43 6.8 Further Applications of Newton’s Laws of Motion

Summary

  • Integrate concepts from kinematics to solve problems using Newton’s laws of motion.

Integrating Concepts: Newton’s Laws of Motion and Kinematics

Biomechanics 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\:=}\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 Fnet = ma or Fnet = 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.

Problems & Exercises

1: 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?

An Achilles tendon is shown in the figure with two forces acting upward, one at an angle of plus twenty degrees, one at minus twenty degrees. F sub one, equal to two hundred newtons, is shown by a vector making an angle twenty degrees toward the right with the vertical, and F sub two, equal to two hundred newtons, is shown making an angle of twenty degrees left from the vertical.
Figure 4. Achilles tendon

2: 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.

A lady is being pulled away from a burning building using a rope. She is in the middle of the rope; her weight is shown by a vector acting vertically downward. Tension, T sub one, acts upward through the left side of the rope, making an angle of fifteen degrees with the vertical. Tension T sub two acts through the right side of the rope, making an angle of ten degrees above the positive x axis.
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)
 

3: 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.

4: 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.

Solutions

Problems & Exercises

2: T1 = 736 N   T2 =  194  N    as  net force is 0 N so using magnitudes only

T1 cos 15o + T2 sin 10o = Weight = mg    and   T1 sin 15o = T2  cos10o

An object of mass m is shown being pulled by two ropes. Tension T sub two acts toward the right at an angle of ten degrees above the horizontal. Another rope makes an angle fifteen degrees to the left of the vertical direction, and tension in the rope is T sub one, shown by a vector arrow. Weight w is acting vertically downward.
Figure 6.

3:   (a)  7.43 m/s (b) 2.97 m  as the acceleration is 650 N / 70.0 kg = 9.29 m/s2

4: (a) 4.20 m/s   7 (b) 29.4 m/s2  7 (c) 4.31 x 103 N

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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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