt{(1.20\textbf{ m/s})^2+(0.750\textbf{ m/s})^2}}[/latex]

yielding

[latex]\boldsymbol{v_{tot}=1.42\textbf{ m/s}}.[/latex]

The direction of the total velocity, θ, is given by:

[latex]\boldsymbol{\theta=\textbf{tan}^{-1}(v_y/v_x)=\textbf{tan}^{-1}(0.750/1.20)}.[/latex]

This equation gives

[latex]\boldsymbol{\theta=32.0^0}.[/latex]

Discussion

Both the magnitude v and the direction θ of the total velocity are consistent with Figure 4. Note that because the velocity of the river is large compared with the velocity of the boat, it is swept rapidly downstream. This result is evidenced by the small angle (only 32.0°) the total velocity has relative to the riverbank.

Example 2: Calculating Velocity: Wind Velocity Causes an Airplane to Drift

Calculate the wind velocity for the situation shown in Figure 5. The plane is known to be moving at 45.0 m/s due north relative to the air mass, while its velocity relative to the ground (its total velocity) is 38.0 m/s in a direction 20.0° west of north.

[caption id="" align="aligncenter" width="325"]Figure 5. An airplane is known to be heading north at 45.0 m/s, though its velocity relative to the ground is 38.0 m/s at an angle west of north. What is the speed and direction of the wind?[/caption]

Strategy

In this problem, somewhat different from the previous example, we know the total velocity v_tot and that it is the sum of two other velocities, v_w (the wind) and v_p (the plane relative to the air mass). The quantity v_p is known, and we are asked to find v_w. None of the velocities are perpendicular, but it is possible to find their components along a common set of perpendicular axes. If we can find the components of v_w, then we can combine them to solve for its magnitude and direction. As shown in Figure 5, we choose a coordinate system with its x-axis due east and its y-axis due north (parallel to v_p). (You may wish to look back at the discussion of the addition of vectors using perpendicular components in Chapter 3.3 Vector Addition and Subtraction: Analytical Methods.)

Solution

Because v_tot is the vector sum of the v_w and v_p, its x- and y-components are the sums of the x- and y-components of the wind and plane velocities. Note that the plane only has vertical component of velocity so v_px = 0 and v_py = v_p. That is,

[latex]\boldsymbol{v_{totx}=v_{wx}}[/latex]

and

[latex]\boldsymbol{v_{toty}=v_{wy}+v_p.}[/latex]

We can use the first of these two equations to find v_wx:

[latex]\boldsymbol{v_{wy}=v_{totx}=v_{tot}\textbf{cos }110^0}.[/latex]

Because v_tot = 38.0 m/s and cos 110° = -0.342 we have

[latex]\boldsymbol{v_{wy}=(38.0\textbf{ m/s})(-0.342)=-13\textbf{ m/s}.}[/latex]

The minus sign indicates motion west which is consistent with the diagram.

Now, to find v_wy we note that

[latex]\boldsymbol{v_{toty}=v_{wy}+v_p}[/latex]

Here v_toty = v_tot sin 110°; thus,

[latex]\boldsymbol{v_{wy}=(38.0\textbf{ m/s})(0.940)-45.0\textbf{ m/s}=-9.29\textbf{ m/s}.}[/latex]

This minus sign indicates motion south which is consistent with the diagram.

Now that the perpendicular components of the wind velocity v_wx and v_wy are known, we can find the magnitude and direction of v_w. First, the magnitude is

[latex]\boldsymbol{v_w=\sqrt{v_{wx}^2+v_{wy}^2}}[/latex]

[latex]\boldsymbol{=(-13.0\textbf{ m/s})^2+(-9.29\textbf{ m/s})^2}\:\:\:\:\:\:[/latex]

so that

[latex]\boldsymbol{v_w=16.0\textbf{ m/s}.}[/latex]

The direction is:

[latex]\boldsymbol{\theta=\textbf{tan}^{-1}(v_{wy}/v_{wx})=\textbf{tan}^{-1}(-9.29/-13.0)}[/latex]

giving

[latex]\boldsymbol{\theta=35.6^0}.[/latex]

Discussion

The wind’s speed and direction are consistent with the significant effect the wind has on the total velocity of the plane, as seen in Figure 5. Because the plane is fighting a strong combination of crosswind and head-wind, it ends up with a total velocity significantly less than its velocity relative to the air mass as well as heading in a different direction.

Note that in both of the last two examples, we were able to make the mathematics easier by choosing a coordinate system with one axis parallel to one of the velocities. We will repeatedly find that choosing an appropriate coordinate system makes problem solving easier. For example, in projectile motion we always use a coordinate system with one axis parallel to gravity.

Relative Velocities and Classical Relativity

When adding velocities, we have been careful to specify that the velocity is relative to some reference frame. These velocities are called relative velocities. For example, the velocity of an airplane relative to an air mass is different from its velocity relative to the ground. Both are quite different from the velocity of an airplane relative to its passengers (which should be close to zero). Relative velocities are one aspect of relativity, which is defined to be the study of how different observers moving relative to each other measure the same phenomenon.

Nearly everyone has heard of relativity and immediately associates it with Albert Einstein (1879–1955), the greatest physicist of the 20th century. Einstein revolutionized our view of nature with his modern theory of relativity, which we shall study in later chapters. The relative velocities in this section are actually aspects of classical relativity, first discussed correctly by Galileo and Isaac Newton. Classical relativity is limited to situations where speeds are less than about 1% of the speed of light—that is, less than 3,000 km/s. Most things we encounter in daily life move slower than this speed.

Let us consider an example of what two different observers see in a situation analyzed long ago by Galileo. Suppose a sailor at the top of a mast on a moving ship drops his binoculars. Where will it hit the deck? Will it hit at the base of the mast, or will it hit behind the mast because the ship is moving forward? The answer is that if air resistance is negligible, the binoculars will hit at the base of the mast at a point directly below its point of release. Now let us consider what two different observers see when the binoculars drop. One observer is on the ship and the other on shore. The binoculars have no horizontal velocity relative to the observer on the ship, and so he sees them fall straight down the mast. (See Figure 6.) To the observer on shore, the binoculars and the ship have the same horizontal velocity, so both move the same distance forward while the binoculars are falling. This observer sees the curved path shown in Figure 6. Although the paths look different to the different observers, each sees the same result—the binoculars hit at the base of the mast and not behind it. To get the correct description, it is crucial to correctly specify the velocities relative to the observer.

[caption id="" align="aligncenter" width="300"]Figure 6. Classical relativity. The same motion as viewed by two different observers. An observer on the moving ship sees the binoculars dropped from the top of its mast fall straight down. An observer on shore sees the binoculars take the curved path, moving forward with the ship. Both observers see the binoculars strike the deck at the base of the mast. The initial horizontal velocity is different relative to the two observers. (The ship is shown moving rather fast to emphasize the effect.)[/caption]

Example 3: Calculating Relative Velocity: An Airline Passenger Drops a Coin

An airline passenger drops a coin while the plane is moving at 260 m/s. What is the velocity of the coin when it strikes the floor 1.50 m below its point of release: (a) Measured relative to the plane? (b) Measured relative to the Earth?

Strategy

Both problems can be solved with the techniques for falling objects and projectiles. In part (a), the initial velocity of the coin is zero relative to the plane, so the motion is that of a falling object (one-dimensional). In part (b), the initial velocity is 260 m/s horizontal relative to the Earth and gravity is vertical, so this motion is a projectile motion. In both parts, it is best to use a coordinate system with vertical and horizontal axes.

Solution for (a)

Using the given information, we note that the initial velocity and position are zero, and the final position is 1.50 m. The final velocity can be found using the equation:

[latex]\boldsymbol{v_y^2=v_{0y}^2-2g(y-y_0).}[/latex]

Substituting known values into the equation, we get

[latex]\boldsymbol{v_y^2=0^2-2(9.80\textbf{ m/s}^2)(-1.50\textbf{ m}-0\textbf{ m})=29.4\textbf{ m}^2/\textbf{ s}^2}[/latex]

yielding

[latex]\boldsymbol{v_y=-5.42\textbf{ m/s}.}[/latex]

We know that the square root of 29.4 has two roots: 5.42 and -5.42. We choose the negative root because we know that the velocity is directed downwards, and we have defined the positive direction to be upwards. There is no initial horizontal velocity relative to the plane and no horizontal acceleration, and so the motion is straight down relative to the plane.

Solution for (b)

Because the initial vertical velocity is zero relative to the ground and vertical motion is independent of horizontal motion, the final vertical velocity for the coin relative to the ground is v_y = -5.42 m/s, the same as found in part (a). In contrast to part (a), there now is a horizontal component of the velocity. However, since there is no horizontal acceleration, the initial and final horizontal velocities are the same and v_x = 260 m/s. The x- and y-components of velocity can be combined to find the magnitude of the final velocity:

[latex]\boldsymbol{v=\sqrt{{v_x}^2+{v_y}^2.}}[/latex]

Thus,

[latex]\boldsymbol{v=\sqrt{(260\textbf{ m/s})^2+(-5.42\textbf{ m/s})^2}}[/latex]

yielding

[latex]\boldsymbol{v=260.06\textbf{ m/s}.}[/latex]

The direction is given by:

[latex]\boldsymbol{\theta=\textbf{tan}^{-1}(v_y/v_x)=\textbf{tan}^{-1}(-5.42/260)}[/latex]

so that

[latex]\boldsymbol{\theta=\textbf{tan}^{-1}(-0.0208)=-1.19^0.}[/latex]

Discussion

In part (a), the final velocity relative to the plane is the same as it would be if the coin were dropped from rest on the Earth and fell 1.50 m. This result fits our experience; objects in a plane fall the same way when the plane is flying horizontally as when it is at rest on the ground. This result is also true in moving cars. In part (b), an observer on the ground sees a much different motion for the coin. The plane is moving so fast horizontally to begin with that its final velocity is barely greater than the initial velocity. Once again, we see that in two dimensions, vectors do not add like ordinary numbers—the final velocity v in part (b) is not (260 - 5.42) m/s; rather, it is 260.06 m/s. The velocity’s magnitude had to be calculated to five digits to see any difference from that of the airplane. The motions as seen by different observers (one in the plane and one on the ground) in this example are analogous to those discussed for the binoculars dropped from the mast of a moving ship, except that the velocity of the plane is much larger, so that the two observers see very different paths. (See Figure 7.) In addition, both observers see the coin fall 1.50 m vertically, but the one on the ground also sees it move forward 144 m (this calculation is left for the reader). Thus, one observer sees a vertical path, the other a nearly horizontal path.

MAKING CONNECTIONS: RELATIVITY AND EINSTEIN

Because Einstein was able to clearly define how measurements are made (some involve light) and because the speed of light is the same for all observers, the outcomes are spectacularly unexpected. Time varies with observer, energy is stored as increased mass, and more surprises await.

PHET EXPLORATIONS: MOTION IN 2D

Try the new "Ladybug Motion 2D" simulation for the latest updated version. Learn about position, velocity, and acceleration vectors. Move the ball with the mouse or let the simulation move the ball in four types of motion (2 types of linear, simple harmonic, circle).

[caption id="" align="aligncenter" width="450"] Figure 8. Motion in 2D[/caption]

Summary

• Velocities in two dimensions are added using the same analytical vector techniques, which are rewritten as
  [latex]\boldsymbol{v_x=v\:\textbf{cos}\:\theta}[/latex]
  [latex]\boldsymbol{v_y=v\:\textbf{sin}\:\theta}[/latex]
  [latex]\boldsymbol{v=\sqrt{v_x^2+v_y^2}}[/latex]
  [latex]\boldsymbol{\theta=\textbf{tan}^{-1}(v_y/v_x).}[/latex]
• Relative velocity is the velocity of an object as observed from a particular reference frame, and it varies dramatically with reference frame.
• Relativity is the study of how different observers measure the same phenomenon, particularly when the observers move relative to one another. Classical relativity is limited to situations where speed is less than about 1% of the speed of light (3000 km/s).

Conceptual Questions

1: What frame or frames of reference do you instinctively use when driving a car? When flying in a commercial jet airplane?

2: A basketball player dribbling down the court usually keeps his eyes fixed on the players around him. He is moving fast. Why doesn’t he need to keep his eyes on the ball?

3: If someone is riding in the back of a pickup truck and throws a softball straight backward, is it possible for the ball to fall straight down as viewed by a person standing at the side of the road? Under what condition would this occur? How would the motion of the ball appear to the person who threw it?

4: The hat of a jogger running at constant velocity falls off the back of his head. Draw a sketch showing the path of the hat in the jogger’s frame of reference. Draw its path as viewed by a stationary observer.

5: A clod of dirt falls from the bed of a moving truck. It strikes the ground directly below the end of the truck. What is the direction of its velocity relative to the truck just before it hits? Is this the same as the direction of its velocity relative to ground just before it hits? Explain your answers.

Problems & Exercises

1: Bryan Allen pedaled a human-powered aircraft across the English Channel from the cliffs of Dover to Cap Gris-Nez on June 12, 1979. (a) He flew for 169 min at an average velocity of 3.53 m/s in a direction 45° south of east. What was his total displacement? (b) Allen encountered a headwind averaging 2.00 m/s almost precisely in the opposite direction of his motion relative to the Earth. What was his average velocity relative to the air? (c) What was his total displacement relative to the air mass?

2: A seagull flies at a velocity of 9.00 m/s straight into the wind. (a) If it takes the bird 20.0 min to travel 6.00 km relative to the Earth, what is the velocity of the wind? (b) If the bird turns around and flies with the wind, how long will he take to return 6.00 km? (c) Discuss how the wind affects the total round-trip time compared to what it would be with no wind.

3: Near the end of a marathon race, the first two runners are separated by a distance of 45.0 m. The front runner has a velocity of 3.50 m/s, and the second a velocity of 4.20 m/s. (a) What is the velocity of the second runner relative to the first? (b) If the front runner is 250 m from the finish line, who will win the race, assuming they run at constant velocity? (c) What distance ahead will the winner be when she crosses the finish line?

4: Verify that the coin dropped by the airline passenger in the Example 3 travels 144 m horizontally while falling 1.50 m in the frame of reference of the Earth.

5: A football quarterback is moving straight backward at a speed of 2.00 m/s when he throws a pass to a player 18.0 m straight downfield. The ball is thrown at an angle of 25.0° relative to the ground and is caught at the same height as it is released. What is the initial velocity of the ball relative to the quarterback ?

6: A ship sets sail from Rotterdam, The Netherlands, heading due north at 7.00 m/s relative to the water. The local ocean current is 1.50 m/s in a direction 40.0° north of east. What is the velocity of the ship relative to the Earth?

7: (a) A jet airplane flying from Darwin, Australia, has an air speed of 260 m/s in a direction 5.0° south of west. It is in the jet stream, which is blowing at 35.0 m/s in a direction 15° south of east. What is the velocity of the airplane relative to the Earth? (b) Discuss whether your answers are consistent with your expectations for the effect of the wind on the plane’s path.

8: (a) In what direction would the ship in Exercise 6 have to travel in order to have a velocity straight north relative to the Earth, assuming its speed relative to the water remains 7.00 m/s? (b) What would its speed be relative to the Earth?

9: (a) Another airplane is flying in a jet stream that is blowing at 45.0 m/s in a direction 20° south of east (as in Exercise 7). Its direction of motion relative to the Earth is 45.0° south of west, while its direction of travel relative to the air is 5.00° south of west. What is the airplane’s speed relative to the air mass? (b) What is the airplane’s speed relative to the Earth?

10: A sandal is dropped from the top of a 15.0-m-high mast on a ship moving at 1.75 m/s due south. Calculate the velocity of the sandal when it hits the deck of the ship: (a) relative to the ship and (b) relative to a stationary observer on shore. (c) Discuss how the answers give a consistent result for the position at which the sandal hits the deck.

11: The velocity of the wind relative to the water is crucial to sailboats. Suppose a sailboat is in an ocean current that has a velocity of 2.20 m/s in a direction 30.0° east of north relative to the Earth. It encounters a wind that has a velocity of 4.50 m/s in a direction of 50.0° south of west relative to the Earth. What is the velocity of the wind relative to the water?

12: The great astronomer Edwin Hubble discovered that all distant galaxies are receding from our Milky Way Galaxy with velocities proportional to their distances. It appears to an observer on the Earth that we are at the center of an expanding universe. Figure 9 illustrates this for five galaxies lying along a straight line, with the Milky Way Galaxy at the center. Using the data from the figure, calculate the velocities: (a) relative to galaxy 2 and (b) relative to galaxy 5. The results mean that observers on all galaxies will see themselves at the center of the expanding universe, and they would likely be aware of relative velocities, concluding that it is not possible to locate the center of expansion with the given information.

13: (a) Use the distance and velocity data in Figure 9 to find the rate of expansion as a function of distance.

(b) If you extrapolate back in time, how long ago would all of the galaxies have been at approximately the same position? The two parts of this problem give you some idea of how the Hubble constant for universal expansion and the time back to the Big Bang are determined, respectively.

14: An athlete crosses a 25-m-wide river by swimming perpendicular to the water current at a speed of 0.5 m/s relative to the water. He reaches the opposite side at a distance 40 m downstream from his starting point. How fast is the water in the river flowing with respect to the ground? What is the speed of the swimmer with respect to a friend at rest on the ground?

15: A ship sailing in the Gulf Stream is heading 25.0° west of north at a speed of 4.00 m/s relative to the water. Its velocity relative to the Earth is 4.80 m/s 5.00° west of north. What is the velocity of the Gulf Stream? (The velocity obtained is typical for the Gulf Stream a few hundred kilometers off the east coast of the United States.)

16: An ice hockey player is moving at 8.00 m/s when he hits the puck toward the goal. The speed of the puck relative to the player is 29.0 m/s. The line between the center of the goal and the player makes a 90.0° angle relative to his path as shown in Figure 10. What angle must the puck’s velocity make relative to the player (in his frame of reference) to hit the center of the goal?

15: Unreasonable Results Suppose you wish to shoot supplies straight up to astronauts in an orbit 36,000 km above the surface of the Earth. (a) At what velocity must the supplies be launched? (b) What is unreasonable about this velocity? (c) Is there a problem with the relative velocity between the supplies and the astronauts when the supplies reach their maximum height? (d) Is the premise unreasonable or is the available equation inapplicable? Explain your answer.

16: Unreasonable Results A commercial airplane has an air speed of 280 m/s due east and flies with a strong tailwind. It travels 3000 km in a direction 5° south of east in 1.50 h. (a) What was the velocity of the plane relative to the ground? (b) Calculate the magnitude and direction of the tailwind’s velocity. (c) What is unreasonable about both of these velocities? (d) Which premise is unreasonable?

17: Construct Your Own Problem Consider an airplane headed for a runway in a cross wind. Construct a problem in which you calculate the angle the airplane must fly relative to the air mass in order to have a velocity parallel to the runway. Among the things to consider are the direction of the runway, the wind speed and direction (its velocity) and the speed of the plane relative to the air mass. Also calculate the speed of the airplane relative to the ground. Discuss any last minute maneuvers the pilot might have to perform in order for the plane to land with its wheels pointing straight down the runway.

Glossary

classical relativity

the study of relative velocities in situations where speeds are less than about 1% of the speed of light—that is, less than 3000 km/s

relative velocity

the velocity of an object as observed from a particular reference frame

relativity

the study of how different observers moving relative to each other measure the same phenomenon

velocity

speed in a given direction

vector addition

the rules that apply to adding vectors together

Solution

Problems & Exercises:

1: (a) [latex]\boldsymbol{35.8\textbf{ km}}[/latex] south of east (b) [latex]\boldsymbol{5.53\textbf{ m/s}}[/latex] south of east (c) [latex]\boldsymbol{56.1\textbf{ km}}[/latex] south of east

3: (a) $$\boldsymbol{0.70\textbf{ m/s faster}}$$ (b) Second runner wins (c) $$\boldsymbol{4.17\textbf{ m}}$$

5: [latex]\boldsymbol{17.0\textbf{ m/s}},\:\boldsymbol{22.1^0}[/latex] 7: (a) [latex]\boldsymbol{230\textbf{ m/s}},\:\boldsymbol{8.0^0}[/latex] south of west (b) The wind should make the plane travel slower and more to the south, which is what was calculated

9: (a) $$\boldsymbol{63.5\textbf{ m/s}}$$ (b) $$\boldsymbol{29.6\textbf{ m/s}}$$

11: [latex]\boldsymbol{6.68\textbf{ m/s}},\:\boldsymbol{53.3^0}[/latex] south of west

13: (a) [latex]\boldsymbol{H_{\textbf{average}}=14.9\frac{\textbf{ km/s}}{\textbf{Mly}}}[/latex] (b) $$\boldsymbol{20.2\textbf{ billion years}}$$

15: [latex]\boldsymbol{1.72\textbf{ m/s}},\:\boldsymbol{42.3^0}[/latex] north of east

]]> 265 194 6 0 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/chapter/4-0-introduction/ Thu, 29 Jun 2017 22:13:26 +0000 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/chapter/4-0-introduction/ [caption id="" align="aligncenter" width="875"]Figure 1. Newton’s laws of motion describe the motion of the dolphin’s path. (credit: Jin Jang)[/caption]

Motion draws our attention. Motion itself can be beautiful, causing us to marvel at the forces needed to achieve spectacular motion, such as that of a dolphin jumping out of the water, or a pole vaulter, or the flight of a bird, or the orbit of a satellite. The study of motion is kinematics, but kinematics only describes the way objects move—their velocity and their acceleration. Dynamics considers the forces that affect the motion of moving objects and systems. Newton’s laws of motion are the foundation of dynamics. These laws provide an example of the breadth and simplicity of principles under which nature functions. They are also universal laws in that they apply to similar situations on Earth as well as in space.

Isaac Newton’s (1642–1727) laws of motion were just one part of the monumental work that has made him legendary. The development of Newton’s laws marks the transition from the Renaissance into the modern era. This transition was characterized by a revolutionary change in the way people thought about the physical universe. For many centuries natural philosophers had debated the nature of the universe based largely on certain rules of logic with great weight given to the thoughts of earlier classical philosophers such as Aristotle (384–322 BC). Among the many great thinkers who contributed to this change were Newton and Galileo.

[caption id="" align="aligncenter" width="340"]Figure 2. Isaac Newton’s monumental work, Philosophiae Naturalis Principia Mathematica, was published in 1687. It proposed scientific laws that are still used today to describe the motion of objects. (credit: Service commun de la documentation de l'Université de Strasbourg)[/caption]

Galileo was instrumental in establishing observation as the absolute determinant of truth, rather than “logical” argument. Galileo’s use of the telescope was his most notable achievement in demonstrating the importance of observation. He discovered moons orbiting Jupiter and made other observations that were inconsistent with certain ancient ideas and religious dogma. For this reason, and because of the manner in which he dealt with those in authority, Galileo was tried by the Inquisition and punished. He spent the final years of his life under a form of house arrest. Because others before Galileo had also made discoveries by observing the nature of the universe, and because repeated observations verified those of Galileo, his work could not be suppressed or denied. After his death, his work was verified by others, and his ideas were eventually accepted by the church and scientific communities.

Galileo also contributed to the formation of what is now called Newton’s first law of motion. Newton made use of the work of his predecessors, which enabled him to develop laws of motion, discover the law of gravity, invent calculus, and make great contributions to the theories of light and color. It is amazing that many of these developments were made with Newton working alone, without the benefit of the usual interactions that take place among scientists today.

It was not until the advent of modern physics early in the 20th century that it was discovered that Newton’s laws of motion produce a good approximation to motion only when the objects are moving at speeds much, much less than the speed of light and when those objects are larger than the size of most molecules (about[latex]\boldsymbol{10^{-9}\textbf{ m}}[/latex]in diameter). These constraints define the realm of classical mechanics, as discussed in Chapter 1 Introduction to the Nature of Science and Physics. At the beginning of the 20^th century, Albert Einstein (1879–1955) developed the theory of relativity and, along with many other scientists, developed quantum theory. This theory does not have the constraints present in classical physics. All of the situations we consider in this chapter, and all those preceding the introduction of relativity in Chapter 28 Special Relativity, are in the realm of classical physics.

MAKING CONNECTIONS: PAST AND PRESENT PHILOSOPHY

The importance of observation and the concept of cause and effect were not always so entrenched in human thinking. This realization was a part of the evolution of modern physics from natural philosophy. The achievements of Galileo, Newton, Einstein, and others were key milestones in the history of scientific thought. Most of the scientific theories that are described in this book descended from the work of these scientists.

]]> 269 266 1 0 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/chapter/4-1-development-of-force-concept/ Thu, 29 Jun 2017 22:13:27 +0000 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/chapter/4-1-development-of-force-concept/

Summary

• Define force.

Dynamics is the study of the forces that cause objects and systems to move. To understand this, we need a working definition of force. Our intuitive definition of force—that is, a push or a pull—is a good place to start. We know that a push or pull has both magnitude and direction (therefore, it is a vector quantity) and can vary considerably in each regard. For example, a cannon exerts a strong force on a cannonball that is launched into the air. In contrast, Earth exerts only a tiny downward pull on a flea. Our everyday experiences also give us a good idea of how multiple forces add. If two people push in different directions on a third person, as illustrated in Figure 1, we might expect the total force to be in the direction shown. Since force is a vector, it adds just like other vectors, as illustrated in Figure 2(a) for two ice skaters. Forces, like other vectors, are represented by arrows and can be added using the familiar head-to-tail method or by trigonometric methods. These ideas were developed in Chapter 3 Two-Dimensional Kinematics.

[caption id="attachment_3535" align="aligncenter" width="300"]Figure 1. Part (a) shows an overhead view of two ice skaters pushing on a third. Forces are vectors and add like other vectors, so the total force on the third skater is in the direction shown. In part (b), we see a free-body diagram representing the forces acting on the third skater.[/caption]

Figure 1(b) is our first example of a free-body diagram, which is a technique used to illustrate all the external forces acting on a body. The body is represented by a single isolated point (or free body), and only those forces acting on the body from the outside (external forces) are shown. (These forces are the only ones shown, because only external forces acting on the body affect its motion. We can ignore any internal forces within the body.) Free-body diagrams are very useful in analyzing forces acting on a system and are employed extensively in the study and application of Newton’s laws of motion.

A more quantitative definition of force can be based on some standard force, just as distance is measured in units relative to a standard distance. One possibility is to stretch a spring a certain fixed distance, as illustrated in Figure 2, and use the force it exerts to pull itself back to its relaxed shape—called a restoring force—as a standard. The magnitude of all other forces can be stated as multiples of this standard unit of force. Many other possibilities exist for standard forces. (One that we will encounter in Chapter 22 Magnetism is the magnetic force between two wires carrying electric current.) Some alternative definitions of force will be given later in this chapter.

[caption id="" align="aligncenter" width="400"]Figure 2. The force exerted by a stretched spring can be used as a standard unit of force. (a) This spring has a length x when undistorted. (b) When stretched a distance Δx, the spring exerts a restoring force, F_restore, which is reproducible. (c) A spring scale is one device that uses a spring to measure force. The force F_restore is exerted on whatever is attached to the hook. Here F_restore has a magnitude of 6 units in the force standard being employed.[/caption]

TAKE-HOME EXPERIMENT: FORCE STANDARDS

To investigate force standards and cause and effect, get two identical rubber bands. Hang one rubber band vertically on a hook. Find a small household item that could be attached to the rubber band using a paper clip, and use this item as a weight to investigate the stretch of the rubber band. Measure the amount of stretch produced in the rubber band with one, two, and four of these (identical) items suspended from the rubber band. What is the relationship between the number of items and the amount of stretch? How large a stretch would you expect for the same number of items suspended from two rubber bands? What happens to the amount of stretch of the rubber band (with the weights attached) if the weights are also pushed to the side with a pencil?

Section Summary

• Dynamics is the study of how forces affect the motion of objects.
• Force is a push or pull that can be defined in terms of various standards, and it is a vector having both magnitude and direction.
• External forces are any outside forces that act on a body. A free-body diagram is a drawing of all external forces acting on a body.

Conceptual Questions

1: Propose a force standard different from the example of a stretched spring discussed in the text. Your standard must be capable of producing the same force repeatedly.

2: What properties do forces have that allow us to classify them as vectors?

Glossary

dynamics

the study of how forces affect the motion of objects and systems

external force

a force acting on an object or system that originates outside of the object or system

free-body diagram

a sketch showing all of the external forces acting on an object or system; the system is represented by a dot, and the forces are represented by vectors extending outward from the dot

force

a push or pull on an object with a specific magnitude and direction; can be represented by vectors; can be expressed as a multiple of a standard force

]]> 272 266 2 0 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/chapter/4-2-hookes-law-originally-section-5-3-elasticity-stress-and-strain/ Thu, 29 Jun 2017 22:13:28 +0000 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/chapter/4-2-hookes-law-originally-section-5-3-elasticity-stress-and-strain/

Summary

• State Hooke’s law.
• Explain Hooke’s law using graphical representation between deformation and applied force.
• Discuss deformations such as changes in length
• Determine the change in length given mass, length and radius.

Forces can affect an object’s shape. If a bulldozer pushes a car into a wall, the car will not move but it will noticeably change shape. A change in shape due to the application of a force is a deformation. Even very small forces are known to cause some deformation. For small deformations, two important characteristics are observed. First, the object returns to its original shape when the force is removed—that is, the deformation is elastic for small deformations. Second, the size of the deformation is proportional to the force—that is, for small deformations, Hooke’s law is obeyed. In equation form, Hooke’s law is given by

[latex]\boldsymbol{F = k\Delta{L},}[/latex]

where ΔL is the amount of deformation (the change in length, for example) produced by the force F, and k is a proportionality constant that depends on the shape and composition of the object and the direction of the force. Note that this force is a function of the deformation ΔL — it is not constant as a kinetic friction force is. Sometimes we use Δx instead of ΔL.  The deformation can be along any axis.  Rearranging this to

[latex]\boldsymbol{\Delta{L}\: = }[/latex][latex size="2"]\boldsymbol{\frac{F}{k}}[/latex]

makes it clear that the deformation is proportional to the applied force. Figure 1 shows the Hooke’s law relationship between the extension ΔL of a spring or of a human bone. For metals or springs, the straight line region in which Hooke’s law pertains is much larger. Bones are brittle and the elastic region is small and the fracture abrupt. Eventually a large enough stress to the material will cause it to break or fracture. Tensile strength is the breaking stress that will cause permanent deformation or fracture of a material.

HOOKE'S LAW

[latex]\boldsymbol{F = k\Delta{L},}[/latex]

where ΔL is the amount of deformation (the change in length, for example) produced by the force F, and k is a proportionality constant that depends on the shape and composition of the object and the direction of the force.

[latex]\boldsymbol{\Delta{L}\:=}[/latex][latex size="2"]\boldsymbol{\frac{F}{k}}[/latex]

The proportionality constant k depends upon a number of factors for the material. For example, a guitar string made of nylon stretches when it is tightened, and the elongation ΔL is proportional to the force applied (at least for small deformations). Thicker nylon strings and ones made of steel stretch less for the same applied force, implying they have a larger k (see Figure 2). Finally, all three strings return to their normal lengths when the force is removed, provided the deformation is small. Most materials will behave in this manner if the deformation is less than about 0.1% or about 1 part in 10³.

STRETCH YOURSELF A LITTLE

How would you go about measuring the proportionality constant k of a rubber band? If a rubber band stretched 3 cm when a 100-g mass was attached to it, then how much would it stretch if two similar rubber bands were attached to the same mass—even if put together in parallel or alternatively if tied together in series?

 

Conceptual Questions

1: How is a bow and arrow like a spring?

Problems & Exercises

1. A mass is suspended from a vertical spring such that it exerts a downwards force of magnitude 5.02 newtons on the spring.  The spring stretches 0.0456 metres when the mass is attached to it.  What is the spring constant in newton/metres?  Hint:  Suspended means hanging in the air not moving.  What is the net force on a mass that is not moving?
2. The shock absorbers on my car have a spring constant of 9876 newton/metres. When a certain person sits in the car they exert a downwards force of 97,900 newtons on the car.  How far "down" does the car move after they sit in it?    (Watch the movie "The Italian Job" and see how  they used this type of physics to figure out which truck was actually carrying the gold.)
3. A spring balance has a spring constant of 34.5 newton/metres and it stretches 3.21 centimetres when an unknown mass is attached to it.  What is the force in newtons exerted on the spring by the unknown mass?
4. If an applied force of 12 newtons stretches a certain spring 4.0 centimetres, how much stretch will occur for an applied for of 18 newtons?

 

Solutions

Problems & Exercises 1) 110 N/m 2) 0.101 m = 10.1 cm

3) 1.11 newtons

4) 6.0 cm       By ratio and proportion    12 N / 4.0 cm = 18 N/ ?    so ? = 18 x 4 / 12 =  

We now consider the type of deformation that will cause a change in length (tension and compression).   There are also sideways forces that cause shear (stress), and changes in volume, but we will not go into detail about them in this course.

Changes in Length—Tension and Compression: Elastic Modulus

A change in length ΔL is produced when a force is applied to a wire or rod parallel to its length L₀, either stretching it (a tension) or compressing it. (See Figure 3.)

[caption id="" align="aligncenter" width="175"]Figure 3. (a) Tension. The rod is stretched a length ΔL when a force is applied parallel to its length. (b) Compression. The same rod is compressed by forces with the same magnitude in the opposite direction. For very small deformations and uniform materials, ΔL is approximately the same for the same magnitude of tension or compression. For larger deformations, the cross-sectional area changes as the rod is compressed or stretched.[/caption]

Experiments have shown that the change in length (ΔL) depends on only a few variables. As already noted, ΔL is proportional to the force F and depends on the substance from which the object is made. Additionally, the change in length is proportional to the original length L₀ and inversely proportional to the cross-sectional area of the wire or rod. For example, a long guitar string will stretch more than a short one, and a thick string will stretch less than a thin one. We can combine all these factors into one equation for ΔL:

[latex]\boldsymbol{\Delta{L}\:=}[/latex][latex size = "2"]\boldsymbol{\frac{1}{Y}\frac{\vec{\textbf{F}}}{A}}[/latex][latex]\boldsymbol{L_0},[/latex]

where ΔL is the change in length, F the applied force, Y is a factor, called the elastic modulus or Young’s modulus, that depends on the substance, A is the cross-sectional area, and L₀ is the original length. Table 3 lists values of Y for several materials—those with a large Y are said to have a large tensile stiffness because they deform less for a given tension or compression.

Material	Young’s modulus (tension–compression)Y[latex]\boldsymbol{(10^9\textbf{N/m}^2)}[/latex]	Shear modulus S[latex]\boldsymbol{(10^9\textbf{N/m}^2)}[/latex]	Bulk modulus B[latex]\boldsymbol{(10^9\textbf{N/m}^2)}[/latex]
Aluminum	70	25	75
Bone – tension	16	80	8
Bone – compression	9
Brass	90	35	75
Brick	15
Concrete	20
Glass	70	20	30
Granite	45	20	45
Hair (human)	10
Hardwood	15	10
Iron, cast	100	40	90
Lead	16	5	50
Marble	60	20	70
Nylon	5
Polystyrene	3
Silk	6
Spider thread	3
Steel	210	80	130
Tendon	1
Acetone			0.7
Ethanol			0.9
Glycerin			4.5
Mercury			25
Water			2.2
Table 3. Elastic Moduli1.

Young’s moduli are not listed for liquids and gases in Table 3 because they cannot be stretched or compressed in only one direction. Note that there is an assumption that the object does not accelerate, so that there are actually two applied forces of magnitude F acting in opposite directions. For example, the strings in Figure 3 are being pulled down by a force of magnitude w and held up by the ceiling, which also exerts a force of magnitude w.

Example 1: The Stretch of a Long Cable

Suspension cables are used to carry gondolas at ski resorts. (See Figure 4) Consider a suspension cable that includes an unsupported span of 3 km. Calculate the amount of stretch in the steel cable. Assume that the cable has a diameter of 5.6 cm and the maximum tension it can withstand is 3.0 × 10⁶ N.

Strategy

The force is equal to the maximum tension, or F = 3.0 × 10⁶ N. The cross-sectional area is  πr²  =  2.46 × 10^-3 m². The equation [latex]\boldsymbol{\Delta{L} = \frac{1}{Y}\frac{F}{A}L_0}[/latex] can be used to find the change in length.

Solution

All quantities are known. Thus,

[latex]\boldsymbol{\Delta{L}\:=(\frac{1}{210\times10^9\textbf{ N/m}^2})(\frac{3.0\times10^6\textbf{ N}}{2.46\times10^{-3}\textbf{ m}^2})(3020\textbf{ m})}[/latex]

[latex]\boldsymbol{=18\textbf{ m}}.[/latex]

Discussion

This is quite a stretch, but only about 0.6% of the unsupported length. Effects of temperature upon length might be important in these environments.

Bones, on the whole, do not fracture due to tension or compression. Rather they generally fracture due to sideways impact or bending, resulting in the bone shearing or snapping. The behaviour of bones under tension and compression is important because it determines the load the bones can carry. Bones are classified as weight-bearing structures such as columns in buildings and trees. Weight-bearing structures have special features; columns in building have steel-reinforcing rods while trees and bones are fibrous. The bones in different parts of the body serve different structural functions and are prone to different stresses. Thus the bone in the top of the femur is arranged in thin sheets separated by marrow while in other places the bones can be cylindrical and filled with marrow or just solid. Overweight people have a tendency toward bone damage due to sustained compressions in bone joints and tendons.

Another biological example of Hooke’s law occurs in tendons. Functionally, the tendon (the tissue connecting muscle to bone) must stretch easily at first when a force is applied, but offer a much greater restoring force for a greater strain. Figure 5 shows a stress-strain relationship for a human tendon. Some tendons have a high collagen content so there is relatively little strain, or length change; others, like support tendons (as in the leg) can change length up to 10%. Note that this stress-strain curve is nonlinear, since the slope of the line changes in different regions. In the first part of the stretch called the toe region, the fibers in the tendon begin to align in the direction of the stress—this is called uncrimping. In the linear region, the fibrils will be stretched, and in the failure region individual fibers begin to break. A simple model of this relationship can be illustrated by springs in parallel: different springs are activated at different lengths of stretch. Examples of this are given in the problems at end of this chapter. Ligaments (tissue connecting bone to bone) behave in a similar way.

[caption id="" align="aligncenter" width="225"]Figure 5. Typical stress-strain curve for mammalian tendon. Three regions are shown: (1) toe region (2) linear region, and (3) failure region.[/caption]

Unlike bones and tendons, which need to be strong as well as elastic, the arteries and lungs need to be very stretchable. The elastic properties of the arteries are essential for blood flow. The pressure in the arteries increases and arterial walls stretch when the blood is pumped out of the heart. When the aortic valve shuts, the pressure in the arteries drops and the arterial walls relax to maintain the blood flow. When you feel your pulse, you are feeling exactly this—the elastic behaviour of the arteries as the blood gushes through with each pump of the heart. If the arteries were rigid, you would not feel a pulse. The heart is also an organ with special elastic properties. The lungs expand with muscular effort when we breathe in but relax freely and elastically when we breathe out. Our skins are particularly elastic, especially for the young. A young person can go from 100 kg to 60 kg with no visible sag in their skins. The elasticity of all organs reduces with age. Gradual physiological aging through reduction in elasticity starts in the early 20s.

Example 2: Calculating Deformation: How Much Does Your Leg Shorten When You Stand on It?

Calculate the change in length of the upper leg bone (the femur) when a 70.0 kg man supports 62.0 kg of his mass on it, assuming the bone to be equivalent to a uniform rod that is 40.0 cm long and 2.00 cm in radius.

Strategy

The force is equal to the weight supported, or

[latex]\boldsymbol{F = mg =(62.0\textbf{ kg})(9.80\textbf{ m/s}^2) = 607.6\textbf{ N}},[/latex]

and the cross-sectional area is  πr²  =  1.257 × 10^-3 m². The equation [latex]\boldsymbol{\Delta{L} = \frac{1}{Y}\frac{F}{A}L_0}[/latex] can be used to find the change in length.

Solution

All quantities except ΔL are known. Note that the compression value for Young’s modulus for bone must be used here. Thus,

[latex]\boldsymbol{\Delta{L}=(\frac{1}{9\times10^9\textbf{ N/m}^2})(\frac{607.6\textbf{ N}}{1.257\times10^{-3}\textbf{ m}^2})(0.400\textbf{ m})}[/latex]

[latex]\boldsymbol{=2\times10^{-5}\textbf{ m}}.[/latex]

Discussion

This small change in length seems reasonable, consistent with our experience that bones are rigid. In fact, even the rather large forces encountered during strenuous physical activity do not compress or bend bones by large amounts. Although bone is rigid compared with fat or muscle, several of the substances listed in Table 3 have larger values of Young’s modulus Y. In other words, they are more rigid.

The equation for change in length is traditionally rearranged and written in the following form:

[latex size="2"]\boldsymbol{\frac{F}{A}}[/latex][latex]\boldsymbol{=Y}[/latex][latex size="2"]\boldsymbol{\frac{\Delta{L}}{L_0}}.[/latex]

The ratio of force to area, [latex]\boldsymbol{\frac{F}{A}},[/latex]  is defined as stress (measured in N/m²), and the ratio of the change in length to length,[latex]\boldsymbol{\frac{\Delta{L}}{L_0}},[/latex]is defined as strain (a unitless quantity). In other words,

[latex]\boldsymbol{\textbf{stress}=Y\times\textbf{strain}}.[/latex]

In this form, the equation is analogous to Hooke’s law, with stress analogous to force and strain analogous to deformation. If we again rearrange this equation to the form

[latex]\boldsymbol{F=YA}[/latex][latex size="2"]\boldsymbol{\frac{\Delta{L}}{L_0}},[/latex]

we see that it is the same as Hooke’s law with a proportionality constant

[latex]\boldsymbol{k\:=}[/latex][latex size="2"]\boldsymbol{\frac{YA}{L_0}}.[/latex]

This general idea—that force and the deformation it causes are proportional for small deformations—applies to changes in length, sideways bending, and changes in volume.

STRESS

The ratio of force to area, [latex]\boldsymbol{\frac{F}{A}},[/latex] is defined as stress measured in N/m².

STRAIN

The ratio of the change in length to length, [latex]\boldsymbol{\frac{\Delta{L}}{L_0}},[/latex] is defined as strain (a unitless quantity). In other words,

[latex]\boldsymbol{\textbf{stress}=Y\times\textbf{strain}}.[/latex]

Section Summary

• Hooke’s law is given by
  [latex]\boldsymbol{F=k\Delta{L}},[/latex]    or  [latex]\boldsymbol{F=k\Delta{x}},[/latex]

  where ΔL is the amount of deformation (the change in length), F is the applied force, and k is a proportionality constant that depends on the shape and composition of the object and the direction of the force. The relationship between the deformation and the applied force can also be written as

  [latex]\boldsymbol{\Delta{L}\:=}[/latex][latex size="2"]\boldsymbol{\frac{1}{Y}\frac{F}{A}}[/latex][latex]\boldsymbol{L_0},[/latex]

  where Y is Young’s modulus, which depends on the substance, A is the cross-sectional area, and L₀ is the original length.

• The ratio of force to area, [latex]\boldsymbol{\frac{F}{A}},[/latex] is defined as stress, measured in N/m².
• The ratio of the change in length to length, [latex]\boldsymbol{\frac{\Delta{L}}{L_0}},[/latex] is defined as strain (a unitless quantity). In other words,
  [latex]\boldsymbol{\textbf{stress}=Y\times\textbf{strain}}.[/latex]

Conceptual Questions

1: The elastic properties of the arteries are essential for blood flow. Explain the importance of this in terms of the characteristics of the flow of blood (pulsating or continuous).

2: What are you feeling when you feel your pulse? Measure your pulse rate for 10 s and for 1 min. Is there a factor of 6 difference?

3: Examine different types of shoes, including sports shoes and thongs. In terms of physics, why are the bottom surfaces designed as they are? What differences will dry and wet conditions make for these surfaces?

4: Would you expect your height to be different depending upon the time of day? Why or why not?

5: Why can a squirrel jump from a tree branch to the ground and run away undamaged, while a human could break a bone in such a fall?

Problems & Exercises

1: During a circus act, one performer swings upside down hanging from a trapeze holding another, also upside-down, performer by the legs. If the upward force on the lower performer is three times her weight, how much do the bones (the femurs) in her upper legs stretch? You may assume each is equivalent to a uniform rod 35.0 cm long and 1.80 cm in radius. Her mass is 60.0 kg.

2: During a wrestling match, a 150 kg wrestler briefly stands on one hand during a maneuver designed to perplex his already moribund adversary. By how much does the upper arm bone shorten in length? The bone can be represented by a uniform rod 38.0 cm in length and 2.10 cm in radius.

3: (a) The “lead” in pencils is a graphite composition with a Young’s modulus of about 1 × 10⁹ N/m². Calculate the change in length of the lead in an automatic pencil if you tap it straight into the pencil with a force of 4.0 N. The lead is 0.50 mm in diameter and 60 mm long. (b) Is the answer reasonable? That is, does it seem to be consistent with what you have observed when using pencils?

4: TV broadcast antennas are the tallest artificial structures on Earth. In 1987, a 72.0-kg physicist placed himself and 400 kg of equipment at the top of one 610-m high antenna to perform gravity experiments. By how much was the antenna compressed, if we consider it to be equivalent to a steel cylinder 0.150 m in radius?

5: (a) By how much does a 65.0-kg mountain climber stretch her 0.800-cm diameter nylon rope when she hangs 35.0 m below a rock outcropping? (b) Does the answer seem to be consistent with what you have observed for nylon ropes? Would it make sense if the rope were actually a bungee cord?

7: As an oil well is drilled, each new section of drill pipe supports its own weight and that of the pipe and drill bit beneath it. Calculate the stretch in a new 6.00 m length of steel pipe that supports 3.00 km of pipe having a mass of 20.0 kg/m and a 100-kg drill bit. The pipe is equivalent in stiffness to a solid cylinder 5.00 cm in diameter.

8: Calculate the force a piano tuner applies to stretch a steel piano wire 8.00 mm, if the wire is originally 0.850 mm in diameter and 1.35 m long.

12: To consider the effect of wires hung on poles, we take data from Chapter 4.7 Example 2, in which tensions in wires supporting a traffic light were calculated. The left wire made an angle 30.0° below the horizontal with the top of its pole and carried a tension of 108 N. The 12.0 m tall hollow aluminum pole is equivalent in stiffness to a 4.50 cm diameter solid cylinder. (a) How far is it bent to the side? (b) By how much is it compressed?

15: This problem returns to the tightrope walker studied in Chapter 4.5 Example 2, who created a tension of 3.94 × 10³ N in a wire making an angle 5.0° below the horizontal with each supporting pole. Calculate how much this tension stretches the steel wire if it was originally 15 m long and 0.50 cm in diameter.

Footnotes

1. 1 Approximate and average values. Young’s moduli Y for tension and compression sometimes differ but are averaged here. Bone has significantly different Young’s moduli for tension and compression.

Glossary

deformation

change in shape due to the application of force

Hooke’s law

proportional relationship between the force F on a material and the deformation ΔL it causes, F = kΔL

tensile strength

the breaking stress that will cause permanent deformation or fraction of a material

stress

ratio of force to area

strain

ratio of change in length to original length

shear deformation

deformation perpendicular to the original length of an object

Solutions

Problems & Exercises 1: [latex]\boldsymbol{1.90\times10^{-3}\textbf{ cm}}[/latex]

3: (a) $$\boldsymbol{1\textbf{ mm}}$$ (b) This does seem reasonable, since the lead does seem to shrink a little when you push on it.

5: (a) $$\boldsymbol{9\textbf{ cm}}$$ (b) This seems reasonable for nylon climbing rope, since it is not supposed to stretch that much.

7: $$\boldsymbol{8.59\textbf{ mm}}$$ 9: [latex]\boldsymbol{1.49\times10^{-7}\textbf{ m}}[/latex]

11: (a) [latex]\boldsymbol{3.99\times10^{-7}\textbf{ m}}[/latex] (b) [latex]\boldsymbol{9.67\times10^{-8}\textbf{ m}}[/latex]

13: [latex]\boldsymbol{4\times10^6\textbf{ N/m}^2}.[/latex] This is about 36 atm, greater than a typical jar can withstand. 15: $$\boldsymbol{1.4\textbf{ cm}}$$

]]> 278 266 3 0 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/chapter/4-3-newtons-first-law-of-motion-inertia/ Thu, 29 Jun 2017 22:13:28 +0000 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/chapter/4-3-newtons-first-law-of-motion-inertia/

Summary

• Define mass and inertia.
• Understand Newton's first law of motion.

Experience suggests that an object at rest will remain at rest if left alone, and that an object in motion tends to slow down and stop unless some effort is made to keep it moving. What Newton’s first law of motion states, however, is the following:

NEWTON'S FIRST LAW OF MOTION

A body at rest remains at rest, or, if in motion, remains in motion at a constant velocity unless acted on by a net external force.

Note the repeated use of the verb “remains.” We can think of this law as preserving the status quo of motion.

Rather than contradicting our experience, Newton’s first law of motion states that there must be a cause (which is a net external force) for there to be any change in velocity (either a change in magnitude or direction). We will define net external force in the next section. An object sliding across a table or floor slows down due to the net force of friction acting on the object. If friction disappeared, would the object still slow down?

The idea of cause and effect is crucial in accurately describing what happens in various situations. For example, consider what happens to an object sliding along a rough horizontal surface. The object quickly grinds to a halt. If we spray the surface with talcum powder to make the surface smoother, the object slides farther. If we make the surface even smoother by rubbing lubricating oil on it, the object slides farther yet. Extrapolating to a frictionless surface, we can imagine the object sliding in a straight line indefinitely. Friction is thus the cause of the slowing (consistent with Newton’s first law). The object would not slow down at all if friction were completely eliminated. Consider an air hockey table. When the air is turned off, the puck slides only a short distance before friction slows it to a stop. However, when the air is turned on, it creates a nearly frictionless surface, and the puck glides long distances without slowing down. Additionally, if we know enough about the friction, we can accurately predict how quickly the object will slow down. Friction is an external force.

Newton’s first law is completely general and can be applied to anything from an object sliding on a table to a satellite in orbit to blood pumped from the heart. Experiments have thoroughly verified that any change in velocity (speed or direction) must be caused by an external force. The idea of generally applicable or universal laws is important not only here—it is a basic feature of all laws of physics. Identifying these laws is like recognizing patterns in nature from which further patterns can be discovered. The genius of Galileo, who first developed the idea for the first law, and Newton, who clarified it, was to ask the fundamental question, “What is the cause?” Thinking in terms of cause and effect is a worldview fundamentally different from the typical ancient Greek approach when questions such as “Why does a tiger have stripes?” would have been answered in Aristotelian fashion, “That is the nature of the beast.” True perhaps, but not a useful insight.

Mass

The property of a body to remain at rest or to remain in motion with constant velocity is called inertia. Newton’s first law is often called the law of inertia. As we know from experience, some objects have more inertia than others. It is obviously more difficult to change the motion of a large boulder than that of a basketball, for example. The inertia of an object is measured by its mass. Roughly speaking, mass is a measure of the amount of “stuff” (or matter) in something. The quantity or amount of matter in an object is determined by the numbers of atoms and molecules of various types it contains. Unlike weight, mass does not vary with location. The mass of an object is the same on Earth, in orbit, or on the surface of the Moon. In practice, it is very difficult to count and identify all of the atoms and molecules in an object, so masses are not often determined in this manner. Operationally, the masses of objects are determined by comparison with the standard kilogram.

Check Your Understanding

1: Which has more mass: a kilogram of cotton balls or a kilogram of gold?

Section Summary

• Newton’s first law of motion states that a body at rest remains at rest, or, if in motion, remains in motion at a constant velocity unless acted on by a net external force. This is also known as the law of inertia.
• Inertia is the tendency of an object to remain at rest or remain in motion. Inertia is related to an object’s mass.
• Mass is the quantity of matter in a substance.

Conceptual Questions

1: How are inertia and mass related?

2: What is the relationship between weight and mass? Which is an intrinsic, unchanging property of a body?

Glossary

inertia

the tendency of an object to remain at rest or remain in motion

law of inertia

see Newton’s first law of motion

mass

the quantity of matter in a substance; measured in kilograms

Newton’s first law of motion

a body at rest remains at rest, or, if in motion, remains in motion at a constant velocity unless acted on by a net external force; also known as the law of inertia

Solutions

Check Your Understanding 1: They are equal. A kilogram of one substance is equal in mass to a kilogram of another substance. The quantities that might differ between them are volume and density.

]]> 279 266 4 0 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/chapter/4-4-newtons-second-law-of-motion-concept-of-a-system/ Thu, 29 Jun 2017 22:13:28 +0000 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/chapter/4-4-newtons-second-law-of-motion-concept-of-a-system/

Summary

• Define net force, external force, and system.
• Understand Newton’s second law of motion.
• Apply Newton’s second law to determine the weight of an object.

Newton’s second law of motion is closely related to Newton’s first law of motion. It mathematically states the cause and effect relationship between force and changes in motion. Newton’s second law of motion is more quantitative and is used extensively to calculate what happens in situations involving a force. Before we can write down Newton’s second law as a simple equation giving the exact relationship of force, mass, and acceleration, we need to sharpen some ideas that have already been mentioned.

First, what do we mean by a change in motion? The answer is that a change in motion is equivalent to a change in velocity. A change in velocity means, by definition, that there is an acceleration. Newton’s first law says that a net external force causes a change in motion; thus, we see that a net external force causes acceleration.

Another question immediately arises. What do we mean by an external force? An intuitive notion of external is correct—an external force acts from outside the system of interest. For example, in Figure 1(a) the system of interest is the wagon plus the child in it. The two forces exerted by the other children are external forces. An internal force acts between elements of the system. Again looking at Figure 1(a), the force the child in the wagon exerts to hang onto the wagon is an internal force between elements of the system of interest. Only external forces affect the motion of a system, according to Newton’s first law. (The internal forces actually cancel, as we shall see in the next section.) You must define the boundaries of the system before you can determine which forces are external. Sometimes the system is obvious, whereas other times identifying the boundaries of a system is more subtle. The concept of a system is fundamental to many areas of physics, as is the correct application of Newton’s laws. This concept will be revisited many times on our journey through physics.

[caption id="attachment_3540" align="aligncenter" width="300"]Figure 1. Different forces exerted on the same mass produce different accelerations. (a) Two children push a wagon with a child in it. Arrows representing all external forces are shown. The system of interest is the wagon and its rider. The weight w of the system and the support of the ground N are also shown for completeness and are assumed to cancel. The vector f represents the friction acting on the wagon, and it acts to the left, opposing the motion of the wagon. (b) All of the external forces acting on the system add together to produce a net force, F_net. The free-body diagram shows all of the forces acting on the system of interest. The dot represents the center of mass of the system. Each force vector extends from this dot. Because there are two forces acting to the right, we draw the vectors collinearly. (c) A larger net external force produces a larger acceleration (a'>a) when an adult pushes the child.[/caption]

Now, it seems reasonable that acceleration should be directly proportional to and in the same direction as the net (total) external force acting on a system. This assumption has been verified experimentally and is illustrated in Figure 1. In part (a), a smaller force causes a smaller acceleration than the larger force illustrated in part (c). For completeness, the vertical forces are also shown; they are assumed to cancel since there is no acceleration in the vertical direction. The vertical forces are the weight w and the support of the ground N, and the horizontal force f represents the force of friction. These will be discussed in more detail in later sections. For now, we will define friction as a force that opposes the motion past each other of objects that are touching. Figure 1(b) shows how vectors representing the external forces add together to produce a net force, F_net.

To obtain an equation for Newton’s second law, we first write the relationship of acceleration and net external force as the proportionality

[latex]\boldsymbol{\textbf{a}\propto F_{\textbf{net}}},[/latex]

where the symbol [latex]\boldsymbol{\propto}[/latex] means “proportional to,” and F_net is the net external force. (The net external force is the vector sum of all external forces and can be determined graphically, using the head-to-tail method, or analytically, using components. The techniques are the same as for the addition of other vectors, and are covered in Chapter 3 Two-Dimensional Kinematics.) This proportionality states what we have said in words—acceleration is directly proportional to the net external force. Once the system of interest is chosen, it is important to identify the external forces and ignore the internal ones. It is a tremendous simplification not to have to consider the numerous internal forces acting between objects within the system, such as muscular forces within the child’s body, let alone the myriad of forces between atoms in the objects, but by doing so, we can easily solve some very complex problems with only minimal error due to our simplification

Now, it also seems reasonable that acceleration should be inversely proportional to the mass of the system. In other words, the larger the mass (the inertia), the smaller the acceleration produced by a given force. And indeed, as illustrated in Figure 2, the same net external force applied to a car produces a much smaller acceleration than when applied to a basketball. The proportionality is written as

[latex]\boldsymbol{\textbf{a}\:\propto}[/latex][latex size="2"]\,\boldsymbol{\frac{1}{m}}[/latex]

where m is the mass of the system. Experiments have shown that acceleration is exactly inversely proportional to mass, just as it is exactly linearly proportional to the net external force.

It has been found that the acceleration of an object depends only on the net external force and the mass of the object. Combining the two proportionalities just given yields Newton's second law of motion.

NEWTON'S SECOND LAW OF MOTION

The acceleration of a system is directly proportional to and in the same direction as the net external force acting on the system, and inversely proportional to its mass.

In equation form, Newton’s second law of motion is

[latex]\boldsymbol{\vec{\textbf{a}}\:=}[/latex][latex size="2"]\,\boldsymbol{\frac{\vec{\textbf{F}}_{\textbf{net}}}{m}}.[/latex]

This is often written in the more familiar form

[latex]\boldsymbol{\vec{\textbf{F}}_{\textbf{net}}=m}\vec{\textbf{a}}.[/latex]

When only the magnitude of force and acceleration are considered, this equation is simply

[latex]\boldsymbol{F_{\textbf{net}}=ma}.[/latex]

Although these last two equations are really the same, the first gives more insight into what Newton’s second law means. The law is a cause and effect relationship among three quantities that is not simply based on their definitions. The validity of the second law is completely based on experimental verification.

Units of Force

$$\boldsymbol{\vec{\textbf{F}}_{\textbf{net}} = m\vec{\textbf{a}}}$$ is used to define the units of force in terms of the three basic units for mass, length, and time. The SI unit of force is called the newton (abbreviated N) and is the force needed to accelerate a 1-kg system at the rate of 1 m/s². That is, $$\boldsymbol{\vec{\textbf{F}}_{\textbf{net}} = m\vec{\textbf{a}}}$$,

[latex]\boldsymbol{1\textbf{ N} = 1\textbf{ kg}\cdotp\textbf{m/s}^2}.[/latex]

While almost the entire world uses the newton for the unit of force, in the United States the most familiar unit of force is the pound (lb), where 1 N = 0.225 lb.

Weight and the Gravitational Force

When an object is dropped, it accelerates toward the center of Earth. Newton’s second law states that a net force on an object is responsible for its acceleration. If air resistance is negligible, the net force on a falling object is the gravitational force, commonly called its weight w. Weight can be denoted as a vector [latex]\vec{\textbf{w}}[/latex] because it has a direction; down is, by definition, the direction of gravity, and hence weight is a downward force. The magnitude of weight is denoted as w. Galileo was instrumental in showing that, in the absence of air resistance, all objects fall with the same acceleration g. Using Galileo’s result and Newton’s second law, we can derive an equation for weight.

Consider an object with mass m falling downward toward Earth. It experiences only the downward force of gravity, which has magnitude w. Newton’s second law states that the magnitude of the net external force on an object is F_net = ma.

Since the object experiences only the downward force of gravity, F_net = w. We know that the acceleration of an object due to gravity is g, or a = g. Substituting these into Newton’s second law gives

WEIGHT

This is the equation for weight—the gravitational force on a mass m:

[latex]\boldsymbol{w=mg}.[/latex]

Since g = 9.80 m/s² on Earth, the weight of a 1.0 kg object on Earth is 9.8 N, as we see:

[latex]\boldsymbol{w=mg=(1.0\textbf{ kg})(9.80\textbf{ m/s}^2)=9.8\textbf{ N}.}[/latex]

Recall that g can take a positive or negative value, depending on the positive direction in the coordinate system. Be sure to take this into consideration when solving problems with weight.

When the net external force on an object is its weight, we say that it is in free-fall. That is, the only force acting on the object is the force of gravity. In the real world, when objects fall downward toward Earth, they are never truly in free-fall because there is always some upward force from the air acting on the object.

The acceleration due to gravity g varies slightly over the surface of Earth, so that the weight of an object depends on location and is not an intrinsic property of the object. Weight varies dramatically if one leaves Earth’s surface. On the Moon, for example, the acceleration due to gravity is only 1.67 m/s². A 1.0-kg mass thus has a weight of 9.8 N on Earth and only about 1.7 N on the Moon.

The broadest definition of weight in this sense is that the weight of an object is the gravitational force on it from the nearest large body, such as Earth, the Moon, the Sun, and so on. This is the most common and useful definition of weight in physics. It differs dramatically, however, from the definition of weight used by NASA and the popular media in relation to space travel and exploration. When they speak of “weightlessness” and “microgravity,” they are really referring to the phenomenon we call “free-fall” in physics. We shall use the above definition of weight, and we will make careful distinctions between free-fall and actual weightlessness.

It is important to be aware that weight and mass are very different physical quantities, although they are closely related. Mass is the quantity of matter (how much “stuff”) and does not vary in classical physics, whereas weight is the gravitational force and does vary depending on gravity. It is tempting to equate the two, since most of our examples take place on Earth, where the weight of an object only varies a little with the location of the object. Furthermore, the terms mass and weight are used interchangeably in everyday language; for example, our medical records often show our “weight” in kilograms, but never in the correct units of newtons.

COMMON MISCONCEPTIONS: MASS VS. WEIGHT

Mass and weight are often used interchangeably in everyday language. However, in science, these terms are distinctly different from one another. Mass is a measure of how much matter is in an object. The typical measure of mass is the kilogram (or the “slug” in English units). Weight, on the other hand, is a measure of the force of gravity acting on an object. Weight is equal to the mass of an object (m) multiplied by the acceleration due to gravity (g). Like any other force, weight is measured in terms of newtons (or pounds in English units).

Assuming the mass of an object is kept intact, it will remain the same, regardless of its location. However, because weight depends on the acceleration due to gravity, the weight of an object can change when the object enters into a region with stronger or weaker gravity. For example, the acceleration due to gravity on the Moon is 1.67 m/s² (which is much less than the acceleration due to gravity on Earth, 9.80 m/s²). If you measured your weight on Earth and then measured your weight on the Moon, you would find that you “weigh” much less, even though you do not look any skinnier. This is because the force of gravity is weaker on the Moon. In fact, when people say that they are “losing weight,” they really mean that they are losing “mass” (which in turn causes them to weigh less).

TAKE-HOME EXPERIMENT: MASS AND WEIGHT

What do bathroom scales measure? When you stand on a bathroom scale, what happens to the scale? It depresses slightly. The scale contains springs that compress in proportion to your weight—similar to rubber bands expanding when pulled. The springs provide a measure of your weight (for an object which is not accelerating). This is a force in newtons (or pounds). In most countries, the measurement is divided by 9.80 to give a reading in mass units of kilograms. The scale measures weight but is calibrated to provide information about mass. While standing on a bathroom scale, push down on a table next to you. What happens to the reading? Why? Would your scale measure the same “mass” on Earth as on the Moon?

Example 1: What Acceleration Can a Person Produce when Pushing a Lawn Mower?

Suppose that the net external force (push minus friction) exerted on a lawn mower is 51 N (about 11 lb) parallel to the ground. The mass of the mower is 24 kg. What is its acceleration?

Strategy

Since F_net and m are given, the acceleration can be calculated directly from Newton’s second law as stated in F_net = m a.

Solution

The magnitude of the acceleration a is [latex]\boldsymbol{a=\frac{F_{\textbf{net}}}{m}}.[/latex] Entering known values gives

[latex]\boldsymbol{a\:=}[/latex][latex size="2"]\boldsymbol{\frac{51\textbf{ N}}{24\textbf{ kg}}}[/latex]

Substituting the units kg⋅m/s² for N yields

[latex]\boldsymbol{a\:=}[/latex][latex size="2"]\boldsymbol{\frac{51\textbf{ kg}\cdotp\textbf{m/s}^2}{24\textbf{ kg}}}[/latex][latex]\boldsymbol{=\:2.1\textbf{ m/s}^2.}[/latex]

Discussion

The direction of the acceleration is the same direction as that of the net force, which is parallel to the ground. There is no information given in this example about the individual external forces acting on the system, but we can say something about their relative magnitudes. For example, the force exerted by the person pushing the mower must be greater than the friction opposing the motion (since we know the mower moves forward), and the vertical forces must cancel if there is to be no acceleration in the vertical direction (the mower is moving only horizontally). The acceleration found is small enough to be reasonable for a person pushing a mower. Such an effort would not last too long because the person’s top speed would soon be reached.

Example 2: What Rocket Thrust Accelerates This Sled?

Prior to manned space flights, rocket sleds were used to test aircraft, missile equipment, and physiological effects on human subjects at high speeds. They consisted of a platform that was mounted on one or two rails and propelled by several rockets. Calculate the magnitude of force exerted by each rocket, called its thrust T, for the four-rocket propulsion system shown in Figure 4. The sled’s initial acceleration is 49 m/s², the mass of the system is 2100 kg, and the force of friction opposing the motion is known to be 650 N.

[caption id="" align="aligncenter" width="325"]Figure 4. A sled experiences a rocket thrust that accelerates it to the right. Each rocket creates an identical thrust T. As in other situations where there is only horizontal acceleration, the vertical forces cancel. The ground exerts an upward force N on the system that is equal in magnitude and opposite in direction to its weight, w. The system here is the sled, its rockets, and rider, so none of the forces between these objects are considered. The arrow representing friction (f) is drawn larger than scale.[/caption]

Strategy

Although there are forces acting vertically and horizontally, we assume the vertical forces cancel since there is no vertical acceleration. This leaves us with only horizontal forces and a simpler one-dimensional problem. Directions are indicated with plus or minus signs, with right taken as the positive direction. See the free-body diagram in the figure.

Solution

Since acceleration, mass, and the force of friction are given, we start with Newton’s second law and look for ways to find the thrust of the engines. Since we have defined the direction of the force and acceleration as acting “to the right,” we need to consider only the magnitudes of these quantities in the calculations. Hence we begin with

[latex]\boldsymbol{F_{\textbf{net}}=ma},[/latex]

where [latex]\boldsymbol{F_{\textbf{net}}}[/latex] is the net force along the horizontal direction. We can see from Figure 4 that the engine thrusts add, while friction opposes the thrust. In equation form, the net external force is

[latex]\boldsymbol{F_{\textbf{net}}=4T-f}[/latex].

Substituting this into Newton’s second law gives

[latex]\boldsymbol{F_{\textbf{net}}=ma=4T-f}.[/latex]

Using a little algebra, we solve for the total thrust 4T:

[latex]\boldsymbol{4\:T=ma+f}.[/latex]

Substituting known values yields

[latex]\boldsymbol{4\:T=ma+f=(2100\textbf{ kg})(49\textbf{ m/s}^2)+650\textbf{ N}}.[/latex]

So the total thrust is

[latex]\boldsymbol{4\:T=1.0\times10^5\textbf{ N}},[/latex]

and the individual thrusts are

[latex]\boldsymbol{T\:=}[/latex][latex size="2"]\boldsymbol{\frac{1.0\times10^5\textbf{ N}}{4}}[/latex][latex]\boldsymbol{=\:2.6\times10^4\textbf{ N}}.[/latex]

Discussion

The numbers are quite large, so the result might surprise you. Experiments such as this were performed in the early 1960s to test the limits of human endurance and the setup designed to protect human subjects in jet fighter emergency ejections. Speeds of 1000 km/h were obtained, with accelerations of 45 g's. (Recall that g, the acceleration due to gravity, is 9.80 m/s².When we say that an acceleration is 45 g's, it is 45 × 9.80 m/s², which is approximately 440 m/s².) While living subjects are not used any more, land speeds of 10,000 km/h have been obtained with rocket sleds. In this example, as in the preceding one, the system of interest is obvious. We will see in later examples that choosing the system of interest is crucial—and the choice is not always obvious.

Newton’s second law of motion is more than a definition; it is a relationship among acceleration, force, and mass. It can help us make predictions. Each of those physical quantities can be defined independently, so the second law tells us something basic and universal about nature. The next section introduces the third and final law of motion.

Section Summary

• Acceleration, a, is defined as a change in velocity, meaning a change in its magnitude or direction, or both.
• An external force is one acting on a system from outside the system, as opposed to internal forces, which act between components within the system.
• Newton’s second law of motion states that the acceleration of a system is directly proportional to and in the same direction as the net external force acting on the system, and inversely proportional to its mass.
• In equation form, Newton’s second law of motion is [latex]\boldsymbol{\textbf{a}=\frac{F_{\textbf{net}}}{m}}.[/latex]
• This is often written in the more familiar form: F_net = m a.
• The weight w of an object is defined as the force of gravity acting on an object of mass m. The object experiences an acceleration due to gravity g:
  [latex]\boldsymbol{w=m\,g}.[/latex]
• If the only force acting on an object is due to gravity, the object is in free fall.
• Friction is a force that opposes the motion past each other of objects that are touching.

Conceptual Questions

1: Which statement is correct? (a) Net force causes motion. (b) Net force causes change in motion. Explain your answer and give an example.

2: Why can we neglect forces such as those holding a body together when we apply Newton’s second law of motion?

3: Explain how the choice of the “system of interest” affects which forces must be considered when applying Newton’s second law of motion.

4: Describe a situation in which the net external force on a system is not zero, yet its speed remains constant.

5: A system can have a nonzero velocity while the net external force on it is zero. Describe such a situation.

6: A rock is thrown straight up. What is the net external force acting on the rock when it is at the top of its trajectory?

7: (a) Give an example of different net external forces acting on the same system to produce different accelerations. (b) Give an example of the same net external force acting on systems of different masses, producing different accelerations. (c) What law accurately describes both effects? State it in words and as an equation.

8: If the acceleration of a system is zero, are no external forces acting on it? What about internal forces? Explain your answers.

9: If a constant, nonzero force is applied to an object, what can you say about the velocity and acceleration of the object?

10: The gravitational force on the basketball in Figure 2 is ignored. When gravity is taken into account, what is the direction of the net external force on the basketball—above horizontal, below horizontal, or still horizontal?

Problems & Exercises

You may assume data taken from illustrations is accurate to three digits.

1: A 63.0-kg sprinter starts a race with an acceleration of 4.20 m/s².What is the net external force on him?

2: If the sprinter from the previous problem accelerates at that rate for 20 m, and then maintains that velocity for the remainder of the 100-m dash, what will be his time for the race?

3: A cleaner pushes a 4.50-kg laundry cart in such a way that the net external force on it is 60.0 N. Calculate the magnitude of its acceleration.

4: Since astronauts in orbit are apparently weightless, a clever method of measuring their masses is needed to monitor their mass gains or losses to adjust diets. One way to do this is to exert a known force on an astronaut and measure the acceleration produced. Suppose a net external force of 50.0 N is exerted and the astronaut’s acceleration is measured to be 0.893 m/s². (a) Calculate her mass. (b) By exerting a force on the astronaut, the vehicle in which they orbit experiences an equal and opposite force. Discuss how this would affect the measurement of the astronaut’s acceleration. Propose a method in which recoil of the vehicle is avoided.

5: In Figure 3, the net external force on the 24-kg mower is stated to be 51 N. If the force of friction opposing the motion is 24 N, what force F (in newtons) is the person exerting on the mower? Suppose the mower is moving at 1.5 m/s when the force F is removed. How far will the mower go before stopping?

6: The same rocket sled drawn in Figure 5 is decelerated at a rate of 196 m/s². What force is necessary to produce this deceleration? Assume that the rockets are off. The mass of the system is 2100 kg.

[caption id="attachment_3549" align="aligncenter" width="300"]Figure 5.[/caption]

7: (a) If the rocket sled shown in Figure 6 starts with only one rocket burning, what is the magnitude of its acceleration? Assume that the mass of the system is 2100 kg, the thrust T is 2.4 × 10⁴ N, and the force of friction opposing the motion is known to be 650 N. (b) Why is the acceleration not one-fourth of what it is with all rockets burning?

[caption id="" align="aligncenter" width="300"]Figure 6.[/caption]

8: What is the deceleration of the rocket sled if it comes to rest in 1.1 s from a speed of 1000 km/h? (Such deceleration caused one test subject to black out and have temporary blindness.)

9: Suppose two children push horizontally, but in exactly opposite directions, on a third child in a wagon. The first child exerts a force of 75.0 N, the second a force of 90.0 N, friction is 12.0 N, and the mass of the third child plus wagon is 23.0 kg. (a) What is the system of interest if the acceleration of the child in the wagon is to be calculated? (b) Draw a free-body diagram, including all forces acting on the system. (c) Calculate the acceleration. (d) What would the acceleration be if friction were 15.0 N?

10: A powerful motorcycle can produce an acceleration of 3.50 m/s² while traveling at 90.0 km/h. At that speed the forces resisting motion, including friction and air resistance, total 400 N. (Air resistance is analogous to air friction. It always opposes the motion of an object.) What is the magnitude of the force the motorcycle exerts backward on the ground to produce its acceleration if the mass of the motorcycle with rider is 245 kg?

11: The rocket sled shown in Figure 7 accelerates at a rate of 49.0 m/s². Its passenger has a mass of 75.0 kg. (a) Calculate the horizontal component of the force the seat exerts against his body. Compare this with his weight by using a ratio. (b) Calculate the direction and magnitude of the total force the seat exerts against his body.

[caption id="attachment_3549" align="aligncenter" width="300"]Figure 7.[/caption]

12: Repeat the previous problem for the situation in which the rocket sled decelerates at a rate of 201 m/s². In this problem, the forces are exerted by the seat and restraining belts.

13: The weight of an astronaut plus his space suit on the Moon is only 250 N. How much do they weigh on Earth? What is the mass on the Moon? On Earth?

14: Suppose the mass of a fully loaded module in which astronauts take off from the Moon is 10,000 kg. The thrust of its engines is 30,000 N. (a) Calculate its the magnitude of acceleration in a vertical takeoff from the Moon. (b) Could it lift off from Earth? If not, why not? If it could, calculate the magnitude of its acceleration.

Glossary

acceleration

the rate at which an object’s velocity changes over a period of time

free-fall

a situation in which the only force acting on an object is the force due to gravity

friction

a force past each other of objects that are touching; examples include rough surfaces and air resistance

net external force

the vector sum of all external forces acting on an object or system; causes a mass to accelerate

Newton’s second law of motion

the net external force F_net on an object with mass m is proportional to and in the same direction as the acceleration of the object, a, and inversely proportional to the mass; defined mathematically as [latex]\boldsymbol{\textbf{a}=\frac{F_{\textbf{net}}}{m}}[/latex]

system

defined by the boundaries of an object or collection of objects being observed; all forces originating from outside of the system are considered external forces

weight

the force w due to gravity acting on an object of mass m; defined mathematically as: w = mg, where g is the magnitude and direction of the acceleration due to gravity

Solutions

Problems & Exercises 1: $$\boldsymbol{265\textbf{ N}}$$ 3: [latex]\boldsymbol{13.3\textbf{ m/s}}[/latex]

7: (a) [latex]\boldsymbol{12\textbf{ m/s}^2}.[/latex] (b) The acceleration is not one-fourth of what it was with all rockets burning because the frictional force is still as large as it was with all rockets burning.

9: (a) The system is the child in the wagon plus the wagon.

(b)

[caption id="" align="aligncenter" width="200"]Figure 8.[/caption]

(c) [latex]\boldsymbol{a=0.130\textbf{ m/s}^2}[/latex] in the direction of the second child’s push. (d) [latex]\boldsymbol{a=0.00\textbf{ m/s}^2}[/latex]

11: (a) [latex]\boldsymbol{3.68\times10^3\textbf{ N}}[/latex]. This force is 5.00 times greater than his weight. (b) [latex]\boldsymbol{3750\textbf{ N};\:11.3^0\textbf{ above horizontal}}[/latex]

13: [latex]\boldsymbol{1.5\times10^3\textbf{ N},\:150\textbf{ kg},\:150\textbf{ kg}}[/latex]

]]> 287 266 5 0 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/chapter/4-5-newtons-third-law-of-motion-symmetry-in-forces/ Thu, 29 Jun 2017 22:13:29 +0000 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/chapter/4-5-newtons-third-law-of-motion-symmetry-in-forces/

Summary

• Understand Newton's third law of motion.
• Apply Newton's third law to define systems and solve problems of motion.

There is a passage in the musical Man of la Mancha that relates to Newton’s third law of motion. Sancho, in describing a fight with his wife to Don Quixote, says, “Of course I hit her back, Your Grace, but she’s a lot harder than me and you know what they say, ‘Whether the stone hits the pitcher or the pitcher hits the stone, it’s going to be bad for the pitcher.’” This is exactly what happens whenever one body exerts a force on another—the first also experiences a force (equal in magnitude and opposite in direction). Numerous common experiences, such as stubbing a toe or throwing a ball, confirm this. It is precisely stated in Newton’s third law of motion.

NEWTON'S THIRD LAW OF MOTION

Whenever one body exerts a force on a second body, the first body experiences a force that is equal in magnitude and opposite in direction to the force that it exerts.

This law represents a certain symmetry in nature: Forces always occur in pairs, and one body cannot exert a force on another without experiencing a force itself. We sometimes refer to this law loosely as “action-reaction,” where the force exerted is the action and the force experienced as a consequence is the reaction. Newton’s third law has practical uses in analyzing the origin of forces and understanding which forces are external to a system.

We can readily see Newton’s third law at work by taking a look at how people move about. Consider a swimmer pushing off from the side of a pool, as illustrated in Figure 1. She pushes against the pool wall with her feet and accelerates in the direction opposite to that of her push. The wall has exerted an equal and opposite force back on the swimmer. You might think that two equal and opposite forces would cancel, but they do not because they act on different systems. In this case, there are two systems that we could investigate: the swimmer or the wall. If we select the swimmer to be the system of interest, as in the figure, then F_{wall on feet} is an external force on this system and affects its motion. The swimmer moves in the direction of F_{wall on feet}. In contrast, the force F_{feet on wall} acts on the wall and not on our system of interest. Thus F_{feet on wall} does not directly affect the motion of the system and does not cancel F_{wall on feet}. Note that the swimmer pushes in the direction opposite to that in which she wishes to move. The reaction to her push is thus in the desired direction.

Other examples of Newton’s third law are easy to find. As a professor paces in front of a whiteboard, she exerts a force backward on the floor. The floor exerts a reaction force forward on the professor that causes her to accelerate forward. Similarly, a car accelerates because the ground pushes forward on the drive wheels in reaction to the drive wheels pushing backward on the ground. You can see evidence of the wheels pushing backward when tires spin on a gravel road and throw rocks backward. In another example, rockets move forward by expelling gas backward at high velocity. This means the rocket exerts a large backward force on the gas in the rocket combustion chamber, and the gas therefore exerts a large reaction force forward on the rocket. This reaction force is called thrust. It is a common misconception that rockets propel themselves by pushing on the ground or on the air behind them. They actually work better in a vacuum, where they can more readily expel the exhaust gases. Helicopters similarly create lift by pushing air down, thereby experiencing an upward reaction force. Birds and airplanes also fly by exerting force on air in a direction opposite to that of whatever force they need. For example, the wings of a bird force air downward and backward in order to get lift and move forward. An octopus propels itself in the water by ejecting water through a funnel from its body, similar to a jet ski. In a situation similar to Sancho’s, professional cage fighters experience reaction forces when they punch, sometimes breaking their hand by hitting an opponent’s body.

Example 1: Getting Up To Speed: Choosing the Correct System

A physics professor pushes a cart of demonstration equipment to a lecture hall, as seen in Figure 2. Her mass is 65.0 kg, the cart’s is 12.0 kg, and the equipment’s is 7.0 kg. Calculate the acceleration produced when the professor exerts a backward force of 150 N on the floor. All forces opposing the motion, such as friction on the cart’s wheels and air resistance, total 24.0 N.

Strategy

Since they accelerate as a unit, we define the system to be the professor, cart, and equipment. This is System 1 in Figure 2. The professor pushes backward with a force F_foot of 150 N. According to Newton’s third law, the floor exerts a forward reaction force F_floor of 150 N on System 1. Because all motion is horizontal, we can assume there is no net force in the vertical direction. The problem is therefore one-dimensional along the horizontal direction. As noted, f opposes the motion and is thus in the opposite direction of F_floor. Note that we do not include the forces F_prof or F_cart because these are internal forces, and we do not include F_foot because it acts on the floor, not on the system. There are no other significant forces acting on System 1. If the net external force can be found from all this information, we can use Newton’s second law to find the acceleration as requested. See the free-body diagram in the figure.

Solution

Newton’s second law is given by

[latex]\boldsymbol{a\:=}[/latex][latex size="2"]\boldsymbol{\frac{F_{\textbf{net}}}{m}.}[/latex]

The net external force on System 1 is deduced from Figure 2 and the discussion above to be

[latex]\boldsymbol{\textbf{F}_{\textbf{net}}=\textbf{F}_{\textbf{floor}}-\textbf{f}=150\textbf{ N}-24.0\textbf{ N}=126\textbf{ N}.}[/latex]

The mass of System 1 is

[latex]\boldsymbol{m=(65.0 + 12.0 + 7.0)\textbf{ kg} = 84\textbf{ kg}.}[/latex]

These values of F_net and m produce an acceleration of

$latex \begin{array}{r @{{}={}}l} \boldsymbol{a} & \boldsymbol{\frac{F_{\textbf{net}}}{m}} \\[1em] \boldsymbol{a} & \boldsymbol{\frac{126 \;\textbf{N}}{84 \;\textbf{kg}} = 1.5 \;\textbf{m} / \;\textbf{s}^2} \end{array} $

Discussion

None of the forces between components of System 1, such as between the professor’s hands and the cart, contribute to the net external force because they are internal to System 1. Another way to look at this is to note that forces between components of a system cancel because they are equal in magnitude and opposite in direction. For example, the force exerted by the professor on the cart results in an equal and opposite force back on her. In this case both forces act on the same system and, therefore, cancel. Thus internal forces (between components of a system) cancel. Choosing System 1 was crucial to solving this problem.

Example 2: Force of the Cart—Choosing a New System

Calculate the force the professor exerts on the cart in Figure 2 using data from the previous example if needed.

Strategy

If we now define the system of interest to be the cart plus equipment (System 2 in Figure 2), then the net external force on System 2 is the force the professor exerts on the cart minus friction. The force she exerts on the cart, F_prof, is an external force acting on System 2. F_prof was internal to System 1, but it is external to System 2 and will enter Newton’s second law for System 2.

Solution

Newton’s second law can be used to find F_prof. Starting with

[latex]\boldsymbol{a\:=}[/latex][latex size="2"]\boldsymbol{\frac{F_{\textbf{net}}}{m}}[/latex]

and noting that the magnitude of the net external force on System 2 is

[latex]\boldsymbol{F_{\textbf{net}}=F_{\textbf{prof}}-f,}[/latex]

we solve for F_prof, the desired quantity:

[latex]\boldsymbol{F_{\textbf{prof}}=F_{\textbf{net}}+f.}[/latex]

The value of f is given, so we must calculate net F_net. That can be done since both the acceleration and mass of System 2 are known. Using Newton’s second law we see that

[latex]\boldsymbol{F_{\textbf{net}}=ma,}[/latex]

where the mass of System 2 is 19.0 kg (m= 12.0 kg + 7.0 kg) and its acceleration was found to be a = 1.5 m/s² in the previous example. Thus,

[latex]\boldsymbol{F_{\textbf{net}}=ma,}[/latex]

[latex]\boldsymbol{F_{\textbf{net}}=(19.0\textbf{ kg})(1.5\textbf{ m/s}^2)=29\textbf{ N}.}[/latex]

Now we can find the desired force:

[latex]\boldsymbol{F_{\textbf{prof}}=F_{\textbf{net}}+f,}[/latex]

[latex]\boldsymbol{F_{\textbf{prof}}=29\textbf{ N}+24.0\textbf{ N}=53\textbf{ N}.}[/latex]

Discussion

It is interesting that this force is significantly less than the 150-N force the professor exerted backward on the floor. Not all of that 150-N force is transmitted to the cart; some of it accelerates the professor.

The choice of a system is an important analytical step both in solving problems and in thoroughly understanding the physics of the situation (which is not necessarily the same thing).

PHET EXPLORATIONS: GRAVITY FORCE LAB

Visualize the gravitational force that two objects exert on each other. Change properties of the objects in order to see how it changes the gravity force. [caption id="" align="aligncenter" width="450"] Figure 3. Gravity Force Lab[/caption]

Section Summary

• Newton’s third law of motion represents a basic symmetry in nature. It states: Whenever one body exerts a force on a second body, the first body experiences a force that is equal in magnitude and opposite in direction to the force that the first body exerts.
• A thrust is a reaction force that pushes a body forward in response to a backward force. Rockets, airplanes, and cars are pushed forward by a thrust reaction force.

Conceptual Questions

1: When you take off in a jet aircraft, there is a sensation of being pushed back into the seat. Explain why you move backward in the seat—is there really a force backward on you? (The same reasoning explains whiplash injuries, in which the head is apparently thrown backward.)

2: A device used since the 1940s to measure the kick or recoil of the body due to heart beats is the “ballistocardiograph.” What physics principle(s) are involved here to measure the force of cardiac contraction? How might we construct such a device?

3: Describe a situation in which one system exerts a force on another and, as a consequence, experiences a force that is equal in magnitude and opposite in direction. Which of Newton’s laws of motion apply?

4: Why does an ordinary rifle recoil (kick backward) when fired? The barrel of a recoilless rifle is open at both ends. Describe how Newton’s third law applies when one is fired. Can you safely stand close behind one when it is fired?

5: An American football lineman reasons that it is senseless to try to out-push the opposing player, since no matter how hard he pushes he will experience an equal and opposite force from the other player. Use Newton’s laws and draw a free-body diagram of an appropriate system to explain how he can still out-push the opposition if he is strong enough.

6: Newton’s third law of motion tells us that forces always occur in pairs of equal and opposite magnitude. Explain how the choice of the “system of interest” affects whether one such pair of forces cancels.

Problems & Exercises

1: What net external force is exerted on a 1100-kg artillery shell fired from a battleship if the shell is accelerated at 2.40 × 10⁴ m/s²? What is the magnitude of the force exerted on the ship by the artillery shell?

2: A brave but inadequate rugby player is being pushed backward by an opposing player who is exerting a force of 800 N on him. The mass of the losing player plus equipment is 90.0 kg, and he is accelerating at 1.20 m/s² backward. (a) What is the force of friction between the losing player’s feet and the grass? (b) What force does the winning player exert on the ground to move forward if his mass plus equipment is 110 kg? (c) Draw a sketch of the situation showing the system of interest used to solve each part. For this situation, draw a free-body diagram and write the net force equation.

Glossary

Newton’s third law of motion

whenever one body exerts a force on a second body, the first body experiences a force that is equal in magnitude and opposite in direction to the force that the first body exerts

thrust

a reaction force that pushes a body forward in response to a backward force; rockets, airplanes, and cars are pushed forward by a thrust reaction force

Solutions

Problems & Exercises

1: Force on shell: [latex]\boldsymbol{2.64\times10^7\textbf{ N}}[/latex] Force exerted on ship = [latex]\boldsymbol{-2.64\times10^7\textbf{ N}},[/latex] by Newton’s third law

]]> 291 266 6 0 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/chapter/4-6-normal-tension-and-other-examples-of-forces/ Thu, 29 Jun 2017 22:13:30 +0000 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/chapter/4-6-normal-tension-and-other-examples-of-forces/

Summary

• Define normal and tension forces.
• Apply Newton's laws of motion to solve problems involving a variety of forces.
• Use trigonometric identities to resolve weight into components.

Forces are given many names, such as push, pull, thrust, lift, weight, friction, and tension. Traditionally, forces have been grouped into several categories and given names relating to their source, how they are transmitted, or their effects. The most important of these categories are discussed in this section, together with some interesting applications. Further examples of forces are discussed later in this text.

Normal Force

Weight (also called force of gravity) is a pervasive force that acts at all times and must be counteracted to keep an object from falling. You definitely notice that you must support the weight of a heavy object by pushing up on it when you hold it stationary, as illustrated in Figure 1(a). But how do inanimate objects like a table support the weight of a mass placed on them, such as shown in Figure 1(b)? When the bag of dog food is placed on the table, the table actually sags slightly under the load. This would be noticeable if the load were placed on a card table, but even rigid objects deform when a force is applied to them. Unless the object is deformed beyond its limit, it will exert a restoring force much like a deformed spring (or trampoline or diving board). The greater the deformation, the greater the restoring force. So when the load is placed on the table, the table sags until the restoring force becomes as large as the weight of the load. At this point the net external force on the load is zero. That is the situation when the load is stationary on the table. The table sags quickly, and the sag is slight so we do not notice it. But it is similar to the sagging of a trampoline when you climb onto it.

[caption id="" align="aligncenter" width="400"]Figure 1. (a) The person holding the bag of dog food must supply an upward force F_hand equal in magnitude and opposite in direction to the weight of the food w. (b) The card table sags when the dog food is placed on it, much like a stiff trampoline. Elastic restoring forces in the table grow as it sags until they supply a force N equal in magnitude and opposite in direction to the weight of the load.[/caption]

We must conclude that whatever supports a load, be it animate or not, must supply an upward force equal to the weight of the load, as we assumed in a few of the previous examples. If the force supporting a load is perpendicular to the surface of contact between the load and its support, this force is defined to be a normal force and here is given the symbol N. (This is not the unit for force N.) The word normal means perpendicular to a surface. The normal force can be less than the object’s weight if the object is on an incline, as you will see in the next example.

COMMON MISCONCEPTIONS: NORMAL FORCE (N) VS. NEWTON (N)

In this section we have introduced the quantity normal force, which is represented by the variable N. This should not be confused with the symbol for the newton, which is also represented by the letter N. These symbols are particularly important to distinguish because the units of a normal force (N) happen to be newtons (N). For example, the normal force N that the floor exerts on a chair might be N = 100 N. One important difference is that normal force is a vector, while the newton is simply a unit. Be careful not to confuse these letters in your calculations! You will encounter more similarities among variables and units as you proceed in physics. Another example of this is the quantity work (W) and the unit watts (W).

Example 1: Weight on an Incline, a Two-Dimensional Problem

Consider the skier on a slope shown in Figure 2. Her mass including equipment is 60.0 kg. (a) What is her acceleration if friction is negligible? (b) What is her acceleration if friction is known to be 45.0 N?

Strategy

This is a two-dimensional problem, since the forces on the skier (the system of interest) are not parallel. The approach we have used in two-dimensional kinematics also works very well here. Choose a convenient coordinate system and project the vectors onto its axes, creating two connected one-dimensional problems to solve. The most convenient coordinate system for motion on an incline is one that has one coordinate parallel to the slope and one perpendicular to the slope. (Remember that motions along mutually perpendicular axes are independent.) We use the symbols ⊥ and [latex]\parallel[/latex] to represent perpendicular and parallel, respectively. This choice of axes simplifies this type of problem, because there is no motion perpendicular to the slope and because friction is always parallel to the surface between two objects. The only external forces acting on the system are the skier’s weight, friction, and the support of the slope, respectively labeled w, and N in Figure 2. N is always perpendicular to the slope, and f is parallel to it. But w is not in the direction of either axis, and so the first step we take is to project it into components along the chosen axes, defining [latex]\boldsymbol{w_{\parallel}}[/latex] to be the component of weight parallel to the slope and [latex]\boldsymbol{w_{\perp}}[/latex] the component of weight perpendicular to the slope. Once this is done, we can consider the two separate problems of forces parallel to the slope and forces perpendicular to the slope.

Solution

The magnitude of the component of the weight parallel to the slope is [latex]\boldsymbol{w_{\parallel} =w\:\textbf{sin}\:(25^o)=mg\:\textbf{sin}\:(25^0)},[/latex] and the magnitude of the component of the weight perpendicular to the slope is [latex]\boldsymbol{w_{\perp}=w\:\textbf{cos}\:(25^0)=mg\:\textbf{cos}\:(25^0)}.[/latex]

(a) Neglecting friction. Since the acceleration is parallel to the slope, we need only consider forces parallel to the slope. (Forces perpendicular to the slope add to zero, since there is no acceleration in that direction.) The forces parallel to the slope are the amount of the skier’s weight parallel to the slope [latex]\boldsymbol{w_{\parallel}}[/latex] and friction f. Using Newton’s second law, with subscripts to denote quantities parallel to the slope,

[latex]\boldsymbol{a_\parallel\: =}[/latex][latex size="2"]\boldsymbol{\frac{F_{\textbf{net}\parallel}}{m}}[/latex]

where [latex]\boldsymbol{F_{\textbf{net}\parallel}=w_{\parallel} =mg\:\textbf{sin}\:(25^0)},[/latex] assuming no friction for this part, so that

[latex]\boldsymbol{a_{\parallel}\: =}[/latex][latex size="2"]\boldsymbol{\frac{F_{\textbf{net}\parallel}}{m}}[/latex][latex]\boldsymbol{=}[/latex][latex size="2"]\boldsymbol{\frac{mg\:\textbf{sin}\:(25^0)}{m}}[/latex][latex]\boldsymbol{\:=g\:\textbf{sin}\:(25^0)}[/latex]

[latex]\boldsymbol{(9.80\textbf{ m/s}^2)(0.4226)=4.14\textbf{ m/s}^2}[/latex]

is the acceleration.

(b) Including friction. We now have a given value for friction, and we know its direction is parallel to the slope and it opposes motion between surfaces in contact. So the net external force is now

[latex]\boldsymbol{F_{\textbf{net}\parallel}=w_{\parallel} -f,}[/latex]

and substituting this into Newton’s second law, [latex]\boldsymbol{a_{\parallel} =\frac{F_{\textbf{net}}}{m},}[/latex] gives

[latex]\boldsymbol{a_{\parallel}\: =}[/latex][latex size="2"]\boldsymbol{\frac{F_{\textbf{net}\parallel}}{m}}[/latex][latex]\boldsymbol{=}[/latex][latex size="2"]\boldsymbol{\frac{w_{\parallel} -f}{m}}[/latex][latex]\boldsymbol{=}[/latex][latex size="2"]\boldsymbol{\frac{mg\:\textbf{sin}(25^0)-f}{m}.}[/latex]

We substitute known values to obtain

[latex]\boldsymbol{a_{\parallel}\: =}[/latex][latex size="2"]\boldsymbol{\frac{(60.0\textbf{ kg})(9.80\textbf{ m/s}^2)(0.4226)-45.0\textbf{ N}}{60.0\textbf{ kg}},}[/latex]

which yields

[latex]\boldsymbol{a_{\parallel} =3.39\textbf{ m/s}^2,}[/latex]

which is the acceleration parallel to the incline when there is 45.0 N of opposing friction.

Discussion

Since friction always opposes motion between surfaces, the acceleration is smaller when there is friction than when there is none. In fact, it is a general result that if friction on an incline is negligible, then the acceleration down the incline is a = g sin θ, regardless of mass. This is related to the previously discussed fact that all objects fall with the same acceleration in the absence of air resistance. Similarly, all objects, regardless of mass, slide down a frictionless incline with the same acceleration (if the angle is the same).

RESOLVING WEIGHT INTO COMPONENTS

[caption id="" align="aligncenter" width="425"]Figure 3. An object rests on an incline that makes an angle θ with the horizontal.[/caption]

When an object rests on an incline that makes an angle θ with the horizontal, the force of gravity acting on the object is divided into two components: a force acting perpendicular to the plane,[latex]\boldsymbol{\textbf{w}_{\perp}},[/latex] and a force acting parallel to the plane, [latex]\boldsymbol{\textbf{w}_{\parallel}}.[/latex] The perpendicular force of weight, [latex]\boldsymbol{\textbf{w}_{\perp}},[/latex] is typically equal in magnitude and opposite in direction to the normal force, N. The force acting parallel to the plane, [latex]\boldsymbol{\textbf{w}_{\parallel}},[/latex] causes the object to accelerate down the incline. The force of friction, f, opposes the motion of the object, so it acts upward along the plane.

It is important to be careful when resolving the weight of the object into components. If the angle of the incline is at an angle θ to the horizontal, then the magnitudes of the weight components are

[latex]\boldsymbol{w_{\parallel} =w\:\textbf{sin}\:(\theta)=mg\:\textbf{sin}\:(\theta)}[/latex]

and

[latex]\boldsymbol{w_{\perp}=w\:\textbf{cos}\:(\theta)=mg\:\textbf{cos}\:(\theta).}[/latex]

Instead of memorizing these equations, it is helpful to be able to determine them from reason. To do this, draw the right triangle formed by the three weight vectors. Notice that the θ of the incline is the same as the angle formed between [latex]\textbf{w}[/latex] and [latex]\boldsymbol{\textbf{w}_{\perp}}.[/latex] Knowing this property, you can use trigonometry to determine the magnitude of the weight components:

$latex \begin{array}{r @{{}={}}l} \boldsymbol{\textbf{cos}(\theta)} & \boldsymbol{\frac{w_{\perp}}{w}} \\[1em] \boldsymbol{w_{\perp}} & \boldsymbol{w\:\textbf{cos}\:(\theta)=mg\:\textbf{cos}\:(\theta)} \end{array} $

$latex \begin{array}{r @{{}={}}l} \boldsymbol{\textbf{sin}(\theta)} & \boldsymbol{\frac{w_{\parallel}}{w}} \\[1em] \boldsymbol{w_{\parallel}} & \boldsymbol{w\:\textbf{sin}\:(\theta) = mg\:\textbf{sin}\:(\theta)} \end{array} $

TAKE-HOME EXPERIMENT: FORCE PARALLEL

To investigate how a force parallel to an inclined plane changes, find a rubber band, some objects to hang from the end of the rubber band, and a board you can position at different angles. How much does the rubber band stretch when you hang the object from the end of the board? Now place the board at an angle so that the object slides off when placed on the board. How much does the rubber band extend if it is lined up parallel to the board and used to hold the object stationary on the board? Try two more angles. What does this show?

Tension

A tension is a force along the length of a medium, especially a force carried by a flexible medium, such as a rope or cable. The word “tension” comes from a Latin word meaning “to stretch.” Not coincidentally, the flexible cords that carry muscle forces to other parts of the body are called tendons. Any flexible connector, such as a string, rope, chain, wire, or cable, can exert pulls only parallel to its length; thus, a force carried by a flexible connector is a tension with direction parallel to the connector. It is important to understand that tension is a pull in a connector. In contrast, consider the phrase: “You can’t push a rope.” The tension force pulls outward along the two ends of a rope.

Consider a person holding a mass on a rope as shown in Figure 4.

[caption id="" align="aligncenter" width="200"]Figure 4. When a perfectly flexible connector (one requiring no force to bend it) such as this rope transmits a force T, that force must be parallel to the length of the rope, as shown. The pull such a flexible connector exerts is a tension. Note that the rope pulls with equal force but in opposite directions on the hand and the supported mass (neglecting the weight of the rope). This is an example of Newton’s third law. The rope is the medium that carries the equal and opposite forces between the two objects. The tension anywhere in the rope between the hand and the mass is equal. Once you have determined the tension in one location, you have determined the tension at all locations along the rope.[/caption]

Tension in the rope must equal the weight of the supported mass, as we can prove using Newton’s second law. If the 5.00-kg mass in the figure is stationary, then its acceleration is zero, and thus F_net = 0. The only external forces acting on the mass are its weight w and the tension T supplied by the rope. Thus,

[latex]\boldsymbol{F_{\textbf{net}}=T-w=0,}[/latex]

where T and w are the magnitudes of the tension and weight and their signs indicate direction, with up being positive here. Thus, just as you would expect, the tension equals the weight of the supported mass:

[latex]\boldsymbol{T=w=mg}[/latex]

For a 5.00-kg mass, then (neglecting the mass of the rope) we see that

[latex]\boldsymbol{T=mg=(5.00\textbf{ kg})(9.80\textbf{ m/s}^2)=49.0\textbf{ N}.}[/latex]

If we cut the rope and insert a spring, the spring would extend a length corresponding to a force of 49.0 N, providing a direct observation and measure of the tension force in the rope.

Flexible connectors are often used to transmit forces around corners, such as in a hospital traction system, a finger joint, or a bicycle brake cable. If there is no friction, the tension is transmitted undiminished. Only its direction changes, and it is always parallel to the flexible connector. This is illustrated in Figure 5 (a) and (b).

[caption id="" align="aligncenter" width="368"] Figure 5. (a) Tendons in the finger carry force T from the muscles to other parts of the finger, usually changing the force’s direction, but not its magnitude (the tendons are relatively friction free). (b) The brake cable on a bicycle carries the tension T from the handlebars to the brake mechanism. Again, the direction but not the magnitude of T is changed.[/caption]

Example 2: What Is the Tension in a Tightrope?

Calculate the tension in the wire supporting the 70.0-kg tightrope walker shown in Figure 6.

Strategy

As you can see in the figure, the wire is not perfectly horizontal (it cannot be!), but is bent under the person’s weight. Thus, the tension on either side of the person has an upward component that can support his weight. As usual, forces are vectors represented pictorially by arrows having the same directions as the forces and lengths proportional to their magnitudes. The system is the tightrope walker, and the only external forces acting on him are his weight w and the two tensions T_L (left tension) and T_R (right tension), as illustrated. It is reasonable to neglect the weight of the wire itself. The net external force is zero since the system is stationary. A little trigonometry can now be used to find the tensions. One conclusion is possible at the outset—we can see from part (b) of the figure that the magnitudes of the tensions T_L and T_R must be equal. This is because there is no horizontal acceleration in the rope, and the only forces acting to the left and right are T_L and T_R. Thus, the magnitude of those forces must be equal so that they cancel each other out.

Whenever we have two-dimensional vector problems in which no two vectors are parallel, the easiest method of solution is to pick a convenient coordinate system and project the vectors onto its axes. In this case the best coordinate system has one axis horizontal and the other vertical. We call the horizontal the x-axis and the vertical the y-axis.

Solution

First, we need to resolve the tension vectors into their horizontal and vertical components. It helps to draw a new free-body diagram showing all of the horizontal and vertical components of each force acting on the system.

Consider the horizontal components of the forces (denoted with a subscript[latex]\boldsymbol{x}[/latex]):

[latex]\boldsymbol{F_{\textbf{net}x}=T_{\textbf{L}x}-T_{\textbf{R}x}.}[/latex]

The net external horizontal force F_netx = 0, since the person is stationary. Thus,

$latex \begin{array}{l @{{}={}} l} \boldsymbol{F_{\textbf{net} x} = 0} & \boldsymbol{T_{\textbf{L} x} - T_{\textbf{R} x}} \\[1em] \boldsymbol{T_{\textbf{L} x}} & \boldsymbol{T_{\textbf{R} x}} \end{array} $

Now, observe Figure 7. You can use trigonometry to determine the magnitude of T_L and T_R. Notice that:

$latex \begin{array}{l @{{}={}} l} \boldsymbol{\textbf{cos} \; (5.0 ^{\circ})} & \boldsymbol{\frac{T_{\textbf{L} x}}{T_{\textbf{L}}}} \\[1em] \boldsymbol{T_{\textbf{L} x}} &\boldsymbol{T_{\textbf{L}} \; \textbf{cos} \; (5.0 ^{\circ})} \\[1em] \boldsymbol{\textbf{cos} \; (5.0 ^{\circ})} & \boldsymbol{\frac{T_{\textbf{R} x}}{T_{\textbf{R}}}} \\[1em] \boldsymbol{T_{\textbf{R} x}} &\boldsymbol{T_{\textbf{R}} \; \textbf{cos} \; (5.0 ^{\circ})} \end{array} $

Equating T_Lx and T_Rx:

[latex]\boldsymbol{T_{\textbf{L}}\textbf{cos}(5.0^0)=T_{\textbf{R}}\textbf{cos}(5.0^0).}[/latex]

Thus,

[latex]\boldsymbol{T_{\textbf{L}}=T_{\textbf{R}}=T,}[/latex]

as predicted. Now, considering the vertical components (denoted by a subscript y), we can solve for T. Again, since the person is stationary, Newton’s second law implies that net F_y = 0. Thus, as illustrated in the free-body diagram in Figure 7,

[latex]\boldsymbol{F_{\textbf{net}y}=T_{\textbf{L}y}+T_{\textbf{R}y}-w=0.}[/latex]

Observing Figure 7, we can use trigonometry to determine the relationship between T_Ly, T_Ry, and T. As we determined from the analysis in the horizontal direction, T_L = T_R = T:

$latex \begin{array}{l @{{}={}} l} \boldsymbol{\textbf{sin} (5.0 ^{\circ})} & \boldsymbol{\frac{T_{\textbf{L} y}}{T_{\textbf{L}}}} \\[1em] \boldsymbol{T_{\textbf{L}y} = T_{\textbf{L}} \;\textbf{sin}(5.0^{\circ})} & \boldsymbol{T \;\textbf{sin} (5.0 ^{\circ})} \\[1em] \boldsymbol{\textbf{sin} (5.0 ^{\circ})} & \boldsymbol{\frac{T_{\textbf{R} y}}{T_{\textbf{R}}}} \\[1em] \boldsymbol{T_{\textbf{R}y} = T_{\textbf{R}} \;\textbf{sin}(5.0^{\circ})} & \boldsymbol{T \;\textbf{sin} (5.0 ^{\circ})} \end{array} $

Now, we can substitute the values for T_Ly and T_Ry, into the net force equation in the vertical direction:

$latex \begin{array}{l @{{}={}} l} \boldsymbol{F_{\textbf{net} y}} & \boldsymbol{T_{\textbf{L} y} + T_{\textbf{L} y} - w = 0} \\[1em] \boldsymbol{F_{\textbf{net} y}} & \boldsymbol{T \; \textbf{sin} \;(5.0 ^{\circ}) + T \; \textbf{sin} \;(5.0 ^{\circ}) - w = 0} \\[1em] \boldsymbol{2 \; T \; \textbf{sin} (5.0^{\circ}) - w} & \boldsymbol{0} \\[1em] \boldsymbol{2 \; T \; \textbf{sin} (5.0^{\circ})} & \boldsymbol{w} \end{array} $

and

[latex]\boldsymbol{T\:=}[/latex][latex size="2"]\boldsymbol{\frac{w}{2\:\textbf{sin}(5.0^0)}}[/latex][latex]\boldsymbol{=}[/latex][latex size="2"]\boldsymbol{\frac{mg}{2\:\textbf{sin}(5.0^0)},}[/latex]

so that

[latex]\boldsymbol{T\:=}[/latex][latex size="2"]\boldsymbol{\frac{(70.0\textbf{ kg})(9.80\textbf{ m/s}^2)}{2(0.0872)}},[/latex]

and the tension is

[latex]\boldsymbol{T=3900\textbf{ N}.}[/latex]

Discussion

Note that the vertical tension in the wire acts as a normal force that supports the weight of the tightrope walker. The tension is almost six times the 686-N weight of the tightrope walker. Since the wire is nearly horizontal, the vertical component of its tension is only a small fraction of the tension in the wire. The large horizontal components are in opposite directions and cancel, and so most of the tension in the wire is not used to support the weight of the tightrope walker.

If we wish to create a very large tension, all we have to do is exert a force perpendicular to a flexible connector, as illustrated in Figure 8. As we saw in the last example, the weight of the tightrope walker acted as a force perpendicular to the rope. We saw that the tension in the roped related to the weight of the tightrope walker in the following way:

[latex]\boldsymbol{T\:=}[/latex][latex size="2"]\boldsymbol{\frac{w}{2\:\textbf{sin}(\theta)}}.[/latex]

We can extend this expression to describe the tension T created when a perpendicular force (F_⊥) is exerted at the middle of a flexible connector:

[latex]\boldsymbol{T\:=}[/latex][latex size="2"]\boldsymbol{\frac{F_{\perp}}{2\:\textbf{sin}(\theta)}.}[/latex]

Note that θ is the angle between the horizontal and the bent connector. In this case, T becomes very large as θ approaches zero. Even the relatively small weight of any flexible connector will cause it to sag, since an infinite tension would result if it were horizontal (i.e., θ = 0 and sinθ = 0). (See Figure 8.)

[caption id="" align="aligncenter" width="550"]Figure 8. We can create a very large tension in the chain by pushing on it perpendicular to its length, as shown. Suppose we wish to pull a car out of the mud when no tow truck is available. Each time the car moves forward, the chain is tightened to keep it as nearly straight as possible. The tension in the chain is given by T = F_⊥ / 2 sin(θ); since θ is small, T is very large. This situation is analogous to the tightrope walker shown in Figure 6, except that the tensions shown here are those transmitted to the car and the tree rather than those acting at the point where F_⊥ is applied.[/caption]

[caption id="" align="aligncenter" width="325"]Figure 9. Unless an infinite tension is exerted, any flexible connector—such as the chain at the bottom of the picture—will sag under its own weight, giving a characteristic curve when the weight is evenly distributed along the length. Suspension bridges—such as the Golden Gate Bridge shown in this image—are essentially very heavy flexible connectors. The weight of the bridge is evenly distributed along the length of flexible connectors, usually cables, which take on the characteristic shape. (credit: Leaflet, Wikimedia Commons)[/caption]

Extended Topic: Real Forces and Inertial Frames

There is another distinction among forces in addition to the types already mentioned. Some forces are real, whereas others are not. Real forces are those that have some physical origin, such as the gravitational pull. Contrastingly, fictitious forces are those that arise simply because an observer is in an accelerating frame of reference, such as one that rotates (like a merry-go-round) or undergoes linear acceleration (like a car slowing down). For example, if a satellite is heading due north above Earth’s northern hemisphere, then to an observer on Earth it will appear to experience a force to the west that has no physical origin. Of course, what is happening here is that Earth is rotating toward the east and moves east under the satellite. In Earth’s frame this looks like a westward force on the satellite, or it can be interpreted as a violation of Newton’s first law (the law of inertia). An inertial frame of reference is one in which all forces are real and, equivalently, one in which Newton’s laws have the simple forms given in this chapter.

Earth’s rotation is slow enough that Earth is nearly an inertial frame. You ordinarily must perform precise experiments to observe fictitious forces and the slight departures from Newton’s laws, such as the effect just described. On the large scale, such as for the rotation of weather systems and ocean currents, the effects can be easily observed.

The crucial factor in determining whether a frame of reference is inertial is whether it accelerates or rotates relative to a known inertial frame. Unless stated otherwise, all phenomena discussed in this text are considered in inertial frames.

All the forces discussed in this section are real forces, but there are a number of other real forces, such as lift and thrust, that are not discussed in this section. They are more specialized, and it is not necessary to discuss every type of force. It is natural, however, to ask where the basic simplicity we seek to find in physics is in the long list of forces. Are some more basic than others? Are some different manifestations of the same underlying force? The answer to both questions is yes, as will be seen in the next (extended) section and in the treatment of modern physics later in the text.

PHET EXPLORATIONS: FORCES IN 1 DIMENSION

Explore the forces at work when you try to push a filing cabinet. Create an applied force and see the resulting friction force and total force acting on the cabinet. Charts show the forces, position, velocity, and acceleration vs. time. View a free-body diagram of all the forces (including gravitational and normal forces). [caption id="" align="aligncenter" width="450"] Figure 10. Forces in 1 Dimension[/caption]

Section Summary

• When objects rest on a surface, the surface applies a force to the object that supports the weight of the object. This supporting force acts perpendicular to and away from the surface. It is called a normal force, N.
• When objects rest on a non-accelerating horizontal surface, the magnitude of the normal force is equal to the weight of the object:
  [latex]\boldsymbol{N=mg}.[/latex]
• When objects rest on an inclined plane that makes an angle θ with the horizontal surface, the weight of the object can be resolved into components that act perpendicular ([latex]\textbf{w}_{\boldsymbol{\perp}}[/latex]) and parallel ([latex]\textbf{w}_{\boldsymbol{\parallel}}[/latex]) to the surface of the plane. These components can be calculated using:
  [latex]\boldsymbol{w_{\parallel} =w\:\textbf{sin}\:(\theta)=mg\:\textbf{sin}\:(\theta)}[/latex]
  [latex]\boldsymbol{w_{\perp}=w\:\textbf{cos}\:(\theta)=mg\:\textbf{cos}\:(\theta).}[/latex]
• The pulling force that acts along a stretched flexible connector, such as a rope or cable, is called tension, T.When a rope supports the weight of an object that is at rest, the tension in the rope is equal to the weight of the object:
  [latex]\boldsymbol{T=mg}.[/latex]
• In any inertial frame of reference (one that is not accelerated or rotated), Newton’s laws have the simple forms given in this chapter and all forces are real forces having a physical origin.

Conceptual Questions

1: If a leg is suspended by a traction setup as shown in Figure 11, what is the tension in the rope?

2: In a traction setup for a broken bone, with pulleys and rope available, how might we be able to increase the force along the tibia using the same weight? (See Figure 11.) (Note that the tibia is the shin bone shown in this image.)

Problems & Exercises

1: Two teams of nine members each engage in a tug of war. Each of the first team’s members has an average mass of 68 kg and exerts an average force of 1350 N horizontally. Each of the second team’s members has an average mass of 73 kg and exerts an average force of 1365 N horizontally. (a) What is magnitude of the acceleration of the two teams? (b) What is the tension in the section of rope between the teams?

2: What force does a trampoline have to apply to a 45.0-kg gymnast to accelerate her straight up at 7.50 m/s²? Note that the answer is independent of the velocity of the gymnast—she can be moving either up or down, or be stationary.

3: (a) Calculate the tension in a vertical strand of spider web if a spider of mass 8.00 × 10^-5 kg hangs motionless on it. (b) Calculate the tension in a horizontal strand of spider web if the same spider sits motionless in the middle of it much like the tightrope walker in Figure 6. The strand sags at an angle of 12° below the horizontal. Compare this with the tension in the vertical strand (find their ratio).

4: Suppose a 60.0-kg gymnast climbs a rope. (a) What is the tension in the rope if he climbs at a constant speed? (b) What is the tension in the rope if he accelerates upward at a rate of 1.50 m/s²?

5: Show that, as stated in the text, a force F_⊥ exerted on a flexible medium at its center and perpendicular to its length (such as on the tightrope wire in Figure 6) gives rise to a tension of magnitude [latex]T=\frac{F_{\perp}}{2\:\textbf{sin}(\theta)}.[/latex]

6: Consider the baby being weighed in Figure 12. (a) What is the mass of the child and basket if a scale reading of 55 N is observed? (b) What is the tension T₁ in the cord attaching the baby to the scale? (c) What is the tension T₂ in the cord attaching the scale to the ceiling, if the scale has a mass of 0.500 kg? (d) Draw a sketch of the situation indicating the system of interest used to solve each part. The masses of the cords are negligible.

Glossary

inertial frame of reference

a coordinate system that is not accelerating; all forces acting in an inertial frame of reference are real forces, as opposed to fictitious forces that are observed due to an accelerating frame of reference

normal force

the force that a surface applies to an object to support the weight of the object; acts perpendicular to the surface on which the object rests

tension

the pulling force that acts along a medium, especially a stretched flexible connector, such as a rope or cable; when a rope supports the weight of an object, the force on the object due to the rope is called a tension force

Solutions

Problems & Exercises 1: (a) [latex]\boldsymbol{0.11\textbf{ m/s}^2}[/latex] (b) [latex]\boldsymbol{1.2\times10^4\textbf{ N}}[/latex]

3: (a) [latex]\boldsymbol{7.84\times10^{-4}\textbf{ N}}[/latex] (b) [latex]\boldsymbol{1.89\times10^{-3}\textbf{ N}}.[/latex]This is 2.41 times the tension in the vertical strand.

5: Newton’s second law applied in vertical direction gives

[latex]\boldsymbol{F_y=F-2T\:\textbf{sin}\:\theta=0}[/latex]

[latex]\boldsymbol{F=2T\:\textbf{sin}\:\theta}[/latex]

[latex]\boldsymbol{T=} [/latex] [latex size="2"] \boldsymbol{\frac{F}{2\:\textbf{sin}\:\theta}}.[/latex]

]]> 304 266 7 0 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/chapter/4-7-problem-solving-strategies/ Thu, 29 Jun 2017 22:13:31 +0000 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/chapter/4-7-problem-solving-strategies/

Summary

• Understand and apply a problem-solving procedure to solve problems using Newton's laws of motion.

Success in problem solving is obviously necessary to understand and apply physical principles, not to mention the more immediate need of passing exams. The basics of problem solving, presented earlier in this text, are followed here, but specific strategies useful in applying Newton’s laws of motion are emphasized. These techniques also reinforce concepts that are useful in many other areas of physics. Many problem-solving strategies are stated outright in the worked examples, and so the following techniques should reinforce skills you have already begun to develop.

Problem-Solving Strategy for Newton’s Laws of Motion

Step 1. As usual, it is first necessary to identify the physical principles involved. Once it is determined that Newton’s laws of motion are involved (if the problem involves forces), it is particularly important to draw a careful sketch of the situation. Such a sketch is shown in Figure 1(a). Then, as in Figure 1(b), use arrows to represent all forces, label them carefully, and make their lengths and directions correspond to the forces they represent (whenever sufficient information exists).

Step 2. Identify what needs to be determined and what is known or can be inferred from the problem as stated. That is, make a list of knowns and unknowns. Then carefully determine the system of interest. This decision is a crucial step, since Newton’s second law involves only external forces. Once the system of interest has been identified, it becomes possible to determine which forces are external and which are internal, a necessary step to employ Newton’s second law. (See Figure 1(c).) Newton’s third law may be used to identify whether forces are exerted between components of a system (internal) or between the system and something outside (external). As illustrated earlier in this chapter, the system of interest depends on what question we need to answer. This choice becomes easier with practice, eventually developing into an almost unconscious process. Skill in clearly defining systems will be beneficial in later chapters as well.

A diagram showing the system of interest and all of the external forces is called a free-body diagram. Only forces are shown on free-body diagrams, not acceleration or velocity. We have drawn several of these in worked examples. Figure 1(c) shows a free-body diagram for the system of interest. Note that no internal forces are shown in a free-body diagram.

Step 3. Once a free-body diagram is drawn, Newton’s second law can be applied to solve the problem. This is done in Figure 1(d) for a particular situation. In general, once external forces are clearly identified in free-body diagrams, it should be a straightforward task to put them into equation form and solve for the unknown, as done in all previous examples. If the problem is one-dimensional—that is, if all forces are parallel—then they add like scalars. If the problem is two-dimensional, then it must be broken down into a pair of one-dimensional problems. This is done by projecting the force vectors onto a set of axes chosen for convenience. As seen in previous examples, the choice of axes can simplify the problem. For example, when an incline is involved, a set of axes with one axis parallel to the incline and one perpendicular to it is most convenient. It is almost always convenient to make one axis parallel to the direction of motion, if this is known.

Applying Newton’s Second Law

Before you write net force equations, it is critical to determine whether the system is accelerating in a particular direction. If the acceleration is zero in a particular direction, then the net force is zero in that direction. Similarly, if the acceleration is nonzero in a particular direction, then the net force is described by the equation: F_net = ma. For example, if the system is accelerating in the horizontal direction, but it is not accelerating in the vertical direction, then you will have the following conclusions:

[latex]\boldsymbol{F_{\textbf{net} \; x}=ma},[/latex]

[latex]\boldsymbol{F_{\textbf{net} \; y}=0}.[/latex]

You will need this information in order to determine unknown forces acting in a system.

Step 4. As always, check the solution to see whether it is reasonable. In some cases, this is obvious. For example, it is reasonable to find that friction causes an object to slide down an incline more slowly than when no friction exists. In practice, intuition develops gradually through problem solving, and with experience it becomes progressively easier to judge whether an answer is reasonable. Another way to check your solution is to check the units. If you are solving for force and end up with units of m/s, then you have made a mistake.

Section Summary

• To solve problems involving Newton’s laws of motion, follow the procedure described:
  1. Draw a sketch of the problem.
  2. Identify known and unknown quantities, and identify the system of interest. Draw a free-body diagram, which is a sketch showing all of the forces acting on an object. The object is represented by a dot, and the forces are represented by vectors extending in different directions from the dot. If vectors act in directions that are not horizontal or vertical, resolve the vectors into horizontal and vertical components and draw them on the free-body diagram.
  3. Write Newton’s second law in the horizontal and vertical directions and add the forces acting on the object. If the object does not accelerate in a particular direction (for example, the x-direction) then F_netx = 0. If the object does accelerate in that direction, F_netx = ma.
  4. Check your answer. Is the answer reasonable? Are the units correct?

Problems & Exercises

1: A 5.00 × 10⁵-kg rocket is accelerating straight up. Its engines produce 1.250 × 10⁷ N of thrust, and air resistance is 4.50 × 10⁶ N. What is the rocket’s acceleration? Explicitly show how you follow the steps in the Problem-Solving Strategy for Newton’s laws of motion.

2: The wheels of a midsize car exert a force of 2100 N backward on the road to accelerate the car in the forward direction. If the force of friction including air resistance is 250 N and the acceleration of the car is 1.80 m/s², what is the mass of the car plus its occupants? Explicitly show how you follow the steps in the Problem-Solving Strategy for Newton’s laws of motion. For this situation, draw a free-body diagram and write the net force equation.

3: Calculate the force a 70.0-kg high jumper must exert on the ground to produce an upward acceleration 4.00 times the acceleration due to gravity. Explicitly show how you follow the steps in the Problem-Solving Strategy for Newton’s laws of motion.

4: When landing after a spectacular somersault, a 40.0-kg gymnast decelerates by pushing straight down on the mat. Calculate the force she must exert if her deceleration is 7.00 times the acceleration due to gravity. Explicitly show how you follow the steps in the Problem-Solving Strategy for Newton’s laws of motion.

5: A freight train consists of two 8.00 × 10⁴-kg engines and 45 cars with average masses of 5.50 × 10⁴ kg. (a) What force must each engine exert backward on the track to accelerate the train at a rate of 5.00 × 10^-2 m/s² if the force of friction is 7.50 × 10⁵ N, assuming the engines exert identical forces? This is not a large frictional force for such a massive system. Rolling friction for trains is small, and consequently trains are very energy-efficient transportation systems. (b) What is the force in the coupling between the 37th and 38th cars (this is the force each exerts on the other), assuming all cars have the same mass and that friction is evenly distributed among all of the cars and engines?

6: Commercial airplanes are sometimes pushed out of the passenger loading area by a tractor. (a) An 1800-kg tractor exerts a force of 1.75 × 10⁴ N backward on the pavement, and the system experiences forces resisting motion that total 2400 N. If the acceleration is 0.150 m/s², what is the mass of the airplane? (b) Calculate the force exerted by the tractor on the airplane, assuming 2200 N of the friction is experienced by the airplane. (c) Draw two sketches showing the systems of interest used to solve each part, including the free-body diagrams for each.

7: A 1100-kg car pulls a boat on a trailer. (a) What total force resists the motion of the car, boat, and trailer, if the car exerts a 1900-N force on the road and produces an acceleration of 0.550 m/s²? The mass of the boat plus trailer is 700 kg. (b) What is the force in the hitch between the car and the trailer if 80% of the resisting forces are experienced by the boat and trailer?

8: (a) Find the magnitudes of the forces F₁ and F₂ that add to give the total force F_tot shown in Figure 2. This may be done either graphically or by using trigonometry. (b) Show graphically that the same total force is obtained independent of the order of addition of F₁ and F₂. (c) Find the direction and magnitude of some other pair of vectors that add to give F_tot. Draw these to scale on the same drawing used in part (b) or a similar picture.

[caption id="" align="aligncenter" width="325"]Figure 2.[/caption]

9: Two children pull a third child on a snow saucer sled exerting forces F₁ and F₂ as shown from above in Figure 3. Find the acceleration of the 49.00-kg sled and child system. Note that the direction of the frictional force is unspecified; it will be in the opposite direction of the sum of F₁ and F₂.

[caption id="" align="aligncenter" width="300"]Figure 3. An overhead view of the horizontal forces acting on a child’s snow saucer sled.[/caption]

10: Suppose your car was mired deeply in the mud and you wanted to use the method illustrated in Figure 4 to pull it out. (a) What force would you have to exert perpendicular to the center of the rope to produce a force of 12,000 N on the car if the angle is 2.00°? In this part, explicitly show how you follow the steps in the Problem-Solving Strategy for Newton’s laws of motion. (b) Real ropes stretch under such forces. What force would be exerted on the car if the angle increases to 7.00° and you still apply the force found in part (a) to its center?

[caption id="" align="aligncenter" width="500"]Figure 4.[/caption]

11: What force is exerted on the tooth in Figure 5 if the tension in the wire is 25.0 N? Note that the force applied to the tooth is smaller than the tension in the wire, but this is necessitated by practical considerations of how force can be applied in the mouth. Explicitly show how you follow steps in the Problem-Solving Strategy for Newton’s laws of motion.

[caption id="" align="aligncenter" width="225"]Figure 5. Braces are used to apply forces to teeth to realign them. Shown in this figure are the tensions applied by the wire to the protruding tooth. The total force applied to the tooth by the wire, F_app, points straight toward the back of the mouth.[/caption]

12: Figure 6 shows Superhero and Trusty Sidekick hanging motionless from a rope. Superhero’s mass is 90.0 kg, while Trusty Sidekick’s is 55.0 kg, and the mass of the rope is negligible. (a) Draw a free-body diagram of the situation showing all forces acting on Superhero, Trusty Sidekick, and the rope. (b) Find the tension in the rope above Superhero. (c) Find the tension in the rope between Superhero and Trusty Sidekick. Indicate on your free-body diagram the system of interest used to solve each part.

[caption id="" align="aligncenter" width="151"]Figure 6. Superhero and Trusty Sidekick hang motionless on a rope as they try to figure out what to do next. Will the tension be the same everywhere in the rope?[/caption]

13: A nurse pushes a cart by exerting a force on the handle at a downward angle 35.0° below the horizontal. The loaded cart has a mass of 28.0 kg, and the force of friction is 60.0 N. (a) Draw a free-body diagram for the system of interest. (b) What force must the nurse exert to move at a constant velocity?

14: Construct Your Own Problem Consider the tension in an elevator cable during the time the elevator starts from rest and accelerates its load upward to some cruising velocity. Taking the elevator and its load to be the system of interest, draw a free-body diagram. Then calculate the tension in the cable. Among the things to consider are the mass of the elevator and its load, the final velocity, and the time taken to reach that velocity.

15: Construct Your Own Problem Consider two people pushing a toboggan with four children on it up a snow-covered slope. Construct a problem in which you calculate the acceleration of the toboggan and its load. Include a free-body diagram of the appropriate system of interest as the basis for your analysis. Show vector forces and their components and explain the choice of coordinates. Among the things to be considered are the forces exerted by those pushing, the angle of the slope, and the masses of the toboggan and children.

16: Unreasonable Results (a) Repeat Exercise 7, but assume an acceleration of 1.20 m/s² is produced. (b) What is unreasonable about the result? (c) Which premise is unreasonable, and why is it unreasonable?

17: Unreasonable Results (a) What is the initial acceleration of a rocket that has a mass of 1.50 × 10⁶ kg at takeoff, the engines of which produce a thrust of 2.00 × 10⁶ N? Do not neglect gravity. (b) What is unreasonable about the result? (This result has been unintentionally achieved by several real rockets.) (c) Which premise is unreasonable, or which premises are inconsistent? (You may find it useful to compare this problem to the rocket problem earlier in this section.)

Solutions

Problems & Exercises 1:

[caption id="" align="aligncenter" width="82"]Figure 7.[/caption]

Using the free-body diagram: [latex]\boldsymbol{F_{\textbf{net}}=T-f-mg=ma},[/latex] so that

$latex \boldsymbol{ a = \frac{T - f - mg}{m} = \frac{1.250 \times 10^7 \; \textbf{N} - 4.50 \times 10^6 \; N - (5.00 \times 10^5 \;\textbf{kg})(9.80 \;\textbf{m/s}^2)}{5.00 \times 10^5 \;\textbf{kg}} = 6.20 \;\textbf{m/s}^2} $

3: [caption id="" align="aligncenter" width="131"]Figure 8.[/caption]

1. Use Newton’s laws of motion

2. Given :[latex]\boldsymbol{a=4.00g=(4.00)(9.80\textbf{ m/s}^2)=39.2\textbf{ m/s}^2\:;\: m=70.0\textbf{ kg}},[/latex]

   Find: [latex]\boldsymbol{F}.[/latex]

3. [latex]\boldsymbol{\sum{F}=+F-w=ma},[/latex] so that [latex]\boldsymbol{F=ma+w=ma+mg=m(a+g)}.[/latex]

   [latex]\boldsymbol{F=(70.0\textbf{ kg})[(39.2\textbf{ m/s}^2)+(9.80\textbf{ m/s}^2)]=3.43\times10^3\textbf{ N}}.[/latex] The force exerted by the high-jumper is actually down on the ground, but [latex]\boldsymbol{F}[/latex] is up from the ground and makes him jump.

4. This result is reasonable, since it is quite possible for a person to exert a force of the magnitude of [latex]\boldsymbol{10^3\textbf{ N}}.[/latex]

5: (a) [latex]\boldsymbol{4.41\times10^5\textbf{ N}}[/latex] (b) [latex]\boldsymbol{1.50\times10^5\textbf{ N}}[/latex]

7: (a) [latex]\boldsymbol{910\textbf{ N}}[/latex] (b) [latex]\boldsymbol{1.11\times10^3\textbf{ N}}[/latex]

9: [latex]\boldsymbol{a=0.139\textbf{ m/s}},\boldsymbol{\theta=12.4^o}[/latex] north of east 11:

1. Use Newton’s laws since we are looking for forces.
2. Draw a free-body diagram:

   [caption id="" align="aligncenter" width="200"]Figure 9.[/caption]

3. The tension is given as [latex]\boldsymbol{T=25.0\textbf{ N}}.[/latex] Find [latex]\boldsymbol{F_{\textbf{app}}}.[/latex] Using Newton’s laws gives: [latex]\boldsymbol{\sum{F}_y=0},[/latex] so that applied force is due to the y-components of the two tensions: [latex]\boldsymbol{F_{\textbf{app}}=2T\textbf{sin}\theta=2(25.0\textbf{ N})\textbf{sin}(15^0)=12.9\textbf{ N}}[/latex]

   The x-components of the tension cancel. [latex]\boldsymbol{\sum{F}_x=0}.[/latex]

4. This seems reasonable, since the applied tensions should be greater than the force applied to the tooth.

]]> 314 266 8 0 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/chapter/4-8-further-applications-of-newtons-laws-of-motion/ Thu, 29 Jun 2017 22:13:32 +0000 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/chapter/4-8-further-applications-of-newtons-laws-of-motion/

Summary

• Apply problem-solving techniques to solve for quantities in more complex systems of forces.
• Integrate concepts from kinematics to solve problems using Newton's laws of motion.

There are many interesting applications of Newton’s laws of motion, a few more of which are presented in this section. These serve also to illustrate some further subtleties of physics and to help build problem-solving skills.

Example 1: Drag Force on a Barge

Suppose two tugboats push on a barge at different angles, as shown in Figure 1. The first tugboat exerts a force of 2.7 × 10⁵ N in the x-direction, and the second tugboat exerts a force of 3.6 × 10⁵ N in the y-direction.

If the mass of the barge is 5.0 × 10⁶ kg and its acceleration is observed to be 7.5 × 10^-2 m/s² in the direction shown, what is the drag force of the water on the barge resisting the motion? (Note: drag force is a frictional force exerted by fluids, such as air or water. The drag force opposes the motion of the object.)

Strategy

The directions and magnitudes of acceleration and the applied forces are given in Figure 1(a). We will define the total force of the tugboats on the barge as F_app so that:

[latex]\boldsymbol{F_{\textbf{app}}=F_x+F_y}[/latex]

Since the barge is flat bottomed, the drag of the water F_D will be in the direction opposite to F_app, as shown in the free-body diagram in Figure 1(b). The system of interest here is the barge, since the forces on it are given as well as its acceleration. Our strategy is to find the magnitude and direction of the net applied force F_app, and then apply Newton’s second law to solve for the drag force F_D.

Solution

Since F_x and F_y are perpendicular, the magnitude and direction of F_app are easily found. First, the resultant magnitude is given by the Pythagorean theorem:

[latex]\boldsymbol{F_{\textbf{app}}=\sqrt{F_x^2+F_y^2}}[/latex]

[latex]\boldsymbol{F_{\textbf{app}}=\sqrt{(2.7\times10^5\textbf{ N})^2+(3.6\times10^5\textbf{ N})^2}=4.5\times10^5\textbf{ N}}.[/latex]

The angle is given by

[latex]\boldsymbol{\theta=\textbf{tan}^{-1}(\frac{F_y}{F_x})}[/latex]

[latex]\boldsymbol{\theta=\textbf{tan}^{-1}(\frac{3.6\times10^5\textbf{ N}}{2.7\times10^5\textbf{ N}})=53^o},[/latex]

which we know, because of Newton’s first law, is the same direction as the acceleration. F_D is in the opposite direction of F_app, since it acts to slow down the acceleration. Therefore, the net external force is in the same direction as F_app, but its magnitude is slightly less than F_app. The problem is now one-dimensional. From Figure 1(b), we can see that

[latex]\boldsymbol{F_{\textbf{net}}=F_{\textbf{app}}-F_{\textbf{D}}}.[/latex]

But Newton’s second law states that

[latex]\boldsymbol{F_{\textbf{net}}=ma}.[/latex]

Thus,

[latex]\boldsymbol{F_{\textbf{app}}-F_{\textbf{D}}=ma}.[/latex]

This can be solved for the magnitude of the drag force of the water F_D in terms of known quantities:

[latex]\boldsymbol{F_{\textbf{D}}=F_{\textbf{app}}-ma}.[/latex]

Substituting known values gives

[latex]\boldsymbol{F_{\textbf{D}}=(4.5\times10^5\textbf{ N})-(5.0\times10^6\textbf{ kg})(7.5\times10^{-2}\textbf{ m/s}^2)=7.5\times10^4\textbf{ N}}.[/latex]

The direction of F_D has already been determined to be in the direction opposite to F_app, or at an angle of 53° south of west.

Discussion

The numbers used in this example are reasonable for a moderately large barge. It is certainly difficult to obtain larger accelerations with tugboats, and small speeds are desirable to avoid running the barge into the docks. Drag is relatively small for a well-designed hull at low speeds, consistent with the answer to this example, where F_D is less than 1/600th of the weight of the ship.

In the earlier example of a tightrope walker we noted that the tensions in wires supporting a mass were equal only because the angles on either side were equal. Consider the following example, where the angles are not equal; slightly more trigonometry is involved.

Example 2: Different Tensions at Different Angles

Consider the traffic light (mass 15.0 kg) suspended from two wires as shown in Figure 2. Find the tension in each wire, neglecting the masses of the wires.

Strategy

The system of interest is the traffic light, and its free-body diagram is shown in Figure 2(c). The three forces involved are not parallel, and so they must be projected onto a coordinate system. The most convenient coordinate system has one axis vertical and one horizontal, and the vector projections on it are shown in part (d) of the figure. There are two unknowns in this problem (T₁ and T₂), so two equations are needed to find them. These two equations come from applying Newton’s second law along the vertical and horizontal axes, noting that the net external force is zero along each axis because acceleration is zero.

Solution

First consider the horizontal or x-axis:

[latex]\boldsymbol{F_{\textbf{net}x}=T_{2x}-T_{1x}=0}.[/latex]

Thus, as you might expect,

[latex]\boldsymbol{T_{1x}=T_{2x}}.[/latex]

This gives us the following relationship between T₁ and T₂:

[latex]\boldsymbol{T_1\textbf{cos}(30^0)=T_2\textbf{cos}(45^0)}.[/latex]

Thus,

[latex]\boldsymbol{T_2=(1.225)T_1}.[/latex]

Note that T₁ and T₂ are not equal in this case, because the angles on either side are not equal. It is reasonable that T₂ ends up being greater than T₁ , because it is exerted more vertically than T₁

Now consider the force components along the vertical or y-axis:

[latex]\boldsymbol{F_{\textbf{net}y}=T_{1y}+T_{2y}-w=0}.[/latex]

This implies

[latex]\boldsymbol{T_{1y}+T_{2y}=w}.[/latex]

Substituting the expressions for the vertical components gives

[latex]\boldsymbol{T_1 \;\textbf{sin}(30^0)+T_2 \;\textbf{sin}(45^0)=w}.[/latex]

There are two unknowns in this equation, but substituting the expression for T₂ in terms of T₁ reduces this to one equation with one unknown:

[latex]\boldsymbol{T_1(0.500)+(1.225T_1)(0.707)=w=mg},[/latex]

which yields

[latex]\boldsymbol{(1.366)T_1=(15.0\textbf{ kg})(9.80\textbf{ m/s}^2)}.[/latex]

Solving this last equation gives the magnitude of T₁ to be

[latex]\boldsymbol{T_1=108\textbf{ N}}.[/latex]

Finally, the magnitude of T₂ is determined using the relationship between them, T₂= 1.225 T₁found above. Thus we obtain

[latex]\boldsymbol{T_2=132\textbf{ N}}.[/latex]

Discussion

Both tensions would be larger if both wires were more horizontal, and they will be equal if and only if the angles on either side are the same (as they were in the earlier example of a tightrope walker).

The bathroom scale is an excellent example of a normal force acting on a body. It provides a quantitative reading of how much it must push upward to support the weight of an object. But can you predict what you would see on the dial of a bathroom scale if you stood on it during an elevator ride? Will you see a value greater than your weight when the elevator starts up? What about when the elevator moves upward at a constant speed: will the scale still read more than your weight at rest? Consider the following example.

Example 3: What Does the Bathroom Scale Read in an Elevator?

Figure 3 shows a 75.0-kg man (weight of about 165 lb) standing on a bathroom scale in an elevator. Calculate the scale reading: (a) if the elevator accelerates upward at a rate of 1.20 m/s², and (b) if the elevator moves upward at a constant speed of 1 m/s.

Strategy

If the scale is accurate, its reading will equal F_p, the magnitude of the force the person exerts downward on it. Figure 3(a) shows the numerous forces acting on the elevator, scale, and person. It makes this one-dimensional problem look much more formidable than if the person is chosen to be the system of interest and a free-body diagram is drawn as in Figure 3(b). Analysis of the free-body diagram using Newton’s laws can produce answers to both parts (a) and (b) of this example, as well as some other questions that might arise. The only forces acting on the person are his weight w and the upward force of the scale F_s. According to Newton’s third law F_p and F_s are equal in magnitude and opposite in direction, so that we need to find F_s in order to find what the scale reads. We can do this, as usual, by applying Newton’s second law,

[latex]\boldsymbol{F_{\textbf{net}}=ma.}[/latex]

From the free-body diagram we see that F_net = F_s - w, so that

[latex]\boldsymbol{F_{\textbf{s}}-w=ma.}[/latex]

Solving for F_s gives an equation with only one unknown:

[latex]\boldsymbol{F_{\textbf{s}}=ma+w,}[/latex]

or, because weight = gravity force downwards = mg, simply

[latex]\boldsymbol{F_{\textbf{s}}=ma+mg.}[/latex]

No assumptions were made about the acceleration, and so this solution should be valid for a variety of accelerations in addition to the ones in this exercise.

Solution for (a)

In this part of the problem, a = 1.20 m/s², so that

[latex]\boldsymbol{F_{\textbf{s}}=(75.0\textbf{ kg})(1.20\textbf{ m/s}^2)+(75.0\textbf{ kg})(9.80\textbf{ m/s}^2) = 825 N,}[/latex]

Discussion for (a)

This is about 185 lb. What would the scale have read if he were stationary? Since his acceleration would be zero, the force of the scale would be equal to his weight:

[latex]\boldsymbol{F_{\textbf{net}}=ma=0=F_{\textbf{s}}-w}[/latex]

[latex]\boldsymbol{F_{\textbf{s}}=w=mg}[/latex]

[latex]\boldsymbol{F_{\textbf{s}}=(75.0\textbf{ kg})(9.80\textbf{ m/s}^2) = 735 N} [/latex]

So, the scale reading in the elevator is greater than his 735-N (165 lb) weight. This means that the scale is pushing up on the person with a force greater than his weight, as it must in order to accelerate him upward. Clearly, the greater the acceleration of the elevator, the greater the scale reading, consistent with what you feel in rapidly accelerating versus slowly accelerating elevators.

Solution for (b)

Now, what happens when the elevator reaches a constant upward velocity? Will the scale still read more than his weight? For any constant velocity—up, down, or stationary—acceleration is zero because [latex]\boldsymbol{a=\frac{\Delta{v}}{\Delta{t}}},[/latex] and Δv = 0.

Thus,

[latex]\boldsymbol{F_{\textbf{s}}=ma+mg=0+mg}[/latex]

Now

[latex]\boldsymbol{F_{\textbf{s}}=(75.0\textbf{ kg})(9.80\textbf{ m/s}^2) = 735 N} [/latex]

Discussion for (b)

The scale reading is 735 N, which equals the person’s weight. This will be the case whenever the elevator has a constant velocity—moving up, moving down, or stationary.

The solution to the previous example also applies to an elevator accelerating downward, as mentioned. When an elevator accelerates downward,[latex]\boldsymbol{a}[/latex]is negative, and the scale reading is less than the weight of the person, until a constant downward velocity is reached, at which time the scale reading again becomes equal to the person’s weight. If the elevator is in free-fall and accelerating downward at g, then the scale reading will be zero and the person will appear to be weightless.

Integrating Concepts: Newton’s Laws of Motion and Kinematics

Physics is most interesting and most powerful when applied to general situations that involve more than a narrow set of physical principles. Newton’s laws of motion can also be integrated with other concepts that have been discussed previously in this text to solve problems of motion. For example, forces produce accelerations, a topic of kinematics, and hence the relevance of earlier chapters. When approaching problems that involve various types of forces, acceleration, velocity, and/or position, use the following steps to approach the problem:

Problem-Solving Strategy

Step 1. Identify which physical principles are involved. Listing the givens and the quantities to be calculated will allow you to identify the principles involved.

Step 2. Solve the problem using strategies outlined in the text. If these are available for the specific topic, you should refer to them. You should also refer to the sections of the text that deal with a particular topic. The following worked example illustrates how these strategies are applied to an integrated concept problem.

Example 4: What Force Must a Soccer Player Exert to Reach Top Speed?

A soccer player starts from rest and accelerates forward, reaching a velocity of 8.00 m/s in 2.50 s. (a) What was his average acceleration? (b) What average force did he exert backward on the ground to achieve this acceleration? The player’s mass is 70.0 kg, and air resistance is negligible.

Strategy

1. To solve an integrated concept problem, we must first identify the physical principles involved and identify the chapters in which they are found. Part (a) of this example considers acceleration along a straight line. This is a topic of kinematics. Part (b) deals with force, a topic of dynamics found in this chapter.
2. The following solutions to each part of the example illustrate how the specific problem-solving strategies are applied. These involve identifying knowns and unknowns, checking to see if the answer is reasonable, and so forth.

Solution for (a)

We are given the initial and final velocities (zero and 8.00 m/s forward); thus, the change in velocity is Δv = 8.00 m/s. We are given the elapsed time, and so Δt = 2.50 s. The unknown is acceleration, which can be found from its definition:

[latex]\boldsymbol{a\:=}[/latex][latex size ="2"]\boldsymbol{\frac{\Delta{v}}{\Delta{t}}}[/latex].

Substituting the known values yields

[latex]\boldsymbol{a=\frac{8.00\textbf{ m/s}}{2.50\textbf{ s}}}[/latex]

[latex]\boldsymbol{=3.20\textbf{ m/s}^2}.\:\:\:\:\:[/latex]

Discussion for (a)

This is an attainable acceleration for an athlete in good condition.

Solution for (b)

Here we are asked to find the average force the player exerts backward to achieve this forward acceleration. Neglecting air resistance, this would be equal in magnitude to the net external force on the player, since this force causes his acceleration. Since we now know the player’s acceleration and are given his mass, we can use Newton’s second law to find the force exerted. That is,

[latex]\boldsymbol{F_{\textbf{net}}=ma.}[/latex]

Substituting the known values of m and a gives

[latex]\boldsymbol{F_{\textbf{net}}=(70.0\textbf{ kg})(3.20\textbf{ m/s}^2)=224N}[/latex]

Discussion for (b)

This is about 50 pounds, a reasonable average force.

This worked example illustrates how to apply problem-solving strategies to situations that include topics from different chapters. The first step is to identify the physical principles involved in the problem. The second step is to solve for the unknown using familiar problem-solving strategies. These strategies are found throughout the text, and many worked examples show how to use them for single topics. You will find these techniques for integrated concept problems useful in applications of physics outside of a physics course, such as in your profession, in other science disciplines, and in everyday life. The following problems will build your skills in the broad application of physical principles.

Summary

• Newton’s laws of motion can be applied in numerous situations to solve problems of motion.
• Some problems will contain multiple force vectors acting in different directions on an object. Be sure to draw diagrams, resolve all force vectors into horizontal and vertical components, and draw a free-body diagram. Always analyze the direction in which an object accelerates so that you can determine whether F_net = ma or F_net = 0.
• The normal force on an object is not always equal in magnitude to the weight of the object. If an object is accelerating, the normal force will be less than or greater than the weight of the object. Also, if the object is on an inclined plane, the normal force will always be less than the full weight of the object.
• Some problems will contain various physical quantities, such as forces, acceleration, velocity, or position. You can apply concepts from kinematics and dynamics in order to solve these problems of motion.

Conceptual Questions

1: To simulate the apparent weightlessness of space orbit, astronauts are trained in the hold of a cargo aircraft that is accelerating downward at[latex]\boldsymbol{g}.[/latex]Why will they appear to be weightless, as measured by standing on a bathroom scale, in this accelerated frame of reference? Is there any difference between their apparent weightlessness in orbit and in the aircraft?

2: A cartoon shows the toupee coming off the head of an elevator passenger when the elevator rapidly stops during an upward ride. Can this really happen without the person being tied to the floor of the elevator? Explain your answer.

Problems & Exercises

1: A flea jumps by exerting a force of 1.20 × 10^-5 N straight down on the ground. A breeze blowing on the flea parallel to the ground exerts a force of 0.500 × 10^-6 N on the flea. Find the direction and magnitude of the acceleration of the flea if its mass is 6.00 × 10^-7 kg. Do not neglect the gravitational force.

2: Two muscles in the back of the leg pull upward on the Achilles tendon, as shown in Figure 4. (These muscles are called the medial and lateral heads of the gastrocnemius muscle.) Find the magnitude and direction of the total force on the Achilles tendon. What type of movement could be caused by this force?

3: A 76.0-kg person is being pulled away from a burning building as shown in Figure 5. Calculate the tension in the two ropes if the person is momentarily motionless. Include a free-body diagram in your solution.

[caption id="" align="aligncenter" width="424"]Figure 5. The force T₂ needed to hold steady the person being rescued from the fire is less than her weight and less than the force T₁ in the other rope, since the more vertical rope supports a greater part of her weight (a vertical force)[/caption]

4: Integrated Concepts A 35.0-kg dolphin decelerates from 12.0 to 7.50 m/s in 2.30 s to join another dolphin in play. What average force was exerted to slow him if he was moving horizontally? (The gravitational force is balanced by the buoyant force of the water.)

5: Integrated Concepts When starting a foot race, a 70.0-kg sprinter exerts an average force of 650 N backward on the ground for 0.800 s. (a) What is his final speed? (b) How far does he travel?   Hint:  Find his acceleration first then use kinematics.

6: Integrated Concepts A large rocket has a mass of 2.00 × 10⁶ kg at takeoff, and its engines produce a thrust of 3.50 × 10⁷ N. (a) Find its initial acceleration if it takes off vertically. (b) How long does it take to reach a velocity of 120 km/h straight up, assuming constant mass and thrust? (c) In reality, the mass of a rocket decreases significantly as its fuel is consumed. Describe qualitatively how this affects the acceleration and time for this motion.

7: Integrated Concepts A basketball player jumps straight up for a ball. To do this, he lowers his body 0.300 m and then accelerates through this distance by forcefully straightening his legs. This player leaves the floor with a vertical velocity sufficient to carry him 0.900 m above the floor. (a) Calculate his velocity when he leaves the floor. (b) Calculate his acceleration while he is straightening his legs. He goes from zero to the velocity found in part (a) in a distance of 0.300 m. (c) Calculate the force he exerts on the floor to do this, given that his mass is 110 kg.

8: Integrated Concepts A 2.50-kg fireworks shell is fired straight up from a mortar and reaches a height of 110 m. (a) Neglecting air resistance (a poor assumption, but we will make it for this example), calculate the shell’s velocity when it leaves the mortar. (b) The mortar itself is a tube 0.450 m long. Calculate the average acceleration of the shell in the tube as it goes from zero to the velocity found in (a). (c) What is the average force on the shell in the mortar? Express your answer in newtons and as a ratio to the weight of the shell.

9: Integrated Concepts Repeat Exercise 8 for a shell fired at an angle 10.0° from the vertical.

10: Integrated Concepts An elevator filled with passengers has a mass of 1700 kg. (a) The elevator accelerates upward from rest at a rate of 1.20 m/s² for 1.50 s. Calculate the tension in the cable supporting the elevator. (b) The elevator continues upward at constant velocity for 8.50 s. What is the tension in the cable during this time? (c) The elevator decelerates at a rate of 0.600{ m/s² for 3.00 s. What is the tension in the cable during deceleration? (d) How high has the elevator moved above its original starting point, and what is its final velocity?

11: Unreasonable Results (a) What is the final velocity of a car originally traveling at 50.0 km/h that decelerates at a rate of 0.400 m/s² for 50.0 s? (b) What is unreasonable about the result? (c) Which premise is unreasonable, or which premises are inconsistent?

12: Unreasonable Results A 75.0-kg man stands on a bathroom scale in an elevator that accelerates from rest to 30.0 m/s in 2.00 s. (a) Calculate the scale reading in newtons and compare it with his weight. (The scale exerts an upward force on him equal to its reading.) (b) What is unreasonable about the result? (c) Which premise is unreasonable, or which premises are inconsistent?

Solutions

Problems & Exercises 1: [latex]\boldsymbol{10.2\textbf{ m/s}^2, 4.67^0\textbf{ from vertical}}[/latex] 3: T₁ = 736 N   T₂ =  194  N    as  net force is 0 N so using magnitudes only T1 cos 15^o + T2 sin 10^o = Weight = mg    and   T1 sin 15^o = T2  cos10^o 

[caption id="" align="aligncenter" width="225"]Figure 6.[/caption]

5:   (a)  7.43 m/s (b) 2.97 m  as the acceleration is 650 N / 70.0 kg = 9.29 m/s²7: (a) [latex]\boldsymbol{4.20\textbf{ m/s}}[/latex]  (b) [latex]\boldsymbol{29.4\textbf{ m/s}^2}[/latex] (c) [latex]\boldsymbol{4.31\times10^3\textbf{ N}}[/latex] 9: (a) $$\boldsymbol{47.1\textbf{ m/s}}$$   (b) [latex]\boldsymbol{2.47\times10^3\textbf{ m/s}^2}[/latex] (c ) [latex]\boldsymbol{6.18\times10^3\textbf{ N}}.[/latex] The average force is 252 times the shell’s weight.

]]> 321 266 9 0 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/chapter/4-9-extended-topic-the-four-basic-forces-an-introduction/ Thu, 29 Jun 2017 22:13:33 +0000 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/chapter/4-9-extended-topic-the-four-basic-forces-an-introduction/

Summary

• Explain the concept of a field
• Explain the four basic forces that underlie the processes in nature.

One of the most remarkable simplifications in physics is that only four distinct forces account for all known phenomena. In fact, nearly all of the forces we experience directly are due to only one basic force, called the electromagnetic force. (The gravitational force is the only force we experience directly that is not electromagnetic.) This is a tremendous simplification of the myriad of apparently different forces we can list, only a few of which were discussed in the previous section. As we will see, the basic forces are all thought to act through the exchange of microscopic carrier particles, and the characteristics of the basic forces are determined by the types of particles exchanged. Action at a distance, such as the gravitational force of Earth on the Moon, is explained by the existence of a force field rather than by “physical contact.”

The four basic forces are the gravitational force, the electromagnetic force, the weak nuclear force, and the strong nuclear force. Their properties are summarized in Table 1. Since the weak and strong nuclear forces act over an extremely short range, the size of a nucleus or less, we do not experience them directly, although they are crucial to the very structure of matter. These forces determine which nuclei are stable and which decay, and they are the basis of the release of energy in certain nuclear reactions. Nuclear forces determine not only the stability of nuclei, but also the relative abundance of elements in nature. The properties of the nucleus of an atom determine the number of electrons it has and, thus, indirectly determine the chemistry of the atom. More will be said of all of these topics in later chapters.

CONCEPT CONNECTIONS: THE FOUR BASIC FORCES

The four basic forces will be encountered in more detail as you progress through the text. The gravitational force is defined in Chapter 6 Uniform Circular Motion and Gravitation, electric force in Chapter 18 Electric Charge and Electric Field, magnetic force in Chapter 22 Magnetism, and nuclear forces in Chapter 31 Radioactivity and Nuclear Physics. On a macroscopic scale, electromagnetism and gravity are the basis for all forces. The nuclear forces are vital to the substructure of matter, but they are not directly experienced on the macroscopic scale.

Force	Approximate Relative Strengths	Range	Attraction/Repulsion	Carrier Particle
Gravitational	[latex]\boldsymbol{10^{-38}}[/latex]	[latex]\boldsymbol{\infty}[/latex]	attractive only	Graviton
Electromagnetic	[latex]\boldsymbol{10^{-2}}[/latex]	[latex]\boldsymbol{\infty}[/latex]	attractive and repulsive	Photon
Weak nuclear	[latex]\boldsymbol{10^{-13}}[/latex]	[latex]\boldsymbol{<10^{-18}\textbf{m}}[/latex]	attractive and repulsive	[latex]\textbf{W}^+,\textbf{W}^-,\textbf{Z}^0[/latex]
Strong nuclear	[latex]\boldsymbol{1}[/latex]	[latex]\boldsymbol{<10^{-15}\textbf{m}}[/latex]	attractive and repulsive	gluons
Table 1. Properties of the Four Basic Forces1.

The gravitational force is surprisingly weak—it is only because gravity is always attractive that we notice it at all. Our weight is the gravitational force due to the entire Earth acting on us. On the very large scale, as in astronomical systems, the gravitational force is the dominant force determining the motions of moons, planets, stars, and galaxies. The gravitational force also affects the nature of space and time. As we shall see later in the study of general relativity, space is curved in the vicinity of very massive bodies, such as the Sun, and time actually slows down near massive bodies.

Electromagnetic forces can be either attractive or repulsive. They are long-range forces, which act over extremely large distances, and they nearly cancel for macroscopic objects. (Remember that it is the net external force that is important.) If they did not cancel, electromagnetic forces would completely overwhelm the gravitational force. The electromagnetic force is a combination of electrical forces (such as those that cause static electricity) and magnetic forces (such as those that affect a compass needle). These two forces were thought to be quite distinct until early in the 19th century, when scientists began to discover that they are different manifestations of the same force. This discovery is a classical case of the unification of forces. Similarly, friction, tension, and all of the other classes of forces we experience directly (except gravity, of course) are due to electromagnetic interactions of atoms and molecules. It is still convenient to consider these forces separately in specific applications, however, because of the ways they manifest themselves.

CONCEPT CONNECTIONS: UNIFYING FORCES

Attempts to unify the four basic forces are discussed in relation to elementary particles later in this text. By “unify” we mean finding connections between the forces that show that they are different manifestations of a single force. Even if such unification is achieved, the forces will retain their separate characteristics on the macroscopic scale and may be identical only under extreme conditions such as those existing in the early universe.

Physicists are now exploring whether the four basic forces are in some way related. Attempts to unify all forces into one come under the rubric of Grand Unified Theories (GUTs), with which there has been some success in recent years. It is now known that under conditions of extremely high density and temperature, such as existed in the early universe, the electromagnetic and weak nuclear forces are indistinguishable. They can now be considered to be different manifestations of one force, called the electroweak force. So the list of four has been reduced in a sense to only three. Further progress in unifying all forces is proving difficult—especially the inclusion of the gravitational force, which has the special characteristics of affecting the space and time in which the other forces exist.

While the unification of forces will not affect how we discuss forces in this text, it is fascinating that such underlying simplicity exists in the face of the overt complexity of the universe. There is no reason that nature must be simple—it simply is.

Action at a Distance: Concept of a Field

All forces act at a distance. This is obvious for the gravitational force. Earth and the Moon, for example, interact without coming into contact. It is also true for all other forces. Friction, for example, is an electromagnetic force between atoms that may not actually touch. What is it that carries forces between objects? One way to answer this question is to imagine that a force field surrounds whatever object creates the force. A second object (often called a test object) placed in this field will experience a force that is a function of location and other variables. The field itself is the “thing” that carries the force from one object to another. The field is defined so as to be a characteristic of the object creating it; the field does not depend on the test object placed in it. Earth’s gravitational field, for example, is a function of the mass of Earth and the distance from its center, independent of the presence of other masses. The concept of a field is useful because equations can be written for force fields surrounding objects (for gravity, this yields w = mg at Earth’s surface), and motions can be calculated from these equations. (See Figure 1.)

CONCEPT CONNECTIONS: FORCE FIELDS

The concept of a force field is also used in connection with electric charge and is presented in Chapter 18 Electric Charge and Electric Field. It is also a useful idea for all the basic forces, as will be seen in Chapter 33 Particle Physics. Fields help us to visualize forces and how they are transmitted, as well as to describe them with precision and to link forces with subatomic carrier particles.

The field concept has been applied very successfully; we can calculate motions and describe nature to high precision using field equations. As useful as the field concept is, however, it leaves unanswered the question of what carries the force. It has been proposed in recent decades, starting in 1935 with Hideki Yukawa’s (1907–1981) work on the strong nuclear force, that all forces are transmitted by the exchange of elementary particles. We can visualize particle exchange as analogous to macroscopic phenomena such as two people passing a basketball back and forth, thereby exerting a repulsive force without touching one another. (See Figure 2.)

This idea of particle exchange deepens rather than contradicts field concepts. It is more satisfying philosophically to think of something physical actually moving between objects acting at a distance. Table 1 lists the exchange or carrier particles, both observed and proposed, that carry the four forces. But the real fruit of the particle-exchange proposal is that searches for Yukawa’s proposed particle found it and a number of others that were completely unexpected, stimulating yet more research. All of this research eventually led to the proposal of quarks as the underlying substructure of matter, which is a basic tenet of GUTs. If successful, these theories will explain not only forces, but also the structure of matter itself. Yet physics is an experimental science, so the test of these theories must lie in the domain of the real world. As of this writing, scientists at the CERN laboratory in Switzerland are starting to test these theories using the world’s largest particle accelerator: the Large Hadron Collider. This accelerator (27 km in circumference) allows two high-energy proton beams, traveling in opposite directions, to collide. An energy of 14 trillion electron volts will be available. It is anticipated that some new particles, possibly force carrier particles, will be found. (See Figure 3.) One of the force carriers of high interest that researchers hope to detect is the Higgs boson. The observation of its properties might tell us why different particles have different masses.

Tiny particles also have wave-like behaviour, something we will explore more in a later chapter. To better understand force-carrier particles from another perspective, let us consider gravity. The search for gravitational waves has been going on for a number of years. Almost 100 years ago, Einstein predicted the existence of these waves as part of his general theory of relativity. Gravitational waves are created during the collision of massive stars, in black holes, or in supernova explosions—like shock waves. These gravitational waves will travel through space from such sites much like a pebble dropped into a pond sends out ripples—except these waves move at the speed of light. A detector apparatus has been built in the U.S., consisting of two large installations nearly 3000 km apart—one in Washington state and one in Louisiana! The facility is called the Laser Interferometer Gravitational-Wave Observatory (LIGO). Each installation is designed to use optical lasers to examine any slight shift in the relative positions of two masses due to the effect of gravity waves. The two sites allow simultaneous measurements of these small effects to be separated from other natural phenomena, such as earthquakes. Initial operation of the detectors began in 2002, and work is proceeding on increasing their sensitivity. Similar installations have been built in Italy (VIRGO), Germany (GEO600), and Japan (TAMA300) to provide a worldwide network of gravitational wave detectors.

International collaboration in this area is moving into space with the joint EU/US project LISA (Laser Interferometer Space Antenna). Earthquakes and other Earthly noises will be no problem for these monitoring spacecraft. LISA will complement LIGO by looking at much more massive black holes through the observation of gravitational-wave sources emitting much larger wavelengths. Three satellites will be placed in space above Earth in an equilateral triangle (with 5,000,000-km sides) (Figure 4). The system will measure the relative positions of each satellite to detect passing gravitational waves. Accuracy to within 10% of the size of an atom will be needed to detect any waves. The launch of this project might be as early as 2018.

“I’m sure LIGO will tell us something about the universe that we didn’t know before. The history of science tells us that any time you go where you haven’t been before, you usually find something that really shakes the scientific paradigms of the day. Whether gravitational wave astrophysics will do that, only time will tell.” —David Reitze, LIGO Input Optics Manager, University of Florida

The ideas presented in this section are but a glimpse into topics of modern physics that will be covered in much greater depth in later chapters.

Summary

• The various types of forces that are categorized for use in many applications are all manifestations of the four basic forces in nature.
• The properties of these forces are summarized in Table 1.
• Everything we experience directly without sensitive instruments is due to either electromagnetic forces or gravitational forces. The nuclear forces are responsible for the submicroscopic structure of matter, but they are not directly sensed because of their short ranges. Attempts are being made to show all four forces are different manifestations of a single unified force.
• A force field surrounds an object creating a force and is the carrier of that force.

Conceptual Questions

1: Explain, in terms of the properties of the four basic forces, why people notice the gravitational force acting on their bodies if it is such a comparatively weak force.

2: What is the dominant force between astronomical objects? Why are the other three basic forces less significant over these very large distances?

3: Give a detailed example of how the exchange of a particle can result in an attractive force. (For example, consider one child pulling a toy out of the hands of another.)

Problems & Exercises

1: (a) What is the strength of the weak nuclear force relative to the strong nuclear force? (b) What is the strength of the weak nuclear force relative to the electromagnetic force? Since the weak nuclear force acts at only very short distances, such as inside nuclei, where the strong and electromagnetic forces also act, it might seem surprising that we have any knowledge of it at all. We have such knowledge because the weak nuclear force is responsible for beta decay, a type of nuclear decay not explained by other forces.

2: (a) What is the ratio of the strength of the gravitational force to that of the strong nuclear force? (b) What is the ratio of the strength of the gravitational force to that of the weak nuclear force? (c) What is the ratio of the strength of the gravitational force to that of the electromagnetic force? What do your answers imply about the influence of the gravitational force on atomic nuclei?

3: What is the ratio of the strength of the strong nuclear force to that of the electromagnetic force? Based on this ratio, you might expect that the strong force dominates the nucleus, which is true for small nuclei. Large nuclei, however, have sizes greater than the range of the strong nuclear force. At these sizes, the electromagnetic force begins to affect nuclear stability. These facts will be used to explain nuclear fusion and fission later in this text.

Footnotes

1. 1 The graviton is a proposed particle, though it has not yet been observed by scientists. See the discussion of gravitational waves later in this section. The particles W⁺, W^-, and Z⁰ are called vector bosons; these were predicted by theory and first observed in 1983. There are eight types of gluons proposed by scientists, and their existence is indicated by meson exchange in the nuclei of atoms.

Glossary

carrier particle

a fundamental particle of nature that is surrounded by a characteristic force field; photons are carrier particles of the electromagnetic force

force field

a region in which a test particle will experience a force

Solutions

Problems & Exercises

1: (a)   1 x 10^-13   (b) 1 x 10^-11

3:   1 x 10²

]]> 326 266 10 0 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/chapter/5-0-introduction/ Thu, 29 Jun 2017 22:13:34 +0000 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/chapter/5-0-introduction/

Describe the forces on the hip joint. What means are taken to ensure that this will be a good movable joint? From the photograph (for an adult) in Figure 1, estimate the dimensions of the artificial device.

It is difficult to categorize forces into various types (aside from the four basic forces discussed in previous chapter). We know that a net force affects the motion, position, and shape of an object. It is useful at this point to look at some particularly interesting and common forces that will provide further applications of Newton’s laws of motion. We have in mind the forces of friction, air or liquid drag, and deformation.

]]> 329 327 1 0 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/chapter/5-1-friction/ Thu, 29 Jun 2017 22:13:36 +0000 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/chapter/5-1-friction/

Summary

• Discuss the general characteristics of friction.
• Describe the various types of friction.
• Calculate the magnitude of static and kinetic friction.

Friction is a force that is around us all the time that opposes relative motion between systems in contact but also allows us to move (which you have discovered if you have ever tried to walk on ice). While a common force, the behavior of friction is actually very complicated and is still not completely understood. We have to rely heavily on observations for whatever understandings we can gain. However, we can still deal with its more elementary general characteristics and understand the circumstances in which it behaves.

FRICTION

Friction is a force that opposes relative motion between systems in contact.

One of the simpler characteristics of friction is that it is parallel to the contact surface between systems and always in a direction that opposes motion or attempted motion of the systems relative to each other. If two systems are in contact and moving relative to one another, then the friction between them is called kinetic friction. For example, friction slows a hockey puck sliding on ice. But when objects are stationary, static friction can act between them; the static friction is usually greater than the kinetic friction between the objects.

KINETIC FRICTION

If two systems are in contact and moving relative to one another, then the friction between them is called kinetic friction.

Imagine, for example, trying to slide a heavy crate across a concrete floor—you may push harder and harder on the crate and not move it at all. This means that the static friction responds to what you do—it increases to be equal to and in the opposite direction of your push. But if you finally push hard enough, the crate seems to slip suddenly and starts to move. Once in motion it is easier to keep it in motion than it was to get it started, indicating that the kinetic friction force is less than the static friction force. If you add mass to the crate, say by placing a box on top of it, you need to push even harder to get it started and also to keep it moving. Furthermore, if you oiled the concrete you would find it to be easier to get the crate started and keep it going (as you might expect).

Figure 1 is a crude pictorial representation of how friction occurs at the interface between two objects. Close-up inspection of these surfaces shows them to be rough. So when you push to get an object moving (in this case, a crate), you must raise the object until it can skip along with just the tips of the surface hitting, break off the points, or do both. A considerable force can be resisted by friction with no apparent motion. The harder the surfaces are pushed together (such as if another box is placed on the crate), the more force is needed to move them. Part of the friction is due to adhesive forces between the surface molecules of the two objects, which explain the dependence of friction on the nature of the substances. Adhesion varies with substances in contact and is a complicated aspect of surface physics. Once an object is moving, there are fewer points of contact (fewer molecules adhering), so less force is required to keep the object moving. At small but nonzero speeds, friction is nearly independent of speed.

The magnitude of the frictional force has two forms: one for static situations (static friction), the other for when there is motion (kinetic friction).

When there is no motion between the objects, the magnitude of static friction f_s is

[latex]\boldsymbol{f_{\textbf{s}}\leq\mu_{\textbf{s}}\:N,}[/latex]

where μ_s is the coefficient of static friction and N is the magnitude of the normal force (the force perpendicular to the surface).

MAGNITUDE OF STATIC FRICTION

Magnitude of static friction f_s is

[latex]\boldsymbol{f_{\textbf{s}}\leq\mu_{\textbf{s}}\:N,}[/latex]

where μ_s is the coefficient of static friction and N is the magnitude of the normal force.

The symbol ≤ means less than or equal to, implying that static friction can have a minimum and a maximum value of μ_sN. Static friction is a responsive force that increases to be equal and opposite to whatever force is exerted, up to its maximum limit. Once the applied force exceeds f_s(max), the object will move. Thus

[latex]\boldsymbol{f_{\textbf{s(max)}}=\mu_{\textbf{s}}N}.[/latex]

Once an object is moving, the magnitude of kinetic friction f_k is given by

[latex]\boldsymbol{f_{\textbf{k}}=\mu_{\textbf{k}}N},[/latex]

where μ_k is the coefficient of kinetic friction. A system in which f_k = μ_kN is described as a system in which friction behaves simply.

MAGNITUDE OF KINETIC FRICTION

The magnitude of kinetic friction f_k is given by

[latex]\boldsymbol{f_{\textbf{k}}=\mu_{\textbf{k}}N},[/latex]

where μ_k is the coefficient of kinetic friction.

As seen in Table 1, the coefficients of kinetic friction are less than their static counterparts. That values of μ in Table 1 are stated to only one or, at most, two digits is an indication of the approximate description of friction given by the above two equations.

[latex]\textbf{System}[/latex]	[latex]\textbf{Static friction,}\boldsymbol{\mu_{\textbf{s}}}[/latex]	[latex]\textbf{Kinetic friction,}\boldsymbol{\mu_{\textbf{k}}}[/latex]
Rubber on dry concrete	1.0	0.7
Rubber on wet concrete	0.7	0.5
Wood on wood	0.5	0.3
Waxed wood on wet snow	0.14	0.1
Metal on wood	0.5	0.3
Steel on steel (dry)	0.6	0.3
Steel on steel (oiled)	0.05	0.03
Teflon on steel	0.04	0.04
Bone lubricated by synovial fluid	0.016	0.015
Shoes on wood	0.9	0.7
Shoes on ice	0.1	0.05
Ice on ice	0.1	0.03
Steel on ice	0.4	0.02
Table 1. Coefficients of Static and Kinetic Friction.

The equations given earlier include the dependence of friction on materials and the normal force. The direction of friction is always opposite that of motion, parallel to the surface between objects, and perpendicular to the normal force. For example, if the crate you try to push (with a force parallel to the floor) has a mass of 100 kg, then the normal force would be equal to its weight, W = mg = (100 kg)(9.80 m/s²) = 980 N, perpendicular to the floor. If the coefficient of static friction is 0.45, you would have to exert a force parallel to the floor greater than f_s(max) = μN = (0.45)(980 N) = 440 N to move the crate. Once there is motion, friction is less and the coefficient of kinetic friction might be 0.30, so that a force of only 290 N (f_k = μ_kN = (0.30)(980 N) = 290 N) would keep it moving at a constant speed. If the floor is lubricated, both coefficients are considerably less than they would be without lubrication. Coefficient of friction is a unit less quantity with a magnitude usually between 0 and 1.0. The coefficient of the friction depends on the two surfaces that are in contact.

TAKE-HOME EXPERIMENT

Find a small plastic object (such as a food container) and slide it on a kitchen table by giving it a gentle tap. Now spray water on the table, simulating a light shower of rain. What happens now when you give the object the same-sized tap? Now add a few drops of (vegetable or olive) oil on the surface of the water and give the same tap. What happens now? This latter situation is particularly important for drivers to note, especially after a light rain shower. Why?

Many people have experienced the slipperiness of walking on ice. However, many parts of the body, especially the joints, have much smaller coefficients of friction—often three or four times less than ice. A joint is formed by the ends of two bones, which are connected by thick tissues. The knee joint is formed by the lower leg bone (the tibia) and the thighbone (the femur). The hip is a ball (at the end of the femur) and socket (part of the pelvis) joint. The ends of the bones in the joint are covered by cartilage, which provides a smooth, almost glassy surface. The joints also produce a fluid (synovial fluid) that reduces friction and wear. A damaged or arthritic joint can be replaced by an artificial joint (Figure 2). These replacements can be made of metals (stainless steel or titanium) or plastic (polyethylene), also with very small coefficients of friction.

Other natural lubricants include saliva produced in our mouths to aid in the swallowing process, and the slippery mucus found between organs in the body, allowing them to move freely past each other during heartbeats, during breathing, and when a person moves. Artificial lubricants are also common in hospitals and doctor’s clinics. For example, when ultrasonic imaging is carried out, the gel that couples the transducer to the skin also serves to to lubricate the surface between the transducer and the skin—thereby reducing the coefficient of friction between the two surfaces. This allows the transducer to mover freely over the skin.

Example 1: Skiing Exercise

A skier with a mass of 62 kg is sliding down a snowy slope. Find the coefficient of kinetic friction for the skier if friction is known to be 45.0 N. Strategy

The magnitude of kinetic friction was given in to be 45.0 N. Kinetic friction is related to the normal force N as f_k = μ_kN; thus, the coefficient of kinetic friction can be found if we can find the normal force of the skier on a slope. The normal force is always perpendicular to the surface, and since there is no motion perpendicular to the surface, the normal force should equal the component of the skier’s weight perpendicular to the slope. (See the skier and free-body diagram in Figure 3.)

That is,

[latex]\boldsymbol{{N}={w}_{\perp}=w\:\textbf{cos}25^0=mg\:\textbf{cos}25^0.}[/latex]

Substituting this into our expression for kinetic friction, we get

[latex]\boldsymbol{f_{\textbf{k}}=\mu_{\textbf{k}}mg\:\textbf{cos}25^0,}[/latex]

which can now be solved for the coefficient of kinetic friction μ_k.

Solution

Solving for μ_k gives

[latex]\boldsymbol{\mu_{\textbf{k}}\:=}[/latex][latex size="2"]\boldsymbol{\frac{f_{\textbf{k}}}{{N}}}[/latex][latex]\boldsymbol{=}[/latex][latex size="2"]\boldsymbol{\frac{f_{\textbf{k}}}{w\textbf{cos}25^0}}[/latex][latex]\boldsymbol{=}[/latex][latex size="2"]\boldsymbol{\frac{f_{\textbf{k}}}{mg\textbf{cos}25^0}}.[/latex]

Substituting known values on the right-hand side of the equation,

[latex]\boldsymbol{\mu_{\textbf{k}}\:=}[/latex][latex size="2"]\boldsymbol{\frac{45.0\textbf{ N}}{(62\textbf{ kg})(9.80\textbf{ m/s}^2)(0.906)}}[/latex][latex]\boldsymbol{=\:0.082.}[/latex]

Discussion

This result is a little smaller than the coefficient listed in Table 1 for waxed wood on snow, but it is still reasonable since values of the coefficients of friction can vary greatly. In situations like this, where an object of mass m slides down a slope that makes an angle θ with the horizontal, friction is given by f_k = μ_kmg cos θ. All objects will slide down a slope with constant acceleration under these circumstances. Proof of this is left for this chapter’s Problems and Exercises.

TAKE-HOME EXPERIMENT

An object will slide down an inclined plane at a constant velocity if the net force on the object is zero. We can use this fact to measure the coefficient of kinetic friction between two objects. As shown in Example 1, the kinetic friction on a slope f_k = μ_kmg cos θ. The component of the weight down the slope is equal to mg sin θ (see the free-body diagram in Figure 3). These forces act in opposite directions, so when they have equal magnitude, the acceleration is zero. Writing these out:

[latex]\boldsymbol{f_{\textbf{k}}=Fg_{\textbf{x}}}[/latex]

[latex]\boldsymbol{\mu_{\textbf{k}}mg \;\textbf{cos}\theta=mg \;\textbf{sin}\theta.}[/latex]

Solving for μ_k, we find that

[latex]\boldsymbol{\mu_{\textbf{k}}\:=}[/latex][latex size="2"]\boldsymbol{\frac{mg \;\textbf{sin}\theta}{mg \;\textbf{cos}\theta}}[/latex][latex]\boldsymbol{=\:\textbf{tan}\theta.}[/latex]

Put a coin on a book and tilt it until the coin slides at a constant velocity down the book. You might need to tap the book lightly to get the coin to move. Measure the angle of tilt relative to the horizontal and find μ_k. Note that the coin will not start to slide at all until an angle greater than θ is attained, since the coefficient of static friction is larger than the coefficient of kinetic friction. Discuss how this may affect the value for μ_k and its uncertainty.

We have discussed that when an object rests on a horizontal surface, there is a normal force supporting it equal in magnitude to its weight. Furthermore, simple friction is always proportional to the normal force.

MAKING CONNECTIONS: SUBMICROSCOPIC EXPLANATIONS OF FRICTION

The simpler aspects of friction dealt with so far are its macroscopic (large-scale) characteristics. Great strides have been made in the atomic-scale explanation of friction during the past several decades. Researchers are finding that the atomic nature of friction seems to have several fundamental characteristics. These characteristics not only explain some of the simpler aspects of friction—they also hold the potential for the development of nearly friction-free environments that could save hundreds of billions of dollars in energy which is currently being converted (unnecessarily) to heat.

Figure 4 illustrates one macroscopic characteristic of friction that is explained by microscopic (small-scale) research. We have noted that friction is proportional to the normal force, but not to the area in contact, a somewhat counterintuitive notion. When two rough surfaces are in contact, the actual contact area is a tiny fraction of the total area since only high spots touch. When a greater normal force is exerted, the actual contact area increases, and it is found that the friction is proportional to this area.

But the atomic-scale view promises to explain far more than the simpler features of friction. The mechanism for how heat is generated is now being determined. In other words, why do surfaces get warmer when rubbed? Essentially, atoms are linked with one another to form lattices. When surfaces rub, the surface atoms adhere and cause atomic lattices to vibrate—essentially creating sound waves that penetrate the material. The sound waves diminish with distance and their energy is converted into heat. Chemical reactions that are related to frictional wear can also occur between atoms and molecules on the surfaces. Figure 5 shows how the tip of a probe drawn across another material is deformed by atomic-scale friction. The force needed to drag the tip can be measured and is found to be related to shear stress, which will be discussed later in this chapter. The variation in shear stress is remarkable (more than a factor of 10¹²) and difficult to predict theoretically, but shear stress is yielding a fundamental understanding of a large-scale phenomenon known since ancient times—friction.

[caption id="" align="aligncenter" width="300"]Figure 5. The tip of a probe is deformed sideways by frictional force as the probe is dragged across a surface. Measurements of how the force varies for different materials are yielding fundamental insights into the atomic nature of friction.[/caption]

PHET EXPLORATIONS: FORCES AND MOTION

Explore the forces at work when you try to push a filing cabinet. Create an applied force and see the resulting friction force and total force acting on the cabinet. Charts show the forces, position, velocity, and acceleration vs. time. Draw a free-body diagram of all the forces (including gravitational and normal forces).

[caption id="" align="aligncenter" width="450"] Figure 6. Forces and Motion[/caption]

Section Summary

• Friction is a contact force between systems that opposes the motion or attempted motion between them. Simple friction is proportional to the normal force N pushing the systems together. (A normal force is always perpendicular to the contact surface between systems.) Friction depends on both of the materials involved. The magnitude of static friction f_s between systems stationary relative to one another is given by
  [latex]\boldsymbol{f_{\textbf{s}}\leq\mu\textbf{N},}[/latex]
  where μ_s is the coefficient of static friction, which depends on both of the materials.
• The kinetic friction force f_k between systems moving relative to one another is given by
  [latex]\boldsymbol{f_{\textbf{k}}=\mu_{\textbf{k}}\textbf{N},}[/latex]
  where μ_kis the coefficient of kinetic friction, which also depends on both materials.

Conceptual Questions

1: Define normal force. What is its relationship to friction when friction behaves simply?

2: The glue on a piece of tape can exert forces. Can these forces be a type of simple friction? Explain, considering especially that tape can stick to vertical walls and even to ceilings.

3: When you learn to drive, you discover that you need to let up slightly on the brake pedal as you come to a stop or the car will stop with a jerk. Explain this in terms of the relationship between static and kinetic friction.

4: When you push a piece of chalk across a chalkboard, it sometimes screeches because it rapidly alternates between slipping and sticking to the board. Describe this process in more detail, in particular explaining how it is related to the fact that kinetic friction is less than static friction. (The same slip-grab process occurs when tires screech on pavement.)

Problems & Exercises

1: A physics major is cooking breakfast when he notices that the frictional force between his steel spatula and his Teflon frying pan is only 0.200 N. Knowing the coefficient of kinetic friction between the two materials, he quickly calculates the normal force. What is it?

2: (a) When rebuilding her car’s engine, a physics major must exert 300 N of force to insert a dry steel piston into a steel cylinder. What is the magnitude of the normal force between the piston and cylinder? (b) What is the magnitude of the force would she have to exert if the steel parts were oiled?

3: (a) What is the maximum frictional force in the knee joint of a person who supports 66.0 kg of her mass on that knee? (b) During strenuous exercise it is possible to exert forces to the joints that are easily ten times greater than the weight being supported. What is the maximum force of friction under such conditions? The frictional forces in joints are relatively small in all circumstances except when the joints deteriorate, such as from injury or arthritis. Increased frictional forces can cause further damage and pain.

4: Suppose you have a 120-kg wooden crate resting on a wood floor. (a) What maximum force can you exert horizontally on the crate without moving it? (b) If you continue to exert this force once the crate starts to slip, what will the magnitude of its acceleration then be?

5: (a) If half of the weight of a small 1.00 × 10³ kg utility truck is supported by its two drive wheels, what is the magnitude of the maximum acceleration it can achieve on dry concrete? (b) Will a metal cabinet lying on the wooden bed of the truck slip if it accelerates at this rate? (c) Solve both problems assuming the truck has four-wheel drive.

6: A team of eight dogs pulls a sled with waxed wood runners on wet snow (mush!). The dogs have average masses of 19.0 kg, and the loaded sled with its rider has a mass of 210 kg. (a) Calculate the magnitude of the acceleration starting from rest if each dog exerts an average force of 185 N backward on the snow. (b) What is the magnitude of the acceleration once the sled starts to move? (c) For both situations, calculate the magnitude of the force in the coupling between the dogs and the sled.

7: Consider the 65.0-kg ice skater being pushed by two others shown in Figure 7. (a) Find the direction and magnitude of F_tot, the total force exerted on her by the others, given that the magnitudes F₁ and F₂ are 26.4 N and 18.6 N, respectively. (b) What is her initial acceleration if she is initially stationary and wearing steel-bladed skates that point in the direction of F_tot? (c) What is her acceleration assuming she is already moving in the direction of F_tot? (Remember that friction always acts in the direction opposite that of motion or attempted motion between surfaces in contact.)

[caption id="attachment_3563" align="aligncenter" width="300"]Figure 7.[/caption]

8: Show that the acceleration of any object down a frictionless incline that makes an angle θ with the horizontal is a = g sin θ. (Note that this acceleration is independent of mass.)

9: Show that the acceleration of any object down an incline where friction behaves simply (that is, where f_k = μ_kN) is a = g (sin θ - μ_kcos θ). Note that the acceleration is independent of mass and reduces to the expression found in the previous problem when friction becomes negligibly small (μ_k = 0).

10: Calculate the deceleration of a snow boarder going up a 5.0°, slope assuming the coefficient of friction for waxed wood on wet snow. The result of Exercise 9 may be useful, but be careful to consider the fact that the snow boarder is going uphill. Explicitly show how you follow the steps in Chapter 4.6 Problem-Solving Strategies.

11: (a) Calculate the acceleration of a skier heading down a 10.0° slope, assuming the coefficient of friction for waxed wood on wet snow. (b) Find the angle of the slope down which this skier could coast at a constant velocity. You can neglect air resistance in both parts, and you will find the result of Exercise 9 to be useful. Explicitly show how you follow the steps in the Chapter 4.6 Problem-Solving Strategies.

12: If an object is to rest on an incline without slipping, then friction must equal the component of the weight of the object parallel to the incline. This requires greater and greater friction for steeper slopes. Show that the maximum angle of an incline above the horizontal for which an object will not slide down is θ = tan^-1 μ_s. You may use the result of the previous problem. Assume that a=0 and that static friction has reached its maximum value.

13: Calculate the maximum deceleration of a car that is heading down a 6° slope (one that makes an angle of 6° with the horizontal) under the following road conditions. You may assume that the weight of the car is evenly distributed on all four tires and that the coefficient of static friction is involved—that is, the tires are not allowed to slip during the deceleration. (Ignore rolling.) Calculate for a car: (a) On dry concrete. (b) On wet concrete. (c) On ice, assuming that μ_s = 0.100, the same as for shoes on ice.

14: Calculate the maximum acceleration of a car that is heading up a 4° slope (one that makes an angle of 4° with the horizontal) under the following road conditions. Assume that only half the weight of the car is supported by the two drive wheels and that the coefficient of static friction is involved—that is, the tires are not allowed to slip during the acceleration. (Ignore rolling.) (a) On dry concrete. (b) On wet concrete. (c) On ice, assuming that μ_s = 0.100, the same as for shoes on ice.

15: Repeat Exercise 14 for a car with four-wheel drive.

16: A freight train consists of two 8.00 × 10⁵-kg engines and 45 cars with average masses of 5.50 × 10⁵ kg. (a) What force must each engine exert backward on the track to accelerate the train at a rate of 5.00 × 10^-2 m/s² if the force of friction is 7.50 × 10⁵ N, assuming the engines exert identical forces? This is not a large frictional force for such a massive system. Rolling friction for trains is small, and consequently trains are very energy-efficient transportation systems. (b) What is the magnitude of the force in the coupling between the 37th and 38th cars (this is the force each exerts on the other), assuming all cars have the same mass and that friction is evenly distributed among all of the cars and engines?

17: Consider the 52.0-kg mountain climber in Figure 8. (a) Find the tension in the rope and the force that the mountain climber must exert with her feet on the vertical rock face to remain stationary. Assume that the force is exerted parallel to her legs. Also, assume negligible force exerted by her arms. (b) What is the minimum coefficient of friction between her shoes and the cliff?

18: A contestant in a winter sporting event pushes a 45.0-kg block of ice across a frozen lake as shown in Figure 9(a). (a) Calculate the minimum force F he must exert to get the block moving. (b) What is the magnitude of its acceleration once it starts to move, if that force is maintained?

19: Repeat Exercise 18 with the contestant pulling the block of ice with a rope over his shoulder at the same angle above the horizontal as shown in Figure 9(b).

Glossary

friction

a force that opposes relative motion or attempts at motion between systems in contact

kinetic friction

a force that opposes the motion of two systems that are in contact and moving relative to one another

static friction

a force that opposes the motion of two systems that are in contact and are not moving relative to one another

magnitude of static friction

f_s ≤ μ_sN, where μ_s is the coefficient of static friction and N is the magnitude of the normal force

magnitude of kinetic friction

f_k = μ_kN, where μ_k is the coefficient of kinetic friction

Solutions

Problems & Exercises 1: [latex]\boldsymbol{5.00\textbf{ N}}[/latex]

4: (a) $$\boldsymbol{588\textbf{ N}}$$ (b) [latex]\boldsymbol{1.96\textbf{ m/s}^2}[/latex]

6: (a) $$\boldsymbol{3.29\textbf{ m/s}^2}$$ (b) $$\boldsymbol{3.52\textbf{ m/s}^2}$$ (c) $$\boldsymbol{980\textbf{ N; }945\textbf{ N}}$$

10: [latex]\boldsymbol{1.83\textbf{ m/s}^2}[/latex]

14: (a) [latex]\boldsymbol{4.20\textbf{ m/s}^2}[/latex] (b) [latex]\boldsymbol{2.74\textbf{ m/s}^2}[/latex] (c) [latex]\boldsymbol{-0.195\textbf{ m/s}^2}[/latex]

16: (a) [latex]\boldsymbol{1.03\times10^6\textbf{ N}}[/latex] (b) [latex]\boldsymbol{3.48\times10^5\textbf{ N}}[/latex]

18: (a) [latex]\boldsymbol{51.0\textbf{ N}}[/latex] (b) [latex]\boldsymbol{0.720\textbf{ m/s}^2}[/latex]

]]> 339 327 2 0 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/chapter/5-2-drag-forces/ Thu, 29 Jun 2017 22:13:37 +0000 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/chapter/5-2-drag-forces/

Summary

• Express mathematically the drag force.
• Discuss the applications of drag force.
• Define terminal velocity.
• Determine the terminal velocity given mass.

Another interesting force in everyday life is the force of drag on an object when it is moving in a fluid (either a gas or a liquid). You feel the drag force when you move your hand through water. You might also feel it if you move your hand during a strong wind. The faster you move your hand, the harder it is to move. You feel a smaller drag force when you tilt your hand so only the side goes through the air—you have decreased the area of your hand that faces the direction of motion. Like friction, the drag force always opposes the motion of an object. Unlike simple friction, the drag force is proportional to some function of the velocity of the object in that fluid. This functionality is complicated and depends upon the shape of the object, its size, its velocity, and the fluid it is in. For most large objects such as bicyclists, cars, and baseballs not moving too slowly, the magnitude of the drag force F_D is found to be proportional to the square of the speed of the object. We can write this relationship mathematically as F_D ∝ v². When taking into account other factors, this relationship becomes

[latex]\boldsymbol{F_{\textbf{D}}\:=}[/latex][latex size="2"]\boldsymbol{\frac{1}{2}}[/latex][latex]\boldsymbol{C\rho\textbf{A}v^2,}[/latex]

where C is the drag coefficient, A is the area of the object facing the fluid, and ρ is the density of the fluid. (Recall that density is mass per unit volume.) This equation can also be written in a more generalized fashion as F_D = bv², where b is a constant equivalent to 0.5CρA. We have set the exponent for these equations as 2 because, when an object is moving at high velocity through air, the magnitude of the drag force is proportional to the square of the speed. As we shall see in a few pages on fluid dynamics, for small particles moving at low speeds in a fluid, the exponent is equal to 1.

DRAG FORCE

Drag force F_D is found to be proportional to the square of the speed of the object. Mathematically

[latex]\boldsymbol{F_{\textbf{D}}\propto{v^2}}[/latex]

[latex]\boldsymbol{F_{\textbf{D}}\:=}[/latex][latex size="2"]\boldsymbol{\frac{1}{2}}[/latex][latex]\boldsymbol{\:C\rho\textbf{A}v^2,}[/latex]

where C is the drag coefficient, A is the area of the object facing the fluid, and ρ is the density of the fluid.

Athletes as well as car designers seek to reduce the drag force to lower their race times. (See Figure 1). “Aerodynamic” shaping of an automobile can reduce the drag force and so increase a car’s gas mileage.

The value of the drag coefficient, C, is determined empirically, usually with the use of a wind tunnel. (See Figure 2).

The drag coefficient can depend upon velocity, but we will assume that it is a constant here. Table 2 lists some typical drag coefficients for a variety of objects. Notice that the drag coefficient is a dimensionless quantity. At highway speeds, over 50% of the power of a car is used to overcome air drag. The most fuel-efficient cruising speed is about 70–80 km/h (about 45–50 mi/h). For this reason, during the 1970s oil crisis in the United States, maximum speeds on highways were set at about 90 km/h (55 mi/h).

[latex]\textbf{Object}[/latex]	[latex]\textbf{C}[/latex]
Airfoil	0.05
Toyota Camry	0.28
Ford Focus	0.32
Honda Civic	0.36
Ferrari Testarossa	0.37
Dodge Ram pickup	0.43
Sphere	0.45
Hummer H2 SUV	0.64
Skydiver (feet first)	0.70
Bicycle	0.90
Skydiver (horizontal)	1.0
Circular flat plate	1.12
Table 2. Drag Coefficient Values Typical values of drag coefficient[latex]\boldsymbol{C}.[/latex].

Substantial research is under way in the sporting world to minimize drag. The dimples on golf balls are being redesigned as are the clothes that athletes wear. Bicycle racers and some swimmers and runners wear full bodysuits. Australian Cathy Freeman wore a full body suit in the 2000 Sydney Olympics, and won the gold medal for the 400 m race. Many swimmers in the 2008 Beijing Olympics wore (Speedo) body suits; it might have made a difference in breaking many world records (See Figure 3). Most elite swimmers (and cyclists) shave their body hair. Such innovations can have the effect of slicing away milliseconds in a race, sometimes making the difference between a gold and a silver medal. One consequence is that careful and precise guidelines must be continuously developed to maintain the integrity of the sport.

Some interesting situations connected to Newton’s second law occur when considering the effects of drag forces upon a moving object. For instance, consider a skydiver falling through air under the influence of gravity. The two forces acting on him are the force of gravity and the drag force (ignoring the buoyant force). The downward force of gravity remains constant regardless of the velocity at which the person is moving. However, as the person’s velocity increases, the magnitude of the drag force increases until the magnitude of the drag force is equal to the gravitational force, thus producing a net force of zero. A zero net force means that there is no acceleration, as given by Newton’s second law. At this point, the person’s velocity remains constant and we say that the person has reached his terminal velocity (vt). Since F_D is proportional to the speed, a heavier skydiver must go faster for F_D to equal his weight. Let’s see how this works out more quantitatively.

At the terminal velocity,

[latex]\boldsymbol{F_{\textbf{net}}=mg-F_{\textbf{D}}=ma=0.}[/latex]

Thus,

[latex]\boldsymbol{mg=F_{\textbf{D}}.}[/latex]

Using the equation for drag force, we have

[latex]\boldsymbol{mg\:=}[/latex][latex size="2"]\boldsymbol{\frac{1}{2}}[/latex][latex]\boldsymbol{\:\rho{CA}v^2.}[/latex]

Solving for the velocity, we obtain

[latex]\boldsymbol{v\:=}[/latex][latex size="2'"]\boldsymbol{\sqrt{\frac{2mg}{\rho{CA}}}}.[/latex]

Assume the density of air is ρ = 1.21 kg/m³. A 75-kg skydiver descending head first will have an area approximately A = 0.18 m² and a drag coefficient of approximately C = 0.70. We find that

[latex]\boldsymbol{v=\sqrt{\frac{2(75\textbf{ kg})(9.80\textbf{ m/s}^2)}{(1.21\textbf{ kg/m}^3)(0.70)(0.18\textbf{ m}^2)}}}[/latex]

[latex]\boldsymbol{=98\textbf{ m/s}}[/latex]

[latex]\boldsymbol{=350\textbf{ km/h}}.[/latex]

This means a skydiver with a mass of 75 kg achieves a maximum terminal velocity of about 350 km/h while traveling in a pike (head first) position, minimizing the area and his drag. In a spread-eagle position, that terminal velocity may decrease to about 200 km/h as the area increases. This terminal velocity becomes much smaller after the parachute opens.

TAKE-HOME EXPERIMENT

This interesting activity examines the effect of weight upon terminal velocity. Gather together some nested coffee filters. Leaving them in their original shape, measure the time it takes for one, two, three, four, and five nested filters to fall to the floor from the same height (roughly 2 m). (Note that, due to the way the filters are nested, drag is constant and only mass varies.) They obtain terminal velocity quite quickly, so find this velocity as a function of mass. Plot the terminal velocity v versus mass. Also plot v² versus mass. Which of these relationships is more linear? What can you conclude from these graphs?

Example 1: A Terminal Velocity

Find the terminal velocity of an 85-kg skydiver falling in a spread-eagle position.

Strategy

At terminal velocity, F_net = 0. Thus the drag force on the skydiver must equal the force of gravity (the person’s weight). Using the equation of drag force, we find [latex]\boldsymbol{mg=\frac{1}{2}\rho{C}\textbf{A}v^2}.[/latex]

Thus the terminal velocity v_t can be written as

[latex]\boldsymbol{v_{\textbf{t}}\:=}[/latex][latex size="2"]\boldsymbol{\sqrt{\frac{2mg}{\rho{CA}}}}.[/latex]

Solution

All quantities are known except the person’s projected area. This is an adult (82 kg) falling spread eagle. We can estimate the frontal area as

[latex]\boldsymbol{A=(2\textbf{ m})(0.35\textbf{ m})=0.70\textbf{ m}^2}.[/latex]

Using our equation for v_t, we find that

[latex]\boldsymbol{v_{\textbf{t}}=\sqrt{\frac{2(85\textbf{ kg})(9.80\textbf{ m/s}^2)}{(1.21\textbf{ kg/m}^3)(1.0)(0.70\textbf{ m}^2)}}}[/latex]

[latex]\boldsymbol{=44\textbf{ m/s}}.[/latex]

Discussion

This result is consistent with the value for v_t mentioned earlier. The 75-kg skydiver going feet first had a v = 98 m/s. He weighed less but had a smaller frontal area and so a smaller drag due to the air.

The size of the object that is falling through air presents another interesting application of air drag. If you fall from a 5-m high branch of a tree, you will likely get hurt—possibly fracturing a bone. However, a small squirrel does this all the time, without getting hurt. You don’t reach a terminal velocity in such a short distance, but the squirrel does.

The following interesting quote on animal size and terminal velocity is from a 1928 essay by a British biologist, J.B.S. Haldane, titled “On Being the Right Size.”

To the mouse and any smaller animal, [gravity] presents practically no dangers. You can drop a mouse down a thousand-yard mine shaft; and, on arriving at the bottom, it gets a slight shock and walks away, provided that the ground is fairly soft. A rat is killed, a man is broken, and a horse splashes. For the resistance presented to movement by the air is proportional to the surface of the moving object. Divide an animal’s length, breadth, and height each by ten; its weight is reduced to a thousandth, but its surface only to a hundredth. So the resistance to falling in the case of the small animal is relatively ten times greater than the driving force.

The above quadratic dependence of air drag upon velocity does not hold if the object is very small, is going very slow, or is in a denser medium than air. Then we find that the drag force is proportional just to the velocity. This relationship is given by Stokes’ law, which states that

[latex]\boldsymbol{F_{\textbf{s}}=6\pi{r}\eta{v}},[/latex]

where r is the radius of the object, η is the viscosity of the fluid, and v is the object’s velocity.

STOKE'S LAW

[latex]\boldsymbol{F_{\textbf{s}}=6\pi{r}\eta{v}},[/latex]

where r is the radius of the object, η is the viscosity of the fluid, and v is the object’s velocity.

Good examples of this law are provided by microorganisms, pollen, and dust particles. Because each of these objects is so small, we find that many of these objects travel unaided only at a constant (terminal) velocity. Terminal velocities for bacteria (size about 1 μm) can be about 2 μm/s. To move at a greater speed, many bacteria swim using flagella (organelles shaped like little tails) that are powered by little motors embedded in the cell. Sediment in a lake can move at a greater terminal velocity (about 5 μm/s), so it can take days to reach the bottom of the lake after being deposited on the surface.

If we compare animals living on land with those in water, you can see how drag has influenced evolution. Fishes, dolphins, and even massive whales are streamlined in shape to reduce drag forces. Birds are streamlined and migratory species that fly large distances often have particular features such as long necks. Flocks of birds fly in the shape of a spear head as the flock forms a streamlined pattern (see Figure 4). In humans, one important example of streamlining is the shape of sperm, which need to be efficient in their use of energy.

GALILEO'S EXPERIMENT

Galileo is said to have dropped two objects of different masses from the Tower of Pisa. He measured how long it took each to reach the ground. Since stopwatches weren’t readily available, how do you think he measured their fall time? If the objects were the same size, but with different masses, what do you think he should have observed? Would this result be different if done on the Moon?

PHET EXPLORATIONS: MASSES & SPRINGS

A realistic mass and spring laboratory. Hang masses from springs and adjust the spring stiffness and damping. You can even slow time. Transport the lab to different planets. A chart shows the kinetic, potential, and thermal energy for each spring.

[caption id="" align="aligncenter" width="450"] Figure 5. Masses & Springs[/caption]

Section Summary

• Drag forces acting on an object moving in a fluid oppose the motion. For larger objects (such as a baseball) moving at a velocity v in air, the drag force is given by
  [latex]\boldsymbol{F_{\textbf{D}}=\frac{1}{2}C\rho {A}v^2},[/latex]
  where C is the drag coefficient (typical values are given in Table 2), A is the area of the object facing the fluid, and ρ is the fluid density.
• For small objects (such as a bacterium) moving in a denser medium (such as water), the drag force is given by Stokes’ law,
  [latex]\boldsymbol{F_{\textbf{s}}=6\pi\eta{rv}},[/latex]
  where r is the radius of the object, η is the fluid viscosity, and v is the object’s velocity.

Conceptual Questions

1: Athletes such as swimmers and bicyclists wear body suits in competition. Formulate a list of pros and cons of such suits.

2: Two expressions were used for the drag force experienced by a moving object in a liquid. One depended upon the speed, while the other was proportional to the square of the speed. In which types of motion would each of these expressions be more applicable than the other one?

3: As cars travel, oil and gasoline leaks onto the road surface. If a light rain falls, what does this do to the control of the car? Does a heavy rain make any difference?

4: Why can a squirrel jump from a tree branch to the ground and run away undamaged, while a human could break a bone in such a fall?

Problems & Exercises

1: The terminal velocity of a person falling in air depends upon the weight and the area of the person facing the fluid. Find the terminal velocity (in meters per second and kilometers per hour) of an 80.0-kg skydiver falling in a pike (headfirst) position with a surface area of 0.140 m².

2: A 60-kg and a 90-kg skydiver jump from an airplane at an altitude of 6000 m, both falling in the pike position. Make some assumption on their frontal areas and calculate their terminal velocities. How long will it take for each skydiver to reach the ground (assuming the time to reach terminal velocity is small)? Assume all values are accurate to three significant digits.

3: A 560-g squirrel with a surface area of 930 cm² falls from a 5.0-m tree to the ground. Estimate its terminal velocity. (Use a drag coefficient for a horizontal skydiver.) What will be the velocity of a 56-kg person hitting the ground, assuming no drag contribution in such a short distance?

4: To maintain a constant speed, the force provided by a car’s engine must equal the drag force plus the force of friction of the road (the rolling resistance). (a) What are the magnitudes of drag forces at 70 km/h and 100 km/h for a Toyota Camry? (Drag area is 0.70 m²) (b) What is the magnitude of drag force at 70 km/h and 100 km/h for a Hummer H2? (Drag area is 2.44 m²) Assume all values are accurate to three significant digits.

5: By what factor does the drag force on a car increase as it goes from 65 to 110 km/h?

6: Calculate the speed a spherical rain drop would achieve falling from 5.00 km (a) in the absence of air drag (b) with air drag. Take the size across of the drop to be 4 mm, the density to be 1.00 × 10³ kg/m³, and the surface area to be πr².

7: Using Stokes’ law, verify that the units for viscosity are kilograms per meter per second.

8: Find the terminal velocity of a spherical bacterium (diameter 2.00 μm) falling in water. You will first need to note that the drag force is equal to the weight at terminal velocity. Take the density of the bacterium to be 1.10 × 10³ kg/m³.

9: Stokes’ law describes seimentation of particles in liquids and can be used to measure viscosity. Particles in liquids achieve terminal velocity quickly. One can measure the time it takes for a particle to fall a certain distance and then use Stokes’ law to calculate the viscosity of the liquid. Suppose a steel ball bearing (density 7.8 × 10³ kg/m³, diameter 3.0 mm) is dropped in a container of motor oil. It takes 12 s to fall a distance of 0.60 m. Calculate the viscosity of the oil.

Glossary

drag force

F_D, found to be proportional to the square of the speed of the object; mathematically

[latex]\boldsymbol{F_{\textbf{D}}\propto{v}^2}[/latex]

[latex]\boldsymbol{F_{\textbf{D}}=\frac{1}{2}C\rho {A}v^2},[/latex]

where C is the drag coefficient, A is the area of the object facing the fluid, and ρ is the density of the fluid

Stokes’ law

F_s = 6πrηv, where r is the radius of the object, η is the viscosity of the fluid, and v is the object’s velocity

Solutions

Problems & Exercises 1: [latex]\boldsymbol{115\textbf{ m/s};\:414\textbf{ km/hr}}[/latex] 3: [latex]\boldsymbol{25\textbf{ m/s};\:9.9\textbf{ m/s}}[/latex] 5: [latex]\boldsymbol{2.9}[/latex] 7: [latex]\boldsymbol{[\eta]\:=}[/latex][latex size="2"]\boldsymbol{\frac{[F_{\textbf{s}}]}{[r][v]}}[/latex][latex]\boldsymbol{=}[/latex][latex size="2"]\boldsymbol{\frac{\textbf{kg}\cdotp\textbf{m/s}^2}{\textbf{m}\cdotp\textbf{m/s}}}[/latex][latex]\boldsymbol{=}[/latex][latex size="2"]\boldsymbol{\frac{\textbf{kg}}{\textbf{m}\cdotp\textbf{s}}}[/latex] 9: [latex]\boldsymbol{0.76\textbf{ kg/m}\cdotp\textbf{s}}[/latex]

]]> 345 327 3 0 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/chapter/6-0-introduction/ Thu, 29 Jun 2017 22:13:38 +0000 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/chapter/6-0-introduction/

Many motions, such as the arc of a bird’s flight or Earth’s path around the Sun, are curved. Recall that Newton’s first law tells us that motion is along a straight line at constant speed unless there is a net external force. We will therefore study not only motion along curves, but also the forces that cause it, including gravitational forces. In some ways, this chapter is a continuation of Chapter 4 Dynamics: Newton's Laws of Motion as we study more applications of Newton’s laws of motion.

This chapter deals with the simplest form of curved motion, uniform circular motion, motion in a circular path at constant speed. Studying this topic illustrates most concepts associated with rotational motion and leads to the study of many new topics we group under the name rotation. Pure rotational motion occurs when points in an object move in circular paths centered on one point. Pure translational motion is motion with no rotation. Some motion combines both types, such as a rotating hockey puck moving along ice.

Glossary

uniform circular motion

the motion of an object in a circular path at constant speed

]]> 348 346 1 0 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/chapter/6-1-rotation-angle-and-angular-velocity/ Thu, 29 Jun 2017 22:13:39 +0000 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/chapter/6-1-rotation-angle-and-angular-velocity/

Summary

• Define arc length, rotation angle, radius of curvature and angular velocity.
• Calculate the angular velocity of a car wheel spin.

In Chapter 2 Kinematics, we studied motion along a straight line and introduced such concepts as displacement, velocity, and acceleration. Chapter 3 Two-Dimensional Kinematics dealt with motion in two dimensions. Projectile motion is a special case of two-dimensional kinematics in which the object is projected into the air, while being subject to the gravitational force, and lands a distance away. In this chapter, we consider situations where the object does not land but moves in a curve. We begin the study of uniform circular motion by defining two angular quantities needed to describe rotational motion.

Rotation Angle

When objects rotate about some axis—for example, when the CD (compact disc) in Figure 1 rotates about its center—each point in the object follows a circular arc. Consider a line from the center of the CD to its edge. Each pit used to record sound along this line moves through the same angle in the same amount of time. The rotation angle is the amount of rotation and is analogous to linear distance. We define the rotation angle Δθ to be the ratio of the arc length to the radius of curvature:

[latex]\boldsymbol{\Delta\theta=}[/latex][latex size="2"]\boldsymbol{\frac{\Delta{s}}{r}}.[/latex]

The arc length Δs is the distance traveled along a circular path as shown in Figure 2 Note that r is the radius of curvature of the circular path.

We know that for one complete revolution, the arc length is the circumference of a circle of radius r. The circumference of a circle is 2πr. Thus for one complete revolution the rotation angle is

[latex]\boldsymbol{\Delta\theta\:=}[/latex][latex size="2"]\boldsymbol{\frac{2\pi{r}}{r}}[/latex][latex]\boldsymbol{=\:2\pi}.[/latex]

This result is the basis for defining the units used to measure rotation angles, Δθ to be radians (rad), defined so that

[latex]\boldsymbol{2\pi\textbf{ rad} = 1\textbf{ revolution}.}[/latex]

A comparison of some useful angles expressed in both degrees and radians is shown in Table 1.

[latex]\textbf{Degree Measures}[/latex]	[latex]\textbf{Radian Measure}[/latex]
[latex]\boldsymbol{30^0}[/latex]	[latex size="1"]\boldsymbol{\frac{\pi}{6}}[/latex]
[latex]\boldsymbol{60^0}[/latex]	[latex size="1"]\boldsymbol{\frac{\pi}{3}}[/latex]
[latex]\boldsymbol{90^0}[/latex]	[latex size="1"]\boldsymbol{\frac{\pi}{2}}[/latex]
[latex]\boldsymbol{120^0}[/latex]	[latex size="1"]\boldsymbol{\frac{2\pi}{3}}[/latex]
[latex]\boldsymbol{135^0}[/latex]	[latex size="1"]\boldsymbol{\frac{3\pi}{4}}[/latex]
[latex]\boldsymbol{180^0}[/latex]	[latex]\boldsymbol{\pi}[/latex]
Table 1. Comparison of Angular Units.

If Δθ = 2π rad, then the CD has made one complete revolution, and every point on the CD is back at its original position. Because there are 360° in a circle or one revolution, the relationship between radians and degrees is thus

[latex]\boldsymbol{2\pi\textbf{rad}=360^0}[/latex]

so that

[latex]\boldsymbol{1\textbf{ rad}\:=}[/latex][latex size="2"]\boldsymbol{\frac{360^0}{2\pi}}[/latex][latex]\boldsymbol{\approx\:57.3^0}.[/latex]

Angular Velocity

How fast is an object rotating? We define angular velocity ω as the rate of change of an angle. In symbols, this is

[latex]\boldsymbol{\omega\:=}[/latex][latex size="2"]\boldsymbol{\frac{\Delta\theta}{\Delta{t}}},[/latex]

where an angular rotation Δθ takes place in a time Δt. The greater the rotation angle in a given amount of time, the greater the angular velocity. The units for angular velocity are radians per second (rad/s).

Angular velocity ω is analogous to linear velocity v. To get the precise relationship between angular and linear velocity, we again consider a pit on the rotating CD. This pit moves an arc length Δs in a time Δt, and so it has a linear velocity

[latex]\boldsymbol{v\:=}[/latex][latex size="2"]\boldsymbol{\frac{\Delta{s}}{\Delta{t}}}.[/latex]

From [latex]\boldsymbol{\Delta\theta=\frac{\Delta{s}}{r}}[/latex] we see that Δs = rΔθ. Substituting this into the expression for v gives

[latex]\boldsymbol{v\:=}[/latex][latex size="2"]\boldsymbol{\frac{r\Delta\theta}{\Delta{t}}}[/latex][latex]\boldsymbol{=\:r\omega}.[/latex]

We write this relationship in two different ways and gain two different insights:

[latex]\boldsymbol{v=r\omega\textbf{ or }\omega\:=}[/latex][latex size="2"]\boldsymbol{\frac{v}{r}}.[/latex]

The first relationship in v = rω or $$\boldsymbol{\omega\:=\frac{v}{r}}$$ states that the linear velocity v is proportional to the distance from the center of rotation, thus, it is largest for a point on the rim (largest r), as you might expect. We can also call this linear speed v of a point on the rim the tangential speed. The second relationship in v = rω or $$\boldsymbol{\omega\:=\frac{v}{r}}[/latex] can be illustrated by considering the tire of a moving car. Note that the speed of a point on the rim of the tire is the same as the speed v of the car. See Figure 4. So the faster the car moves, the faster the tire spins—large v means a large ω, because v = rω. Similarly, a larger-radius tire rotating at the same angular velocity (ω) will produce a greater linear speed (v) for the car.

[caption id="" align="aligncenter" width="300"]Figure 4. A car moving at a velocity v to the right has a tire rotating with an angular velocity ω. The speed of the tread of the tire relative to the axle is v, the same as if the car were jacked up. Thus the car moves forward at linear velocity v=rω, where r is the tire radius. A larger angular velocity for the tire means a greater velocity for the car.[/caption]

Example 1: How Fast Does a Car Tire Spin?

Calculate the angular velocity of a 0.300 m radius car tire when the car travels at 15.0 m/s (about 54 km/h). See Figure 4.

Strategy

Because the linear speed of the tire rim is the same as the speed of the car, we have v = 15.0 m/s. The radius of the tire is given to be r = 0.300 m. Knowing v and r, we can use the second relationship in v = rω, [latex]\boldsymbol{\omega=\frac{v}{r}}[/latex] to calculate the angular velocity.

Solution

To calculate the angular velocity, we will use the following relationship:

[latex]\boldsymbol{\omega\:=}[/latex][latex size="2"]\boldsymbol{\frac{v}{r}}.[/latex]

Substituting the knowns,

[latex]\boldsymbol{\omega\:=}[/latex][latex size="2"]\boldsymbol{\frac{15.0\textbf{ m/s}}{0.300\textbf{ m}}}[/latex][latex]\boldsymbol{=\:50.0\textbf{ rad/s}}.[/latex]

Discussion

When we cancel units in the above calculation, we get 50.0/s. But the angular velocity must have units of rad/s. Because radians are actually unitless (radians are defined as a ratio of distance), we can simply insert them into the answer for the angular velocity. Also note that if an earth mover with much larger tires, say 1.20 m in radius, were moving at the same speed of 15.0 m/s, its tires would rotate more slowly. They would have an angular velocity

[latex]\boldsymbol{\omega=(15.0\textbf{ m/s})/(1.20\textbf{ m})=12.5\textbf{ rad/s}}.[/latex]

Both[latex]\boldsymbol{\omega}[/latex]and[latex]\boldsymbol{v}[/latex]have directions (hence they are angular and linear velocities, respectively). Angular velocity has only two directions with respect to the axis of rotation—it is either clockwise or counterclockwise. Linear velocity is tangent to the path, as illustrated in Figure 5.

TAKE-HOME EXPERIMENT

Tie an object to the end of a string and swing it around in a horizontal circle above your head (swing at your wrist). Maintain uniform speed as the object swings and measure the angular velocity of the motion. What is the approximate speed of the object? Identify a point close to your hand and take appropriate measurements to calculate the linear speed at this point. Identify other circular motions and measure their angular velocities.

PHET EXPLORATIONS: LADYBUG REVOLUTION

Join the ladybug in an exploration of rotational motion. Rotate the merry-go-round to change its angle, or choose a constant angular velocity or angular acceleration. Explore how circular motion relates to the bug's x,y position, velocity, and acceleration using vectors or graphs.

Section Summary

• Uniform circular motion is motion in a circle at constant speed. The rotation angle Δθ is defined as the ratio of the arc length to the radius of curvature:
  [latex]\boldsymbol{\Delta\theta\:=}[/latex][latex size="2"]\boldsymbol{\frac{\Delta{s}}{r}},[/latex]

  where arc length Δs is distance traveled along a circular path and r is the radius of curvature of the circular path. The quantity Δθ is measured in units of radians (rad), for which

  [latex]\boldsymbol{2\pi\textbf{ rad}=360^0=1\textbf{ revolution}}.[/latex]
• The conversion between radians and degrees is 1 rad = 57.3°.
• Angular velocity ω is the rate of change of an angle,
  [latex]\boldsymbol{\omega\:=}[/latex][latex size="2"]\boldsymbol{\frac{\Delta\theta}{\Delta{t}}},[/latex]

  where a rotation Δθ takes place in a time Δt. The units of angular velocity are radians per second (rad/s). Linear velocity v and angular velocity ω are related by

  [latex]\boldsymbol{v=r\omega\textbf{ or }\omega\:=}[/latex][latex size="2"]\boldsymbol{\frac{v}{r}}.[/latex]

Conceptual Questions

1: There is an analogy between rotational and linear physical quantities. What rotational quantities are analogous to distance and velocity?

Problems & Exercises

1: Semi-trailer trucks have an odometer on one hub of a trailer wheel. The hub is weighted so that it does not rotate, but it contains gears to count the number of wheel revolutions—it then calculates the distance traveled. If the wheel has a 1.15 m diameter and goes through 200,000 rotations, how many kilometers should the odometer read?

2: Microwave ovens rotate at a rate of about 6 rev/min. What is this in revolutions per second? What is the angular velocity in radians per second?

3: An automobile with 0.260 m radius tires travels 80,000 km before wearing them out. How many revolutions do the tires make, neglecting any backing up and any change in radius due to wear?

4: (a) What is the period of rotation of Earth in seconds? (b) What is the angular velocity of Earth? (c) Given that Earth has a radius of 6.4 × 10⁶ m at its equator, what is the linear velocity at Earth’s surface?

5: A baseball pitcher brings his arm forward during a pitch, rotating the forearm about the elbow. If the velocity of the ball in the pitcher’s hand is 35.0 m/s and the ball is 0.300 m from the elbow joint, what is the angular velocity of the forearm?

6: In lacrosse, a ball is thrown from a net on the end of a stick by rotating the stick and forearm about the elbow. If the angular velocity of the ball about the elbow joint is 30.0 rad/s and the ball is 1.30 m from the elbow joint, what is the velocity of the ball?

7: A truck with 0.420-m-radius tires travels at 32.0 m/s. What is the angular velocity of the rotating tires in radians per second? What is this in rev/min?

8: Integrated Concepts

When kicking a football, the kicker rotates his leg about the hip joint.

(a) If the velocity of the tip of the kicker’s shoe is 35.0 m/s and the hip joint is 1.05 m from the tip of the shoe, what is the shoe tip’s angular velocity?

(b) The shoe is in contact with the initially stationary 0.500 kg football for 20.0 ms. What average force is exerted on the football to give it a velocity of 20.0 m/s?

(c) Find the maximum range of the football, neglecting air resistance.

9: Construct Your Own Problem

Consider an amusement park ride in which participants are rotated about a vertical axis in a cylinder with vertical walls. Once the angular velocity reaches its full value, the floor drops away and friction between the walls and the riders prevents them from sliding down. Construct a problem in which you calculate the necessary angular velocity that assures the riders will not slide down the wall. Include a free body diagram of a single rider. Among the variables to consider are the radius of the cylinder and the coefficients of friction between the riders’ clothing and the wall.

Glossary

arc length

Δs, the distance traveled by an object along a circular path

pit

a tiny indentation on the spiral track moulded into the top of the polycarbonate layer of CD

rotation angle

the ratio of the arc length to the radius of curvature on a circular path:

[latex]\boldsymbol{\Delta\theta\:=}[/latex][latex size="2"]\boldsymbol{\frac{\Delta{s}}{r}}[/latex]

radius of curvature

radius of a circular path

radians

a unit of angle measurement

angular velocity

ω, the rate of change of the angle with which an object moves on a circular path

Solutions

Problems & Exercises 1: [latex]\boldsymbol{723\textbf{ km}}[/latex] 3: [latex]\boldsymbol{5\times10^7\textbf{ rotations}}[/latex] 5: [latex]\boldsymbol{117\textbf{ rad/s}}[/latex]

7: [latex]\boldsymbol{76.2\textbf{ rad/s }728\textbf{ rpm}}[/latex]

8: (a) [latex]\boldsymbol{33.3\textbf{ rad/s}}[/latex] (b) [latex]\boldsymbol{500\textbf{ N}}[/latex] (c) [latex]\boldsymbol{40.8\textbf{ m}}[/latex]

]]> 355 346 2 0 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/chapter/6-2-centripetal-acceleration/ Thu, 29 Jun 2017 22:13:39 +0000 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/chapter/6-2-centripetal-acceleration/

Summary

• Establish the expression for centripetal acceleration.
• Explain the centrifuge.

We know from kinematics that acceleration is a change in velocity, either in its magnitude or in its direction, or both. In uniform circular motion, the direction of the velocity changes constantly, so there is always an associated acceleration, even though the magnitude of the velocity might be constant. You experience this acceleration yourself when you turn a corner in your car. (If you hold the wheel steady during a turn and move at constant speed, you are in uniform circular motion.) What you notice is a sideways acceleration because you and the car are changing direction. The sharper the curve and the greater your speed, the more noticeable this acceleration will become. In this section we examine the direction and magnitude of that acceleration.

Figure 1 shows an object moving in a circular path at constant speed. The direction of the instantaneous velocity is shown at two points along the path. Acceleration is in the direction of the change in velocity, which points directly toward the center of rotation (the center of the circular path). This pointing is shown with the vector diagram in the figure. We call the acceleration of an object moving in uniform circular motion (resulting from a net external force) the centripetal acceleration (a_c); centripetal means “toward the center” or “center seeking.”

The direction of centripetal acceleration is toward the center of curvature, but what is its magnitude? Note that the triangle formed by the velocity vectors and the one formed by the radii r and Δs are similar. Both the triangles ABC and PQR are isosceles triangles (two equal sides). The two equal sides of the velocity vector triangle are the speeds v₁ = v₂ = v. Using the properties of two similar triangles, we obtain

[latex size="2"]\boldsymbol{\frac{\Delta{v}}{v}}[/latex][latex]\boldsymbol{=}[/latex][latex size="2"]\boldsymbol{\frac{\Delta{s}}{r}}.[/latex]

Acceleration is Δv/Δt, and so we first solve this expression for Δv:

[latex]\boldsymbol{\Delta{v}\:=}[/latex][latex size="2"]\boldsymbol{\frac{v}{r}}[/latex][latex]\boldsymbol{\Delta{s}}.[/latex]

Then we divide this by Δt, yielding

[latex size="2"]\boldsymbol{\frac{\Delta{v}}{\Delta{t}}}[/latex][latex]\boldsymbol{=}[/latex][latex size="2"]\boldsymbol{\frac{v}{r}\times\frac{\Delta{s}}{\Delta{t}}}.[/latex]

Finally, noting that Δv/Δt = a_c and that Δs/Δt = v, the linear or tangential speed, we see that the magnitude of the centripetal acceleration is

[latex]\boldsymbol{a_{\textbf{c}}\:=}[/latex][latex size="2"]\boldsymbol{\frac{v^2}{r}},[/latex]

which is the acceleration of an object in a circle of radius r at a speed v. So, centripetal acceleration is greater at high speeds and in sharp curves (smaller radius), as you have noticed when driving a car. But it is a bit surprising that a_c is proportional to speed squared, implying, for example, that it is four times as hard to take a curve at 100 km/h than at 50 km/h. A sharp corner has a small radius, so that a_c is greater for tighter turns, as you have probably noticed.

It is also useful to express a_c in terms of angular velocity. Substituting v = rω into the above expression, we find a_c = (rω)²/r = rω². We can express the magnitude of centripetal acceleration using either of two equations:

[latex]\boldsymbol{a_{\textbf{c}}\:=}[/latex][latex size="2"]\boldsymbol{\frac{v^2}{r}}[/latex][latex]\boldsymbol{;\:a_{\textbf{c}}=r\omega^2}.[/latex]

Recall that the direction of a_c is toward the center. You may use whichever expression is more convenient, as illustrated in examples below.

A centrifuge (see Figure 2b) is a rotating device used to separate specimens of different densities. High centripetal acceleration significantly decreases the time it takes for separation to occur, and makes separation possible with small samples. Centrifuges are used in a variety of applications in science and medicine, including the separation of single cell suspensions such as bacteria, viruses, and blood cells from a liquid medium and the separation of macromolecules, such as DNA and protein, from a solution. Centrifuges are often rated in terms of their centripetal acceleration relative to acceleration due to gravity (g); maximum centripetal acceleration of several hundred thousand g is possible in a vacuum. Human centrifuges, extremely large centrifuges, have been used to test the tolerance of astronauts to the effects of accelerations larger than that of Earth’s gravity.

Example 1: How Does the Centripetal Acceleration of a Car Around a Curve Compare with That Due to Gravity?

What is the magnitude of the centripetal acceleration of a car following a curve of radius 500 m at a speed of 25.0 m/s (about 90 km/h)? Compare the acceleration with that due to gravity for this fairly gentle curve taken at highway speed. See Figure 2(a).

Strategy

Because v and r are given, the first expression in [latex]\boldsymbol{a_{\textbf{c}}=\frac{v^2}{r};\:a_{\textbf{c}}=r\omega^2}[/latex] is the most convenient to use.

Solution

Entering the given values of v = 25.0 m/s and r = 500 m into the first expression for a_c gives

[latex]\boldsymbol{a_{\textbf{c}}\:=}[/latex][latex size="2"]\boldsymbol{\frac{v^2}{r}}[/latex][latex]\boldsymbol{=}[/latex][latex size="2"]\boldsymbol{\frac{(25.0\textbf{ m/s})^2}{500\textbf{ m}}}[/latex][latex]\boldsymbol{=\:1.25\textbf{ m/s}^2}.[/latex]

Discussion

To compare this with the acceleration due to gravity (g = 9.80 m/s²), we take the ratio of a_c /g = (1.25 m/s²)/(9.80m/s²) = 0.128. Thus, a_c = 0.128 g and is noticeable especially if you were not wearing a seat belt.

[caption id="" align="aligncenter" width="198"]Figure 2. (a) The car following a circular path at constant speed is accelerated perpendicular to its velocity, as shown. The magnitude of this centripetal acceleration is found in Example 1. (b) A particle of mass in a centrifuge is rotating at constant angular velocity . It must be accelerated perpendicular to its velocity or it would continue in a straight line. The magnitude of the necessary acceleration is found in Example 2.[/caption]

Example 2: How Big Is The Centripetal Acceleration in an Ultracentrifuge?

Calculate the centripetal acceleration of a point 7.50 cm from the axis of an ultracentrifuge spinning at 7.5 × 10⁴ rev/min. Determine the ratio of this acceleration to that due to gravity. See Figure 2(b).

Strategy

The term rev/min stands for revolutions per minute. By converting this to radians per second, we obtain the angular velocity ω. Because r is given, we can use the second expression in the equation [latex]\boldsymbol{a_{\textbf{c}}=\frac{v^2}{r};\:a_{\textbf{c}}=r\omega^2}[/latex] to calculate the centripetal acceleration.

Solution

To convert 7.5 × 10⁴ rev/min to radians per second, we use the facts that one revolution is 2π rad and one minute is 60.0 s. Thus,

[latex]\boldsymbol{\omega\:=7.50\times10^4}[/latex][latex size="2"]\boldsymbol{\frac{\textbf{rev}}{\textbf{min}}\times\frac{2\pi\textbf{ rad}}{1\textbf{ rev}}\times\frac{1\textbf{ min}}{60.0\textbf{ s}}}[/latex][latex]\boldsymbol{=\:7854\textbf{ rad/s}}.[/latex]

Now the centripetal acceleration is given by the second expression in [latex]\boldsymbol{a_{\textbf{c}}=\frac{v^2}{r};\:a_{\textbf{c}}=r\omega^2}[/latex] as

[latex]\boldsymbol{a_{\textbf{c}}=r\omega^2}.[/latex]

Converting 7.50 cm to meters and substituting known values gives

[latex]\boldsymbol{a_{\textbf{c}}=(0.0750\textbf{ m})(7854\textbf{ rad/s})^2=4.63\times10^6\textbf{ m/s}^2}.[/latex]

Note that the unitless radians are discarded in order to get the correct units for centripetal acceleration. Taking the ratio of a_c to g yields

[latex size="2"]\boldsymbol{\frac{a_{\textbf{c}}}{g}}[/latex][latex]\boldsymbol{=}[/latex][latex size="2"]\boldsymbol{\frac{4.63\times10^6}{9.80}}[/latex][latex]\boldsymbol{=4.72\times10^5}.[/latex]

Discussion

This last result means that the centripetal acceleration is 472,000 times as strong as g. It is no wonder that such high ω centrifuges are called ultracentrifuges. The extremely large accelerations involved greatly decrease the time needed to cause the sedimentation of blood cells or other materials.

Of course, a net external force is needed to cause any acceleration, just as Newton proposed in his second law of motion. So a net external force is needed to cause a centripetal acceleration. In Chapter 6.3 Centripetal Force, we will consider the forces involved in circular motion.

PHET EXPLORATIONS: LADYBUG MOTION 2D

Learn about position, velocity and acceleration vectors. Move the ladybug by setting the position, velocity or acceleration, and see how the vectors change. Choose linear, circular or elliptical motion, and record and playback the motion to analyze the behavior.

[caption id="" align="aligncenter" width="450"] Figure 3. Ladybug Motion in 2D[/caption]

Section Summary

• Centripetal acceleration a_c is the acceleration experienced while in uniform circular motion. It always points toward the center of rotation. It is perpendicular to the linear velocity[latex]\boldsymbol{v}[/latex]and has the magnitude

[latex]\boldsymbol{a_{\textbf{c}}\:=}[/latex][latex size="2"]\boldsymbol{\frac{v^2}{r}}[/latex][latex]\boldsymbol{;\:a_{\textbf{c}}=r\omega^2}.[/latex]

• The unit of centripetal acceleration is m/s².

Conceptual Questions

1: Can centripetal acceleration change the speed of circular motion? Explain.

Problems & Exercises

1: A fairground ride spins its occupants inside a flying saucer-shaped container. If the horizontal circular path the riders follow has an 8.00 m radius, at how many revolutions per minute will the riders be subjected to a centripetal acceleration whose magnitude is 1.50 times that due to gravity?

2: A runner taking part in the 200 m dash must run around the end of a track that has a circular arc with a radius of curvature of 30 m. If he completes the 200 m dash in 23.2 s and runs at constant speed throughout the race, what is the magnitude of his centripetal acceleration as he runs the curved portion of the track?

3: Taking the age of Earth to be about 4 × 10⁹ years and assuming its orbital radius of 1.5 × 10¹¹ has not changed and is circular, calculate the approximate total distance Earth has traveled since its birth (in a frame of reference stationary with respect to the Sun).

4: The propeller of a World War II fighter plane is 2.30 m in diameter.

(a) What is its angular velocity in radians per second if it spins at 1200 rev/min?

(b) What is the linear speed of its tip at this angular velocity if the plane is stationary on the tarmac?

(c) What is the centripetal acceleration of the propeller tip under these conditions? Calculate it in meters per second squared and convert to multiples of g.

5: An ordinary workshop grindstone has a radius of 7.50 cm and rotates at 6500 rev/min.

(a) Calculate the magnitude of the centripetal acceleration at its edge in meters per second squared and convert it to multiples of[latex]\boldsymbol{g}.[/latex]

(b) What is the linear speed of a point on its edge?

6: Helicopter blades withstand tremendous stresses. In addition to supporting the weight of a helicopter, they are spun at rapid rates and experience large centripetal accelerations, especially at the tip.

(a) Calculate the magnitude of the centripetal acceleration at the tip of a 4.00 m long helicopter blade that rotates at 300 rev/min.

(b) Compare the linear speed of the tip with the speed of sound (taken to be 340 m/s).

7: Olympic ice skaters are able to spin at about 5 rev/s.

(a) What is their angular velocity in radians per second?

(b) What is the centripetal acceleration of the skater’s nose if it is 0.120 m from the axis of rotation?

(c) An exceptional skater named Dick Button was able to spin much faster in the 1950s than anyone since—at about 9 rev/s. What was the centripetal acceleration of the tip of his nose, assuming it is at 0.120 m radius?

(d) Comment on the magnitudes of the accelerations found. It is reputed that Button ruptured small blood vessels during his spins.

8: What percentage of the acceleration at Earth’s surface is the acceleration due to gravity at the position of a satellite located 300 km above Earth?

9: Verify that the linear speed of an ultracentrifuge is about 0.50 km/s, and Earth in its orbit is about 30 km/s by calculating:

(a) The linear speed of a point on an ultracentrifuge 0.100 m from its center, rotating at 50,000 rev/min.

(b) The linear speed of Earth in its orbit about the Sun (use data from the text on the radius of Earth’s orbit and approximate it as being circular).

10: A rotating space station is said to create “artificial gravity”—a loosely-defined term used for an acceleration that would be crudely similar to gravity. The outer wall of the rotating space station would become a floor for the astronauts, and centripetal acceleration supplied by the floor would allow astronauts to exercise and maintain muscle and bone strength more naturally than in non-rotating space environments. If the space station is 200 m in diameter, what angular velocity would produce an “artificial gravity” of 9.80 m/s² at the rim?

11: At takeoff, a commercial jet has a 60.0 m/s speed. Its tires have a diameter of 0.850 m.

(a) At how many rev/min are the tires rotating?

(b) What is the centripetal acceleration at the edge of the tire?

(c) With what force must a determined 1.00 × 10^-15 kg bacterium cling to the rim?

(d) Take the ratio of this force to the bacterium’s weight.

12: Integrated Concepts

Riders in an amusement park ride shaped like a Viking ship hung from a large pivot are rotated back and forth like a rigid pendulum. Sometime near the middle of the ride, the ship is momentarily motionless at the top of its circular arc. The ship then swings down under the influence of gravity.

(a) Assuming negligible friction, find the speed of the riders at the bottom of its arc, given the system's center of mass travels in an arc having a radius of 14.0 m and the riders are near the center of mass.

(b) What is the centripetal acceleration at the bottom of the arc?

(c) Draw a free body diagram of the forces acting on a rider at the bottom of the arc.

(d) Find the force exerted by the ride on a 60.0 kg rider and compare it to her weight.

(e) Discuss whether the answer seems reasonable.

13: Unreasonable Results

A mother pushes her child on a swing so that his speed is 9.00 m/s at the lowest point of his path. The swing is suspended 2.00 m above the child’s center of mass.

(a) What is the magnitude of the centripetal acceleration of the child at the low point?

(b) What is the magnitude of the force the child exerts on the seat if his mass is 18.0 kg?

(c) What is unreasonable about these results?

(d) Which premises are unreasonable or inconsistent?

Glossary

centripetal acceleration

the acceleration of an object moving in a circle, directed toward the center

ultracentrifuge

a centrifuge optimized for spinning a rotor at very high speeds

Solutions

Problems & Exercises 1: $$\boldsymbol{12.9\textbf{ rev/min}}$$ 3: [latex]\boldsymbol{4\times10^{21}\textbf{ m}}[/latex]

5: (a) [latex]\boldsymbol{3.47\times10^4\textbf{ m/s}^2},\boldsymbol{\:3.55\times10^3\textbf{ g}}[/latex] (b) [latex]\boldsymbol{51.1\textbf{ m/s}}[/latex]

7: (a) [latex]\boldsymbol{31.4\textbf{ rad/s}}[/latex] (b) [latex]\boldsymbol{118\textbf{ m/s}}[/latex] (c) [latex]\boldsymbol{384\textbf{ m/s}}[/latex] (d) The centripetal acceleration felt by Olympic skaters is 12 times larger than the acceleration due to gravity. That’s quite a lot of acceleration in itself. The centripetal acceleration felt by Button’s nose was 39.2 times larger than the acceleration due to gravity. It is no wonder that he ruptured small blood vessels in his spins.

9: (a) $$\boldsymbol{0.524\textbf{ km/s}}$$ (b) $$\boldsymbol{29.7\textbf{ km/s}}$$

11: (a) [latex]\boldsymbol{1.35\times10^3\textbf{ rpm}}[/latex] (b) [latex]\boldsymbol{8.47\times10^3\textbf{ m/s}^2}[/latex] (c) [latex]\boldsymbol{8.47\times10^{-12}\textbf{ N}}[/latex] (d) [latex]\boldsymbol{865}[/latex]

12: (a) [latex]\boldsymbol{16.6\textbf{ m/s}}[/latex] (b) [latex]\boldsymbol{19.6\textbf{ m/s}^2}[/latex] (c)

[caption id="" align="aligncenter" width="150"]Figure 4.[/caption]

(d) [latex]\boldsymbol{1.76\times10^3\textbf{ N}\text{ or }3.00\:\omega},[/latex] that is, the normal force (upward) is three times her weight. (e) This answer seems reasonable, since she feels like she’s being forced into the chair MUCH stronger than just by gravity.

13: (a) [latex]\boldsymbol{40.5\textbf{ m/s}^2}[/latex] (b) $$\boldsymbol{905\textbf{ N}}$$ (c) The force in part (b) is very large. The acceleration in part (a) is too much, about 4 g. (d) The speed of the swing is too large. At the given velocity at the bottom of the swing, there is enough kinetic energy to send the child all the way over the top, ignoring friction.

]]> 360 346 3 0 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/chapter/6-3-centripetal-force/ Thu, 29 Jun 2017 22:13:42 +0000 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/chapter/6-3-centripetal-force/

Summary

• Calculate coefficient of friction on a car tire.
• Calculate ideal speed and angle of a car on a turn.

Any force or combination of forces can cause a centripetal or radial acceleration. Just a few examples are the tension in the rope on a tether ball, the force of Earth’s gravity on the Moon, friction between roller skates and a rink floor, a banked roadway’s force on a car, and forces on the tube of a spinning centrifuge.

Any net force causing uniform circular motion is called a centripetal force. The direction of a centripetal force is toward the center of curvature, the same as the direction of centripetal acceleration. According to Newton’s second law of motion, net force is mass times acceleration: net F = ma. For uniform circular motion, the acceleration is the centripetal acceleration— a = a_c. Thus, the magnitude of centripetal force F_c is

[latex]\boldsymbol{F_{\textbf{c}}=ma_{\textbf{c}}}.[/latex]

By using the expressions for centripetal acceleration a_c from [latex]\boldsymbol{a_{\textbf{c}}=\frac{v^2}{r};\:a_{\textbf{c}}=r\omega^2},[/latex] we get two expressions for the centripetal force F_c in terms of mass, velocity, angular velocity, and radius of curvature:

[latex]\boldsymbol{F_{\textbf{c}}=\:m}[/latex][latex size="2"]\boldsymbol{\frac{v^2}{r}}[/latex][latex]\boldsymbol{;\:F_{\textbf{c}}=mr\omega^2}.[/latex]

You may use whichever expression for centripetal force is more convenient. Centripetal force F_c is always perpendicular to the path and pointing to the center of curvature, because a_c is perpendicular to the velocity and pointing to the center of curvature.

Note that if you solve the first expression for r, you get

[latex]\boldsymbol{r\:=}[/latex][latex size="2"]\boldsymbol{\frac{mv^2}{F_{\textbf{c}}}}.[/latex]

This implies that for a given mass and velocity, a large centripetal force causes a small radius of curvature—that is, a tight curve.

[caption id="" align="aligncenter" width="200"]Figure 1. The frictional force supplies the centripetal force and is numerically equal to it. Centripetal force is perpendicular to velocity and causes uniform circular motion. The larger the F_c, the smaller the radius of curvature r and the sharper the curve. The second curve has the same v, but a larger F_c produces a smaller r'.[/caption]

Example 1: What Coefficient of Friction Do Care Tires Need on a Flat Curve?

(a) Calculate the centripetal force exerted on a 900 kg car that negotiates a 500 m radius curve at 25.0 m/s.

(b) Assuming an unbanked curve, find the minimum static coefficient of friction, between the tires and the road, static friction being the reason that keeps the car from slipping (see Figure 2).

Strategy and Solution for (a)

We know that [latex]\boldsymbol{{F}_{\textbf{c}}=\frac{mv^2}{r}}.[/latex]Thus,

[latex]\boldsymbol{{F}_{\textbf{c}}\:=}[/latex][latex size="2"]\boldsymbol{\frac{mv^2}{r}}[/latex][latex]\boldsymbol{=}[/latex][latex size="2"]\boldsymbol{\frac{(900\textbf{ kg})(25.0\textbf{ m/s})^2}{(500\textbf{ m})}}[/latex][latex]\boldsymbol{=\:1125\textbf{ N}}.[/latex]

Strategy for (b)

Figure 2 shows the forces acting on the car on an unbanked (level ground) curve. Friction is to the left, keeping the car from slipping, and because it is the only horizontal force acting on the car, the friction is the centripetal force in this case. We know that the maximum static friction (at which the tires roll but do not slip) is μ_sN, where μ_s is the static coefficient of friction and N is the normal force. The normal force equals the car’s weight on level ground, so that N = mg. Thus the centripetal force in this situation is

[latex]\boldsymbol{F_{\textbf{c}}=f=\mu_{\textbf{s}}\textbf{N}=\mu_{\textbf{s}}mg}.[/latex]

Now we have a relationship between centripetal force and the coefficient of friction. Using the first expression for F_c from the equation

[latex]\begin{array}{c} \boldsymbol{F_{\textbf{c}}=m\frac{v^2}{r}} \\ \boldsymbol{F_{\textbf{c}}=mr\omega^2} \end{array}[/latex][latex size="4"]\rbrace[/latex],

[latex]\boldsymbol{m}[/latex][latex size="2"]\boldsymbol{\frac{v^2}{r}}[/latex][latex]\boldsymbol{=}[/latex][latex]\boldsymbol{\mu_{\textbf{s}}mg.}[/latex]

We solve this for μ_s, noting that mass cancels, and obtain

[latex]\boldsymbol{\mu_{\textbf{s}}\:=}[/latex][latex size="2"]\boldsymbol{\frac{v^2}{rg}}.[/latex]

Solution for (b)

Substituting the knowns,

[latex]\boldsymbol{\mu_{\textbf{s}}\:=}[/latex][latex size="2"]\boldsymbol{\frac{(25.0\textbf{ m/s})^2}{(500\textbf{ m})(9.80\textbf{ m/s}^2)}}[/latex][latex]\boldsymbol{=\:0.13}.[/latex]

(Because coefficients of friction are approximate, the answer is given to only two digits.)

Discussion

We could also solve part (a) using the first expression in [latex]\begin{array}{l} \boldsymbol{F_{\textbf{c}}=m\frac{v^2}{r}} \\ \boldsymbol{F_{\textbf{c}}=mr\omega^2} \end{array}[/latex][latex size="4"]\rbrace[/latex], because m, v, and r are given. The coefficient of friction found in part (b) is much smaller than is typically found between tires and roads. The car will still negotiate the curve if the coefficient is greater than 0.13, because static friction is a responsive force, being able to assume a value less than but no more than μ_sN. A higher coefficient would also allow the car to negotiate the curve at a higher speed, but if the coefficient of friction is less, the safe speed would be less than 25 m/s. Note that mass cancels, implying that in this example, it does not matter how heavily loaded the car is to negotiate the turn. Mass cancels because friction is assumed proportional to the normal force, which in turn is proportional to mass. If the surface of the road were banked, the normal force would be less as will be discussed below.

Let us now consider banked curves, where the slope of the road helps you negotiate the curve. See Figure 3. The greater the angle θ, the faster you can take the curve. Race tracks for bikes as well as cars, for example, often have steeply banked curves. In an “ideally banked curve,” the angle θ is such that you can negotiate the curve at a certain speed without the aid of friction between the tires and the road. We will derive an expression for θ for an ideally banked curve and consider an example related to it.

For ideal banking, the net external force equals the horizontal centripetal force in the absence of friction. The components of the normal force N in the horizontal and vertical directions must equal the centripetal force and the weight of the car, respectively. In cases in which forces are not parallel, it is most convenient to consider components along perpendicular axes—in this case, the vertical and horizontal directions.

Figure 3 shows a free body diagram for a car on a frictionless banked curve. If the angle θ is ideal for the speed and radius, then the net external force will equal the necessary centripetal force. The only two external forces acting on the car are its weight w and the normal force of the road N. (A frictionless surface can only exert a force perpendicular to the surface—that is, a normal force.) These two forces must add to give a net external force that is horizontal toward the center of curvature and has magnitude mv²/r. Because this is the crucial force and it is horizontal, we use a coordinate system with vertical and horizontal axes. Only the normal force has a horizontal component, and so this must equal the centripetal force—that is,

[latex]\boldsymbol{N\textbf{sin} \;\theta\:=}[/latex][latex size="2"]\boldsymbol{\frac{mv^2}{r}}.[/latex]

Because the car does not leave the surface of the road, the net vertical force must be zero, meaning that the vertical components of the two external forces must be equal in magnitude and opposite in direction. From the figure, we see that the vertical component of the normal force is N cos θ, and the only other vertical force is the car’s weight. These must be equal in magnitude; thus,

[latex]\boldsymbol{N\textbf{cos} \;\theta=mg}.[/latex]

Now we can combine the last two equations to eliminate N and get an expression for θ, as desired. Solving the second equation for N = mg/(cos θ), and substituting this into the first yields

[latex]\boldsymbol{mg}[/latex][latex size="2"]\boldsymbol{\frac{\textbf{sin} \;\theta}{\textbf{cos} \;\theta}}[/latex][latex]\boldsymbol{=}[/latex][latex size="2"]\boldsymbol{\frac{mv^2}{r}}[/latex]

[latex]\boldsymbol{mg \;\textbf{tan}(\theta)\:=}[/latex][latex size="2"]\boldsymbol{\frac{mv^2}{r}}[/latex]

[latex]\boldsymbol{\textbf{tan}\theta\:=}[/latex][latex size="2"]\boldsymbol{\frac{v^2}{rg}.}[/latex]

Taking the inverse tangent gives

[latex]\boldsymbol{\theta=\textbf{tan}^{-1}}[/latex][latex size="2"]\boldsymbol{(\frac{v^2}{rg})}[/latex][latex]\textbf{(ideally banked curve, no friction).}[/latex]

This expression can be understood by considering how θ depends on v and r. A large θ will be obtained for a large v and a small r. That is, roads must be steeply banked for high speeds and sharp curves. Friction helps, because it allows you to take the curve at greater or lower speed than if the curve is frictionless. Note that θ does not depend on the mass of the vehicle.

[caption id="" align="aligncenter" width="300"]Figure 3. The car on this banked curve is moving away and turning to the left.[/caption]

Example 2: What Is the Ideal Speed to Take a Steeply Banked Tight Curve?

Curves on some test tracks and race courses, such as the Daytona International Speedway in Florida, are very steeply banked. This banking, with the aid of tire friction and very stable car configurations, allows the curves to be taken at very high speed. To illustrate, calculate the speed at which a 100 m radius curve banked at 65.0° should be driven if the road is frictionless.

Strategy

We first note that all terms in the expression for the ideal angle of a banked curve except for speed are known; thus, we need only rearrange it so that speed appears on the left-hand side and then substitute known quantities.

Solution

Starting with

[latex]\boldsymbol{\textbf{tan}\theta\:=}[/latex][latex size="2"]\boldsymbol{\frac{v^2}{rg}}[/latex]

we get

[latex]\boldsymbol{v=(rg\:\textbf{tan}\:\theta)^{1/2}}.[/latex]

Noting that tan 65.0º = 2.14, we obtain

[latex]\begin{array}{lcl} \boldsymbol{v} & = & \boldsymbol{[(100\textbf{ m})(9.80\textbf{ m/s}^2)(2.14)]^{1/2}} \\ \boldsymbol{} & = & \boldsymbol{45.8}\textbf{ m/s.} \end{array}[/latex]

Discussion

This is just about 165 km/h, consistent with a very steeply banked and rather sharp curve. Tire friction enables a vehicle to take the curve at significantly higher speeds.

Calculations similar to those in the preceding examples can be performed for a host of interesting situations in which centripetal force is involved—a number of these are presented in this chapter’s Problems and Exercises.

TAKE-HOME EXPERIMENT

Ask a friend or relative to swing a golf club or a tennis racquet. Take appropriate measurements to estimate the centripetal acceleration of the end of the club or racquet. You may choose to do this in slow motion.

PHET EXPLORATIONS: GRAVITY AND ORBITS

Move the sun, earth, moon and space station to see how it affects their gravitational forces and orbital paths. Visualize the sizes and distances between different heavenly bodies, and turn off gravity to see what would happen without it! [caption id="" align="aligncenter" width="450"] Figure 4. Gravity and Orbits[/caption]

Section Summary

• Centripetal force F_c is any force causing uniform circular motion. It is a “center-seeking” force that always points toward the center of rotation. It is perpendicular to linear velocity v and has magnitude

[latex]\boldsymbol{F_{\textbf{c}}=ma_{\textbf{c}}},[/latex]

which can also be expressed as

[latex]\boldsymbol{F_{\textbf{c}}=m\frac{v^2}{r}}[/latex]

or

[latex]\boldsymbol{F_{\textbf{c}}=mr\omega^2}[/latex]

Conceptual Questions

Conceptual Questions

1: If you wish to reduce the stress (which is related to centripetal force) on high-speed tires, would you use large- or small-diameter tires? Explain.

2: Define centripetal force. Can any type of force (for example, tension, gravitational force, friction, and so on) be a centripetal force? Can any combination of forces be a centripetal force?

3: If centripetal force is directed toward the center, why do you feel that you are ‘thrown’ away from the center as a car goes around a curve? Explain.

4: Race car drivers routinely cut corners as shown in Figure 5. Explain how this allows the curve to be taken at the greatest speed.

5: A number of amusement parks have rides that make vertical loops like the one shown in Figure 6. For safety, the cars are attached to the rails in such a way that they cannot fall off. If the car goes over the top at just the right speed, gravity alone will supply the centripetal force. What other force acts and what is its direction if:

(a) The car goes over the top at faster than this speed?

(b)The car goes over the top at slower than this speed?

[caption id="" align="aligncenter" width="300"]Figure 6. Amusement rides with a vertical loop are an example of a form of curved motion.[/caption]

7: What is the direction of the force exerted by the car on the passenger as the car goes over the top of the amusement ride pictured in Figure 6 under the following circumstances:

(a) The car goes over the top at such a speed that the gravitational force is the only force acting?

(b) The car goes over the top faster than this speed?

(c) The car goes over the top slower than this speed?

8: As a skater forms a circle, what force is responsible for making her turn? Use a free body diagram in your answer.

9: Suppose a child is riding on a merry-go-round at a distance about halfway between its center and edge. She has a lunch box resting on wax paper, so that there is very little friction between it and the merry-go-round. Which path shown in Figure 7 will the lunch box take when she lets go? The lunch box leaves a trail in the dust on the merry-go-round. Is that trail straight, curved to the left, or curved to the right? Explain your answer.

10: Do you feel yourself thrown to either side when you negotiate a curve that is ideally banked for your car’s speed? What is the direction of the force exerted on you by the car seat?

11: Suppose a mass is moving in a circular path on a frictionless table as shown in figure. In the Earth’s frame of reference, there is no centrifugal force pulling the mass away from the centre of rotation, yet there is a very real force stretching the string attaching the mass to the nail. Using concepts related to centripetal force and Newton’s third law, explain what force stretches the string, identifying its physical origin.

Problems & Exercises

1: (a) A 22.0 kg child is riding a playground merry-go-round that is rotating at 40.0 rev/min. What centripetal force must she exert to stay on if she is 1.25 m from its center?

(b) What centripetal force does she need to stay on an amusement park merry-go-round that rotates at 3.00 rev/min if she is 8.00 m from its center?

(c) Compare each force with her weight.

2: Calculate the centripetal force on the end of a 100 m (radius) wind turbine blade that is rotating at 0.5 rev/s. Assume the mass is 4 kg.

3: What is the ideal banking angle for a gentle turn of 1.20 km radius on a highway with a 105 km/h speed limit (about 65 mi/h), assuming everyone travels at the limit?

4: What is the ideal speed to take a 100 m radius curve banked at a 20.0° angle?

5: (a) What is the radius of a bobsled turn banked at 75.0° and taken at 30.0 m/s, assuming it is ideally banked?

(b) Calculate the centripetal acceleration.

(c) Does this acceleration seem large to you?

6: Part of riding a bicycle involves leaning at the correct angle when making a turn, as seen in Figure 9. To be stable, the force exerted by the ground must be on a line going through the center of gravity. The force on the bicycle wheel can be resolved into two perpendicular components—friction parallel to the road (this must supply the centripetal force), and the vertical normal force (which must equal the system’s weight).

(a) Show that θ (as defined in the figure) is related to the speed v and radius of curvature r of the turn in the same way as for an ideally banked roadway—that is,  θ = tan^-1v²/rg.

(b) Calculate θ for a 12.0 m/s turn of radius 30.0 m (as in a race).

7: A large centrifuge, like the one shown in Figure 10(a), is used to expose aspiring astronauts to accelerations similar to those experienced in rocket launches and atmospheric reentries.

(a) At what angular velocity is the centripetal acceleration 10 g if the rider is 15.0 m from the center of rotation?

(b) The rider’s cage hangs on a pivot at the end of the arm, allowing it to swing outward during rotation as shown in Figure 10(b). At what angle θ below the horizontal will the cage hang when the centripetal acceleration is 10 g? (Hint: The arm supplies centripetal force and supports the weight of the cage. Draw a free body diagram of the forces to see what the angle θ should be.)

8: Integrated Concepts

If a car takes a banked curve at less than the ideal speed, friction is needed to keep it from sliding toward the inside of the curve (a real problem on icy mountain roads). (a) Calculate the ideal speed to take a 100 m radius curve banked at 15.0º. (b) What is the minimum coefficient of friction needed for a frightened driver to take the same curve at 20.0 km/h?

9: Modern roller coasters have vertical loops like the one shown in Figure 11. The radius of curvature is smaller at the top than on the sides so that the downward centripetal acceleration at the top will be greater than the acceleration due to gravity, keeping the passengers pressed firmly into their seats. What is the speed of the roller coaster at the top of the loop if the radius of curvature there is 15.0 m and the downward acceleration of the car is 1.50 g?

10: Unreasonable Results

(a) Calculate the minimum coefficient of friction needed for a car to negotiate an unbanked 50.0 m radius curve at 30.0 m/s.

(b) What is unreasonable about the result?

(c) Which premises are unreasonable or inconsistent?

Glossary

centripetal force

any net force causing uniform circular motion

ideal banking

the sloping of a curve in a road, where the angle of the slope allows the vehicle to negotiate the curve at a certain speed without the aid of friction between the tires and the road; the net external force on the vehicle equals the horizontal centripetal force in the absence of friction

ideal speed

the maximum safe speed at which a vehicle can turn on a curve without the aid of friction between the tire and the road

ideal angle

the angle at which a car can turn safely on a steep curve, which is in proportion to the ideal speed

banked curve

the curve in a road that is sloping in a manner that helps a vehicle negotiate the curve

Solutions

Problems & Exercises

1: (a) $$\boldsymbol{483\textbf{ N}}$$ (b) $$\boldsymbol{17.4\textbf{ N}}$$ (c) 2.24 times her weight, 0.0807 times her weight

3: [latex]\boldsymbol{4.14^0}[/latex]

5: (a) $$\boldsymbol{24.6\textbf{ m}}$$ (b) [latex]\boldsymbol{36.6\textbf{ m/s}^2}[/latex] (c) [latex]\boldsymbol{a_{\textbf{c}}=3.73\textbf{ g}.}[/latex] This does not seem too large, but it is clear that bobsledders feel a lot of force on them going through sharply banked turns.

7: (a) $$\boldsymbol{2.56\textbf{ rad/s}}$$ (b) [latex]\boldsymbol{5.71^0}[/latex]

8: (a) $$\boldsymbol{16.2\textbf{ m/s}}$$ (b) $$\boldsymbol{0.234}$$

10: (a) $$\boldsymbol{1.84}$$ (b) A coefficient of friction this much greater than 1 is unreasonable . (c) The assumed speed is too great for the tight curve.

]]> 372 346 4 0 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/chapter/6-4-fictitious-forces-and-non-inertial-frames-the-coriolis-force/ Thu, 29 Jun 2017 22:13:43 +0000 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/chapter/6-4-fictitious-forces-and-non-inertial-frames-the-coriolis-force/

Summary

• Discuss the inertial frame of reference.
• Discuss the non-inertial frame of reference.
• Describe the effects of the Coriolis force.

What do taking off in a jet airplane, turning a corner in a car, riding a merry-go-round, and the circular motion of a tropical cyclone have in common? Each exhibits fictitious forces—unreal forces that arise from motion and may seem real, because the observer’s frame of reference is accelerating or rotating.

When taking off in a jet, most people would agree it feels as if you are being pushed back into the seat as the airplane accelerates down the runway. Yet a physicist would say that you tend to remain stationary while the seat pushes forward on you, and there is no real force backward on you. An even more common experience occurs when you make a tight curve in your car—say, to the right. You feel as if you are thrown (that is, forced) toward the left relative to the car. Again, a physicist would say that you are going in a straight line but the car moves to the right, and there is no real force on you to the left. Recall Newton’s first law.

We can reconcile these points of view by examining the frames of reference used. Let us concentrate on people in a car. Passengers instinctively use the car as a frame of reference, while a physicist uses Earth. The physicist chooses Earth because it is very nearly an inertial frame of reference—one in which all forces are real (that is, in which all forces have an identifiable physical origin). In such a frame of reference, Newton’s laws of motion take the form given in Chapter 4 Dynamics: Newton's Laws of Motion The car is a non-inertial frame of reference because it is accelerated to the side. The force to the left sensed by car passengers is a fictitious force having no physical origin. There is nothing real pushing them left—the car, as well as the driver, is actually accelerating to the right.

Let us now take a mental ride on a merry-go-round—specifically, a rapidly rotating playground merry-go-round. You take the merry-go-round to be your frame of reference because you rotate together. In that non-inertial frame, you feel a fictitious force, named centrifugal force (not to be confused with centripetal force), trying to throw you off. You must hang on tightly to counteract the centrifugal force. In Earth’s frame of reference, there is no force trying to throw you off. Rather you must hang on to make yourself go in a circle because otherwise you would go in a straight line, right off the merry-go-round.

This inertial effect, carrying you away from the center of rotation if there is no centripetal force to cause circular motion, is put to good use in centrifuges (see Figure 3). A centrifuge spins a sample very rapidly, as mentioned earlier in this chapter. Viewed from the rotating frame of reference, the fictitious centrifugal force throws particles outward, hastening their sedimentation. The greater the angular velocity, the greater the centrifugal force. But what really happens is that the inertia of the particles carries them along a line tangent to the circle while the test tube is forced in a circular path by a centripetal force.

Let us now consider what happens if something moves in a frame of reference that rotates. For example, what if you slide a ball directly away from the center of the merry-go-round, as shown in Figure 4? The ball follows a straight path relative to Earth (assuming negligible friction) and a path curved to the right on the merry-go-round’s surface. A person standing next to the merry-go-round sees the ball moving straight and the merry-go-round rotating underneath it. In the merry-go-round’s frame of reference, we explain the apparent curve to the right by using a fictitious force, called the Coriolis force, that causes the ball to curve to the right. The fictitious Coriolis force can be used by anyone in that frame of reference to explain why objects follow curved paths and allows us to apply Newton’s Laws in non-inertial frames of reference.

[caption id="" align="aligncenter" width="250"]Figure 4. Looking down on the counterclockwise rotation of a merry-go-round, we see that a ball slid straight toward the edge follows a path curved to the right. The person slides the ball toward point B, starting at point A. Both points rotate to the shaded positions (A’ and B’) shown in the time that the ball follows the curved path in the rotating frame and a straight path in Earth’s frame.[/caption]

Up until now, we have considered Earth to be an inertial frame of reference with little or no worry about effects due to its rotation. Yet such effects do exist—in the rotation of weather systems, for example. Most consequences of Earth’s rotation can be qualitatively understood by analogy with the merry-go-round. Viewed from above the North Pole, Earth rotates counterclockwise, as does the merry-go-round in Figure 4. As on the merry-go-round, any motion in Earth’s northern hemisphere experiences a Coriolis force to the right. Just the opposite occurs in the southern hemisphere; there, the force is to the left. Because Earth’s angular velocity is small, the Coriolis force is usually negligible, but for large-scale motions, such as wind patterns, it has substantial effects.

The Coriolis force causes hurricanes in the northern hemisphere to rotate in the counterclockwise direction, while the tropical cyclones (what hurricanes are called below the equator) in the southern hemisphere rotate in the clockwise direction. The terms hurricane, typhoon, and tropical storm are regionally-specific names for tropical cyclones, storm systems characterized by low pressure centers, strong winds, and heavy rains. Figure 5 helps show how these rotations take place. Air flows toward any region of low pressure, and tropical cyclones contain particularly low pressures. Thus winds flow toward the center of a tropical cyclone or a low-pressure weather system at the surface. In the northern hemisphere, these inward winds are deflected to the right, as shown in the figure, producing a counterclockwise circulation at the surface for low-pressure zones of any type. Low pressure at the surface is associated with rising air, which also produces cooling and cloud formation, making low-pressure patterns quite visible from space. Conversely, wind circulation around high-pressure zones is clockwise in the northern hemisphere but is less visible because high pressure is associated with sinking air, producing clear skies.

The rotation of tropical cyclones and the path of a ball on a merry-go-round can just as well be explained by inertia and the rotation of the system underneath. When non-inertial frames are used, fictitious forces, such as the Coriolis force, must be invented to explain the curved path. There is no identifiable physical source for these fictitious forces. In an inertial frame, inertia explains the path, and no force is found to be without an identifiable source. Either view allows us to describe nature, but a view in an inertial frame is the simplest and truest, in the sense that all forces have real origins and explanations.

Section Summary

• Rotating and accelerated frames of reference are non-inertial.
• Fictitious forces, such as the Coriolis force, are needed to explain motion in such frames.

Conceptual Questions

1: When a toilet is flushed or a sink is drained, the water (and other material) begins to rotate about the drain on the way down. Assuming no initial rotation and a flow initially directly straight toward the drain, explain what causes the rotation and which direction it has in the northern hemisphere. (Note that this is a small effect and in most toilets the rotation is caused by directional water jets.) Would the direction of rotation reverse if water were forced up the drain?

2: Is there a real force that throws water from clothes during the spin cycle of a washing machine? Explain how the water is removed.

3: In one amusement park ride, riders enter a large vertical barrel and stand against the wall on its horizontal floor. The barrel is spun up and the floor drops away. Riders feel as if they are pinned to the wall by a force something like the gravitational force. This is a fictitious force sensed and used by the riders to explain events in the rotating frame of reference of the barrel. Explain in an inertial frame of reference (Earth is nearly one) what pins the riders to the wall, and identify all of the real forces acting on them.

4: Action at a distance, such as is the case for gravity, was once thought to be illogical and therefore untrue. What is the ultimate determinant of the truth in physics, and why was this action ultimately accepted?

5: Two friends are having a conversation. Anna says a satellite in orbit is in freefall because the satellite keeps falling toward Earth. Tom says a satellite in orbit is not in freefall because the acceleration due to gravity is not 9.80 m/s². Who do you agree with and why?

6: A non-rotating frame of reference placed at the center of the Sun is very nearly an inertial one. Why is it not exactly an inertial frame?

Glossary

fictitious force

a force having no physical origin

centrifugal force

a fictitious force that tends to throw an object off when the object is rotating in a non-inertial frame of reference

Coriolis force

the fictitious force causing the apparent deflection of moving objects when viewed in a rotating frame of reference

non-inertial frame of reference

an accelerated frame of reference

]]> 378 346 5 0 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/chapter/6-5-newtons-universal-law-of-gravitation/ Thu, 29 Jun 2017 22:13:45 +0000 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/chapter/6-5-newtons-universal-law-of-gravitation/

Summary

• Explain Earth’s gravitational force.
• Describe the gravitational effect of the Moon on Earth.
• Discuss weightlessness in space.
• Examine the Cavendish experiment

What do aching feet, a falling apple, and the orbit of the Moon have in common? Each is caused by the gravitational force. Our feet are strained by supporting our weight—the force of Earth’s gravity on us. An apple falls from a tree because of the same force acting a few meters above Earth’s surface. And the Moon orbits Earth because gravity is able to supply the necessary centripetal force at a distance of hundreds of millions of meters. In fact, the same force causes planets to orbit the Sun, stars to orbit the center of the galaxy, and galaxies to cluster together. Gravity is another example of underlying simplicity in nature. It is the weakest of the four basic forces found in nature, and in some ways the least understood. It is a force that acts at a distance, without physical contact, and is expressed by a formula that is valid everywhere in the universe, for masses and distances that vary from the tiny to the immense.

Sir Isaac Newton was the first scientist to precisely define the gravitational force, and to show that it could explain both falling bodies and astronomical motions. See Figure 1. But Newton was not the first to suspect that the same force caused both our weight and the motion of planets. His forerunner Galileo Galilei had contended that falling bodies and planetary motions had the same cause. Some of Newton’s contemporaries, such as Robert Hooke, Christopher Wren, and Edmund Halley, had also made some progress toward understanding gravitation. But Newton was the first to propose an exact mathematical form and to use that form to show that the motion of heavenly bodies should be conic sections—circles, ellipses, parabolas, and hyperbolas. This theoretical prediction was a major triumph—it had been known for some time that moons, planets, and comets follow such paths, but no one had been able to propose a mechanism that caused them to follow these paths and not others.

The gravitational force is relatively simple. It is always attractive, and it depends only on the masses involved and the distance between them. Stated in modern language, Newton’s universal law of gravitation states that every particle in the universe attracts every other particle with a force along a line joining them. The force is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

MISCONCEPTION ALERT

The magnitude of the force on each object (one has larger mass than the other) is the same, consistent with Newton’s third law.

The bodies we are dealing with tend to be large. To simplify the situation we assume that the body acts as if its entire mass is concentrated at one specific point called the center of mass (CM), which will be further explored in Chapter 8 Linear Momentum and Collisions. For two bodies having masses m₁ and m₂ with a distance r between their centers of mass, the equation for Newton’s universal law of gravitation is

[latex]\boldsymbol{F=G}[/latex][latex size="2"]\boldsymbol{\frac{m_1 m_2}{r^2}},[/latex]

where F is the magnitude of the gravitational force and G is a proportionality factor called the gravitational constant. G is a universal gravitational constant—that is, it is thought to be the same everywhere in the universe. It has been measured experimentally to be

[latex]\boldsymbol{G=6.674\times10^{-11}}[/latex][latex size="2"]\boldsymbol{\frac{\textbf{N}\cdotp\textbf{m}^2}{\textbf{kg}^2}}[/latex]

in SI units. Note that the units of G are such that a force in newtons is obtained from [latex]\boldsymbol{F=G\frac{m_1 m_2}{r^2}},[/latex] when considering masses in kilograms and distance in meters. For example, two 1.000 kg masses separated by 1.000 m will experience a gravitational attraction of 6.674 × 10^-11 N. This is an extraordinarily small force. The small magnitude of the gravitational force is consistent with everyday experience. We are unaware that even large objects like mountains exert gravitational forces on us. In fact, our body weight is the force of attraction of the entire Earth on us with a mass of 6 × 10²⁴ kg.

Recall that the acceleration due to gravity g is about 9.80 m/s² on Earth. We can now determine why this is so. The weight of an object mg is the gravitational force between it and Earth. Substituting m₁g for F in Newton’s universal law of gravitation gives

[latex]\boldsymbol{mg=G}[/latex][latex size="2"]\boldsymbol{\frac{m_1 m_2}{r^2}},[/latex]

where m₁ is the mass of the object, m₂ is the mass of Earth, and r is the distance to the center of Earth (the distance between the centers of mass of the object and Earth). See Figure 3. The mass m₁ of the object cancels, leaving an equation for g:

[latex]\boldsymbol{g=G}[/latex][latex size="2"]\boldsymbol{\frac{m_2}{r^2}}.[/latex]

Substituting known values for Earth’s mass and radius (to three significant figures),

[latex]\boldsymbol{g\:=}[/latex][latex size="2"]\boldsymbol{(}[/latex][latex]\boldsymbol{6.67\times10^{-11}}[/latex][latex size="2"]\boldsymbol{\frac{\textbf{ N}\cdotp\textbf{m}^2}{\textbf{kg}^2}\:)}[/latex][latex]\boldsymbol{\times}[/latex][latex size="2"]\boldsymbol{\frac{5.98\times10^{24}\textbf{ kg}}{(6.38\times10^6\textbf{ m})^2}},[/latex]

and we obtain a value for the acceleration of a falling body:

[latex]\boldsymbol{g=9.80\textbf{ m/s}^2}.[/latex]

This is the expected value and is independent of the body’s mass. Newton’s law of gravitation takes Galileo’s observation that all masses fall with the same acceleration a step further, explaining the observation in terms of a force that causes objects to fall—in fact, in terms of a universally existing force of attraction between masses.

TAKE-HOME EXPERIMENT

Take a marble, a ball, and a spoon and drop them from the same height. Do they hit the floor at the same time? If you drop a piece of paper as well, does it behave like the other objects? Explain your observations.

MAKING CONNECTIONS

Attempts are still being made to understand the gravitational force. As we shall see in Chapter 33 Particle Physics, modern physics is exploring the connections of gravity to other forces, space, and time. General relativity alters our view of gravitation, leading us to think of gravitation as bending space and time.

In the following example, we make a comparison similar to one made by Newton himself. He noted that if the gravitational force caused the Moon to orbit Earth, then the acceleration due to gravity should equal the centripetal acceleration of the Moon in its orbit. Newton found that the two accelerations agreed “pretty nearly.”

Example 1: Earth's Gravitational Force Is the Centripetal Force Making the Moon Move in a Curved Path

(a) Find the acceleration due to Earth’s gravity at the distance of the Moon.

(b) Calculate the centripetal acceleration needed to keep the Moon in its orbit (assuming a circular orbit about a fixed Earth), and compare it with the value of the acceleration due to Earth’s gravity that you have just found.

Strategy for (a)

This calculation is the same as the one finding the acceleration due to gravity at Earth’s surface, except that r is the distance from the center of Earth to the center of the Moon. The radius of the Moon’s nearly circular orbit is 3.84 × 10⁸ m.

Solution for (a)

Substituting known values into the expression for g found above, remembering that m₂ is the mass of Earth not the Moon, yields

[latex]\boldsymbol{g=G\frac{m_2}{r^2}=(6.67\times10^{-11}\frac{\textbf{N}\cdotp\textbf{m}^2}{kg^2})\times\frac{5.98\times10^{24}\textbf{ kg}}{(3.84\times10^8\textbf{ m})^2}}[/latex]

[latex]\boldsymbol{=2.70\times10^{-3}\textbf{ m/s}^2}.[/latex]

Strategy for (b)

Centripetal acceleration can be calculated using either form of

[latex]\begin{array}{lcl} \boldsymbol{a_{\textbf{c}}} & = & \boldsymbol{\frac{v^2}{r}} \\ \boldsymbol{a_{\textbf{c}}} & = & \boldsymbol{r\omega^2} \end{array}[/latex][latex size="4"]\rbrace[/latex]

We choose to use the second form:

[latex]\boldsymbol{a_{\textbf{c}}=r\omega^2},[/latex]

where[latex]\boldsymbol{\omega}[/latex]is the angular velocity of the Moon about Earth.

Solution for (b)

Given that the period (the time it takes to make one complete rotation) of the Moon’s orbit is 27.3 days, (d) and using

[latex]\boldsymbol{1\textbf{ d}\times24}[/latex][latex size="2"]\boldsymbol{\frac{\textbf{hr}}{\textbf{d}}}[/latex][latex]\boldsymbol{\times\:60}[/latex][latex size="2"]\boldsymbol{\frac{\textbf{min}}{\textbf{hr}}}[/latex][latex]\boldsymbol{\times\:60}[/latex][latex size="2"]\boldsymbol{\frac{\textbf{s}}{\textbf{min}}}[/latex][latex]\boldsymbol{=86,400\textbf{ s}}[/latex]

we see that

[latex]\boldsymbol{\omega\:=}[/latex][latex size="2"]\boldsymbol{\frac{\Delta\theta}{\Delta{t}}}[/latex][latex]\boldsymbol{\:=}[/latex][latex size="2"]\boldsymbol{\frac{2\pi\textbf{ rad}}{(27.3\textbf{ d})(86,400\textbf{ s/d})}}[/latex][latex]\boldsymbol{=2.66\times10^{-6}}[/latex][latex size="2"]\boldsymbol{\frac{\textbf{rad}}{\textbf{s}}}.[/latex]

The centripetal acceleration is

[latex]\begin{array}{lcl} \boldsymbol{a_{\textbf{c}}} & = & \boldsymbol{r\omega^2=(3.84\times10^8\textbf{ m})(2.66\times10^{-6}\textbf{ rad/s})^2} \\ {} & = & \boldsymbol{2.72\times10^{-3}\textbf{ m/s}^2} \end{array}[/latex]

The direction of the acceleration is toward the center of the Earth.

Discussion

The centripetal acceleration of the Moon found in (b) differs by less than 1% from the acceleration due to Earth’s gravity found in (a). This agreement is approximate because the Moon’s orbit is slightly elliptical, and Earth is not stationary (rather the Earth-Moon system rotates about its center of mass, which is located some 1700 km below Earth’s surface). The clear implication is that Earth’s gravitational force causes the Moon to orbit Earth.

Why does Earth not remain stationary as the Moon orbits it? This is because, as expected from Newton’s third law, if Earth exerts a force on the Moon, then the Moon should exert an equal and opposite force on Earth (see Figure 4). We do not sense the Moon’s effect on Earth’s motion, because the Moon’s gravity moves our bodies right along with Earth but there are other signs on Earth that clearly show the effect of the Moon’s gravitational force as discussed in Chapter 6.6 Satellites and Kepler's Laws: An Argument for Simplicity.

Tides

Ocean tides are one very observable result of the Moon’s gravity acting on Earth. Figure 5 is a simplified drawing of the Moon’s position relative to the tides. Because water easily flows on Earth’s surface, a high tide is created on the side of Earth nearest to the Moon, where the Moon’s gravitational pull is strongest. Why is there also a high tide on the opposite side of Earth? The answer is that Earth is pulled toward the Moon more than the water on the far side, because Earth is closer to the Moon. So the water on the side of Earth closest to the Moon is pulled away from Earth, and Earth is pulled away from water on the far side. As Earth rotates, the tidal bulge (an effect of the tidal forces between an orbiting natural satellite and the primary planet that it orbits) keeps its orientation with the Moon. Thus there are two tides per day (the actual tidal period is about 12 hours and 25.2 minutes), because the Moon moves in its orbit each day as well).

The Sun also affects tides, although it has about half the effect of the Moon. However, the largest tides, called spring tides, occur when Earth, the Moon, and the Sun are aligned. The smallest tides, called neap tides, occur when the Sun is at a 90° angle to the Earth-Moon alignment.

Tides are not unique to Earth but occur in many astronomical systems. The most extreme tides occur where the gravitational force is the strongest and varies most rapidly, such as near black holes (see Figure 7). A few likely candidates for black holes have been observed in our galaxy. These have masses greater than the Sun but have diameters only a few kilometers across. The tidal forces near them are so great that they can actually tear matter from a companion star.

”Weightlessness” and Microgravity

In contrast to the tremendous gravitational force near black holes is the apparent gravitational field experienced by astronauts orbiting Earth. What is the effect of “weightlessness” upon an astronaut who is in orbit for months? Or what about the effect of weightlessness upon plant growth? Weightlessness doesn’t mean that an astronaut is not being acted upon by the gravitational force. There is no “zero gravity” in an astronaut’s orbit. The term just means that the astronaut is in free-fall, accelerating with the acceleration due to gravity. If an elevator cable breaks, the passengers inside will be in free fall and will experience weightlessness. You can experience short periods of weightlessness in some rides in amusement parks.

Microgravity refers to an environment in which the apparent net acceleration of a body is small compared with that produced by Earth at its surface. Many interesting biology and physics topics have been studied over the past three decades in the presence of microgravity. Of immediate concern is the effect on astronauts of extended times in outer space, such as at the International Space Station. Researchers have observed that muscles will atrophy (waste away) in this environment. There is also a corresponding loss of bone mass. Study continues on cardiovascular adaptation to space flight. On Earth, blood pressure is usually higher in the feet than in the head, because the higher column of blood exerts a downward force on it, due to gravity. When standing, 70% of your blood is below the level of the heart, while in a horizontal position, just the opposite occurs. What difference does the absence of this pressure differential have upon the heart?

Some findings in human physiology in space can be clinically important to the management of diseases back on Earth. On a somewhat negative note, spaceflight is known to affect the human immune system, possibly making the crew members more vulnerable to infectious diseases. Experiments flown in space also have shown that some bacteria grow faster in microgravity than they do on Earth. However, on a positive note, studies indicate that microbial antibiotic production can increase by a factor of two in space-grown cultures. One hopes to be able to understand these mechanisms so that similar successes can be achieved on the ground. In another area of physics space research, inorganic crystals and protein crystals have been grown in outer space that have much higher quality than any grown on Earth, so crystallography studies on their structure can yield much better results.

Plants have evolved with the stimulus of gravity and with gravity sensors. Roots grow downward and shoots grow upward. Plants might be able to provide a life support system for long duration space missions by regenerating the atmosphere, purifying water, and producing food. Some studies have indicated that plant growth and development are not affected by gravity, but there is still uncertainty about structural changes in plants grown in a microgravity environment.

The Cavendish Experiment: Then and Now

As previously noted, the universal gravitational constant G is determined experimentally. This definition was first done accurately by Henry Cavendish (1731–1810), an English scientist, in 1798, more than 100 years after Newton published his universal law of gravitation. The measurement of G is very basic and important because it determines the strength of one of the four forces in nature. Cavendish’s experiment was very difficult because he measured the tiny gravitational attraction between two ordinary-sized masses (tens of kilograms at most), using apparatus like that in Figure 9. Remarkably, his value for G differs by less than 1% from the best modern value.

One important consequence of knowing G was that an accurate value for Earth’s mass could finally be obtained. This was done by measuring the acceleration due to gravity as accurately as possible and then calculating the mass of Earth m₂ from the relationship Newton’s universal law of gravitation gives

[latex]\boldsymbol{m_1g=G}[/latex][latex size="2"]\boldsymbol{\frac{m_1 m_2}{r^2}},[/latex]

where m₁ is the mass of the object, m₂ is the mass of Earth, and r is the distance to the center of Earth (the distance between the centers of mass of the object and Earth). See Figure 2. The mass m₁ of the object cancels, leaving an equation for g:

[latex]\boldsymbol{g=G}[/latex][latex size="2"]\boldsymbol{\frac{m_2}{r^2}}.[/latex]

Rearranging to solve for m₂ yields

[latex]\boldsymbol{m_2\:=}[/latex][latex size="2"]\boldsymbol{\frac{gr^2}{G}}.[/latex]

So m₂ can be calculated because all quantities on the right, including the radius of Earth r, are known from direct measurements. We shall see in Chapter 6.6 Satellites and Kepler's Laws: An Argument for Simplicity that knowing G also allows for the determination of astronomical masses. Interestingly, of all the fundamental constants in physics, G is by far the least well determined.

The Cavendish experiment is also used to explore other aspects of gravity. One of the most interesting questions is whether the gravitational force depends on substance as well as mass—for example, whether one kilogram of lead exerts the same gravitational pull as one kilogram of water. A Hungarian scientist named Roland von Eötvös pioneered this inquiry early in the 20th century. He found, with an accuracy of five parts per billion, that the gravitational force does not depend on the substance. Such experiments continue today, and have improved upon Eötvös’ measurements. Cavendish-type experiments such as those of Eric Adelberger and others at the University of Washington, have also put severe limits on the possibility of a fifth force and have verified a major prediction of general relativity—that gravitational energy contributes to rest mass. Ongoing measurements there use a torsion balance and a parallel plate (not spheres, as Cavendish used) to examine how Newton’s law of gravitation works over sub-millimeter distances. On this small-scale, do gravitational effects depart from the inverse square law? So far, no deviation has been observed.

[caption id="" align="aligncenter" width="250"]Figure 9. Cavendish used an apparatus like this to measure the gravitational attraction between the two suspended spheres (m) and the two on the stand (M) by observing the amount of torsion (twisting) created in the fiber. Distance between the masses can be varied to check the dependence of the force on distance. Modern experiments of this type continue to explore gravity.[/caption]

Section Summary

• Newton’s universal law of gravitation: Every particle in the universe attracts every other particle with a force along a line joining them. The force is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. In equation form, this is
  [latex]\boldsymbol{F=G}[/latex][latex size="2"]\boldsymbol{\frac{m_1 m_2}{r^2}},[/latex]

  where F is the magnitude of the gravitational force. G is the gravitational constant, given by

  [latex]\boldsymbol{G=6.674\times10^{-11}\textbf{ N}\cdotp\textbf{m}^2\textbf{/kg}^2}[/latex]
• Newton’s law of gravitation applies universally.

Conceptual Questions

1: Action at a distance, such as is the case for gravity, was once thought to be illogical and therefore untrue. What is the ultimate determinant of the truth in physics, and why was this action ultimately accepted?

2: Two friends are having a conversation. Anna says a satellite in orbit is in freefall because the satellite keeps falling toward Earth. Tom says a satellite in orbit is not in freefall because the acceleration due to gravity is not 9.80 m/s². Who do you agree with and why?

3: Draw a free body diagram for a satellite in an elliptical orbit showing why its speed increases as it approaches its parent body and decreases as it moves away.

4: Newton’s laws of motion and gravity were among the first to convincingly demonstrate the underlying simplicity and unity in nature. Many other examples have since been discovered, and we now expect to find such underlying order in complex situations. Is there proof that such order will always be found in new explorations?

Problems & Exercises

1: (a) Calculate Earth’s mass given the acceleration due to gravity at the North Pole is 9.830 m/s² and the radius of the Earth is 6371 km from center to pole.

(b) Compare this with the accepted value of 5.979 × 10²⁴ kg.

2: (a) Calculate the magnitude of the acceleration due to gravity on the surface of Earth due to the Moon.

(b) Calculate the magnitude of the acceleration due to gravity at Earth due to the Sun.

(c) Take the ratio of the Moon’s acceleration to the Sun’s and comment on why the tides are predominantly due to the Moon in spite of this number.

3: (a) What is the acceleration due to gravity on the surface of the Moon?

(b) On the surface of Mars? The mass of Mars is 6.418 × 10²³ kg and its radius is 3.38 × 10⁶ m.

4: (a) Calculate the acceleration due to gravity on the surface of the Sun.

(b) By what factor would your weight increase if you could stand on the Sun? (Never mind that you cannot.)

5: The Moon and Earth rotate about their common center of mass, which is located about 4700 km from the center of Earth. (This is 1690 km below the surface.)

(a) Calculate the magnitude of the acceleration due to the Moon’s gravity at that point.

(b) Calculate the magnitude of the centripetal acceleration of the center of Earth as it rotates about that point once each lunar month (about 27.3 d) and compare it with the acceleration found in part (a). Comment on whether or not they are equal and why they should or should not be.

6: Solve part (b) of Example 1 using a_c = v²/r.

7: Astrology, that unlikely and vague pseudoscience, makes much of the position of the planets at the moment of one’s birth. The only known force a planet exerts on Earth is gravitational.

(a) Calculate the magnitude of the gravitational force exerted on a 4.20 kg baby by a 100 kg father 0.200 m away at birth (he is assisting, so he is close to the child).

(b) Calculate the magnitude of the force on the baby due to Jupiter if it is at its closest distance to Earth, some 6.29 × 10¹¹ m away. How does the force of Jupiter on the baby compare to the force of the father on the baby? Other objects in the room and the hospital building also exert similar gravitational forces. (Of course, there could be an unknown force acting, but scientists first need to be convinced that there is even an effect, much less that an unknown force causes it.)

8: The existence of the dwarf planet Pluto was proposed based on irregularities in Neptune’s orbit. Pluto was subsequently discovered near its predicted position. But it now appears that the discovery was fortuitous, because Pluto is small and the irregularities in Neptune’s orbit were not well known. To illustrate that Pluto has a minor effect on the orbit of Neptune compared with the closest planet to Neptune:

(a) Calculate the acceleration due to gravity at Neptune due to Pluto when they are 4.50 × 10¹² m apart, as they are at present. The mass of Pluto is 1.4 × 10²² kg.

(b) Calculate the acceleration due to gravity at Neptune due to Uranus, presently about 2.50 × 10¹² m apart, and compare it with that due to Pluto. The mass of Uranus is 8.62 × 10²⁵ kg.

9: (a) The Sun orbits the Milky Way galaxy once each 2.60 × 10⁸ y, with a roughly circular orbit averaging 3.00 × 10⁴ light years in radius. (A light year is the distance traveled by light in 1 y.) Calculate the centripetal acceleration of the Sun in its galactic orbit. Does your result support the contention that a nearly inertial frame of reference can be located at the Sun?

(b) Calculate the average speed of the Sun in its galactic orbit. Does the answer surprise you?

10: Unreasonable Result

A mountain 10.0 km from a person exerts a gravitational force on him equal to 2.00% of his weight.

(a) Calculate the mass of the mountain.

(b) Compare the mountain’s mass with that of Earth.

(c) What is unreasonable about these results?

(d) Which premises are unreasonable or inconsistent? (Note that accurate gravitational measurements can easily detect the effect of nearby mountains and variations in local geology.)

Glossary

gravitational constant, G

a proportionality factor used in the equation for Newton’s universal law of gravitation; it is a universal constant—that is, it is thought to be the same everywhere in the universe

center of mass

the point where the entire mass of an object can be thought to be concentrated

microgravity

an environment in which the apparent net acceleration of a body is small compared with that produced by Earth at its surface

Newton’s universal law of gravitation

every particle in the universe attracts every other particle with a force along a line joining them; the force is directly proportional to the product of their masses and inversely proportional to the square of the distance between them

Solutions

Problems & Exercises

1: (a) [latex]\boldsymbol{5.979\times10^{24}\textbf{ kg}}[/latex] (b) This is identical to the best value to three significant figures.

3: (a) [latex]\boldsymbol{1.62\textbf{ m/s}^2}[/latex] (b) [latex]\boldsymbol{3.75\textbf{ m/s}^2}[/latex]

5: (a) [latex]\boldsymbol{3.42\times10^{-5}\textbf{ m/s}^2}[/latex] (b) [latex]\boldsymbol{3.34\times10^{-5}\textbf{ m/s}^2}[/latex] The values are nearly identical. One would expect the gravitational force to be the same as the centripetal force at the core of the system.

7: (a) [latex]\boldsymbol{7.01\times10^{-7}\textbf{ N}}[/latex] (b) [latex]\boldsymbol{1.35\times10^{-6}\textbf{ N},\:0.521}[/latex]

9: (a) [latex]\boldsymbol{1.66\times10^{-10}\textbf{ m/s}^2}[/latex] (b) [latex]\boldsymbol{2.17\times10^5\textbf{ m/s}}[/latex]

10: (a) [latex]\boldsymbol{2.937\times10^{17}\textbf{ kg}}[/latex] (b) [latex]\boldsymbol{4.91\times10^{-8}}[/latex] of the Earth’s mass. (c) The mass of the mountain and its fraction of the Earth’s mass are too great. (d) The gravitational force assumed to be exerted by the mountain is too great.

]]> 388 346 6 0 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/chapter/6-6-satellites-and-keplers-laws-an-argument-for-simplicity/ Thu, 29 Jun 2017 22:13:46 +0000 https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/chapter/6-6-satellites-and-keplers-laws-an-argument-for-simplicity/

Summary

• State Kepler’s laws of planetary motion.
• Derive the third Kepler’s law for circular orbits.
• Discuss the Ptolemaic model of the universe.
• Define a planet according to the I.A.U. the International Astronomical Union
• Explain why Pluto is not a planet

Examples of gravitational orbits abound. Hundreds of artificial satellites orbit Earth together with thousands of pieces of debris. The Moon’s orbit about Earth has intrigued humans from time immemorial. The orbits of planets, asteroids, meteors, and comets about the Sun are no less interesting. If we look further, we see almost unimaginable numbers of stars, galaxies, and other celestial objects orbiting one another and interacting through gravity.

All these motions are governed by gravitational force, and it is possible to describe them to various degrees of precision. Precise descriptions of complex systems must be made with large computers. However, we can describe an important class of orbits without the use of computers, and we shall find it instructive to study them. These orbits have the following characteristics:

1. A small mass m₁ orbits a much larger mass m₂.This allows us to view the motion as if m₂ were stationary—in fact, as if from an inertial frame of reference placed on m₂—without significant error. Mass m₁ is the satellite of m₂, if the orbit is gravitationally bound.
2. The system is isolated from other masses. This allows us to neglect any small effects due to outside masses.

The conditions are satisfied, to good approximation, by Earth’s satellites (including the Moon), by objects orbiting the Sun, and by the satellites of other planets. Historically, planets were studied first, and there is a classical set of three laws, called Kepler’s laws of planetary motion, that describe the orbits of all bodies satisfying the two previous conditions (not just planets in our solar system). These descriptive laws are named for the German astronomer Johannes Kepler (1571–1630), who devised them after careful study (over some 20 years) of a large amount of meticulously recorded observations of planetary motion done by Tycho Brahe (1546–1601). Such careful collection and detailed recording of methods and data are hallmarks of good science. Data constitute the evidence from which new interpretations and meanings can be constructed.

Kepler’s Laws of Planetary Motion

Kepler’s First Law

The orbit of each planet about the Sun is an ellipse with the Sun at one focus.

Kepler’s Second Law

Each planet moves so that an imaginary line drawn from the Sun to the planet sweeps out equal areas in equal times (see Figure 2).

Kepler’s Third Law

The ratio of the squares of the periods of any two planets about the Sun is equal to the ratio of the cubes of their average distances from the Sun. In equation form, this is

[latex size="2"]\boldsymbol{\frac{T_1^2}{T_2^2}}[/latex][latex]\boldsymbol{=}[/latex][latex size="2"]\boldsymbol{\frac{r_1^3}{r_2^3}},[/latex]

where T is the period (time for one orbit) and r is the average radius. This equation is valid only for comparing two small masses orbiting the same large one. Most importantly, this is a descriptive equation only, giving no information as to the cause of the equality.

Note again that while, for historical reasons, Kepler’s laws are stated for planets orbiting the Sun, they are actually valid for all bodies satisfying the two previously stated conditions.

Example 1: Find the Time for One Orbit of an Earth Satellite

Given that the Moon orbits Earth each 27.3 d and that it is an average distance of 3.84 × 10⁸ m from the center of Earth, calculate the period of an artificial satellite orbiting at an average altitude of 1500 km above Earth’s surface.

Strategy

The period, or time for one orbit, is related to the radius of the orbit by Kepler’s third law, given in mathematical form in [latex]\boldsymbol{\frac{T_1^2}{T_2^2}=\frac{r_1^3}{r_2^3}}.[/latex] Let us use the subscript 1 for the Moon and the subscript 2 for the satellite. We are asked to find T₂. The given information tells us that the orbital radius of the Moon is r₁ = 3.84 × 10⁸ m, and that the period of the Moon is T₁ = 27.3 d. The height of the artificial satellite above Earth’s surface is given, and so we must add the radius of Earth (6380 km) to get r₂ = (1500+6380) km = 7880 km. Now all quantities are known, and so T₂ can be found.

Solution

Kepler’s third law is

[latex size="2"]\boldsymbol{\frac{T_1^2}{T_2^2}}[/latex][latex]\boldsymbol{=}[/latex][latex size="2"]\boldsymbol{\frac{r_1^3}{r_2^3}}.[/latex]

To solve for T₂, we cross-multiply and take the square root, yielding

[latex]\boldsymbol{T_2^2=T_1^2}[/latex][latex size="2"]\boldsymbol{(\frac{r_2}{r_1})^3}[/latex]

[latex]\boldsymbol{T_2=T_1}[/latex][latex size="2"]\boldsymbol{(\frac{r_2}{r_1})^{3/2}}.[/latex]

Substituting known values yields

[latex]\begin{array}{lcl} \boldsymbol{T_2} & = & \boldsymbol{27.3\textbf{ d}\times\frac{24.0\textbf{ h}}{\textbf{d}}\times(\frac{7880\textbf{ km}}{3.84\times10^5\textbf{ km}})^{3/2}} \\ {} & = & \boldsymbol{1.93\textbf{ h}} \end{array}[/latex]

Discussion This is a reasonable period for a satellite in a fairly low orbit. It is interesting that any satellite at this altitude will orbit in the same amount of time. This fact is related to the condition that the satellite’s mass is small compared with that of Earth.

People immediately search for deeper meaning when broadly applicable laws, like Kepler’s, are discovered. It was Newton who took the next giant step when he proposed the law of universal gravitation. While Kepler was able to discover what was happening, Newton discovered that gravitational force was the cause.

Derivation of Kepler’s Third Law for Circular Orbits

We shall derive Kepler’s third law, starting with Newton’s laws of motion and his universal law of gravitation. The point is to demonstrate that the force of gravity is the cause for Kepler’s laws (although we will only derive the third one).

Let us consider a circular orbit of a small mass m₁ around a large mass m₂, satisfying the two conditions stated at the beginning of this section. Gravity supplies the centripetal force to mass m₁. Starting with Newton’s second law applied to circular motion,

[latex]\boldsymbol{F_{\textbf{net}}=m_1a_{\textbf{c}}=m_1}[/latex][latex size="2"]\boldsymbol{\frac{v^2}{r}}.[/latex]

The net external force on mass m₁ is gravity, and so we substitute the force of gravity for F_net:

[latex]\boldsymbol{G}[/latex][latex size="2"]\boldsymbol{\frac{m_1 m_2}{r^2}}[/latex][latex]\boldsymbol{=m_1}[/latex][latex size="2"]\boldsymbol{\frac{v^2}{r}}.[/latex]

The mass m₁ cancels, yielding

[latex]\boldsymbol{G}[/latex][latex size="2"]\boldsymbol{\frac{m_2}{r}}[/latex][latex]\boldsymbol{=\:v^2}.[/latex]

The fact that m₁ cancels out is another aspect of the oft-noted fact that at a given location all masses fall with the same acceleration. Here we see that at a given orbital radius r, all masses orbit at the same speed. (This was implied by the result of the preceding worked example.) Now, to get at Kepler’s third law, we must get the period T into the equation. By definition, period T is the time for one complete orbit. Now the average speed v is the circumference divided by the period—that is,

[latex]\boldsymbol{v\:=}[/latex][latex size="2"]\boldsymbol{\frac{2\pi{r}}{T}}.[/latex]

Substituting this into the previous equation gives

[latex]\boldsymbol{G}[/latex][latex size="2"]\boldsymbol{\frac{m_2}{r}}[/latex][latex]\boldsymbol{=}[/latex][latex size="2"]\boldsymbol{\frac{4\pi^2r^2}{T^2}}.[/latex]

Solving for T²yields

[latex]\boldsymbol{T^2\:=}[/latex][latex size="2"]\boldsymbol{\frac{4\pi^2}{G m_2}}[/latex][latex]\boldsymbol{r^3}.[/latex]

Using subscripts 1 and 2 to denote two different satellites, and taking the ratio of the last equation for satellite 1 to satellite 2 yields

[latex size="2"]\boldsymbol{\frac{T_1^2}{T_2^2}}[/latex][latex]\boldsymbol{=}[/latex][latex size="2"]\boldsymbol{\frac{r_1^3}{r_2^3}}.[/latex]

This is Kepler’s third law. Note that Kepler’s third law is valid only for comparing satellites of the same parent body, because only then does the mass of the parent body m₂ cancel.

Now consider what we get if we solve [latex]\boldsymbol{T^2=\frac{4\pi^2}{Gm_2}r^3}.[/latex] We obtain a relationship that can be used to determine the mass m₂ of a parent body from the orbits of its satellites:

[latex size="2"]\boldsymbol{\frac{r^3}{T^2}}[/latex][latex]\boldsymbol{=}[/latex][latex size="2"]\boldsymbol{\frac{G}{4\pi^2}}[/latex][latex]\boldsymbol{m_2}.[/latex]

If r and T are known for a satellite, then the mass m₂ of the parent can be calculated. This principle has been used extensively to find the masses of heavenly bodies that have satellites. Furthermore, the ratio r³/T² should be a constant for all satellites of the same parent body (because r³/T² = Gm₂/4π²). (See Table 2).

It is clear from Table 2 that the ratio of r³/T² is constant, at least to the third digit, for all listed satellites of the Sun, and for those of Jupiter. Small variations in that ratio have two causes—uncertainties in the r  and T data, and perturbations of the orbits due to other bodies. Interestingly, those perturbations can be—and have been—used to predict the location of new planets and moons. This is another verification of Newton’s universal law of gravitation.

MAKING CONNECTIONS

Newton’s universal law of gravitation is modified by Einstein’s general theory of relativity, as we shall see in Chapter 33 Particle Physics. Newton’s gravity is not seriously in error—it was and still is an extremely good approximation for most situations. Einstein’s modification is most noticeable in extremely large gravitational fields, such as near black holes. However, general relativity also explains such phenomena as small but long-known deviations of the orbit of the planet Mercury from classical predictions.

The Case for Simplicity

The development of the universal law of gravitation by Newton played a pivotal role in the history of ideas. While it is beyond the scope of this text to cover that history in any detail, we note some important points. The definition of planet set in 2006 by the International Astronomical Union (IAU) states that in the solar system, a planet is a celestial body that:

1. is in orbit around the Sun,
2. has sufficient mass to assume hydrostatic equilibrium and
3. has cleared the neighborhood around its orbit.

A non-satellite body fulfilling only the first two of the above criteria is classified as “dwarf planet.”

In 2006, Pluto was demoted to a ‘dwarf planet’ after scientists revised their definition of what constitutes a “true” planet.

Parent	Satellite	Average orbital radius r(km)	Period T(y)	r3 / T2 (km3 / y2)
Earth	Moon	[latex]\boldsymbol{3.84\times10^5}[/latex]	0.07481	[latex]\boldsymbol{1.01\times10^{19}}[/latex]
Sun	Mercury	[latex]\boldsymbol{5.79\times10^7}[/latex]	0.2409	[latex]\boldsymbol{3.34\times10^{24}}[/latex]
	Venus	[latex]\boldsymbol{1.082\times10^8}[/latex]	0.6150	[latex]\boldsymbol{3.35\times10^{24}}[/latex]
	Earth	[latex]\boldsymbol{1.496\times10^8}[/latex]	1.000	[latex]\boldsymbol{3.35\times10^{24}}[/latex]
	Mars	[latex]\boldsymbol{2.279\times10^8}[/latex]	1.881	[latex]\boldsymbol{3.35\times10^{24}}[/latex]
	Jupiter	[latex]\boldsymbol{7.783\times10^8}[/latex]	11.86	[latex]\boldsymbol{3.35\times10^{24}}[/latex]
	Saturn	[latex]\boldsymbol{1.427\times10^9}[/latex]	29.46	[latex]\boldsymbol{3.35\times10^{24}}[/latex]
	Neptune	[latex]\boldsymbol{4.497\times10^9}[/latex]	164.8	[latex]\boldsymbol{3.35\times10^{24}}[/latex]
	Pluto	[latex]\boldsymbol{5.90\times10^9}[/latex]	248.3	[latex]\boldsymbol{3.33\times10^{24}}[/latex]
Jupiter	Io	[latex]\boldsymbol{4.22\times10^5}[/latex]	0.00485 (1.77 d)	[latex]\boldsymbol{3.19\times10^{21}}[/latex]
	Europa	[latex]\boldsymbol{6.71\times10^5}[/latex]	0.00972 (3.55 d)	[latex]\boldsymbol{3.20\times10^{21}}[/latex]
	Ganymede	[latex]\boldsymbol{1.07\times10^6}[/latex]	0.0196 (7.16 d)	[latex]\boldsymbol{3.19\times10^{21}}[/latex]
	Callisto	[latex]\boldsymbol{1.88\times10^6}[/latex]	0.0457 (16.19 d)	[latex]\boldsymbol{3.20\times10^{21}}[/latex]
Table 2. Orbital Data and Kepler’s Third Law.

The universal law of gravitation is a good example of a physical principle that is very broadly applicable. That single equation for the gravitational force describes all situations in which gravity acts. It gives a cause for a vast number of effects, such as the orbits of the planets and moons in the solar system. It epitomizes the underlying unity and simplicity of physics.

Before the discoveries of Kepler, Copernicus, Galileo, Newton, and others, the solar system was thought to revolve around Earth as shown in Figure 3(a). This is called the Ptolemaic view, for the Greek philosopher who lived in the second century AD. This model is characterized by a list of facts for the motions of planets with no cause and effect explanation. There tended to be a different rule for each heavenly body and a general lack of simplicity.

Figure 3(b) represents the modern or Copernican model. In this model, a small set of rules and a single underlying force explain not only all motions in the solar system, but all other situations involving gravity. The breadth and simplicity of the laws of physics are compelling. As our knowledge of nature has grown, the basic simplicity of its laws has become ever more evident.

Section Summary

• Kepler’s laws are stated for a small mass m₁ orbiting a larger mass m₂ in near-isolation. Kepler’s laws of planetary motion are then as follows:

  Kepler’s first law

  The orbit of each planet about the Sun is an ellipse with the Sun at one focus.

  Kepler’s second law

  Each planet moves so that an imaginary line drawn from the Sun to the planet sweeps out equal areas in equal times.

  Kepler’s third law

  The ratio of the squares of the periods of any two planets about the Sun is equal to the ratio of the cubes of their average distances from the Sun:

  [latex size="2"]\boldsymbol{\frac{T_1^2}{T_2^2}}[/latex][latex]\boldsymbol{=}[/latex][latex size="2"]\boldsymbol{\frac{r_1^3}{r_2^3}},[/latex]

  where T is the period (time for one orbit) and r is the average radius of the orbit.

• The period and radius of a satellite’s orbit about a larger body m₂ are related by
  [latex]\boldsymbol{T_2\:=}[/latex][latex size="2"]\boldsymbol{\frac{4\pi^2}{Gm_2}}[/latex][latex]\boldsymbol{r^3}[/latex]

  or

  [latex size="2"]\boldsymbol{\frac{r^3}{T^2}}[/latex][latex]\boldsymbol{=}[/latex][latex size="2"]\boldsymbol{\frac{G}{4\pi^2}}[/latex][latex]\boldsymbol{m_2}.[/latex]

Conceptual Questions

1: In what frame(s) of reference are Kepler’s laws valid? Are Kepler’s laws purely descriptive, or do they contain causal information?

Problems & Exercises

1: A geosynchronous Earth satellite is one that has an orbital period of precisely 1 day. Such orbits are useful for communication and weather observation because the satellite remains above the same point on Earth (provided it orbits in the equatorial plane in the same direction as Earth’s rotation). Calculate the radius of such an orbit based on the data for the moon in Table 2.

2: Calculate the mass of the Sun based on data for Earth’s orbit and compare the value obtained with the Sun’s actual mass.

3: Find the mass of Jupiter based on data for the orbit of one of its moons, and compare your result with its actual mass.

4: Find the ratio of the mass of Jupiter to that of Earth based on data in Table 2.

5: Astronomical observations of our Milky Way galaxy indicate that it has a mass of about 8.0 × 10¹¹ solar masses. A star orbiting on the galaxy’s periphery is about 6.0 × 10⁴ light years from its center. (a) What should the orbital period of that star be? (b) If its period is 6.0 × 10⁷ instead, what is the mass of the galaxy? Such calculations are used to imply the existence of “dark matter” in the universe and have indicated, for example, the existence of very massive black holes at the centers of some galaxies.

6: Integrated Concepts

Space debris left from old satellites and their launchers is becoming a hazard to other satellites. (a) Calculate the speed of a satellite in an orbit 900 km above Earth’s surface. (b) Suppose a loose rivet is in an orbit of the same radius that intersects the satellite’s orbit at an angle of 90° relative to Earth. What is the velocity of the rivet relative to the satellite just before striking it? (c) Given the rivet is 3.00 mm in size, how long will its collision with the satellite last? (d) If its mass is 0.500 g, what is the average force it exerts on the satellite? (e) How much energy in joules is generated by the collision? (The satellite’s velocity does not change appreciably, because its mass is much greater than the rivet’s.)

7: Unreasonable Results

(a) Based on Kepler’s laws and information on the orbital characteristics of the Moon, calculate the orbital radius for an Earth satellite having a period of 1.00 h. (b) What is unreasonable about this result? (c) What is unreasonable or inconsistent about the premise of a 1.00 h orbit?

8: Construct Your Own Problem

On February 14, 2000, the NEAR spacecraft was successfully inserted into orbit around Eros, becoming the first artificial satellite of an asteroid. Construct a problem in which you determine the orbital speed for a satellite near Eros. You will need to find the mass of the asteroid and consider such things as a safe distance for the orbit. Although Eros is not spherical, calculate the acceleration due to gravity on its surface at a point an average distance from its center of mass. Your instructor may also wish to have you calculate the escape velocity from this point on Eros.

Solutions

Problems & Exercises 2: [latex]\boldsymbol{1.98\times10^{30}\textbf{ kg}}[/latex] 4: [latex]\boldsymbol{\frac{M_{\textbf{J}}}{M_{\textbf{E}}}=316}[/latex]

6: (a) [latex]\boldsymbol{7.4\times10^3\textbf{ m/s}}[/latex] (b) [latex]\boldsymbol{1.05\times10^3\textbf{ m/s}}[/latex] (c) [latex]\boldsymbol{2.86\times10^{-7}\textbf{ s}}[/latex] (d) [latex]\boldsymbol{1.84\times10^7\textbf{ N}}[/latex] (e) [latex]\boldsymbol{2.76\times10^4\textbf{ J}}[/latex]

7: (a) [latex]\boldsymbol{5.08\times10^3\textbf{ km}}[/latex] (b) This radius is unreasonable because it is less than the radius of earth. (c) The premise of a one-hour orbit is inconsistent with the known radius of the earth.

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