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If the weight of the fluid can be neglected, the pressure in a fluid is the same throughout its volume. We used that approximation in our discussion of bulk stress and strain in Section 11-6. But often the fluid’s weight is not negligible. Atmospheric pressure is less at high altitude than at sea level, which is why an airplane cabin has to be pressurized when flying at 35,000 feet. When you dive into deep water, your ears tell you that the pressure increases rapidly with increasing depth below the surface.

We can derive a general relation between the pressure p at any point in a fluid at rest and the elevation y of the point. We’ll assume that the density p and the acceleration due to gravity g are the same throughout the fluid. If the fluid is in equilibrium, every volume element is in equilibrium. Consider a thin element of fluid, with height dy (Fig.14-2). The bottom and top surfaces each have area A, and they are at elevations y and y + dy above some reference level where y = O


What are the other forces on this fluid element? Call the pressure at the bottom surface p; the total y-component of upward force on this surface is pA. The pressure at the top surface is p + dp, and the total y-component of (downward) force on the top surface is -(p + dp)A. The fluid element is in equilibrium, so the total y-component.

This equation shows that when y increases, p decreases; that is, as we move upward in the fluid, pressure decreases, as we expect. If p, and P2 are the pressures at elevations y, and Y2′ respectively, and if p and g are constant, then

It’s often convenient to express Eq. (14-5) in terms of the depth below the surface of a fluid (Fig. 14-3). Take point 1 at any level in the fluid and let p represent the pressure at this point. Take point 2 at the surface of the fluid, where the pressure is Po (subscript zero for zero depth). The depth of point 1below the surface is h =Y2 – y” and Eq. (14-5) becomes the pressure p at a depth h is greater than the pressure Po at the surface by an amount pgh. Note that the pressure is the same at any two points at the same level in the fluid  The shape of the container does not matter (Fig. 14-4). Equation (14-6) shows that if we increase the pressure Po at the top surface, possibly by using a piston that fits tightly inside the container to push down on the fluid surface, the pressure p at any depth increases by exactly the same amount. This fact was recognized in 1653 by the French scientist Blaise Pascal (1623-1662) and is called Pascal’s law: Pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and the walls of the containing vessel. The hydraulic lift shown schematically in Fig. 14-5 illustrates Pascal’s law. A piston with small cross-section area A, exerts a force F, on the surface of a liquid such as oil. The applied pressure p = F/A, is transmitted through the connecting pipe to a larger piston of area A2• The applied pressure is the same in both cylinders, so the hydraulic lift is a force-multiplying device with a multiplication factor equal to the ratio of the areas of the two pistons. Dentist’s chairs, car lifts and jacks, many elevators, and hydraulic brakes all use this principle a very small difference. But between sea level and the summit of
Mount Everest (8882 m) the density of air changes by nearly a factor of three, and in this case we cannot use Eq. (14-6). Liquids, by contrast, are nearly in compressible, and it is usually a very good approximation to regard their density as independent of pressure. A pressure of several hundred atmospheres will cause only a few percent increase in the density of most liquids..

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