11.12

Concepts & Calculations

In the behavior of fluids pressure plays an important role. As we have seen in this chapter, pressure is the magnitude of the force acting perpendicular to a surface divided by the area of the surface. Pressure should not be confused, however, with the force itself. The idea of a force is introduced in Chapter 4. The next example serves to emphasize that pressure and force are different concepts. In addition, it reviews the techniques that Chapters 4 and 9 discuss for analyzing objects that are in equilibrium.

| Example 18 | | | Pressure and Force

Figure 11-42a shows a rear view of a loaded two-wheeled wheelbarrow on a horizontal surface. It has balloon tires and a weight of , which is uniformly distributed. The left tire has a contact area with the ground of , whereas the right tire is underinflated and has a contact area of . Find the force and the pressure that each tire applies to the ground.

Figure 11-42 (a) Rear view of a wheelbarrow with balloon tires. (b) Free-body diagram of the wheelbarrow, showing its weight and the forces and that the ground applies, respectively, to the left and right tires.

Concept Questions and Answers Force is a vector. Therefore, both a direction and a magnitude are needed to specify it. Are both a direction and a magnitude needed to specify a pressure?

Answer No. Only a magnitude is needed to specify a pressure because pressure is a scalar quantity, not a vector quantity. The definition of pressure is the magnitude of the force acting perpendicular to a surface divided by the area of the surface. This definition does not include directional information.

The problem asks for the force that each tire applies to the ground. How is this force related to the force that the ground applies to each tire?

Answer As Newton’s law of action–reaction indicates, the force that a tire applies to the ground has the same magnitude but opposite direction as the force that the ground applies to the tire.

Do the left and right tires apply the same force to the ground?

Answer Yes. Since the weight is uniformly distributed, each tire supports half the load. Therefore, each tire applies the same force to the ground.

Do the left and right tires apply the same pressure to the ground?

Answer No. Pressure involves the contact area as well as the magnitude of the force. Since the force magnitudes are the same but the contact areas are different, the pressures are different, too.

Solution Since the wheelbarrow is at rest, it is in equilibrium and the net force acting on it must be zero, as Chapters 4 and 9 discuss. Referring to the free-body diagram in Figure 11-42b and taking upward as the positive direction, we have

where FL and FR are, respectively, the forces that the ground applies to the left and right tires. At equilibrium the net torque acting on the wheelbarrow must also be zero, as Chapter 9 discusses. In the free-body diagram the weight acts at the center of gravity, which is at the center of the wheelbarrow. Using this point as the axis of rotation for torques, we can see that the forces FL and FR have the same lever arm , which is half the width of the wheelbarrow. For this axis, the weight W has a zero lever arm and, hence, a zero torque. Therefore, it does not contribute to the net torque. Remembering that counterclockwise torques are positive and setting the net torque equal to zero, we have

Substituting this result into the force equation gives

The value of 342 N is the magnitude of the upward force that the ground applies to each tire. Using Newton’s action–reaction law, we conclude that each tire applies to the ground a . Using this value and the given contact areas, we find that the left and right tires apply the following pressures PL and PR to the ground:

One manifestation of fluid pressure is the buoyant force that is applied to any object immersed in a fluid. Section 11.6 discusses this force, as well as Archimedes’ principle, which specifies a convenient way to determine it. Example 19 focuses on the essence of this principle.

| Concepts & Calculations Example 19 | | | The Buoyant Force

A father (weight ) and his daughter (weight ) are spending the day at the lake. They are each sitting on a beach ball that is just submerged beneath the water (see Figure 11-43). Ignoring the weight of the air within the balls and the parts of their legs that are underwater, find the radius of each ball.

Figure 11-43 The two bathers are sitting on different-sized beach balls that are just submerged beneath the water.

Concept Questions and Answers Each person and beach ball is in equilibrium, being stationary and having no acceleration. Thus, the net force acting on each is zero. What balances the downward-acting weight in each case?

Answer The downward-acting weight is balanced by the upward-acting buoyant force FB that the water applies to the ball.

In which case is the buoyant force greater?

Answer The buoyant force acting on the father’s beach ball is greater, since it must balance his greater weight.

In the situation described, what determines the magnitude of the buoyant force?

Answer According to Archimedes’ principle, the magnitude of the buoyant force equals the weight of the fluid that the ball displaces. Since the ball is completely submerged, it displaces a volume of water that equals the ball’s volume. The weight of this volume of water is the magnitude of the buoyant force.

Which beach ball has the larger radius?

Answer The father’s ball has the larger volume and the larger radius. This follows because a larger buoyant force acts on that ball. For the buoyant force to be larger, that ball must displace a greater volume of water, according to Archimedes’ principle. Therefore, the volume of that ball is larger, since the balls are completely submerged.

Solution For an object in equilibrium, the net force acting on the object must be zero. Therefore, taking upward to be the positive direction, we have

Archimedes’ principle specifies that the magnitude of the buoyant force is the weight of the water displaced by the ball. Using the definition of density given in Equation 11.1, the mass of the displaced water is , where is the density of water (see Table 11-1) and V is the volume displaced. Since all of the ball is submerged, , assuming that the ball remains spherical. The weight of the displaced water is . With this value for the buoyant force, the force equation becomes

Solving for the radius, we find that

As expected, the radius of the father’s beach ball is greater.

At the end of the problem set for this chapter, you will find homework problems that contain both conceptual and quantitative parts. These problems are grouped under the heading Concepts & Calculations.

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