ORMAT 8
Equilibrium, Elasticity, and Fracture
Looking Ahead
The goal of Chapter 25 is to learn about the magnetic field, and how magnetic fields exert forces on currents and moving charges. In this chapter, you will learn to:
• Understand basic magnetic phenomena.
Looking Back
This chapter uses what you have learned about circular motion, torque, and electric fields and electric dipoles. Please review:
• Sections 7.2–7.3 Forces and circular motion.
A large suspension bridge is a remarkable structure, a precise blend of engineering and imagination. In its construction, its designers must ensure its stability and endurance. To do so, they must understand how to balance the many forces acting on it so that it remains motionless. They must recognize how the bridge responds to the enormous forces acting on it—how its parts stretch or flex under the bridge’s load. And the designers must certainly be able to calculate the maximum forces that can be applied to bridge members, such as beams or cables, before such members fail.
In this chapter, we study how any structure—from a bridge to a tree to a knee joint—must be designed in order to support itself and external loads. We will see under what conditions an object under load will remain motionless, and how it responds to changes in the forces acting on it. Is the structure stable if outside forces disturb it, or does it come crashing down? How do its members bend and stretch as forces act? And are the members strong enough to resist such bending, or is their ultimate strength exceeded, with disastrous results?
8.1 Torque and Static Equilibrium
In the preceding chapters, we have made an in-depth study of motion and its causes. In many disciplines, it is just important to understand under what conditions objects do not move. In structural engineering, buildings and dams must be designed such that they remain motionless, even when huge forces act on them. In sports science, a correct stationary position is often the starting point for a successful athletic event. And a study of, say, the knee joint must understand the enormous forces that the joint must sustain, even when the subject is standing at rest.
Recall from Section 5.2 that when any object is at rest, we say that it is in static equilibrium. In that section we also found the condition necessary for a particle to be in static equilibrium: the net force on the particle must be zero, so that it does not accelerate. Such a situation is shown in Figure 8.1a, where the two forces applied to the block balance, and the block can remain at rest. But in Chapter 7 we moved beyond the particle model to study extended objects that could rotate as well as move translationally. For an extended object, having is by itself not enough to ensure static equilibrium. In Figure 8.1b, the same two forces are applied, but this time off center. The net force is still zero, so the block will not undergo an overall acceleration. But the block will begin to rotate, because now there is a net torque applied to it. So there is a second condition for static equilibrium: the net torque on the object must be zero.
FIG_StaticEquilibrium
Figure 8.1 A block with no net force acting on it may still be out of equilibrium.
For an object to be in static equilibrium, then, it must have both zero net force acting and zero net torque acting on it. If we write the net force in component form, these conditions are
Conditions for static equilibrium
STOP TO THINK 8.1 Which of the objects below could be in static equilibrium?
Example 8.1 Lifting a weight
Weightlifting can exert extraordinary forces on the body’s joints. In the strict curl event, a standing athlete lifts the barbell using only the forearms. The record weight lifted in the strict curl is over 200 pounds (about 900 N). Figure 8.2a shows the arrangement of the arm bones and the main lifting muscles when the forearm is horizontal. With the arms held stationary in this position, what is the tension in the lifting biceps muscle when 900 N is lifted?
FIG_StrictCurl
Figure 8.2 Forces and torques in holding a barbell.
Prepare Figure 8.2b shows a simplified model of the arm, indicating the forces acting on the forearm. is the tension force due to the muscle, is the downward force of the weight, and is the force of the elbow joint on the forearm.
Solve For the forearm to be in static equilibrium, the net force and net torque on it must be zero. Setting the net force to zero gives
We know the magnitude of the weight force, = 450N (assuming that each arm supports half the 900N total weight). We don’t know either of the other two forces and nor does the force equation above give us enough information to find them. Using the fact that the torque must be zero gives us that extra information. Choosing the elbow joint to be the axis of rotation, we have
Note that makes no contribution to the torque, since it acts directly at the pivot. We can solve the above equation for to find
.
Assess We see that we need to use the fact that both the net force and the net torque are zero for an object in equilibrium. The short distance from the muscle to the elbow joint means that the force of the muscle has to be very large in order to counter the torque generated by a force applied at the end of the forearm.
Choosing the Pivot Point
In Example 8.1 above, we calculated the net torque using the elbow joint as the axis of rotation or pivot point. But we have learned that the torque on an object depends on which point is chosen as the axis of rotation. Was there something special about our choice of the elbow joint as the axis?
Consider the hammer shown in Figure 8.3, supported on a pegboard by two pegs A and B. Because it’s in static equilibrium, the net torque around the pivot at A must be zero: the clockwise torque due to the weight is exactly balanced by the counterclockwise torque due to the force of peg B. But if instead we take B as the pivot, the net torque is still zero. The counterclockwise torque due to (with a large force but small moment arm) balances the clockwise torque due to (with a small force but large moment arm). Indeed, for an object in static equilibrium, the net torque about every axis must be zero. This means that when you calculate the torque, you can pick any point you wish as an axis. Even if we use the arbitrary point C, which is not a real pivot point for the hammer, we would find the torque to be zero.
FIG_TorqueIndependentOfAxis
Figure 8.3 The net torque on an object in static equilibrium is zero when calculated about any axis.
Although any choice of axis will work, some choices are better because they simplify the calculations. Often, there is a “natural” axis of rotation in the problem, an axis about which rotation would occur if the object were not in static equilibrium. Example 8.1 is of this type, with the elbow joint as a natural axis of rotation.
If no point naturally suggests itself as an axis, you should look for a point on the object where a force acts. Good choices are where either several forces act at once, or where a force whose magnitude you don’t know acts. Such points are good choices because any force acting at the axis does not contribute to the torque. By choosing the axis where a force acts, that force will not appear in the torque equation.
Don’t agonize too much about this choice of axis. If you can’t decide, pick any point on the object where a force acts. You will still be able to solve the problem.
Problem-Solving Strategy 25.1 Static equilibrium problems
Prepare Model the object as a simple shape. Draw a pictorial representation that shows all forces and distances. List known information.
· Pick an axis or pivot about which the torques will be calculated.
· Determine the moment arms of all forces about this pivot point.
· Determine the sign of each torque about this pivot point.
Solve The mathematical representation is based on the fact that an object in static equilibrium has no net force and no net torque.
and
· Write equations for.
· Solve the resulting equations.
Assess Check that your result is reasonable and answers the question.
Example 8.2 A board on sawhorses
A board weighing 100 N sits on two sawhorses, as shown in Figure 8.4a. What are the magnitudes of the two forces and of the sawhorses acting on the board?
FIG_Sawhorses
Figure 8.4 A board sitting on two sawhorses.
Prepare The board and the forces acting on it are shown in Figure 8.4b. and are the forces on the board due to the sawhorses, and is the weight of the board, acting at the center of gravity.
As discussed above, a good choice for the pivot is a point where an unknown force acts, since then the unknown force contributes nothing to the torque. Two such points are Point A in Figure 8.4b, where force acts, or point B, where acts. Either will work; let’s choose point A for this example.
With this choice of pivot, we see that the moment arm for which acts halfway down the board, is Since tends to rotate the board clockwise, its torque is negative. The moment arm for is the distance and, since tends to rotate the board counterclockwise, its torque is positive.
Solve Since the board is in static equilibrium, the net force and the net torque must both be zero. The forces acting have only y components, so we needn’t worry about the x components. So we have
from which we can solve for:
Next we compute the torque around axis A, and set it to zero. We have
from which we find
From the force equation above, we also have
Assess It seems reasonable that since more of the board sits over the right sawhorse.
Example 8.3 Choosing a different axis
Let’s repeat Example 8.2, but with the axis at a different point. Let’s put the axis at the center of gravity where the weight acts. Will this change our results?
Prepare In Figure 8.5 we define new distances and from the axis to where the forces and act. Moment arm and force produce a negative torque, while and produce a positive torque.
FIG_SawhorsesDifferentAxis
Figure 8.5 A board on two sawhorses, with axis at the center of gravity.
Solve The force equation still reads , from which we have, as before,
Now we need to calculate the net torque about the center of gravity. The weight acting at the pivot, now does not contribute to the torque. We have
If we insert the expression for from the force equation into the torque result, we get
.
We can solve this for to get
as before. Again, we can use the force equation to find that = 25 N.
Assess Even with a very different choice of axis from Example 8.2, we get the same result for the forces.
STOP TO THINK 8.2 A uniform beam is suspended from a wire, and has a pivot on its left end. In in which direction is the force of the pivot on the beam?
A. B. C. D. E.
An interesting application of the principle of static equilibrium is the determination of the position of the center of gravity of the human body. As we’ll see later, the position of a person’s center of gravity plays an important role in many aspects of sports and athletics. Because the human body is highly flexible, the position of the center of gravity is quite variable, and depends on just how the body is posed.
FIG_ReactionBoard
Figure 8.6 Board-and-scale method of finding a subject's center of gravity.
The horizontal position of the body’s center of gravity can be located accurately from simple measurements with a reaction board and a scale. Figure 8.6 shows the method. The subject lies or stands on the board in the desired posture. The board has a pivot at its left end, while its right end is supported on a spring scale. The spring scale pushes up on the board with a force the magnitude F of this force is the scale reading. The following example shows how the position of the subject’s center of gravity can be using this method.
Example 8.4 Finding the center of gravity of the human body
A subject, weighing 730 N, lies on a reaction board that is 2.5 m long and weighs 60 N. The scale on the right reads 300 N. Find the distance from the pivot to the subject’s center of gravity.
FIG_ReactionBoardExample
Figure 8.7 Finding the center of gravity of the human body.
Prepare The forces and distances in the problem are drawn in Figure 8.7. We’ll calculate all torques around the pivot at the left end of the board. Then the torque due to is positive, and those due to and are negative. The torque due to which acts at the pivot, is zero.