# Statics and Torque

## Applications of Statics, Including Problem-Solving Strategies

### Learning Objectives

By the end of this section, you will be able to:
1. Discuss the applications of Statics in real life.
2. State and discuss various problem-solving strategies in Statics.
Statics can be applied to a variety of situations, ranging from raising a drawbridge to bad posture and back strain. We begin with a discussion of problem-solving strategies specifically used for statics. Since statics is a special case of Newton’s laws, both the general problem-solving strategies and the special strategies for Newton’s laws, discussed in Problem-Solving Strategies, still apply.
Now let us apply this problem-solving strategy for the pole vaulter shown in the three figures below. The pole is uniform and has a mass of 5.00 kg. In , the pole’s cg lies halfway between the vaulter’s hands. It seems reasonable that the force exerted by each hand is equal to half the weight of the pole, or 24.5 N. This obviously satisfies the first condition for equilibrium  . The second condition  is also satisfied, as we can see by choosing the cg to be the pivot point. The weight exerts no torque about a pivot point located at the cg, since it is applied at that point and its lever arm is zero. The equal forces exerted by the hands are equidistant from the chosen pivot, and so they exert equal and opposite torques. Similar arguments hold for other systems where supporting forces are exerted symmetrically about the cg. For example, the four legs of a uniform table each support one-fourth of its weight.
In , a pole vaulter holding a pole with its cg halfway between his hands is shown. Each hand exerts a force equal to half the weight of the pole, . (b) The pole vaulter moves the pole to his left, and the forces that the hands exert are no longer equal. See . If the pole is held with its cg to the left of the person, then he must push down with his right hand and up with his left. The forces he exerts are larger here because they are in opposite directions and the cg is at a long distance from either hand.
Similar observations can be made using a meter stick held at different locations along its length.

If the pole vaulter holds the pole as shown in , the situation is not as simple. The total force he exerts is still equal to the weight of the pole, but it is not evenly divided between his hands. (If , then the torques about the cg would not be equal since the lever arms are different.) Logically, the right hand should support more weight, since it is closer to the cg. In fact, if the right hand is moved directly under the cg, it will support all the weight. This situation is exactly analogous to two people carrying a load; the one closer to the cg carries more of its weight. Finding the forces  and  is straightforward, as the next example shows.
If the pole vaulter holds the pole from near the end of the pole (), the direction of the force applied by the right hand of the vaulter reverses its direction.
If the pole vaulter holds the pole as he might at the start of a run, shown in , the forces change again. Both are considerably greater, and one force reverses direction.

### Test Prep for AP Courses

### Summary

1. Statics can be applied to a variety of situations, ranging from raising a drawbridge to bad posture and back strain. We have discussed the problem-solving strategies specifically useful for statics. Statics is a special case of Newton’s laws, both the general problem-solving strategies and the special strategies for Newton’s laws, discussed in Problem-Solving Strategies, still apply.

### Conceptual Questions

### Problems & Exercises


# Statics and Torque

## Simple Machines

### Learning Objectives

By the end of this section, you will be able to:
1. Describe different simple machines.
2. Calculate the mechanical advantage.
Simple machines are devices that can be used to multiply or augment a force that we apply – often at the expense of a distance through which we apply the force. The word for “machine” comes from the Greek word meaning “to help make things easier.” Levers, gears, pulleys, wedges, and screws are some examples of machines. Energy is still conserved for these devices because a machine cannot do more work than the energy put into it. However, machines can reduce the input force that is needed to perform the job. The ratio of output to input force magnitudes for any simple machine is called its mechanical advantage (MA).
One of the simplest machines is the lever, which is a rigid bar pivoted at a fixed place called the fulcrum. Torques are involved in levers, since there is rotation about a pivot point. Distances from the physical pivot of the lever are crucial, and we can obtain a useful expression for the MA in terms of these distances.
 shows a lever type that is used as a nail puller. Crowbars, seesaws, and other such levers are all analogous to this one.

 is the input force and

 is the output force. There are three vertical forces acting on the nail puller (the system of interest) – these are

 and

.

 is the reaction force back on the system, equal and opposite to

. (Note that

 is not a force on the system.)

 is the normal force upon the lever, and its torque is zero since it is exerted at the pivot. The torques due to

 and  must be equal to each other  if the nail is not moving, to satisfy the second condition for equilibrium

     . (In order for the nail to actually move, the torque due to  must be ever-so-slightly greater than torque due to  .) Hence,
Notice that  is the distance from the pivot point to the point where the input force  is applied, and  (not labeled on the diagram) is the distance from the pivot point to the point where the output force  is applied. The distances   and  are the perpendicular components of the distances from where the input and output forces are applied to  the pivot, as shown in the figure. Rearranging the last equation gives
What interests us most here is that the magnitude of the force exerted by the nail puller, , is much greater than the magnitude of the input force applied to the puller at the other end, . For the nail puller,
This equation is true for levers in general. For the nail puller, the MA is certainly greater than one. The longer the handle on the nail puller, the greater the force you can exert with it.
Two other types of levers that differ slightly from the nail puller are a wheelbarrow and a shovel, shown in . All these lever types are similar in that only three forces are involved – the input force, the output force, and the force on the pivot – and thus their MAs are given by  and , with distances being measured relative to the physical pivot. The wheelbarrow and shovel differ from the nail puller because both the input and output forces are on the same side of the pivot.
In the case of the wheelbarrow, the output force or load is between the pivot (the wheel’s axle) and the input or applied force. In the case of the shovel, the input force is between the pivot (at the end of the handle) and the load, but the input lever arm is shorter than the output lever arm. In this case, the MA is less than one.
Another very simple machine is the inclined plane. Pushing a cart up a plane is easier than lifting the same cart straight up to the top using a ladder, because the applied force is less. However, the work done in both cases (assuming the work done by friction is negligible) is the same. Inclined lanes or ramps were probably used during the construction of the Egyptian pyramids to move large blocks of stone to the top.
A crank is a lever that can be rotated

   about its pivot, as shown in . Such a machine may not look like a lever, but the physics of its actions remain the same. The MA for a crank is simply the ratio of the radii . Wheels and gears have this simple expression for their MAs too. The MA can be greater than 1, as it is for the crank, or less than 1, as it is for the simplified car axle driving the wheels, as shown. If the axle’s radius is

 and the wheel’s radius is

, then

 and the axle would have to exert a force of

 on the wheel to enable it to exert a force of

 on the ground.
An ordinary pulley has an MA of 1; it only changes the direction of the force and not its magnitude. Combinations of pulleys, such as those illustrated in , are used to multiply force. If the pulleys are friction-free, then the force output is approximately an integral multiple of the tension in the cable. The number of cables pulling directly upward on the system of interest, as illustrated in the figures given below, is approximately the MA of the pulley system. Since each attachment applies an external force in approximately the same direction as the others, they add, producing a total force that is nearly an integral multiple of the input force

  .

### Test Prep for AP Courses

### Section Summary

1. Simple machines are devices that can be used to multiply or augment a force that we apply – often at the expense of a distance through which we have to apply the force.
2. The ratio of output to input forces for any simple machine is called its mechanical advantage
3. A few simple machines are the lever, nail puller, wheelbarrow, crank, etc.

### Conceptual Questions

### Problems & Exercises


# Statics and Torque

## Forces and Torques in Muscles and Joints

### Learning Objectives

By the end of this section, you will be able to:
1. Explain the forces exerted by muscles.
2. State how a bad posture causes back strain.
3. Discuss the benefits of skeletal muscles attached close to joints.
4. Discuss various complexities in the real system of muscles, bones, and joints.
Muscles, bones, and joints are some of the most interesting applications of statics. There are some surprises. Muscles, for example, exert far greater forces than we might think.  shows a forearm holding a book and a schematic diagram of an analogous lever system. The schematic is a good approximation for the forearm, which looks more complicated than it is, and we can get some insight into the way typical muscle systems function by analyzing it.
Muscles can only contract, so they occur in pairs. In the arm, the biceps muscle is a flexor—that is, it closes the limb. The triceps muscle is an extensor that opens the limb. This configuration is typical of skeletal muscles, bones, and joints in humans and other vertebrates. Most skeletal muscles exert much larger forces within the body than the limbs apply to the outside world. The reason is clear once we realize that most muscles are attached to bones via tendons close to joints, causing these systems to have mechanical advantages much less than one. Viewing them as simple machines, the input force is much greater than the output force, as seen in .
In the above example of the biceps muscle, the angle between the forearm and upper arm is 90°. If this angle changes, the force exerted by the biceps muscle also changes. In addition, the length of the biceps muscle changes. The force the biceps muscle can exert depends upon its length; it is smaller when it is shorter than when it is stretched.
Very large forces are also created in the joints. In the previous example, the downward force  exerted by the humerus at the elbow joint equals 407 N, or 6.38 times the total weight supported. (The calculation of

 is straightforward and is left as an end-of-chapter problem.) Because muscles can contract, but not expand beyond their resting length, joints and muscles often exert forces that act in opposite directions and thus subtract. (In the above example, the upward force of the muscle minus the downward force of the joint equals the weight supported—that is,

, approximately equal to the weight supported.) Forces in muscles and joints are largest when their load is a long distance from the joint, as the book is in the previous example.
In racquet sports such as tennis the constant extension of the arm during game play creates large forces in this way. The mass times the lever arm  of a tennis racquet is an important factor, and many players use the heaviest racquet they can handle. It is no wonder that joint deterioration and damage to the tendons in the elbow, such as “tennis elbow,” can result from repetitive motion, undue torques, and possibly poor racquet selection in such sports. Various tried techniques for holding and using a racquet or bat or stick not only increases sporting prowess but can minimize fatigue and long-term damage to the body. For example, tennis balls correctly hit at the “sweet spot” on the racquet will result in little vibration or impact force being felt in the racquet and the body—less torque as explained in Collisions of Extended Bodies in Two Dimensions. Twisting the hand to provide top spin on the ball or using an extended rigid elbow in a backhand stroke can also aggravate the tendons in the elbow.
Training coaches and physical therapists use the knowledge of relationships between forces and torques in the treatment of muscles and joints. In physical therapy, an exercise routine can apply a particular force and torque which can, over a period of time, revive muscles and joints. Some exercises are designed to be carried out under water, because this requires greater forces to be exerted, further strengthening muscles. However, connecting tissues in the limbs, such as tendons and cartilage as well as joints are sometimes damaged by the large forces they carry. Often, this is due to accidents, but heavily muscled athletes, such as weightlifters, can tear muscles and connecting tissue through effort alone.
The back is considerably more complicated than the arm or leg, with various muscles and joints between vertebrae, all having mechanical advantages less than 1. Back muscles must, therefore, exert very large forces, which are borne by the spinal column. Discs crushed by mere exertion are very common. The jaw is somewhat exceptional—the masseter muscles that close the jaw have a mechanical advantage greater than 1 for the back teeth, allowing us to exert very large forces with them. A cause of stress headaches is persistent clenching of teeth where the sustained large force translates into fatigue in muscles around the skull.
 shows how bad posture causes back strain. In part (a), we see a person with good posture. Note that her upper body’s cg is directly above the pivot point in the hips, which in turn is directly above the base of support at her feet. Because of this, her upper body’s weight exerts no torque about the hips. The only force needed is a vertical force at the hips equal to the weight supported. No muscle action is required, since the bones are rigid and transmit this force from the floor. This is a position of unstable equilibrium, but only small forces are needed to bring the upper body back to vertical if it is slightly displaced. Bad posture is shown in part (b); we see that the upper body’s cg is in front of the pivot in the hips. This creates a clockwise torque around the hips that is counteracted by muscles in the lower back. These muscles must exert large forces, since they have typically small mechanical advantages. (In other words, the perpendicular lever arm for the muscles is much smaller than for the cg.) Poor posture can also cause muscle strain for people sitting at their desks using computers. Special chairs are available that allow the body’s CG to be more easily situated above the seat, to reduce back pain. Prolonged muscle action produces muscle strain. Note that the cg of the entire body is still directly above the base of support in part (b) of . This is compulsory; otherwise the person would not be in equilibrium. We lean forward for the same reason when carrying a load on our backs, to the side when carrying a load in one arm, and backward when carrying a load in front of us, as seen in .
You have probably been warned against lifting objects with your back. This action, even more than bad posture, can cause muscle strain and damage discs and vertebrae, since abnormally large forces are created in the back muscles and spine.
What are the benefits of having most skeletal muscles attached so close to joints? One advantage is speed because small muscle contractions can produce large movements of limbs in a short period of time. Other advantages are flexibility and agility, made possible by the large numbers of joints and the ranges over which they function. For example, it is difficult to imagine a system with biceps muscles attached at the wrist that would be capable of the broad range of movement we vertebrates possess.
There are some interesting complexities in real systems of muscles, bones, and joints. For instance, the pivot point in many joints changes location as the joint is flexed, so that the perpendicular lever arms and the mechanical advantage of the system change, too. Thus the force the biceps muscle must exert to hold up a book varies as the forearm is flexed. Similar mechanisms operate in the legs, which explain, for example, why there is less leg strain when a bicycle seat is set at the proper height. The methods employed in this section give a reasonable description of real systems provided enough is known about the dimensions of the system. There are many other interesting examples of force and torque in the body—a few of these are the subject of end-of-chapter problems.

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### Section Summary

1. Statics plays an important part in understanding everyday strains in our muscles and bones.
2. Many lever systems in the body have a mechanical advantage of significantly less than one, as many of our muscles are attached close to joints.
3. Someone with good posture stands or sits in such a way that the person's center of gravity lies directly above the pivot point in the hips, thereby avoiding back strain and damage to disks.

### Conceptual Questions

### Problems & Exercises


# Rotational Motion and Angular Momentum

## Introduction to Rotational Motion and Angular Momentum

Why do tornadoes spin at all? And why do tornadoes spin so rapidly? The answer is that air masses that produce tornadoes are themselves rotating, and when the radii of the air masses decrease, their rate of rotation increases. An ice skater increases her spin in an exactly analogous manner as seen in . The skater starts her rotation with outstretched limbs and increases her spin by pulling them in toward her body. The same physics describes the exhilarating spin of a skater and the wrenching force of a tornado.
Clearly, force, energy, and power are associated with rotational motion. These and other aspects of rotational motion are covered in this chapter. We shall see that all important aspects of rotational motion either have already been defined for linear motion or have exact analogs in linear motion. First, we look at angular acceleration—the rotational analog of linear acceleration.


# Rotational Motion and Angular Momentum

## Angular Acceleration

### Learning Objectives

By the end of this section, you will be able to:
1. Describe uniform circular motion.
2. Explain non-uniform circular motion.
3. Calculate angular acceleration of an object.
4. Observe the link between linear and angular acceleration.
Uniform Circular Motion and Gravitation discussed only uniform circular motion, which is motion in a circle at constant speed and, hence, constant angular velocity. Recall that angular velocity  was defined as the time rate of change of angle :
where  is the angle of rotation as seen in . The relationship between angular velocity  and linear velocity  was also defined in Rotation Angle and Angular Velocity as
or
where  is the radius of curvature, also seen in . According to the sign convention, the counter clockwise direction is considered as positive direction and clockwise direction as negative
Angular velocity is not constant when a skater pulls in her arms, when a child starts up a merry-go-round from rest, or when a computer’s hard disk slows to a halt when switched off. In all these cases, there is an angular acceleration, in which  changes. The faster the change occurs, the greater the angular acceleration. Angular acceleration  is defined as the rate of change of angular velocity. In equation form, angular acceleration is expressed as follows:
where  is the change in angular velocity and  is the change in time. The units of angular acceleration are , or . If  increases, then  is positive. If  decreases, then  is negative.
If the bicycle in the preceding example had been on its wheels instead of upside-down, it would first have accelerated along the ground and then come to a stop. This connection between circular motion and linear motion needs to be explored. For example, it would be useful to know how linear and angular acceleration are related. In circular motion, linear acceleration is tangent to the circle at the point of interest, as seen in . Thus, linear acceleration is called tangential acceleration .
Linear or tangential acceleration refers to changes in the magnitude of velocity but not its direction. We know from Uniform Circular Motion and Gravitation that in circular motion centripetal acceleration,

, refers to changes in the direction of the velocity but not its magnitude. An object undergoing circular motion experiences centripetal acceleration, as seen in . Thus,

 and

 are perpendicular and independent of one another. Tangential acceleration

 is directly related to the angular acceleration

 and is linked to an increase or decrease in the velocity, but not its direction.
Now we can find the exact relationship between linear acceleration  and angular acceleration . Because linear acceleration is proportional to a change in the magnitude of the velocity, it is defined (as it was in One-Dimensional Kinematics) to be
For circular motion, note that , so that
The radius  is constant for circular motion, and so . Thus,
By definition, . Thus,
or
These equations mean that linear acceleration and angular acceleration are directly proportional. The greater the angular acceleration is, the larger the linear (tangential) acceleration is, and vice versa. For example, the greater the angular acceleration of a car’s drive wheels, the greater the acceleration of the car. The radius also matters. For example, the smaller a wheel, the smaller its linear acceleration for a given angular acceleration .
So far, we have defined three rotational quantities— , and . These quantities are analogous to the translational quantities , and .  displays rotational quantities, the analogous translational quantities, and the relationships between them.

### Section Summary

1. Uniform circular motion is the motion with a constant angular velocity .
2. In non-uniform circular motion, the velocity changes with time and the rate of change of angular velocity (i.e. angular acceleration) is .
3. Linear or tangential acceleration refers to changes in the magnitude of velocity but not its direction, given as .
4. For circular motion, note that , so that

5. The radius r is constant for circular motion, and so . Thus,

6. By definition, . Thus,
    or

### Conceptual Questions

### Problems & Exercises


# Rotational Motion and Angular Momentum

## Kinematics of Rotational Motion

### Learning Objectives

By the end of this section, you will be able to:
1. Observe the kinematics of rotational motion.
2. Derive rotational kinematic equations.
3. Evaluate problem solving strategies for rotational kinematics.
Just by using our intuition, we can begin to see how rotational quantities like , , and  are related to one another. For example, if a motorcycle wheel has a large angular acceleration for a fairly long time, it ends up spinning rapidly and rotates through many revolutions. In more technical terms, if the wheel’s angular acceleration  is large for a long period of time , then the final angular velocity  and angle of rotation  are large. The wheel’s rotational motion is exactly analogous to the fact that the motorcycle’s large translational acceleration produces a large final velocity, and the distance traveled will also be large.
Kinematics is the description of motion. The kinematics of rotational motion describes the relationships among rotation angle, angular velocity, angular acceleration, and time. Let us start by finding an equation relating , , and . To determine this equation, we recall a familiar kinematic equation for translational, or straight-line, motion:
Note that in rotational motion , and we shall use the symbol  for tangential or linear acceleration from now on. As in linear kinematics, we assume  is constant, which means that angular acceleration  is also a constant, because . Now, let us substitute  and  into the linear equation above:
The radius  cancels in the equation, yielding
where  is the initial angular velocity. This last equation is a kinematic relationship among , , and  —that is, it describes their relationship without reference to forces or masses that may affect rotation. It is also precisely analogous in form to its translational counterpart.
Starting with the four kinematic equations we developed in One-Dimensional Kinematics, we can derive the following four rotational kinematic equations (presented together with their translational counterparts):
In these equations, the subscript 0 denotes initial values (, , and  are initial values), and the average angular velocity  and average velocity  are defined as follows:
The equations given above in  can be used to solve any rotational or translational kinematics problem in which  and  are constant.
There is translational motion even for something spinning in place, as the following example illustrates.  shows a fly on the edge of a rotating microwave oven plate. The example below calculates the total distance it travels.

### Section Summary

1. Kinematics is the description of motion.
2. The kinematics of rotational motion describes the relationships among rotation angle, angular velocity, angular acceleration, and time.
3. Starting with the four kinematic equations we developed in the One-Dimensional Kinematics, we can derive the four rotational kinematic equations (presented together with their translational counterparts) seen in .
4. In these equations, the subscript 0 denotes initial values ( and  are initial values), and the average angular velocity  and average velocity  are defined as follows:

### Problems & Exercises


# Rotational Motion and Angular Momentum

## Dynamics of Rotational Motion: Rotational Inertia

### Learning Objectives

By the end of this section, you will be able to:
1. Understand the relationship between force, mass and acceleration.
2. Study the turning effect of force.
3. Study the analogy between force and torque, mass and moment of inertia, and linear acceleration and angular acceleration.
If you have ever spun a bike wheel or pushed a merry-go-round, you know that force is needed to change angular velocity as seen in . In fact, your intuition is reliable in predicting many of the factors that are involved. For example, we know that a door opens slowly if we push too close to its hinges. Furthermore, we know that the more massive the door, the more slowly it opens. The first example implies that the farther the force is applied from the pivot, the greater the angular acceleration; another implication is that angular acceleration is inversely proportional to mass. These relationships should seem very similar to the familiar relationships among force, mass, and acceleration embodied in Newton’s second law of motion. There are, in fact, precise rotational analogs to both force and mass.
To develop the precise relationship among force, mass, radius, and angular acceleration, consider what happens if we exert a force  on a point mass  that is at a distance  from a pivot point, as shown in . Because the force is perpendicular to , an acceleration  is obtained in the direction of . We can rearrange this equation such that  and then look for ways to relate this expression to expressions for rotational quantities. We note that , and we substitute this expression into , yielding
Recall that torque is the turning effectiveness of a force. In this case, because  is perpendicular to , torque is simply . So, if we multiply both sides of the equation above by , we get torque on the left-hand side. That is,

or
This last equation is the rotational analog of Newton’s second law (), where torque is analogous to force, angular acceleration is analogous to translational acceleration, and  is analogous to mass (or inertia). The quantity  is called the rotational inertia or moment of inertia of a point mass  a distance  from the center of rotation.

### Rotational Inertia and Moment of Inertia

Before we can consider the rotation of anything other than a point mass like the one in , we must extend the idea of rotational inertia to all types of objects. To expand our concept of rotational inertia, we define the moment of inertia

 of an object to be the sum of

 for all the point masses of which it is composed. That is,

. Here

 is analogous to

 in translational motion. Because of the distance

, the moment of inertia for any object depends on the chosen axis. Actually, calculating

 is beyond the scope of this text except for one simple case—that of a hoop, which has all its mass at the same distance from its axis. A hoop’s moment of inertia around its axis is therefore

, where

 is its total mass and  its radius. (We use  and  for an entire object to distinguish them from  and  for point masses.) In all other cases, we must consult  (note that the table is piece of artwork that has shapes as well as formulae) for formulas for  that have been derived from integration over the continuous body. Note that  has units of mass multiplied by distance squared (), as we might expect from its definition.
The general relationship among torque, moment of inertia, and angular acceleration is
or
where net  is the total torque from all forces relative to a chosen axis. For simplicity, we will only consider torques exerted by forces in the plane of the rotation. Such torques are either positive or negative and add like ordinary numbers. The relationship in  is the rotational analog to Newton’s second law and is very generally applicable. This equation is actually valid for any torque, applied to any object, relative to any axis.
As we might expect, the larger the torque is, the larger the angular acceleration is. For example, the harder a child pushes on a merry-go-round, the faster it accelerates. Furthermore, the more massive a merry-go-round, the slower it accelerates for the same torque. The basic relationship between moment of inertia and angular acceleration is that the larger the moment of inertia, the smaller is the angular acceleration. But there is an additional twist. The moment of inertia depends not only on the mass of an object, but also on its distribution of mass relative to the axis around which it rotates. For example, it will be much easier to accelerate a merry-go-round full of children if they stand close to its axis than if they all stand at the outer edge. The mass is the same in both cases, but the moment of inertia is much larger when the children are at the edge.

### Test Prep for AP Courses

### Section Summary

1. The farther the force is applied from the pivot, the greater is the angular acceleration; angular acceleration is inversely proportional to mass.
2. If we exert a force  on a point mass  that is at a distance  from a pivot point and because the force is perpendicular to , an acceleration  is obtained in the direction of . We can rearrange this equation such that
                and then look for ways to relate this expression to expressions for rotational quantities. We note that

3. Torque is the turning effectiveness of a force. In this case, because  is perpendicular to , torque is simply . If we multiply both sides of the equation above by , we get torque on the left-hand side. That is,

    or

4. The moment of inertia  of an object is the sum of  for all the point masses  of which it is composed. That is,

5. The general relationship among torque, moment of inertia, and angular acceleration is
    or

### Conceptual Questions

### Problems & Exercises


# Rotational Motion and Angular Momentum

## Rotational Kinetic Energy: Work and Energy Revisited

### Learning Objectives

By the end of this section, you will be able to:
1. Derive the equation for rotational work.
2. Calculate rotational kinetic energy.
3. Demonstrate the Law of Conservation of Energy.
In this module, we will learn about work and energy associated with rotational motion.  shows a worker using an electric grindstone propelled by a motor. Sparks are flying, and noise and vibration are created as layers of steel are pared from the pole. The stone continues to turn even after the motor is turned off, but it is eventually brought to a stop by friction. Clearly, the motor had to work to get the stone spinning. This work went into heat, light, sound, vibration, and considerable rotational kinetic energy.
Work must be done to rotate objects such as grindstones or merry-go-rounds. Work was defined in Uniform Circular Motion and Gravitation for translational motion, and we can build on that knowledge when considering work done in rotational motion. The simplest rotational situation is one in which the net force is exerted perpendicular to the radius of a disk (as shown in ) and remains perpendicular as the disk starts to rotate. The force is parallel to the displacement, and so the net work done is the product of the force times the arc length traveled:
To get torque and other rotational quantities into the equation, we multiply and divide the right-hand side of the equation by , and gather terms:
We recognize that  and , so that
This equation is the expression for rotational work. It is very similar to the familiar definition of translational work as force multiplied by distance. Here, torque is analogous to force, and angle is analogous to distance. The equation  is valid in general, even though it was derived for a special case.
To get an expression for rotational kinetic energy, we must again perform some algebraic manipulations. The first step is to note that , so that
Now, we solve one of the rotational kinematics equations for . We start with the equation
Next, we solve for :
Substituting this into the equation for net  and gathering terms yields
This equation is the work-energy theorem for rotational motion only. As you may recall, net work changes the kinetic energy of a system. Through an analogy with translational motion, we define the term  to be rotational kinetic energy  for an object with a moment of inertia  and an angular velocity :
The expression for rotational kinetic energy is exactly analogous to translational kinetic energy, with  being analogous to  and  to . Rotational kinetic energy has important effects. Flywheels, for example, can be used to store large amounts of rotational kinetic energy in a vehicle, as seen in .
Helicopter pilots are quite familiar with rotational kinetic energy. They know, for example, that a point of no return will be reached if they allow their blades to slow below a critical angular velocity during flight. The blades lose lift, and it is impossible to immediately get the blades spinning fast enough to regain it. Rotational kinetic energy must be supplied to the blades to get them to rotate faster, and enough energy cannot be supplied in time to avoid a crash. Because of weight limitations, helicopter engines are too small to supply both the energy needed for lift and to replenish the rotational kinetic energy of the blades once they have slowed down. The rotational kinetic energy is put into them before takeoff and must not be allowed to drop below this crucial level. One possible way to avoid a crash is to use the gravitational potential energy of the helicopter to replenish the rotational kinetic energy of the blades by losing altitude and aligning the blades so that the helicopter is spun up in the descent. Of course, if the helicopter’s altitude is too low, then there is insufficient time for the blade to regain lift before reaching the ground.

### How Thick Is the Soup? Or Why Don’t All Objects Roll Downhill at the Same Rate?

One of the quality controls in a tomato soup factory consists of rolling filled cans down a ramp. If they roll too fast, the soup is too thin. Why should cans of identical size and mass roll down an incline at different rates? And why should the thickest soup roll the slowest?
The easiest way to answer these questions is to consider energy. Suppose each can starts down the ramp from rest. Each can starting from rest means each starts with the same gravitational potential energy , which is converted entirely to , provided each rolls without slipping. , however, can take the form of  or , and total  is the sum of the two. If a can rolls down a ramp, it puts part of its energy into rotation, leaving less for translation. Thus, the can goes slower than it would if it slid down. Furthermore, the thin soup does not rotate, whereas the thick soup does, because it sticks to the can. The thick soup thus puts more of the can’s original gravitational potential energy into rotation than the thin soup, and the can rolls more slowly, as seen in .
Assuming no losses due to friction, there is only one force doing work—gravity. Therefore the total work done is the change in kinetic energy. As the cans start moving, the potential energy is changing into kinetic energy. Conservation of energy gives
More specifically,
or
So, the initial  is divided between translational kinetic energy and rotational kinetic energy; and the greater  is, the less energy goes into translation. If the can slides down without friction, then  and all the energy goes into translation; thus, the can goes faster.

### Test Prep for AP Courses

### Section Summary

1. The rotational kinetic energy  for an object with a moment of inertia  and an angular velocity  is given by

2. Helicopters store large amounts of rotational kinetic energy in their blades. This energy must be put into the blades before takeoff and maintained until the end of the flight. The engines do not have enough power to simultaneously provide lift and put significant rotational energy into the blades.
3. Work and energy in rotational motion are completely analogous to work and energy in translational motion.
4. The equation for the work-energy theorem for rotational motion is,

### Conceptual Questions

### Problems & Exercises


# Rotational Motion and Angular Momentum

## Angular Momentum and Its Conservation

### Learning Objectives

By the end of this section, you will be able to:
1. Understand the analogy between angular momentum and linear momentum.
2. Observe the relationship between torque and angular momentum.
3. Apply the law of conservation of angular momentum.
Why does Earth keep on spinning? What started it spinning to begin with? And how does an ice skater manage to spin faster and faster simply by pulling her arms in? Why does she not have to exert a torque to spin faster? Questions like these have answers based in angular momentum, the rotational analog to linear momentum.
By now the pattern is clear—every rotational phenomenon has a direct translational analog. It seems quite reasonable, then, to define angular momentum  as
This equation is an analog to the definition of linear momentum as . Units for linear momentum are  while units for angular momentum are . As we would expect, an object that has a large moment of inertia , such as Earth, has a very large angular momentum. An object that has a large angular velocity , such as a centrifuge, also has a rather large angular momentum.
When you push a merry-go-round, spin a bike wheel, or open a door, you exert a torque. If the torque you exert is greater than opposing torques, then the rotation accelerates, and angular momentum increases. The greater the net torque, the more rapid the increase in . The relationship between torque and angular momentum is
This expression is exactly analogous to the relationship between force and linear momentum, . The equation  is very fundamental and broadly applicable. It is, in fact, the rotational form of Newton’s second law.

### Conservation of Angular Momentum

We can now understand why Earth keeps on spinning. As we saw in the previous example, . This equation means that, to change angular momentum, a torque must act over some period of time. Because Earth has a large angular momentum, a large torque acting over a long time is needed to change its rate of spin. So what external torques are there? Tidal friction exerts torque that is slowing Earth’s rotation, but tens of millions of years must pass before the change is very significant. Recent research indicates the length of the day was 18 h some 900 million years ago. Only the tides exert significant retarding torques on Earth, and so it will continue to spin, although ever more slowly, for many billions of years.
What we have here is, in fact, another conservation law. If the net torque is zero, then angular momentum is constant or conserved. We can see this rigorously by considering  for the situation in which the net torque is zero. In that case,
implying that
If the change in angular momentum  is zero, then the angular momentum is constant; thus,
or
These expressions are the law of conservation of angular momentum. Conservation laws are as scarce as they are important.
An example of conservation of angular momentum is seen in , in which an ice skater is executing a spin. The net torque on her is very close to zero, because there is relatively little friction between her skates and the ice and because the friction is exerted very close to the pivot point. (Both  and  are small, and so  is negligibly small.) Consequently, she can spin for quite some time. She can do something else, too. She can increase her rate of spin by pulling her arms and legs in. Why does pulling her arms and legs in increase her rate of spin? The answer is that her angular momentum is constant, so that
Expressing this equation in terms of the moment of inertia,
where the primed quantities refer to conditions after she has pulled in her arms and reduced her moment of inertia. Because  is smaller, the angular velocity  must increase to keep the angular momentum constant. The change can be dramatic, as the following example shows.

There are several other examples of objects that increase their rate of spin because something reduced their moment of inertia. Tornadoes are one example. Storm systems that create tornadoes are slowly rotating. When the radius of rotation narrows, even in a local region, angular velocity increases, sometimes to the furious level of a tornado. Earth is another example. Our planet was born from a huge cloud of gas and dust, the rotation of which came from turbulence in an even larger cloud. Gravitational forces caused the cloud to contract, and the rotation rate increased as a result. (See .)
In case of human motion, one would not expect angular momentum to be conserved when a body interacts with the environment as its foot pushes off the ground. Astronauts floating in space aboard the International Space Station have no angular momentum relative to the inside of the ship if they are motionless. Their bodies will continue to have this zero value no matter how they twist about as long as they do not give themselves a push off the side of the vessel.

### Test Prep for AP Courses

### Section Summary

1. Every rotational phenomenon has a direct translational analog , likewise angular momentum  can be defined as
2. This equation is an analog to the definition of linear momentum as . The relationship between torque and angular momentum is

3. Angular momentum, like energy and linear momentum, is conserved. This universally applicable law is another sign of underlying unity in physical laws. Angular momentum is conserved when net external torque is zero, just as linear momentum is conserved when the net external force is zero.

### Conceptual Questions

### Problems & Exercises


# Rotational Motion and Angular Momentum

## Collisions of Extended Bodies in Two Dimensions

### Learning Objectives

By the end of this section, you will be able to:
1. Observe collisions of extended bodies in two dimensions.
2. Examine collision at the point of percussion.
Bowling pins are sent flying and spinning when hit by a bowling ball—angular momentum as well as linear momentum and energy have been imparted to the pins. (See ). Many collisions involve angular momentum. Cars, for example, may spin and collide on ice or a wet surface. Baseball pitchers throw curves by putting spin on the baseball. A tennis player can put a lot of top spin on the tennis ball which causes it to dive down onto the court once it crosses the net. We now take a brief look at what happens when objects that can rotate collide.
Consider the relatively simple collision shown in , in which a disk strikes and adheres to an initially motionless stick nailed at one end to a frictionless surface. After the collision, the two rotate about the nail. There is an unbalanced external force on the system at the nail. This force exerts no torque because its lever arm  is zero. Angular momentum is therefore conserved in the collision. Kinetic energy is not conserved, because the collision is inelastic. It is possible that momentum is not conserved either because the force at the nail may have a component in the direction of the disk’s initial velocity. Let us examine a case of rotation in a collision in .
The above example has other implications. For example, what would happen if the disk hit very close to the nail? Obviously, a force would be exerted on the nail in the forward direction. So, when the stick is struck at the end farthest from the nail, a backward force is exerted on the nail, and when it is hit at the end nearest the nail, a forward force is exerted on the nail. Thus, striking it at a certain point in between produces no force on the nail. This intermediate point is known as the percussion point.
An analogous situation occurs in tennis as seen in . If you hit a ball with the end of your racquet, the handle is pulled away from your hand. If you hit a ball much farther down, for example, on the shaft of the racquet, the handle is pushed into your palm. And if you hit the ball at the racquet’s percussion point (what some people call the “sweet spot”), then little or no force is exerted on your hand, and there is less vibration, reducing chances of a tennis elbow. The same effect occurs for a baseball bat.

### Test Prep for AP Courses

### Section Summary

1. Angular momentum  is analogous to linear momentum and is given by .
2. Angular momentum is changed by torque, following the relationship

3. Angular momentum is conserved if the net torque is zero

           or

    . This equation is known as the law of conservation of angular momentum, which may be conserved in collisions.

### Conceptual Questions

### Problems & Exercises


# Rotational Motion and Angular Momentum

## Gyroscopic Effects: Vector Aspects of Angular Momentum

### Learning Objectives

By the end of this section, you will be able to:
1. Describe the right-hand rule to find the direction of angular velocity, momentum, and torque.
2. Explain the gyroscopic effect.
3. Study how Earth acts like a gigantic gyroscope.
Angular momentum is a vector and, therefore, has direction as well as magnitude. Torque affects both the direction and the magnitude of angular momentum. What is the direction of the angular momentum of a rotating object like the disk in ? The figure shows the right-hand rule used to find the direction of both angular momentum and angular velocity. Both  and  are vectors—each has direction and magnitude. Both can be represented by arrows. The right-hand rule defines both to be perpendicular to the plane of rotation in the direction shown. Because angular momentum is related to angular velocity by , the direction of  is the same as the direction of . Notice in the figure that both point along the axis of rotation.

Now, recall that torque changes angular momentum as expressed by
This equation means that the direction of  is the same as the direction of the torque  that creates it. This result is illustrated in , which shows the direction of torque and the angular momentum it creates.
Let us now consider a bicycle wheel with a couple of handles attached to it, as shown in . (This device is popular in demonstrations among physicists, because it does unexpected things.) With the wheel rotating as shown, its angular momentum is to the woman's left. Suppose the person holding the wheel tries to rotate it as in the figure. Her natural expectation is that the wheel will rotate in the direction she pushes it—but what happens is quite different. The forces exerted create a torque that is horizontal toward the person, as shown in (a). This torque creates a change in angular momentum  in the same direction, perpendicular to the original angular momentum , thus changing the direction of  but not the magnitude of .  shows how  and  add, giving a new angular momentum with direction that is inclined more toward the person than before. The axis of the wheel has thus moved perpendicular to the forces exerted on it, instead of in the expected direction.

This same logic explains the behavior of gyroscopes.  shows the two forces acting on a spinning gyroscope. The torque produced is perpendicular to the angular momentum, thus the direction of the torque is changed, but not its magnitude. The gyroscope precesses around a vertical axis, since the torque is always horizontal and perpendicular to . If the gyroscope is not spinning, it acquires angular momentum in the direction of the torque (), and it rotates around a horizontal axis, falling over just as we would expect.
Earth itself acts like a gigantic gyroscope. Its angular momentum is along its axis and points at Polaris, the North Star. But Earth is slowly precessing (once in about 26,000 years) due to the torque of the Sun and the Moon on its nonspherical shape.

### Test Prep for AP Courses

### Section Summary

1. Torque is perpendicular to the plane formed by  and  and is the direction your right thumb would point if you curled the fingers of your right hand in the direction of . The direction of the torque is thus the same as that of the angular momentum it produces.
2. The gyroscope precesses around a vertical axis, since the torque is always horizontal and perpendicular to . If the gyroscope is not spinning, it acquires angular momentum in the direction of the torque (), and it rotates about a horizontal axis, falling over just as we would expect.
3. Earth itself acts like a gigantic gyroscope. Its angular momentum is along its axis and points at Polaris, the North Star.

### Conceptual Questions

### Problem Exercises


# Fluid Statics

## Introduction to Fluid Statics

Much of what we value in life is fluid: a breath of fresh winter air; the hot blue flame in our gas cooker; the water we drink, swim in, and bathe in; the blood in our veins. What exactly is a fluid? Can we understand fluids with the laws already presented, or will new laws emerge from their study? The physical characteristics of static or stationary fluids and some of the laws that govern their behavior are the topics of this chapter. Fluid Dynamics and Its Biological and Medical Applications explores aspects of fluid flow.


# Fluid Statics

## What Is a Fluid?

### Learning Objectives

By the end of this section, you will be able to:
1. State the common phases of matter.
2. Explain the physical characteristics of solids, liquids, and gases.
3. Describe the arrangement of atoms in solids, liquids, and gases.
Matter most commonly exists as a solid, liquid, gas, or plasma; these states are known as the common phases of matter. Solids have a definite shape and a specific volume, liquids have a definite volume but their shape changes depending on the container in which they are held, gases have neither a definite shape nor a specific volume as their molecules move to fill the container in which they are held, and plasmas also have neither definite shape nor volume. (See .) Liquids, gases, and plasmas are considered to be fluids because they yield to shearing forces, whereas solids resist them. Note that the extent to which fluids yield to shearing forces (and hence flow easily and quickly) depends on a quantity called the viscosity which is discussed in detail in Viscosity and Laminar Flow; Poiseuille’s Law. We can understand the phases of matter and what constitutes a fluid by considering the forces between atoms that make up matter in the three phases.

Atoms in solids are in close contact, with forces between them that allow the atoms to vibrate but not to change positions with neighboring atoms. (These forces can be thought of as springs that can be stretched or compressed, but not easily broken.) Thus a solid resists all types of stress. A solid cannot be easily deformed because the atoms that make up the solid are not able to move about freely. Solids also resist compression, because their atoms form part of a lattice structure in which the atoms are a relatively fixed distance apart. Under compression, the atoms would be forced into one another. Most of the examples we have studied so far have involved solid objects which deform very little when stressed.
In contrast, liquids deform easily when stressed and do not spring back to their original shape once the force is removed because the atoms are free to slide about and change neighbors—that is, they flow (so they are a type of fluid), with the molecules held together by their mutual attraction. When a liquid is placed in a container with no lid on, it remains in the container (providing the container has no holes below the surface of the liquid!). Because the atoms are closely packed, liquids, like solids, resist compression.
Atoms in gases and charged particles in plasmas are separated by distances that are large compared with the size of the particles. The forces between the particles are therefore very weak, except when they collide with one another. Gases and plasmas thus not only flow (and are therefore considered to be fluids) but they are relatively easy to compress because there is much space and little force between the particles. When placed in an open container gases, unlike liquids, will escape. The major distinction is that gases are easily compressed, whereas liquids are not. Plasmas are difficult to contain because they have so much energy. When discussing how substances flow, we shall generally refer to both gases and liquids simply as fluids, and make a distinction between them only when they behave differently.

### Section Summary

1. A fluid is a state of matter that yields to sideways or shearing forces. Liquids and gases are both fluids. Fluid statics is the physics of stationary fluids.

### Conceptual Questions


# Fluid Statics

## Density

### Learning Objectives

By the end of this section, you will be able to:
1. Define density.
2. Calculate the mass of a reservoir from its density.
3. Compare and contrast the densities of various substances.
Which weighs more, a ton of feathers or a ton of bricks? This old riddle plays with the distinction between mass and density. A ton is a ton, of course; but bricks have much greater density than feathers, and so we are tempted to think of them as heavier. (See .)
Density, as you will see, is an important characteristic of substances. It is crucial, for example, in determining whether an object sinks or floats in a fluid. Density is the mass per unit volume of a substance or object. In equation form, density is defined as
where the Greek letter  (rho) is the symbol for density,  is the mass, and  is the volume occupied by the substance.
In the riddle regarding the feathers and bricks, the masses are the same, but the volume occupied by the feathers is much greater, since their density is much lower. The SI unit of density is , representative values are given in . The metric system was originally devised so that water would have a density of , equivalent to . Thus the basic mass unit, the kilogram, was first devised to be the mass of 1000 mL of water, which has a volume of 1000 cm3.
As you can see by examining , the density of an object may help identify its composition. The density of gold, for example, is about 2.5 times the density of iron, which is about 2.5 times the density of aluminum. Density also reveals something about the phase of the matter and its substructure. Notice that the densities of liquids and solids are roughly comparable, consistent with the fact that their atoms are in close contact. The densities of gases are much less than those of liquids and solids, because the atoms in gases are separated by large amounts of empty space.

### Test Prep for AP Courses

### Section Summary

1. Density is the mass per unit volume of a substance or object. In equation form, density is defined as

2. The SI unit of density is .

### Conceptual Questions

### Problems & Exercises


# Fluid Statics

## Pressure

### Learning Objectives

By the end of this section, you will be able to:
1. Define pressure.
2. Explain the relationship between pressure and force.
3. Calculate force given pressure and area.
You have no doubt heard the word pressure being used in relation to blood (high or low blood pressure) and in relation to the weather (high- and low-pressure weather systems). These are only two of many examples of pressures in fluids. Pressure  is defined as
where  is a force applied to an area  that is perpendicular to the force.
A given force can have a significantly different effect depending on the area over which the force is exerted, as shown in . The SI unit for pressure is the pascal, where
In addition to the pascal, there are many other units for pressure that are in common use. In meteorology, atmospheric pressure is often described in units of millibar (mb), where
Pounds per square inch  is still sometimes used as a measure of tire pressure, and millimeters of mercury (mm Hg) is still often used in the measurement of blood pressure. Pressure is defined for all states of matter but is particularly important when discussing fluids.

The force exerted on the end of the tank is perpendicular to its inside surface. This direction is because the force is exerted by a static or stationary fluid. We have already seen that fluids cannot withstand shearing (sideways) forces; they cannot exert shearing forces, either. Fluid pressure has no direction, being a scalar quantity. The forces due to pressure have well-defined directions: they are always exerted perpendicular to any surface. (See the tire in , for example.) Finally, note that pressure is exerted on all surfaces. Swimmers, as well as the tire, feel pressure on all sides. (See .)

### Test Prep for AP Courses

### Section Summary

1. Pressure is the force per unit perpendicular area over which the force is applied. In equation form, pressure is defined as

2. The SI unit of pressure is pascal and .

### Conceptual Questions

### Problems & Exercises


# Fluid Statics

## Variation of Pressure with Depth in a Fluid

### Learning Objectives

By the end of this section, you will be able to:
1. Define pressure in terms of weight.
2. Explain the variation of pressure with depth in a fluid.
3. Calculate density given pressure and altitude.
If your ears have ever popped on a plane flight or ached during a deep dive in a swimming pool, you have experienced the effect of depth on pressure in a fluid. At the Earth’s surface, the air pressure exerted on you is a result of the weight of air above you. This pressure is reduced as you climb up in altitude and the weight of air above you decreases. Under water, the pressure exerted on you increases with increasing depth. In this case, the pressure being exerted upon you is a result of both the weight of water above you and that of the atmosphere above you. You may notice an air pressure change on an elevator ride that transports you many stories, but you need only dive a meter or so below the surface of a pool to feel a pressure increase. The difference is that water is much denser than air, about 775 times as dense.
Consider the container in . Its bottom supports the weight of the fluid in it. Let us calculate the pressure exerted on the bottom by the weight of the fluid. That pressure is the weight of the fluid  divided by the area  supporting it (the area of the bottom of the container):
We can find the mass of the fluid from its volume and density:
The volume of the fluid  is related to the dimensions of the container. It is
where  is the cross-sectional area and  is the depth. Combining the last two equations gives
If we enter this into the expression for pressure, we obtain
The area cancels, and rearranging the variables yields
This value is the pressure due to the weight of a fluid. The equation has general validity beyond the special conditions under which it is derived here. Even if the container were not there, the surrounding fluid would still exert this pressure, keeping the fluid static. Thus the equation  represents the pressure due to the weight of any fluid of average density  at any depth  below its surface. For liquids, which are nearly incompressible, this equation holds to great depths. For gases, which are quite compressible, one can apply this equation as long as the density changes are small over the depth considered.  illustrates this situation.
Atmospheric pressure is another example of pressure due to the weight of a fluid, in this case due to the weight of air above a given height. The atmospheric pressure at the Earth’s surface varies a little due to the large-scale flow of the atmosphere induced by the Earth’s rotation (this creates weather “highs” and “lows”).  However, the average pressure at sea level is given by the standard atmospheric pressure , measured to be
This relationship means that, on average, at sea level, a column of air above  of the Earth’s surface has a weight of , equivalent to . (See .)
What do you suppose is the total pressure at a depth of 10.3 m in a swimming pool? Does the atmospheric pressure on the water’s surface affect the pressure below? The answer is yes. This seems only logical, since both the water’s weight and the atmosphere’s weight must be supported. So the total pressure at a depth of 10.3 m is 2 atm—half from the water above and half from the air above. We shall see in Pascal’s Principle that fluid pressures always add in this way.

### Section Summary

1. Pressure is the weight of the fluid  divided by the area  supporting it (the area of the bottom of the container):

2. Pressure due to the weight of a liquid is given by
    where

### Conceptual Questions

### Problems & Exercises


# Fluid Statics

## Pascal’s Principle

### Learning Objectives

By the end of this section, you will be able to:
1. Define pressure.
2. State Pascal’s principle.
3. Understand applications of Pascal’s principle.
4. Derive relationships between forces in a hydraulic system.
Pressure is defined as force per unit area. Can pressure be increased in a fluid by pushing directly on the fluid? Yes, but it is much easier if the fluid is enclosed. The heart, for example, increases blood pressure by pushing directly on the blood in an enclosed system (valves closed in a chamber). If you try to push on a fluid in an open system, such as a river, the fluid flows away. An enclosed fluid cannot flow away, and so pressure is more easily increased by an applied force.
What happens to a pressure in an enclosed fluid? Since atoms in a fluid are free to move about, they transmit the pressure to all parts of the fluid and to the walls of the container. Remarkably, the pressure is transmitted undiminished. This phenomenon is called Pascal’s principle, because it was first clearly stated by the French philosopher and scientist Blaise Pascal (1623–1662): A change in pressure applied to an enclosed fluid is transmitted undiminished to all portions of the fluid and to the walls of its container.
Pascal’s principle, an experimentally verified fact, is what makes pressure so important in fluids. Since a change in pressure is transmitted undiminished in an enclosed fluid, we often know more about pressure than other physical quantities in fluids. Moreover, Pascal’s principle implies that the total pressure in a fluid is the sum of the pressures from different sources. We shall find this fact—that pressures add—very useful.
Blaise Pascal had an interesting life in that he was home-schooled by his father who removed all of the mathematics textbooks from his house and forbade him to study mathematics until the age of 15. This, of course, raised the boy’s curiosity, and by the age of 12, he started to teach himself geometry. Despite this early deprivation, Pascal went on to make major contributions in the mathematical fields of probability theory, number theory, and geometry. He is also well known for being the inventor of the first mechanical digital calculator, in addition to his contributions in the field of fluid statics.

###     Application of Pascal’s Principle

One of the most important technological applications of Pascal’s principle is found in a hydraulic system, which is an enclosed fluid system used to exert forces. The most common hydraulic systems are those that operate car brakes. Let us first consider the simple hydraulic system shown in .

### Relationship Between Forces in a Hydraulic System

We can derive a relationship between the forces in the simple hydraulic system shown in  by applying Pascal’s principle. Note first that the two pistons in the system are at the same height, and so there will be no difference in pressure due to a difference in depth. Now the pressure due to  acting on area  is simply , as defined by . According to Pascal’s principle, this pressure is transmitted undiminished throughout the fluid and to all walls of the container. Thus, a pressure  is felt at the other piston that is equal to . That is .
But since , we see that .
This equation relates the ratios of force to area in any hydraulic system, providing the pistons are at the same vertical height and that friction in the system is negligible. Hydraulic systems can increase or decrease the force applied to them. To make the force larger, the pressure is applied to a larger area. For example, if a 100-N force is applied to the left cylinder in  and the right one has an area five times greater, then the force out is 500 N. Hydraulic systems are analogous to simple levers, but they have the advantage that pressure can be sent through tortuously curved lines to several places at once.
A simple hydraulic system, such as a simple machine, can increase force but cannot do more work than done on it. Work is force times distance moved, and the wheel cylinder moves through a smaller distance than the pedal cylinder. Furthermore, the more wheels added, the smaller the distance each moves. Many hydraulic systems—such as power brakes and those in bulldozers—have a motorized pump that actually does most of the work in the system. The movement of the legs of a spider is achieved partly by hydraulics. Using hydraulics, a jumping spider can create a force that makes it capable of jumping 25 times its length!

### Section Summary

1. Pressure is force per unit area.
2. A change in pressure applied to an enclosed fluid is transmitted undiminished to all portions of the fluid and to the walls of its container.
3. A hydraulic system is an enclosed fluid system used to exert forces.

### Conceptual Questions

### Problems & Exercises


# Fluid Statics

## Gauge Pressure, Absolute Pressure, and Pressure Measurement

### Learning Objectives

By the end of this section, you will be able to:
1. Define gauge pressure and absolute pressure.
2. Understand the working of aneroid and open-tube barometers.
If you limp into a gas station with a nearly flat tire, you will notice the tire gauge on the airline reads nearly zero when you begin to fill it. In fact, if there were a gaping hole in your tire, the gauge would read zero, even though atmospheric pressure exists in the tire. Why does the gauge read zero? There is no mystery here. Tire gauges are simply designed to read zero at atmospheric pressure and positive when pressure is greater than atmospheric.
Similarly, atmospheric pressure adds to blood pressure in every part of the circulatory system. (As noted in Pascal’s Principle, the total pressure in a fluid is the sum of the pressures from different sources—here, the heart and the atmosphere.) But atmospheric pressure has no net effect on blood flow since it adds to the pressure coming out of the heart and going back into it, too. What is important is how much greater blood pressure is than atmospheric pressure. Blood pressure measurements, like tire pressures, are thus made relative to atmospheric pressure.
In brief, it is very common for pressure gauges to ignore atmospheric pressure—that is, to read zero at atmospheric pressure. We therefore define gauge pressure to be the pressure relative to atmospheric pressure. Gauge pressure is positive for pressures above atmospheric pressure, and negative for pressures below it.
In fact, atmospheric pressure does add to the pressure in any fluid not enclosed in a rigid container. This happens because of Pascal’s principle. The total pressure, or absolute pressure, is thus the sum of gauge pressure and atmospheric pressure:  where  is absolute pressure,  is gauge pressure, and  is atmospheric pressure. For example, if your tire gauge reads 34 psi (pounds per square inch), then the absolute pressure is 34 psi plus 14.7 psi ( in psi), or 48.7 psi (equivalent to 336 kPa).
For reasons we will explore later, in most cases the absolute pressure in fluids cannot be negative. Fluids push rather than pull, so the smallest absolute pressure is zero. (A negative absolute pressure is a pull.) Thus the smallest possible gauge pressure is  (this makes  zero). There is no theoretical limit to how large a gauge pressure can be.
There are a host of devices for measuring pressure, ranging from tire gauges to blood pressure cuffs. Pascal’s principle is of major importance in these devices. The undiminished transmission of pressure through a fluid allows precise remote sensing of pressures. Remote sensing is often more convenient than putting a measuring device into a system, such as a person’s artery.
 shows one of the many types of mechanical pressure gauges in use today. In all mechanical pressure gauges, pressure results in a force that is converted (or transduced) into some type of readout.
An entire class of gauges uses the property that pressure due to the weight of a fluid is given by  Consider the U-shaped tube shown in , for example. This simple tube is called a manometer. In (a), both sides of the tube are open to the atmosphere. Atmospheric pressure therefore pushes down on each side equally so its effect cancels. If the fluid is deeper on one side, there is a greater pressure on the deeper side, and the fluid flows away from that side until the depths are equal.
Let us examine how a manometer is used to measure pressure. Suppose one side of the U-tube is connected to some source of pressure  such as the toy balloon in (b) or the vacuum-packed peanut jar shown in (c). Pressure is transmitted undiminished to the manometer, and the fluid levels are no longer equal. In (b),  is greater than atmospheric pressure, whereas in (c),  is less than atmospheric pressure. In both cases,  differs from atmospheric pressure by an amount , where  is the density of the fluid in the manometer. In (b),  can support a column of fluid of height , and so it must exert a pressure  greater than atmospheric pressure (the gauge pressure  is positive). In (c), atmospheric pressure can support a column of fluid of height , and so  is less than atmospheric pressure by an amount  (the gauge pressure  is negative). A manometer with one side open to the atmosphere is an ideal device for measuring gauge pressures. The gauge pressure is  and is found by measuring .
Mercury manometers are often used to measure arterial blood pressure. An inflatable cuff is placed on the upper arm as shown in . By squeezing the bulb, the person making the measurement exerts pressure, which is transmitted undiminished to both the main artery in the arm and the manometer. When this applied pressure exceeds blood pressure, blood flow below the cuff is cut off. The person making the measurement then slowly lowers the applied pressure and listens for blood flow to resume. Blood pressure pulsates because of the pumping action of the heart, reaching a maximum, called systolic pressure, and a minimum, called diastolic pressure, with each heartbeat. Systolic pressure is measured by noting the value of  when blood flow first begins as cuff pressure is lowered. Diastolic pressure is measured by noting  when blood flows without interruption. The typical blood pressure of a young adult raises the mercury to a height of 120 mm at systolic and 80 mm at diastolic. This is commonly quoted as 120 over 80, or 120/80. The first pressure is representative of the maximum output of the heart; the second is due to the elasticity of the arteries in maintaining the pressure between beats. The density of the mercury fluid in the manometer is 13.6 times greater than water, so the height of the fluid will be 1/13.6 of that in a water manometer. This reduced height can make measurements difficult, so mercury manometers are used to measure larger pressures, such as blood pressure. The density of mercury is such that .
A barometer is a device that measures atmospheric pressure. A mercury barometer is shown in . This device measures atmospheric pressure, rather than gauge pressure, because there is a nearly pure vacuum above the mercury in the tube. The height of the mercury is such that . When atmospheric pressure varies, the mercury rises or falls, giving important clues to weather forecasters. The barometer can also be used as an altimeter, since average atmospheric pressure varies with altitude. Mercury barometers and manometers are so common that units of mm Hg are often quoted for atmospheric pressure and blood pressures.  gives conversion factors for some of the more commonly used units of pressure.

### Section Summary

1. Gauge pressure is the pressure relative to atmospheric pressure.
2. Absolute pressure is the sum of gauge pressure and atmospheric pressure.
3. Aneroid gauge measures pressure using a bellows-and-spring arrangement connected to the pointer of a calibrated scale.
4. Open-tube manometers have U-shaped tubes and one end is always open. It is used to measure pressure.
5. A mercury barometer is a device that measures atmospheric pressure.

### Conceptual Questions

### Problems & Exercises


# Fluid Statics

## Archimedes’ Principle

### Learning Objectives

By the end of this section, you will be able to:
1. Define buoyant force.
2. State Archimedes’ principle.
3. Understand why objects float or sink.
4. Understand the relationship between density and Archimedes’ principle.
When you rise from lounging in a warm bath, your arms feel strangely heavy. This is because you no longer have the buoyant support of the water. Where does this buoyant force come from? Why is it that some things float and others do not? Do objects that sink get any support at all from the fluid? Is your body buoyed by the atmosphere, or are only helium balloons affected? (See .)
Answers to all these questions, and many others, are based on the fact that pressure increases with depth in a fluid. This means that the upward force on the bottom of an object in a fluid is greater than the downward force on the top of the object. There is a net upward, or buoyant force on any object in any fluid. (See .) If the buoyant force is greater than the object’s weight, the object will rise to the surface and float. If the buoyant force is less than the object’s weight, the object will sink. If the buoyant force equals the object’s weight, the object will remain suspended at that depth. The buoyant force is always present whether the object floats, sinks, or is suspended in a fluid.
Just how great is this buoyant force? To answer this question, think about what happens when a submerged object is removed from a fluid, as in .

The space it occupied is filled by fluid having a weight . This weight is supported by the surrounding fluid, and so the buoyant force must equal , the weight of the fluid displaced by the object. It is a tribute to the genius of the Greek mathematician and inventor Archimedes (ca. 287–212 B.C.) that he stated this principle long before concepts of force were well established. Stated in words, Archimedes’ principle is as follows: The buoyant force on an object equals the weight of the fluid it displaces. In equation form, Archimedes’ principle is
where  is the buoyant force and  is the weight of the fluid displaced by the object. Archimedes’ principle is valid in general, for any object in any fluid, whether partially or totally submerged.
Humm … High-tech body swimsuits were introduced in 2008 in preparation for the Beijing Olympics. One concern (and international rule) was that these suits should not provide any buoyancy advantage. How do you think that this rule could be verified?

### Floating and Sinking

Drop a lump of clay in water. It will sink. Then mold the lump of clay into the shape of a boat, and it will float. Because of its shape, the boat displaces more water than the lump and experiences a greater buoyant force. The same is true of steel ships.

### Density and Archimedes’ Principle

Density plays a crucial role in Archimedes’ principle. The average density of an object is what ultimately determines whether it floats. If its average density is less than that of the surrounding fluid, it will float. This is because the fluid, having a higher density, contains more mass and hence more weight in the same volume. The buoyant force, which equals the weight of the fluid displaced, is thus greater than the weight of the object. Likewise, an object denser than the fluid will sink.
The extent to which a floating object is submerged depends on how the object’s density is related to that of the fluid. In , for example, the unloaded ship has a lower density and less of it is submerged compared with the same ship loaded. We can derive a quantitative expression for the fraction submerged by considering density. The fraction submerged is the ratio of the volume submerged to the volume of the object, or
The volume submerged equals the volume of fluid displaced, which we call . Now we can obtain the relationship between the densities by substituting  into the expression. This gives
where  is the average density of the object and  is the density of the fluid. Since the object floats, its mass and that of the displaced fluid are equal, and so they cancel from the equation, leaving

We use this last relationship to measure densities. This is done by measuring the fraction of a floating object that is submerged—for example, with a hydrometer. It is useful to define the ratio of the density of an object to a fluid (usually water) as specific gravity:
where  is the average density of the object or substance and  is the density of water at 4.00°C. Specific gravity is dimensionless, independent of whatever units are used for . If an object floats, its specific gravity is less than one. If it sinks, its specific gravity is greater than one. Moreover, the fraction of a floating object that is submerged equals its specific gravity. If an object’s specific gravity is exactly 1, then it will remain suspended in the fluid, neither sinking nor floating. Scuba divers try to obtain this state so that they can hover in the water. We measure the specific gravity of fluids, such as battery acid, radiator fluid, and urine, as an indicator of their condition. One device for measuring specific gravity is shown in .
There are many obvious examples of lower-density objects or substances floating in higher-density fluids—oil on water, a hot-air balloon, a bit of cork in wine, an iceberg, and hot wax in a “lava lamp,” to name a few. Less obvious examples include lava rising in a volcano and mountain ranges floating on the higher-density crust and mantle beneath them. Even seemingly solid Earth has fluid characteristics.

### More Density Measurements

One of the most common techniques for determining density is shown in .

An object, here a coin, is weighed in air and then weighed again while submerged in a liquid. The density of the coin, an indication of its authenticity, can be calculated if the fluid density is known. This same technique can also be used to determine the density of the fluid if the density of the coin is known. All of these calculations are based on Archimedes’ principle.
Archimedes’ principle states that the buoyant force on the object equals the weight of the fluid displaced. This, in turn, means that the object appears to weigh less when submerged; we call this measurement the object’s apparent weight. The object suffers an apparent weight loss equal to the weight of the fluid displaced. Alternatively, on balances that measure mass, the object suffers an apparent mass loss equal to the mass of fluid displaced. That is
or
The next example illustrates the use of this technique.
This brings us back to Archimedes’ principle and how it came into being. As the story goes, the king of Syracuse gave Archimedes the task of determining whether the royal crown maker was supplying a crown of pure gold. The purity of gold is difficult to determine by color (it can be diluted with other metals and still look as yellow as pure gold), and other analytical techniques had not yet been conceived. Even ancient peoples, however, realized that the density of gold was greater than that of any other then-known substance. Archimedes purportedly agonized over his task and had his inspiration one day while at the public baths, pondering the support the water gave his body. He came up with his now-famous principle, saw how to apply it to determine density, and ran naked down the streets of Syracuse crying “Eureka!” (Greek for “I have found it”). Similar behavior can be observed in contemporary physicists from time to time!

### Section Summary

1. Buoyant force is the net upward force on any object in any fluid. If the buoyant force is greater than the object’s weight, the object will rise to the surface and float. If the buoyant force is less than the object’s weight, the object will sink. If the buoyant force equals the object’s weight, the object will remain suspended at that depth. The buoyant force is always present whether the object floats, sinks, or is suspended in a fluid.
2. Archimedes’ principle states that the buoyant force on an object equals the weight of the fluid it displaces.
3. Specific gravity is the ratio of the density of an object to a fluid (usually water).

### Conceptual Questions

### Problem Exercises


# Fluid Statics

## Cohesion and Adhesion in Liquids: Surface Tension and Capillary Action

### Learning Objectives

By the end of this section, you will be able to:
1. Understand cohesive and adhesive forces.
2. Define surface tension.
3. Understand capillary action.

### Cohesion and Adhesion in Liquids

Children blow soap bubbles and play in the spray of a sprinkler on a hot summer day. (See .) An underwater spider keeps his air supply in a shiny bubble he carries wrapped around him. A technician draws blood into a small-diameter tube just by touching it to a drop on a pricked finger. A premature infant struggles to inflate her lungs. What is the common thread? All these activities are dominated by the attractive forces between atoms and molecules in liquids—both within a liquid and between the liquid and its surroundings.
Attractive forces between molecules of the same type are called cohesive forces. Liquids can, for example, be held in open containers because cohesive forces hold the molecules together. Attractive forces between molecules of different types are called adhesive forces. Such forces cause liquid drops to cling to window panes, for example. In this section we examine effects directly attributable to cohesive and adhesive forces in liquids.

### Surface Tension

Cohesive forces between molecules cause the surface of a liquid to contract to the smallest possible surface area. This general effect is called surface tension. Molecules on the surface are pulled inward by cohesive forces, reducing the surface area. Molecules inside the liquid experience zero net force, since they have neighbors on all sides.
The model of a liquid surface acting like a stretched elastic sheet can effectively explain surface tension effects. For example, some insects can walk on water (as opposed to floating in it) as we would walk on a trampoline—they dent the surface as shown in (a). (b) shows another example, where a needle rests on a water surface. The iron needle cannot, and does not, float, because its density is greater than that of water. Rather, its weight is supported by forces in the stretched surface that try to make the surface smaller or flatter. If the needle were placed point down on the surface, its weight acting on a smaller area would break the surface, and it would sink.

Surface tension is proportional to the strength of the cohesive force, which varies with the type of liquid. Surface tension  is defined to be the force F per unit length  exerted by a stretched liquid membrane:
 lists values of  for some liquids. For the insect of (a), its weight  is supported by the upward components of the surface tension force: , where  is the circumference of the insect’s foot in contact with the water.  shows one way to measure surface tension. The liquid film exerts a force on the movable wire in an attempt to reduce its surface area. The magnitude of this force depends on the surface tension of the liquid and can be measured accurately.
Surface tension is the reason why liquids form bubbles and droplets. The inward surface tension force causes bubbles to be approximately spherical and raises the pressure of the gas trapped inside relative to atmospheric pressure outside. It can be shown that the gauge pressure     inside a spherical bubble is given by
where  is the radius of the bubble. Thus the pressure inside a bubble is greatest when the bubble is the smallest. Another bit of evidence for this is illustrated in . When air is allowed to flow between two balloons of unequal size, the smaller balloon tends to collapse, filling the larger balloon.
Our lungs contain hundreds of millions of mucus-lined sacs called alveoli, which are very similar in size, and about 0.1 mm in diameter. (See .) You can exhale without muscle action by allowing surface tension to contract these sacs. Medical patients whose breathing is aided by a positive pressure respirator have air blown into the lungs, but are generally allowed to exhale on their own. Even if there is paralysis, surface tension in the alveoli will expel air from the lungs. Since pressure increases as the radii of the alveoli decrease, an occasional deep cleansing breath is needed to fully reinflate the alveoli. Respirators are programmed to do this and we find it natural, as do our companion dogs and cats, to take a cleansing breath before settling into a nap.
The tension in the walls of the alveoli results from the membrane tissue and a liquid on the walls of the alveoli containing a long lipoprotein that acts as a surfactant (a surface-tension reducing substance). The need for the surfactant results from the tendency of small alveoli to collapse and the air to fill into the larger alveoli making them even larger (as demonstrated in ). During inhalation, the lipoprotein molecules are pulled apart and the wall tension increases as the radius increases (increased surface tension). During exhalation, the molecules slide back together and the surface tension decreases, helping to prevent a collapse of the alveoli. The surfactant therefore serves to change the wall tension so that small alveoli don’t collapse and large alveoli are prevented from expanding too much. This tension change is a unique property of these surfactants, and is not shared by detergents (which simply lower surface tension). (See .)
If water gets into the lungs, the surface tension is too great and you cannot inhale. This is a severe problem in resuscitating drowning victims. A similar problem occurs in newborn infants who are born without this surfactant—their lungs are very difficult to inflate. This condition is known as hyaline membrane disease and is a leading cause of death for infants, particularly in premature births. Some success has been achieved in treating hyaline membrane disease by spraying a surfactant into the infant’s breathing passages. Emphysema produces the opposite problem with alveoli. Alveolar walls of emphysema victims deteriorate, and the sacs combine to form larger sacs. Because pressure produced by surface tension decreases with increasing radius, these larger sacs produce smaller pressure, reducing the ability of emphysema victims to exhale. A common test for emphysema is to measure the pressure and volume of air that can be exhaled.

### Adhesion and Capillary Action

Why is it that water beads up on a waxed car but does not on bare paint? The answer is that the adhesive forces between water and wax are much smaller than those between water and paint. Competition between the forces of adhesion and cohesion are important in the macroscopic behavior of liquids. An important factor in studying the roles of these two forces is the angle  between the tangent to the liquid surface and the surface. (See .) The contact angle  is directly related to the relative strength of the cohesive and adhesive forces. The larger the strength of the cohesive force relative to the adhesive force, the larger  is, and the more the liquid tends to form a droplet. The smaller  is, the smaller the relative strength, so that the adhesive force is able to flatten the drop.  lists contact angles for several combinations of liquids and solids.
One important phenomenon related to the relative strength of cohesive and adhesive forces is capillary action—the tendency of a fluid to be raised or suppressed in a narrow tube, or capillary tube. This action causes blood to be drawn into a small-diameter tube when the tube touches a drop.
If a capillary tube is placed vertically into a liquid, as shown in , capillary action will raise or suppress the liquid inside the tube depending on the combination of substances. The actual effect depends on the relative strength of the cohesive and adhesive forces and, thus, the contact angle  given in the table. If  is less than , then the fluid will be raised; if  is greater than , it will be suppressed. Mercury, for example, has a very large surface tension and a large contact angle with glass. When placed in a tube, the surface of a column of mercury curves downward, somewhat like a drop. The curved surface of a fluid in a tube is called a meniscus. The tendency of surface tension is always to reduce the surface area. Surface tension thus flattens the curved liquid surface in a capillary tube. This results in a downward force in mercury and an upward force in water, as seen in .

Capillary action can move liquids horizontally over very large distances, but the height to which it can raise or suppress a liquid in a tube is limited by its weight. It can be shown that this height  is given by
If we look at the different factors in this expression, we might see how it makes good sense. The height is directly proportional to the surface tension , which is its direct cause. Furthermore, the height is inversely proportional to tube radius—the smaller the radius , the higher the fluid can be raised, since a smaller tube holds less mass. The height is also inversely proportional to fluid density , since a larger density means a greater mass in the same volume. (See .)
How does sap get to the tops of tall trees? (Recall that a column of water can only rise to a height of 10 m  when there is a vacuum at the top—see .) The question has not been completely resolved, but it appears that it is pulled up like a chain held together by cohesive forces. As each molecule of sap enters a leaf and evaporates (a process called transpiration), the entire chain is pulled up a notch. So a negative pressure created by water evaporation must be present to pull the sap up through the xylem vessels. In most situations, fluids can push but can exert only negligible pull, because the cohesive forces seem to be too small to hold the molecules tightly together. But in this case, the cohesive force of water molecules provides a very strong pull.  shows one device for studying negative pressure. Some experiments have demonstrated that negative pressures sufficient to pull sap to the tops of the tallest trees can be achieved.

### Section Summary

1. Attractive forces between molecules of the same type are called cohesive forces.
2. Attractive forces between molecules of different types are called adhesive forces.
3. Cohesive forces between molecules cause the surface of a liquid to contract to the smallest possible surface area. This general effect is called surface tension.
4. Capillary action is the tendency of a fluid to be raised or suppressed in a narrow tube, or capillary tube which is due to the relative strength of cohesive and adhesive forces.

### Conceptual Questions

### Problems & Exercises