9.2

Rigid Objects in Equilibrium

CONCEPTS AT A GLANCE If a rigid body is in equilibrium, neither its linear motion nor its rotational motion changes. This lack of change leads to certain equations that apply for rigid-body equilibrium, as the Concepts-at-a-Glance chart in Figure 9-5 emphasizes. For instance, an object whose linear motion is not changing has no acceleration . Therefore, the net force applied to the object must be zero, since and . For two-dimensional motion the x and y components of the net force are separately zero: and (Equations 4.9a and 4.9b). In calculating the net force, we include only forces from external agents, or external forces.* In addition to linear motion, we must consider rotational motion, which also does not change under equilibrium conditions. This means that the net external torque acting on the object must be zero, because torque is what causes rotational motion to change. Using the symbol to represent the net external torque (the sum of all positive and negative torques), we have

| (9.2)

We define rigid-body equilibrium, then, in the following way.

Figure 9-5 CONCEPTS AT A GLANCE A body is in equilibrium when the sum of the external forces is zero and when the sum of the external torques is zero. These conditions apply to the pelican, since it is nicely balanced on top of the cactus and is in equilibrium. (Macduff Everton/The Image Bank/Getty Images)

| Equilibrium of a Rigid Body
| A rigid body is in equilibrium if it has zero translational acceleration and zero angular acceleration. In equilibrium, the sum of the externally applied forces is zero, and the sum of the externally applied torques is zero:

| (4.9a and 4.9b)

| (9.2)

The reasoning strategy for analyzing the forces and torques acting on a body in equilibrium is given below. The first four steps of the strategy are essentially the same as those outlined in Section 4.11, where only forces are considered. Steps 5 and 6 have been added to account for any external torques that may be present. Example 3 illustrates how this reasoning strategy is applied to a diving board.

| Reasoning Strategy

Applying the Conditions of Equilibrium to a Rigid Body
1. | Select the object to which the equations for equilibrium are to be applied.
2. | Draw a free-body diagram that shows all the external forces acting on the object.
3. | Choose a convenient set of x, y axes and resolve all forces into components that lie along these axes.
4. | Apply the equations that specify the balance of forces at equilibrium: and .
5. | Select a convenient axis of rotation. Identify the point where each external force acts on the object, and calculate the torque produced by each force about the axis of rotation. Set the sum of the torques about this axis equal to zero: .
6. | Solve the equations in steps 4 and 5 for the desired unknown quantities.

| Example 3 | | | A Diving Board

A woman whose weight is 530 N is poised at the right end of a diving board with a length of 3.90 m. The board has negligible weight and is bolted down at the left end, while being supported 1.40 m away by a fulcrum, as Figure 9-6a shows. Find the forces and that the bolt and the fulcrum, respectively, exert on the board.

Figure 9-6 (a) A diver stands at the end of a diving board. (b) The free-body diagram for the diving board. The box at the upper left shows the positive x and y directions for the forces, as well as the positive direction (counterclockwise) for the torques.

Reasoning Part b of the figure shows the free-body diagram of the diving board. Three forces act on the board: , , and the force due to the diver’s weight . In choosing the directions of and , we have used our intuition: points downward because the bolt must pull in that direction to counteract the tendency of the board to rotate clockwise about the fulcrum; points upward, because the board pushes downward against the fulcrum, which, in reaction, pushes upward on the board. Since the board is stationary, it is in equilibrium.
Solution Since the board is in equilibrium, the sum of the vertical forces must be zero:

| (4.9b)

Similarly, the sum of the torques must be zero, . For calculating torques, we select an axis that passes through the left end of the board and is perpendicular to the page. (We will see shortly that this choice is arbitrary.) The force produces no torque since it passes through the axis and, therefore, has a zero lever arm, while creates a counterclockwise (positive) torque, and produces a clockwise (negative) torque. The free-body diagram shows the lever arms for the torques:

| (9.2)

Solving this equation for F2 yields

This value for F2, along with W = 530 N, can be substituted into Equation 4.9b above to show that .

In Example 3 the sum of the external torques is calculated using an axis that passes through the left end of the diving board. However, the choice of the axis is completely arbitrary, because if an object is in equilibrium, it is in equilibrium with respect to any axis whatsoever. Thus, the sum of the external torques is zero, no matter where the axis is placed. One usually chooses the location so that the lines of action of one or more of the unknown forces pass through the axis. Such a choice simplifies the torque equation, because the torques produced by these forces are zero. For instance, in Example 3 the torque due to the force F1 does not appear in Equation 9.2, because the lever arm of this force is zero.

Problem solving insight

Need more practice?

Interactive LearningWare9.2
A woman whose weight is 530 N is poised at the right end of a diving board with a length of 3.90 m. The board has negligible weight and is bolted down at the left end, while being supported 1.40 m away by a fulcrum, as Figure 9-6a shows. Find the forces and that the bolt and the fulcrum, respectively, exert on the board.
This Interactive LearningWare tutorial is identical to Example 3. However, the solution is carried out differently to emphasize that it does not matter where you choose the axis of rotation. For an interactive solution, go to

In a calculation of torque, the lever arm of the force must be determined relative to the axis of rotation. In Example 3 the lever arms are obvious, but sometimes a little care is needed in determining them, as in the next example.

| Example 4 | | | Fighting a Fire

In Figure 9-7a an 8.00-m ladder of weight WL = 355 N leans against a smooth vertical wall. The term “smooth” means that the wall can exert only a normal force directed perpendicular to the wall and cannot exert a frictional force parallel to it. A firefighter, whose weight is WF = 875 N, stands 6.30 m up from the bottom of the ladder. Assume that the ladder’s weight acts at the ladder’s center and neglect the hose’s weight. Find the forces that the wall and the ground exert on the ladder.

Figure 9-7 (a) A ladder leaning against a smooth wall. (b) The free-body diagram for the ladder. (c) Three of the forces that act on the ladder and their lever arms. The axis of rotation is at the lower end of the ladder and is perpendicular to the page.

Reasoning Part b of the figure shows the free-body diagram of the ladder. The following forces act on the ladder:

1. | Its weight
2. | A force due to the weight of the firefighter
3. | The force applied to the top of the ladder by the wall and directed perpendicular to the wall
4. | The forces and , which are the horizontal and vertical components of the force exerted by the ground on the bottom of the ladder

The ground, unlike the wall, is not smooth, so that the force is produced by static friction and prevents the ladder from slipping. The force is the normal force applied to the ladder by the ground. The ladder is in equilibrium, so the sum of these forces and the sum of the torques produced by them must be zero.
Solution Since the net force acting on the ladder is zero, we have

| (4.9a)

| (4.9b)

Equation 4.9a cannot be solved as it stands, because it contains two unknown variables. However, another equation can be obtained from the fact that the net torque acting on an object in equilibrium is zero. In calculating torques, we use an axis at the left end of the ladder, directed perpendicular to the page, as Figure 9-7c indicates. This axis is convenient, because and produce no torques about it, their lever arms being zero. Consequently, these forces will not appear in the equation representing the balance of torques. The lever arms for the remaining forces are shown in part c as red dashed lines. The following list summarizes these forces, their lever arms, and the torques:

Force | Lever Arm | Torque
P |  |

Setting the sum of the torques equal to zero gives

| (9.2)

Solving this equation for P gives

Substituting P = 727 N into Equation 4.9a indicates that .

Problem solving insight Sometimes it is necessary to use trigonometry to determine the lever arms from the distances given in a problem, as here in Example 4.

To a large extent the directions of the forces acting on an object in equilibrium can be deduced using intuition. Sometimes, however, the direction of an unknown force is not obvious, and it is inadvertently drawn reversed in the free-body diagram. This kind of mistake causes no difficulty. Choosing the direction of an unknown force backward in the free-body diagram simply means that the value determined for the force will be a negative number, as the next example illustrates.

Problem solving insight

| Example 5 | | | Bodybuilding

A bodybuilder holds a dumbbell of weight as in Figure 9-8a. His arm is horizontal and weighs Wa = 31.0 N. The deltoid muscle is assumed to be the only muscle acting and is attached to the arm as shown. The maximum force that the deltoid muscle can supply has a magnitude of 1840 N. Figure 9-8b shows the distances that locate where the various forces act on the arm. What is the weight of the heaviest dumbbell that can be held, and what are the horizontal and vertical force components, and , that the shoulder joint applies to the left end of the arm?

Figure 9-8 (a) The fully extended, horizontal arm of a bodybuilder supports a dumbbell. (b) The free-body diagram for the arm. (c) Three of the forces that act on the arm and their lever arms. The axis of rotation at the left end of the arm is perpendicular to the page. Force vectors are not to scale.

The physics of bodybuilding.

Reasoning Figure 9-8b is the free-body diagram for the arm. Note that is directed to the right, because the deltoid muscle pulls the arm in toward the shoulder joint, and the joint pushes back in accordance with Newton’s third law. The direction of the force , however, is less obvious, and we are alert for the possibility that the direction chosen in the free-body diagram is backward. If so, the value obtained for will be negative.
Solution The arm is in equilibrium, so the net force acting on it is zero:

| (4.9a)

| (4.9b)

Equation 4.9b cannot be solved at this point, because it contains two unknowns, Sy and Wd . However, since the arm is in equilibrium, the torques acting on the arm must balance, and this fact provides another equation. To calculate torques, we choose an axis through the left end of the arm and perpendicular to the page. With this axis, the torques due to and are zero, because the line of action of each force passes through the axis and the lever arm of each force is zero. The list below summarizes the remaining forces, their lever arms (see Figure 9-8c), and the torques.

Force | Lever Arm | Torque
Wa = 31.0 N |  |
Wd |  |
M = 1840 N |  |

The condition specifying a zero net torque is

| (9.2)

Solving this equation for Wd yields

Substituting this value for Wd into Equation 4.9b above and solving for Sy gives us . The minus sign indicates that the choice of direction for Sy in the free-body diagram is wrong. In reality, Sy has a magnitude of 297 N but is directed downward, not upward.

Problem solving insight When a force is negative, such as the vertical force Sy = −297 N in this example, it means that the direction of the force is opposite to that chosen originally.

Check Your Understanding 2

Three forces act on each of the thin, square sheets shown in the drawing. In parts A and B of the drawing, the force labeled acts at the center of the sheet. The forces can have different magnitudes (F or 2F) and can be applied at different points on an object. Determine in which drawing (a) the translational acceleration is equal to zero, but the angular acceleration is not equal to zero, (b) the translational acceleration is not equal to zero, but the angular acceleration is equal to zero, and (c) the object is in equilibrium. (The answers are given at the end of the book.)

Background: Understanding the concept of equilibrium is essential for answering this question. Think about what must be true concerning the net force and the net torque acting on an object in equilibrium.
For similar questions (including calculational counterparts), consult Self-Assessment Test 9.1. This test is described at the end of Section 9.3.

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