4.13

Concepts & Calculations

Newton’s three laws of motion provide the basis for understanding the effect of forces on the motion of an object, as we have seen. The second law is especially important, because it provides the quantitative relationship between force and acceleration. The examples in this section serve as a review of the essential features of this relationship.

| Concepts & Calculations Example 20 | | | Velocity, Acceleration, and Newton’s Second Law of Motion

Figure 4-38 shows two forces, and , acting on a spacecraft, where the plus signs indicate that the forces are directed along the +x axis. A third force also acts on the spacecraft but is not shown in the drawing. The craft is moving with a constant velocity of +850 m/s. Find the magnitude and direction of .

Figure 4-38 Two horizontal forces, and , act on the spacecraft. A third force also acts but is not shown.

Concept Questions and Answers Suppose the spacecraft were stationary. What would be the direction of ?

Answer If the spacecraft were stationary, its acceleration would be zero. According to Newton’s second law, the acceleration of an object is proportional to the net force acting on it. Thus, the net force must also be zero. But the net force is the vector sum of the three forces in this case. Therefore, the force must have a direction such that it balances to zero the forces and . Since and point along the +x axis in Figure 4-38, must then point along the −x axis.

When the spacecraft is moving at a constant velocity of +850 m/s, what is the direction of ?

Answer Since the velocity is constant, the acceleration is still zero. As a result, everything we said in the stationary case applies again here. The net force is zero, and the force must point along the −x axis in Figure 4-38.

Solution Since the velocity is constant, the acceleration is zero. The net force must also be zero, so that

Solving for F3 yields

The minus sign in the answer means that points opposite to the sum of and , or along the −x axis in Figure 4-38. The force
has a magnitude of 8000 N, which is the magnitude of the sum of the forces and . The answer is independent of the velocity of the spacecraft, as long as that velocity remains constant.

| Concepts & Calculations Example 21 | | | The Importance of Mass

On earth a block has a weight of 88 N. This block is sliding on a horizontal surface on the moon, where the acceleration due to gravity is 1.60 m/s2. As Figure 4-39a shows, the block is being pulled by a horizontal rope in which the tension is T = 24 N. The coefficient of kinetic friction between the block and the surface is . Determine the acceleration of the block.

Figure 4-39 (a) A block is sliding on a horizontal surface on the moon. The tension in the rope is . (b) The freebody diagram for the block, including a kinetic frictional force .

Concept Questions and Answers Which of Newton’s laws of motion provides a way to determine the acceleration of the block?

Answer Newton’s second law allows us to calculate the acceleration as , where is the net force acting in the horizontal direction and m is the mass of the block.

This problem deals with a situation on the moon, but the block’s mass on the moon is not given. Instead, the block’s earth-weight is given. Why can the earth-weight be used to obtain a value for the block’s mass that applies on the moon?

Answer Since the block’s earth-weight Wearth is related to the block’s mass according to Wearth = mgearth, we can use Wearth = 88 N and gearth = 9.80 m/s2 to obtain m. But mass is an intrinsic property of the block and does not depend on whether it is on the earth or on the moon. Therefore, the value obtained for m applies on the moon as well as on the earth.

Does the net force equal the tension T ?

Answer No. The net force is the vector sum of all the external forces acting in the horizontal direction. It includes the kinetic frictional force fk as well as the tension T.

Solution Figure 4-39b shows the free-body diagram for the block. The net force along the x axis is , where T is the magnitude of the tension in the rope and fk is the magnitude of the kinetic frictional force. According to Equation 4.8, fk is related to the magnitude FN of the normal force by , where is the coefficient of kinetic friction. The acceleration ax of the block is given by Newton’s second law as

We can obtain an expression for FN by noting that the block does not move in the y direction, so . Therefore, the net force along the y direction must also be zero. An examination of the free-body diagram reveals that , so that . The acceleration in the x direction becomes

Using the earth-weight of the block to determine its mass, we find

The acceleration of the block is, then,

Copyright © 2007 John Wiley & Sons, Inc. All rights reserved.