{"id":88,"date":"2020-12-14T20:22:07","date_gmt":"2020-12-15T01:22:07","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/bcengrphys3\/?post_type=chapter&p=88"},"modified":"2021-01-02T13:21:58","modified_gmt":"2021-01-02T18:21:58","slug":"examples","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/bcengrphys3\/chapter\/examples\/","title":{"raw":"Examples","rendered":"Examples"},"content":{"raw":" \r\n
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Example 1.1 Unit Conversion<\/p>\r\n\r\n<\/header>\r\n

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Now we can set up our unit conversion. We will write the units that we have and then multiply them by the conversion factor so that the units cancel out, as shown:<\/p>\r\n\r\n

[latex]\\bf{80 \\;\\rule[0.5ex]{1.0em}{0.1ex}\\hspace{-1.0em}\\textbf{m}\\times} \\bf{\\frac{1\\textbf{ km}}{1000\\;\\rule[0.25ex]{0.75em}{0.1ex}\\hspace{-0.75em}\\textbf{m}}}[\/latex][latex]\\bf{= 0.080\\textbf{ km}}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n \r\n
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Example 1.2 Convert km\/h to m\/s and ft\/s<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nConvert 3 km\/h to a) m\/s and then b) ft\/s.\r\n\r\nSolution\r\n\r\nRemember that 1 k = 1000 or 1 km = 1000 m.\r\n\r\nRemember that 1 hour = 60 minutes and 1 minute = 60 seconds so 1 hour = 3600 seconds.\r\n\r\na)\u00a0 [latex] 3\u00a0 \\: \\: {km\/h} = \\frac { 3 \\: km}{h} ( \\frac {1000 \\: m} { \\: km} ) ( \\frac {1 \\: h} {3600\u00a0 \\: s}) = \\frac {3000 \\: } {3600 \\: s}\u00a0 \\: = \\: 0.8333333 \\: {m\/s} [\/latex]\r\n\r\n= 0.833 m\/s\u00a0 to three significant figures.\r\n\r\nb) Now to convert to ft\/s there is two ways to do this.\u00a0 Remember that 1 inch is exactly 2.54 cm so 1 foot = 0.3048 m.\r\n\r\n[latex] 0.833 \\: {m\/s} = ( \\frac {0.833 \\: m} {s} ) \\: ( \\frac {1 \\: {ft}} {0.3048 \\: m})\u00a0 = \\: 2.73293963 \\: {ft\/s} [\/latex]\r\n\r\n= 2.73 ft\/s to three significant figures.\r\n\r\nor do this all in one step\r\n\r\n[latex] 3\u00a0 \\: \\: {km\/h} = \\frac { 3 \\: km}{h} ( \\frac {1000 \\: m} { \\: km} ) ( \\frac {1 \\: h} {3600\u00a0 \\: s})\u00a0 ( \\frac {1\u00a0 \\:\u00a0 {ft} } {0.3048 \\: m}\u00a0 ) \\: = \\: 2.734033 \\: {ft\/s} [\/latex]\r\n\r\n= 2.73 ft\/s to three significant figures.\r\n\r\nYou can see that it does not matter in which order you do the conversion as long as you round off to the three significant figures at the end.\r\n\r\n \r\n\r\n<\/div>\r\n<\/div>\r\n
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Example 1.3 average speed<\/p>\r\n\r\n<\/header>\r\n

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Suppose that you drive the 10.0 km from your university to home in 20.0 min. Calculate your average speed (a) in kilometers per hour (km\/h) and (b) in meters per second (m\/s). (Note: Average speed is distance traveled divided by time of travel.)<\/p>\r\n

Strategy<\/strong><\/p>\r\n

First we calculate the average speed using the given units. Then we can get the average speed into the desired units by picking the correct conversion factor and multiplying by it. The correct conversion factor is the one that cancels the unwanted unit and leaves the desired unit in its place.<\/p>\r\n

Solution for (a)<\/strong><\/p>\r\n

(1) Calculate average speed. Average speed is distance traveled divided by time of travel. (Take this definition as a given for now\u2014average speed and other motion concepts will be covered in a later module.) In equation form,<\/p>\r\n\r\n

[latex]{average \\: speed}=\u00a0 \\frac {distance}{time}[\/latex].<\/strong><\/div>\r\n

(2) Substitute the given values for distance and time.<\/p>\r\n\r\n

[latex]\\bf{\\textbf{average speed} =}[\/latex][latex size=\"2\"]\\bf{\\frac {10.0\\textbf{ km}} {20.0\\textbf{ min}}}[\/latex][latex]\\bf{=0.500}[\/latex][latex size=\"2\"]\\bf{\\frac {\\textbf{ km}} {\\textbf{ min}}}[\/latex]<\/div>\r\n

(3) Convert km\/min to km\/h: multiply by the conversion factor that will cancel minutes and leave hours. That conversion factor is 60 min\/hr. Thus,<\/p>\r\n\r\n

[latex]\\bf{\\textbf{average speed}=0.500}[\/latex][latex size=\"2\"]\\frac{\\textbf{km}}{\\textbf{min}}[\/latex][latex]\\bf{\\times}[\/latex][latex size=\"2\"]\\bf{\\frac{60\\textbf{ min}}{1\\textbf{ h}}}[\/latex][latex]\\bf{=30.0}[\/latex][latex size=\"2\"]\\bf{\\frac{\\textbf{km}}{\\textbf{h}}}[\/latex].<\/div>\r\n

Discussion for (a)<\/strong><\/p>\r\n

To check your answer, consider the following:<\/p>\r\n

(1) Be sure that you have properly cancelled the units in the unit conversion. If you have written the unit conversion factor upside down, the units will not cancel properly in the equation. If you accidentally get the ratio upside down, then the units will not cancel; rather, they will give you the wrong units as follows:<\/p>\r\n\r\n

[latex size=\"2\"]\\bf{\\frac{km}{min}}[\/latex][latex]\\bf{\\times}[\/latex][latex size=\"2\"]\\bf{\\frac{1\\textbf{ hr}}{60\\textbf{ min}}}[\/latex][latex]\\bf{=}[\/latex][latex size=\"2\"]\\bf{\\frac{1\\textbf{ km}\\cdot\\textbf{hr}}{60\\textbf{ min}^2}}[\/latex],<\/strong><\/div>\r\n

which are obviously not the desired units of km\/h.<\/p>\r\n

(2) Check that the units of the final answer are the desired units. The problem asked us to solve for average speed in units of km\/h and we have indeed obtained these units.<\/p>\r\n

(3) Check the significant figures. Because each of the values given in the problem has three significant figures, the answer should also have three significant figures. The answer 30.0 km\/h does indeed have three significant figures, so this is appropriate. Note that the significant figures in the conversion factor are not relevant because an hour is defined<\/em> to be 60 minutes, so the precision of the conversion factor is perfect.<\/p>\r\n

(4) Next, check whether the answer is reasonable. Let us consider some information from the problem\u2014if you travel 10 km in a third of an hour (20 min), you would travel three times that far in an hour. The answer does seem reasonable.<\/p>\r\n

Solution for (b)<\/strong><\/p>\r\n

There are several ways to convert the average speed into meters per second.<\/p>\r\n

(1) Start with the answer to (a) and convert km\/h to m\/s. Two conversion factors are needed\u2014one to convert hours to seconds, and another to convert kilometers to meters.<\/p>\r\n

(2) Multiplying by these yields<\/p>\r\n\r\n

[latex]\\bf{\\textbf{Average speed}=30.0}[\/latex][latex size=\"2\"]\\bf{\\frac{km}{h}}[\/latex][latex]\\bf{\\times}[\/latex][latex size=\"2\"]\\bf{\\frac{1\\textbf{ h}}{3,600\\textbf{ s}}}[\/latex][latex]\\bf{\\times}[\/latex][latex size=\"2\"]\\bf{\\frac{1,000\\textbf{ m}} {1\\textbf{ km}}}[\/latex],<\/div>\r\n
[latex]\\bf{\\textbf{Average speed} = 8.33}[\/latex][latex size=\"2\"]\\bf{ \\frac {m} {s}}[\/latex].<\/div>\r\n

Discussion for (b)<\/strong><\/p>\r\n

If we had started with 0.500 km\/min, we would have needed different conversion factors, but the answer would have been the same: 8.33 m\/s.<\/p>\r\n

You may have noted that the answers in the worked example just covered were given to three digits. Why? When do you need to be concerned about the number of digits in something you calculate? Why not write down all the digits your calculator produces?<\/p>\r\nSpecial thanks to OpenStax College Physics for the inspiration for these examples. You can download this excellent open educational resource at https:\/\/openstax.org\/books\/college-physics\/<\/a>\r\n\r\n<\/div>\r\n

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Examples: Equilbrium<\/p>\r\n\r\n<\/header>\r\n

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<\/div>\r\nA 76.0-kg person is being pulled away from a burning building as shown in the image below. Calculate the tension in the two ropes if the person is momentarily motionless. Include a free-body diagram in your solution.<\/span>\r\n\r\n<\/div>\r\n
\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"339\"]\"A The force T2<\/sub><\/strong> needed to hold steady the person being rescued from the fire is less than her weight and less than the force T1<\/sub><\/strong> in the other rope, since the more vertical rope supports a greater part of her weight (a vertical force). This CCBY image is from OpenStax College Physics. You can access the complete textbook https:\/\/openstax.org\/details\/books\/college-physics[\/caption]\r\n\r\n
\u00a0<\/strong><\/figcaption><\/figure>\r\n \r\n\r\nAnswer\r\n\r\nT1<\/sub> = 736 N \u00a0 T2<\/sub> = \u00a0194 \u00a0N\u00a0 \u00a0as net force is 0 N so using magnitudes only\r\n\r\nT1 cos 15\u00ba<\/sup> + T2 sin 10\u00b0<\/sup> = Weight = mg \u00a0 \u00a0and\r\n\r\nT1 sin 15\u00b0<\/sup>\u00a0= T2 \u00a0cos10\u00b0<\/sup>\r\n
\"An<\/figure>\r\n
<\/div>\r\nThis CCBY example is from OpenStax College Physics. You can access the whole book at https:\/\/openstax.org\/details\/books\/college-physics.<\/a>\r\n\r\n<\/div>\r\n<\/div>\r\n
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Examples<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nMore examples\r\n