Lesson: Chapter - 6

Problem Solving with Newton’s Laws

Dynamics problem solving in physics class usually involves difficult calculations that take into account a number of vectors on a free-body diagram. subject test Physics won’t expect you to make any difficult calculations, and the test will usually include the free-body diagrams that you need. Your task will usually be to interpret free-body diagrams rather than to draw them.

Example 1

The Three Stooges are dragging a 10 kg sled across a frozen lake. Moe pulls with force M, Larry pulls with force L, and Curly pulls with force C. If the sled is moving in the direction, and both Moe and Larry are exerting a force of 10 N, what is the x magnitude of the force Curly is exerting? Assuming that friction is negligible, what is the acceleration of the sled? (Note: sin 30 = cos 60 = 0.500 and sin 60 = cos 30 = 0.866.)

The figure above gives us a free-body diagram that shows us the direction in which all forces are acting, but we should be careful to note that vectors in the diagram are not drawn to scale: we cannot estimate the magnitude of C simply by comparing it to M and L.

What is the magnitude of the force Curly is exerting?

Since we know that the motion of the sled is in the x direction, the net force, M + L + C, must also be in the x direction. And since the sled is not moving in the y direction, the y-component of the net force must be zero. Because the y-component of Larry’s force is zero, this implies:

M₉ + c_y = 0

where M_y is the y-component of M and C_y is the y-component of C. We also know:

M_y = M sin ? = (10 N) = sin60 = 8.660 N

C_y = C sin? = C sin (-30) = -0.500C

If we substitute these two equations for M_y and C_y into the equation M_y + C_y = 0 , we have:

8.660 N - 0.500C = 0

C = 17.32 N

What is the acceleration of the sled?

According to Newton’s Second Law, the acceleration of the sled is a = Fm. We know the sled has a mass of 10 kg, so we just need to calculate the magnitude of the

M_x + L_x + C_x = (10 N) cos 60 + 10N + (17.32 N) cos(-30)

0.500(10N) + 10N + 0.886(17.32 N)

30.000 N

Now that we have calculated the magnitude of the net force acting on the sled, a simple calculation can give us the sled’s acceleration:

We have been told that the sled is moving in the x direction, so the acceleration is also in the x direction.

This example problem illustrates the importance of vector components. For the subject test, you will need to break vectors into components on any problem that deals with vectors that are not all parallel or perpendicular. As with this example, however, the subject test will always provide you with the necessary trigonometric values.

Example 2

Each of the following free-body diagrams shows the instantaneous forces, F, acting on a particle and the particle’s instantaneous velocity, v. All forces represented in the diagrams are of the same magnitude.

1. In which diagram is neither the speed nor the direction of the particle being changed?
2. In which diagram is the speed but not the direction of the particle being changed?
3. In which diagram is the direction but not the speed of the particle being changed?
4. In which diagram are both the speed and direction of the particle being changed?

The answer to question 1 is B. The two forces in that diagram cancel each other out, so the net force on the particle is zero. The velocity of a particle only changes under the influence of a net force. The answer to question 2 is C. The net force is in the same direction as the particle’s motion, so the particle continues to accelerate in the same direction. The answer to question 3 is A. Because the force is acting perpendicular to the particle’s velocity, it does not affect the particle’s speed, but rather acts to pull the particle in a circular orbit. Note, however, that the speed of the particle only remains constant if the force acting on the particle remains perpendicular to it. As the direction of the particle changes, the direction of the force must also change to remain perpendicular to the velocity. This rule is the essence of circular motion, which we will examine in more detail later in this book. The answer to question 4 is D. The net force on the particle is in the opposite direction of the particle’s motion, so the particle slows down, stops, and then starts accelerating in the opposite direction.

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