9.3

Center of Gravity

Often, it is important to know the torque produced by the weight of an extended body. In Examples 4 and 5 and Examples 4 and 5, for instance, it is necessary to determine the torques caused by the weight of the ladder and the arm, respectively. In both cases the weight is considered to act at a definite point for the purpose of calculating the torque. This point is called the center of gravity (abbreviated “cg”).

| Definition of Center of Gravity
| The center of gravity of a rigid body is the point at which its weight can be considered to act when the torque due to the weight is being calculated.

When an object has a symmetrical shape and its weight is distributed uniformly, the center of gravity lies at its geometrical center. For instance, Figure 9-9 shows a thin, uniform, horizontal rod of length L attached to a vertical wall by a hinge. The center of gravity of the rod is located at the geometrical center. The lever arm for the weight is L/2, and the magnitude of the torque is W(L/2). In a similar fashion, the center of gravity of any symmetrically shaped and uniform object, such as a sphere, disk, cube, or cylinder, is located at its geometrical center. However, this does not mean that the center of gravity must lie within the object itself. The center of gravity of a compact disc recording, for instance, lies at the center of the hole in the disc and is, therefore, “outside” the object.

Figure 9-9 A thin, uniform, horizontal rod of length L is attached to a vertical wall by a hinge. The center of gravity of the rod is at its geometrical center.

Suppose we have a group of objects, with known weights and centers of gravity, and it is necessary to know the center of gravity for the group as a whole. As an example, Figure 9-10a shows a group composed of two parts: a horizontal uniform board (weight ) and a uniform box (weight ) near the left end of the board. The center of gravity can be determined by calculating the net torque created by the board and box about an axis that is picked arbitrarily to be at the right end of the board. Part a of the figure shows the weights and and their corresponding lever arms x1 and x2. The net torque is . It is also possible to calculate the net torque by treating the total weight as if it were located at the center of gravity and had the lever arm xcg, as part b of the drawing indicates: . The two values for the net torque must be the same, so that

This expression can be solved for xcg, which locates the center of gravity relative to the axis:

| (9.3)

The notation “+ …” indicates that Equation 9.3 can be extended to account for any number of weights distributed along a horizontal line. Figure 9-10c illustrates that the group can be balanced by a single external force (due to the index finger), if the line of action of the force passes through the center of gravity, and if the force is equal in magnitude, but opposite in direction, to the weight of the group. Example 6 demonstrates how to calculate the center of gravity for the human arm.

Figure 9-10 (a) A box rests near the left end of a horizontal board. (b) The total weight () acts at the center of gravity of the group. (c) The group can be balanced by applying an external force (due to the index finger) at the center of gravity.

| Example 6 | | | The Center of Gravity of an Arm

The horizontal arm illustrated in Figure 9-11 is composed of three parts: the upper arm (weight W1 = 17 N), the lower arm (W2 = 11 N), and the hand (W3 = 4.2 N). The drawing shows the center of gravity of each part, measured with respect to the shoulder joint. Find the center of gravity of the entire arm, relative to the shoulder joint.

Figure 9-11 The three parts of a human arm, and the weight and center of gravity for each.

Reasoning and Solution The coordinate xcg of the center of gravity is given by

| (9.3)

The center of gravity plays an important role in determining whether a group of objects remains in equilibrium as the weight distribution within the group changes. A change in the weight distribution causes a change in the position of the center of gravity, and if the change is too great, the group will not remain in equilibrium. Conceptual Example 7 discusses a shift in the center of gravity that led to an embarrassing result.

| Conceptual Example 7 | | | Overloading a Cargo Plane

Figure 9-12a shows a stationary cargo plane with its front landing gear 9 meters off the ground. This accident occurred because the plane was overloaded toward the rear. How did a shift in the center of gravity of the loaded plane cause the accident?

Figure 9-12 (a) This stationary cargo plane is sitting on its tail at Los Angeles International Airport, after being overloaded toward the rear. (© AP/Wide World Photos) (b) In a correctly loaded plane, the center of gravity is between the front and the rear landing gears. (c) When the plane is overloaded toward the rear, the center of gravity shifts behind the rear landing gear, and

Reasoning and Solution Figure 9-12b shows a drawing of a correctly loaded plane, with the center of gravity located between the front and the rear landing gears. The weight of the plane and cargo acts downward at the center of gravity, and the normal forces and act upward at the front and at the rear landing gear, respectively. With respect to an axis at the rear landing gear, the counterclockwise torque due to balances the clockwise torque due to , and the plane remains in equilibrium. Figure 9-12c shows the plane with too much cargo loaded toward the rear, just after the plane has begun to rotate counterclockwise. Because of the overloading, the center of gravity has shifted behind the rear landing gear. The torque due to is now counterclockwise and is not balanced by any clockwise torque. Due to the unbalanced counterclockwise torque, the plane rotates until its tail hits the ground, which applies an upward force to the tail. The clockwise torque due to this upward force balances the counterclockwise torque due to , and the plane comes again into an equilibrium state, this time with the front landing gear 9 meters off the ground.
Related Homework: Problems 19 and 64

As we have seen in Conceptual Example 7, the center of gravity plays an important role in the equilibrium orientation of airplanes. It also plays a similar role in the design of vehicles that can be safely driven with minimal risk to the passengers. Example 8 discusses this role in the context of sport utility vehicles.

A N A L Y Z I N G M U L T I P L E - C O N C E P T P R O B L E M S

| Example 8 | | | The Static Stability Factor and Rollover

Figure 9-13a shows a sport utility vehicle (SUV) that is moving away from you and negotiating a horizontal turn. The radius of the turn is 16 m, and its center is on the right in the drawing. The center of gravity of the vehicle is 0.94 m above the ground and, as an approximation, is assumed to be located midway between the wheels on the left and right sides. The separation between these wheels is the track width and is 1.7 m. What is the greatest speed at which the SUV can negotiate the turn without rolling over?

Figure 9-13 (a) A sport utility vehicle (SUV) is shown moving away from you, following a road curving to the right. The radius of the turn is r. The SUV’s center of gravity is located at a height h above the ground, and its wheels are separated by a distance d. (b) This free-body diagram shows the SUV at the instant it begins to roll over toward the outside of the turn. The forces acting on it are its weight m, the total normal force (acting only on the left-side tires), and the total force of static friction (also acting only on the left-side tires). (c) The lever arms are for an axis passing through the center of gravity and perpendicular to the page.

The physics of the Static Stability Factor and rollover.

Reasoning The free-body diagram in Figure 9-13b shows the SUV at the instant it begins to roll over toward the outside of the turn, which is on the left side of the drawing. At this moment the right-side wheels have just lost contact with the ground, so no forces are acting on them. The forces acting on the SUV are its weight m, the total normal force (acting only on the left-side tires), and the total force of static friction (also acting only on the left-side tires). Since the SUV is moving around the turn, it has a centripetal acceleration and must be experiencing a centripetal force. The static frictional force alone provides the centripetal force. The speed v at which the SUV (mass m) negotiates the turn of radius r is related to the magnitude Fc of the centripetal force by (Equation 5.3). After applying this relation, we will consider rollover by analyzing the torques acting on the SUV with respect to an axis through the center of gravity and perpendicular to the drawing. The normal force will play a role in this analysis, and it will be evaluated by using the fact that it must balance the weight of the vehicle.
Knowns and Unknowns We are given the following data:

Description | Symbol | Value | Comment
Radius of turn | r | 16 m |
Location of center of gravity | h | 0.94 m | Height above ground.
Track width | d | 1.7 m | Distance between left- and right-side wheels.
Unknown Variable |  |  |
Speed of SUV | v | ? |

Modeling the Problem

| Step 1 Speed and Centripetal Force
| According to Equation 5.3, the magnitude of the centripetal force (provided solely by the static frictional force of magnitude fs) is related to the speed v of the car by

Solving for the speed gives Equation 1 at the right. To use this result, we must have a value for fs, which we obtain in Step 2.

| (1)

| Step 2 Torques
| Figure 9-13b shows the free-body diagram for the SUV at the instant rollover begins, when the total normal force and the total static frictional force are acting only on the left-side wheels, the right-side wheels having just lost contact with the ground. At this moment, the sum of the torques due to these two forces is zero for an axis through the center of gravity and perpendicular to the page. Each torque has a magnitude given by Equation 9.1 as the magnitude of the force times the lever arm for the force, and part c of the drawing shows the lever arms. No lever arm is shown for the weight, because it passes through the axis and, therefore, contributes no torque. The force produces a positive torque, since it causes a counterclockwise rotation about the chosen axis, while the force produces a negative torque, since it causes a clockwise rotation. Thus, we have

This result for fs can be substituted into Equation 1, as illustrated at the right. We now proceed to Step 3, to obtain a value for FN.

| Step 3 The Normal Force
| The SUV does not accelerate in the vertical direction, so that the normal force must balance the car’s weight mg, or

The substitution of this result into Equation 2 is shown at the right.

Solution The results of each step can be combined algebraically to show that

The final expression is independent of the mass of the SUV, since m has been eliminated algebraically. The greatest speed at which the SUV can negotiate the turn without rolling over is

At this speed, the SUV will negotiate the turn just on the verge of rolling over. In the final result and beneath the term , we have included the label SSF, which stands for static stability factor. The SSF provides one measure of how susceptible a vehicle is to rollover. Higher values of the SSF are better, because they lead to larger values for v, the greatest speed at which the vehicle can negotiate a turn without rolling over. Note that greater values for the track width d (more widely separated wheels) and smaller values for the height h (center of gravity closer to the ground) lead to higher values of the SSF. A Formula One race car, for example, with its low center of gravity and large track width, is much less prone to rolling over than is an SUV, when both are driven at the same speed around the same curve.
Related Homework: Problem 14

The center of gravity of an object with an irregular shape and a nonuniform weight distribution can be found by suspending the object from two different points P1 and P2, one at a time. Figure 9-14a shows the object at the moment of release, when its weight , acting at the center of gravity, has a nonzero lever arm relative to the axis shown in the drawing. At this instant the weight produces a torque about the axis. The tension force applied to the object by the suspension cord produces no torque because its line of action passes through the axis. Hence, in part a there is a net torque applied to the object, and the object begins to rotate. Friction eventually brings the object to rest as in part b, where the center of gravity lies directly below the point of suspension. In such an orientation, the line of action of the weight passes through the axis, so there is no longer any net torque. In the absence of a net torque the object remains at rest. By suspending the object from a second point P2 (see Figure 9-14c), a second line through the object can be established, along which the center of gravity must also lie. The center of gravity, then, must be at the intersection of the two lines.

Figure 9-14 The center of gravity (cg) of an object can be located by suspending the object from two different points, P1 and P2, one at a time.

The center of gravity is closely related to the center-of-mass concept discussed in 7.5. To see why they are related, let’s replace each occurrence of the weight in Equation 9.3 by W = mg, where m is the mass of a given object and g is the acceleration due to gravity at the location of the object. Suppose that g has the same value everywhere that the objects are located. Then it can be algebraically eliminated from each term on the right side of Equation 9.3. The resulting equation, which contains only masses and distances, is the same as Equation 7.10, which defines the location of the center of mass. Thus, the two points are identical. For ordinary-sized objects, like cars and boats, the center of gravity coincides with the center of mass.

Test your understanding of the key ideas presented in Sections 9.1–9.3:

· | Forces, Lever Arms, and Torques
· | Rigid Bodies in Equilibrium
· | Center of Gravity

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