11.3

Pressure and Depth in a Static Fluid

CONCEPTS AT A GLANCE The deeper an underwater swimmer goes, the more strongly the water pushes on his body and the greater is the pressure that he experiences. To determine the relation between pressure and depth, we turn to Newton’s second law. The Concepts-at-a-Glance chart in Figure 11-5 outlines our approach. This chart is an expanded version of the charts in Figures 4.28 and 10.4. In using the second law, we will focus on two external forces that act on the fluid. One is the gravitational force—that is, the weight of the fluid. The other is the collisional force that is responsible for fluid pressure, as Section 11.2 discusses. Since the fluid is at rest, its acceleration is zero , and it is in equilibrium. By applying the second law in the form , we will derive a relation between pressure and depth. This relation is especially important because it leads to Pascal’s principle (Section 11.5) and Archimedes’ principle (Section 11.6), both of which are essential in describing the properties of static fluids.

Figure 11-5 CONCEPTS AT A GLANCE The gravitational force and the collisional forces exerted by a fluid are used with Newton’s second law to determine the relation between pressure and depth in a static fluid. The deeper that this scuba diver goes, the greater is the water pressure on her. (Stephen Frink/Getty Images)

Figure 11-6 shows a container of fluid and focuses attention on one column of the fluid. The free-body diagram in the figure shows all the vertical forces acting on the column. On the top face (), the fluid pressure P1 generates a downward force whose magnitude is P1A. Similarly, on the bottom face, the pressure P2 generates an upward force of magnitude P2A. The pressure P2 is greater than the pressure P1 because the bottom face supports the weight of more fluid than the upper one does. In fact, the excess weight supported by the bottom face is exactly the weight of the fluid within the column. As the free-body diagram indicates, this weight is mg, where m is the mass of the fluid and g is the magnitude of the acceleration due to gravity. Since the column is in equilibrium, we can set the sum of the vertical forces equal to zero and find that

The mass m is related to the density and the volume V of the column by (Equation 11.1). Since the volume is the cross-sectional area A times the vertical dimension h, we have . With this substitution, the condition for equilibrium becomes . The area A can be eliminated algebraically from this expression, with the result that

| (11.4)

Figure 11-6 (a) A container of fluid in which one column of the fluid is outlined. The fluid is at rest. (b) The free-body diagram, showing the vertical forces acting on the column.

Equation 11.4 indicates that if the pressure P1 is known at a higher level, the larger pressure P2 at a deeper level can be calculated by adding the increment . In determining the pressure increment , we assumed that the density is the same at any vertical distance h or, in other words, the fluid is incompressible. The assumption is reasonable for liquids, since the bottom layers can support the upper layers with little compression. In a gas, however, the lower layers are compressed markedly by the weight of the upper layers, with the result that the density varies with vertical distance. For example, the density of our atmosphere is larger near the earth’s surface than it is at higher altitudes. When applied to gases, the relation can be used only when h is small enough that any variation in is negligible.
A significant feature of Equation 11.4 is that the pressure increment is affected by the vertical distance h, but not by any horizontal distance within the fluid. Conceptual Example 3 helps to clarify this feature.

| Conceptual Example 3 | | | The Hoover Dam

Lake Mead is the largest wholly artificial reservoir in the United States and was formed after the completion of the Hoover Dam in 1936. As Figure 11-7a suggests, the water in the reservoir backs up behind the dam for a considerable distance (about 200 km or 120 miles). Suppose that all the water were removed, except for a relatively narrow vertical column in contact with the dam. Figure 11-7b shows a side view of this hypothetical situation, in which the water against the dam has the same depth as in Figure 11-7a. Would the Hoover Dam still be needed to contain the water in this hypothetical reservoir, or could a much less massive structure do the job?

Figure 11-7 (a) The Hoover Dam in Nevada and Lake Mead behind it. (b) A hypothetical reservoir formed by removing most of the water from Lake Mead. Conceptual Example 3 compares the dam needed for this hypothetical reservoir with the Hoover Dam. (Adam G. Sylvester/Photo Researchers, Inc.)

Reasoning and Solution Since our hypothetical reservoir contains much less water than Lake Mead, it is tempting to say that a much less massive structure than the Hoover Dam would be needed. Not so, however. Hoover Dam would still be needed to contain our hypothetical reservoir. To see why, imagine a small square on the inner face of the dam, located beneath the water. The magnitude of the force on this square is the product of its area and the pressure of the water. But the pressure depends on the depth—that is, on the vertical distance h in Equation 11.4. The horizontal distance of the water behind the dam does not appear in this equation and, therefore, has no effect on the pressure. Consequently, at a given depth the small square experiences a force that is the same in both parts of Figure 11-7. Certainly the dam experiences greater forces generated by greater water pressures at greater depths. But no matter where on the inner face of the dam the square is, the force that the water applies to it depends only on the depth, not on the amount of water backed up behind the dam. Thus, the dam for our imaginary reservoir would sustain the same forces that the Hoover Dam sustains and would need to be equally large.

The next example deals further with the relationship between pressure and depth given by Equation 11.4.

| Example 4 | | | The Swimming Hole

Figure 11-8 shows the cross section of a swimming hole. Points A and B are both located at a distance of below the surface of the water. Find the pressure at each of these two points.

Figure 11-8 The pressures at points A and B are the same, since both points are located at the same vertical distance of 5.50 m beneath the surface of the water.

Reasoning The pressure at point B is the same as that at point A, since both are located at the same vertical distance beneath the surface and only the vertical distance h affects the pressure increment in Equation 11.4. To understand this important feature more clearly, consider the path in Figure 11-8. The pressure decreases on the way up along the vertical segment and increases by the same amount on the way back down along segment . Since no change in pressure occurs along the horizontal segment , the pressure is the same at A and B.
Solution The pressure acting on the surface of the water is the atmospheric pressure of . Using this value as P1 in Equation 11.4, we can determine a value for the pressure P2 at either point A or B, both of which are located 5.50 m under the water. Table 11-1 gives the density of water as .

Figure 11-9 shows an irregularly shaped container of liquid. Reasoning similar to that used in Example 4 leads to the conclusion that the pressure is the same at points A, B, C, and D, since each is at the same vertical distance h beneath the surface. In effect, the arteries in our bodies constitute an irregularly shaped “container” for the blood. The next example examines the blood pressure at different places in this “container.”

Figure 11-9 Since points A, B, C, and D are at the same distance h beneath the liquid surface, the pressure at each of them is the same. (Richard Megna/ Fundamental Photographs)

Problem solving insight The pressure at any point in a fluid depends on the vertical distance h of the point beneath the surface. However, for a given vertical distance, the pressure is the same, no matter where the point is located horizontally in the fluid.

| Example 5 | | | Blood Pressure

Blood in the arteries is flowing, but as a first approximation, the effects of this flow can be ignored and the blood treated as a static fluid. Estimate the amount by which the blood pressure P2 in the anterior tibial artery at the foot exceeds the blood pressure P1 in the aorta at the heart when a person is (a) reclining horizontally as in Figure 11-10a and (b) standing as in Figure 11-10b.

Figure 11-10 The blood pressure in the feet can exceed that in the heart, depending on whether a person is (a) reclining horizontally or (b) standing.

The physics of blood pressure.

Reasoning and Solution

a. | When the body is horizontal, there is little or no vertical separation between the feet and the heart. Since ,

| (11.4)

b. | When an adult is standing up, the vertical separation between the feet and the heart is about 1.35 m, as Figure 11-10b indicates. Table 11-1 gives the density of blood as 1060 kg/m3, so that

Sometimes fluid pressure places limits on how a job can be done. Conceptual Example 6 illustrates how fluid pressure restricts the height to which water can be pumped.

| Conceptual Example 6 | | | Pumping Water

Figure 11-11 shows two methods for pumping water from a well. In one method, the pump is submerged in the water at the bottom of the well, while in the other, it is located at ground level. If the well is shallow, either technique can be used. However, if the well is very deep, only one of the methods works. Which is it?

Figure 11-11 A water pump can be placed at the bottom of a well or at ground level. Conceptual Example 6 discusses the two placements.

Reasoning and Solution To answer this question, we need to examine the nature of the job done by each pump. The pump at the bottom of the well literally pushes water up the pipe. For a very deep well, the column of water becomes very tall, and the pressure at the bottom of the pipe becomes large, due to the pressure increment in Equation 11.4. However, as long as the pump can push with sufficient strength to overcome the large pressure, it can shove the next increment of water into the pipe, so the method can be used for very deep wells. In contrast, the pump at ground level does not push water at all. It removes air from the pipe. As the pump reduces the air pressure within the pipe (see point A), the greater air pressure outside (see point B) pushes water up the pipe. But even the strongest pump can only remove all of the air. Once the air is completely removed, an increase in pump strength does not increase the height to which the water is pushed by the external air pressure. Thus, a ground-level pump can only cause water to rise to a certain maximum height, so it cannot be used for very deep wells. (Incidentally, the pump at ground level works just the way you do when you drink through a straw. You draw some of the air out of the straw, and the external air pressure pushes the liquid up into it.)
Related Homework: Problems 25,77

The physics of pumping water from a well.

Check Your Understanding 1

A scuba diver is swimming under water, and the graph shows a plot of the water pressure acting on the diver as a function of time. In each of the three regions, A → B, B → C, and C → D, does the depth of the diver increase, decrease, or remain constant? (The answer is given at the end of the book.)
Background: The relationship between pressure and depth in a fluid is the focus of this question.
For similar questions (including calculational counterparts), consult Self-Assessment Test 11.1. This test is described at the end of Section 11.6.

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