PP 9.3-9.4_1 - Applications Physics of Hydrostatic Pressure...

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Applications
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Physics of Hydrostatic Pressure and Force A thin horizontal plate of area A m 2 is submerged in a fluid of density ρ kg/m 3 , at a depth of d m below the surface.
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The pressure P on the plate is defined to be the force per unit square: The cylinder of fluid directly above the plate has volume V = Ad , so its mass is m = ρV = ρAd . Hence the force exerted on the plate by the fluid is: gAd mg F ρ = = where g is the acceleration due to gravity gd A F P = =
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The Principle of Direction Invariance At any point in a liquid the pressure is the same in all directions . Thus the pressure in all directions at a depth d in a fluid of density ρ is gd P ρ =
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Example A dam has the shape of a trapezoid. The height is 20m, the width is 50m at the top, and 30m at the bottom. The water level is 4m below the top of the dam. Find the force on the dam due to hydrostatic pressure.
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The difficulty here is that the pressure on the dam increases with depth. So we need to accumulate together the forces on the dam wall at each depth. Fix a vertical x -axis, with the origin at the surface of the water. The water depth is 16m.
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Approximate the pressure on the dam as follows: Divide [0 , 16] into equal subintervals, each of length Δ x = 16 /n Pick points x i * in successive subintervals The i th approximating rectangle has height Δ x , width w i and area A i = w i Δ x .
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Using similar triangles w i = 46 - x i * .
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For small x , the pressure on the i th rectangle, P i , is almost constant, so P i 1000 gx i * Hence the hydrostatic force,
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PP 9.3-9.4_1 - Applications Physics of Hydrostatic Pressure...

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