A swiming pool is 20 ft wide and 40 ft longand its bottom is an inclined plane, the shallow end having a depth of 3 ft and the deep end, 9ft. If the pool is full of water, find the hydrostatic force on the bottom of the pool.
The 40 ft length travels along the length of the angled portion.
Can anyone explain how to solve this? I have the answer, 303,356 lbs, but I do not understand how it was done. I have tried different ways, mostly involving similar triangles, but I cannot get the answer.
Hydrostatic pressure is P = rho g h, where h is the depth of the pool. Hydrostatic force is the integral of pressure over the area. Call x the variable that goes lengthwise across the pool, so that x goes from 0 to 40ft. Then h(x) = 3 + 3x / 20 (x and h are in ft). Call y the variable that goes widthwise across the pool, so that y goes from 0 to 20ft. Call u the variable that goes down the incline of the pool, so that u goes from 0 to 40.4475ft. x and u are related by u = 1.0112x. We want to integrate P over dudy, since that’s the area element of the bottom surface of the pool.
F = Int(rho g h du dy)
rho and g are contants, and h does not vary with y, so we can pull some of this out to get:
F = rho g delta-y Int(h du)
The integral is not hard if we change du = 1.0112dx. Replacing the value for h that we have and integrating over x = 0…40ft gives a value over the integral of 242.685ft[sup]2[/sup]. delta-y is 20ft. For water, rho g = 62.428 lb/ft[sup]3[/sup]. Put it together and what’ve you got? 303006.6 lb. Close enough?