Calculus - Hydrostatic Pressure

Please check my work:

Find the hydrostatic pressure on one end of a water trough full of water, the end of which is a trapezoid with given dimensions: top of trapezoid = 20 feet, sides of trapezoid both = 8 feet, bottom of trapezoid = 12 feet.

Depth of water = 8 feet
Density of water = 62.4 lb/ft^3
gravity = 32.15 ft/s^2

a/(8-xi*) = (4ft)/(8ft)
a = (8-xi*)/2 = 4 - (xi/2)
Wi = 2(6+a) = 2(6+4-(1/2)xi*)=20 - xi*
Ai = Wi*delta x = (20-xi*)delta x
Pi = rho*g*d
Pi = (62.4 lb/ft^3)(32.15 ft/s^2)xi
Fi = Pi*Ai
Fi = (62.4 b/ft^3)(32.1/s^2)xi*(20-xi*) delta x
Fi = Integral from 0 to 8 of:
(62.4)(32.15)x(20-x)dx
Fi = 2006.16*Integral 0 to 8 of: (20x-x^2)dx
=2006.16[10x^2-(x^3)/3] evaluated at 0 and 8
=9.42 x 10^5 (what are the units here? lb/(ft^3*s^2)???

Your density of water in lbs/ft^3 is the weight of water at sea level. It is not mass density. So no need to multiply by 32 ft/sec to turn into a force. That is one error.

I don't follow your work, I probably need a diagram to see what your x and dx is. But I know the area of the end is appx 16*6 or appx 100 ft squared (engineering estimate)and the average depth is about 3 ft, so the area*density water*depth is about E2*62*3 appx 18K lbs, which is about half of your adjusted answer (adjusted by dividing by 32). Recheck your calcs.

also, your depth is wrong. If the slanted sides are 8 ft, the depth is about six feet.

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asked by COFFEE
  1. Hmm. . . Let me think about this one.

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    posted by Josh
  2. 16049.28

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