Variation of Pressure with Depth in a Fluid Quiz

PDF page 495: 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 Figure 11.8. 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):

PDF page 495: 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 Figure 11.8. 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):

PDF page 495: 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 Figure 11.8. 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):

PDF page 496: FIGURE 11.8 The bottom of this container supports the entire weight of the fluid in it. The vertical sides cannot exert an upward force on

PDF page 496: FIGURE 11.8 The bottom of this container supports the entire weight of the fluid in it. The vertical sides cannot exert an upward force on

PDF page 496: the fluid (since it cannot withstand a shearing force), and so the bottom must support it all.

PDF page 495: We can find the mass of the fluid from its volume and density: 11.13 The volume of the fluid is related to the dimensions of the container. It is 11.14 where is the cross-sectional area and is the depth. Combining the last two equations gives 11.15 If we enter this into the expression for pressure, we obtain

PDF page 495: We can find the mass of the fluid from its volume and density: 11.13 The volume of the fluid is related to the dimensions of the container. It is 11.14 where is the cross-sectional area and is the depth. Combining the last two equations gives 11.15 If we enter this into the expression for pressure, we obtain

PDF page 495: The area cancels, and rearranging the variables yields 11.17 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. Example 11.4 illustrates this situation.

PDF page 495: The area cancels, and rearranging the variables yields 11.17 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. Example 11.4 illustrates this situation.

PDF page 496: In Example 11.1, we calculated the mass of water in a large reservoir. We will now consider the pressure and force acting on the dam retaining water. (See Figure 11.9.) The dam is 500 m wide, and the water is 80.0 m deep at the dam. (a) What is the average pressure on the dam due to the water? (b) Calculate the force exerted against the dam and compare it with the weight of water in the dam (previously found to be ). Strategy for (a) The average pressure due to the weight of the water is the pressure at the average depth of 40.0 m, since pressure increases linearly with depth. Solution for (a) The average pressure due to the weight of a fluid is 11.18 Entering the density of water from Table 11.1 and taking to be the average depth of 40.0 m, we obtain

PDF page 496: In Example 11.1, we calculated the mass of water in a large reservoir. We will now consider the pressure and force acting on the dam retaining water. (See Figure 11.9.) The dam is 500 m wide, and the water is 80.0 m deep at the dam. (a) What is the average pressure on the dam due to the water? (b) Calculate the force exerted against the dam and compare it with the weight of water in the dam (previously found to be ). Strategy for (a) The average pressure due to the weight of the water is the pressure at the average depth of 40.0 m, since pressure increases linearly with depth. Solution for (a) The average pressure due to the weight of a fluid is 11.18 Entering the density of water from Table 11.1 and taking to be the average depth of 40.0 m, we obtain

PDF page 496: In Example 11.1, we calculated the mass of water in a large reservoir. We will now consider the pressure and force acting on the dam retaining water. (See Figure 11.9.) The dam is 500 m wide, and the water is 80.0 m deep at the dam. (a) What is the average pressure on the dam due to the water? (b) Calculate the force exerted against the dam and compare it with the weight of water in the dam (previously found to be ). Strategy for (a) The average pressure due to the weight of the water is the pressure at the average depth of 40.0 m, since pressure increases linearly with depth. Solution for (a) The average pressure due to the weight of a fluid is 11.18 Entering the density of water from Table 11.1 and taking to be the average depth of 40.0 m, we obtain

PDF page 497: the water behind the dam. 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 11.22