42 6.7 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).

(a) A sketch is shown of a man hanging from a vine. (b) The forces acting on the person, shown by vector arrows, are tension T, pointing upward at the hand of the man, F sub T, from the same point but in a downward direction, and weight W, acting downward from his stomach. (c) In figure (c) we define only the man as the system of interest. Tension T is acting upward from his hand. The weight W acts in a downward direction. In a free-body diagram W is shown by an arrow acting downward and T is shown by an arrow acting vertically upward. (d) Tension T is shown by an arrow vertically upward and another vector, weight W, is shown by an arrow vertically downward, both having the same lengths. It is indicated that T is equal to minus W.
Figure 1. (a) A sketch of Tarzan hanging from a vine. (b) Arrows are used to represent all forces. T is the tension in the vine above Tarzan, FT is the force he exerts on the vine, and w is his weight. All other forces, such as the nudge of a breeze, are assumed negligible. (c) Suppose we are given the ape man’s mass and asked to find the tension in the vine. We then define the system of interest as shown and draw a free-body diagram. FT is no longer shown, because it is not a force acting on the system of interest; rather, FT acts on the outside world. (d) Showing only the arrows, the head-to-tail method of addition is used. It is apparent that T = -w, if Tarzan is stationary.

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: Fnet = 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 Fnetx = 0. If the object does accelerate in that direction, Fnetx = ma.
    4. Check your answer. Is the answer reasonable? Are the units correct?

Problems & Exercises

1: 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.

2: 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.

3: (a) Find the magnitudes of the forces F1 and F2 that add to give the total force Ftot 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 Ftot. Draw these to scale on the same drawing used in part (b) or a similar picture.

A right triangle is shown made up of three vectors. The first vector, F sub one, is along the triangle’s base toward the right; the second vector, F sub two, is along the perpendicular side pointing upward; and the third vector, F sub tot, is along the hypotenuse pointing up the incline. The magnitude of F sub tot is twenty newtons. In a free-body diagram, F sub one is shown by an arrow pointing right and F sub two is shown by an arrow acting vertically upward.
Figure 2.

4: 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.

An overhead view of a child sitting on a snow saucer sled. Two forces, F sub one equal to ten newtons and F sub two equal to eight newtons, are acting toward the right. F sub one makes an angle of forty-five degrees from the x axis and F sub two makes an angle of thirty degrees from the x axis in a clockwise direction. A friction force f is equal to seven point five newtons, shown by a vector pointing in negative x direction. In the free-body diagram, F sub one and F sub two are shown by arrows toward the right, making a forty-five degree angle above the horizontal and a thirty-degree angle below the horizontal respectively. The friction force f is shown by an arrow along the negative x axis.
Figure 3. An overhead view of the horizontal forces acting on a child’s snow saucer sled.

5: 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.

Two caped superheroes hang on a rope suspended vertically from a bar.
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?

6: 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?

7: 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.

Solutions

Problems & Exercises

1:

Two forces are acting on an object of mass m: F, shown by an arrow pointing upward, and its weight w, shown by an arrow pointing downward. Acceleration a is represented by a vector arrow pointing upward. The figure depicts the forces acting on a high jumper.
Figure 8.
  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]

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Biomechanics of Human Movement Copyright © August 22, 2016 by OpenStax is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.

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