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 net 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 4.37 shows two forces, and , acting on a spacecraft; the plus signs indicate that the forces are directed along the 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 . Find the magnitude and direction of .
Solution
Since the velocity is constant, the acceleration is zero. The net force must also be zero, so that
Solving for yields
The minus sign in the answer means that points opposite to the sum of and , or along the axis in Figure 4.37. 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.
 Suppose the spacecraft were stationary. What would be the direction of ?Answer: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 axis in Figure 4.37, must then point along the axis.
 When the spacecraft is moving at a constant velocity of , what is the direction of ?Answer: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 axis in Figure 4.37.
 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 . As Figure 4.38a shows, the block is being pulled by a horizontal rope in which the tension is . The coefficient of kinetic friction between the block and the surface is . Determine the acceleration of the block.
Solution
Figure 4.38b 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 is the magnitude of the kinetic frictional force. According to Equation 4.8, is related to the magnitude of the normal force by , where is the coefficient of kinetic friction. The acceleration of the block is given by Newton's second law as
We can obtain an expression for 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,