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Hydrostatic Pressure

by courtrigrad
Tags: hydrostatic, pressure
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courtrigrad
#1
Sep14-06, 09:50 PM
P: 1,237
An aquarium 5 ft long, 2 ft wide, and 3 ft deep is full of water. Find (a) the hydrostatic pressure on the bottom of the aquarium, (b) the hydrostatic force on the bottom, and (c) the hydrostatic force on one end of the aquarium.

For (a) the hydrostatic pressure is [tex] P = \rho gd [/tex] which is [tex]187.5 \frac{lb}{ft^{3}} [/tex].
For (b), the hydrostatic force is [tex] F = PA = (187.5 \frac{lb}{ft^{3}})(10 ft) = 1870\frac{lb}{ft^{2}} [/tex]

Now for (c), this is where I become stuck. I know that [tex] P = 62.5(x_{i}) [/tex]. So the water is pushing against the bottom of the aquarium, which is a rectangle. I took a partition of that side, and found the area to be [tex] 5P(\Delta x) = (5\Delta X)62.5x_{i} [/tex]. But I saw the answer, and they have that the area of the partition is [tex] 2\Delta x [/tex]. This in turn messes up the calculation of the integral. What am I doing wrong?

P.S: What exactly does this mean: we choose [tex] x_{i}\in [x_{i-1} , x_{i}] [/tex]? What does [tex] x_{i} [/tex] represent?

Thanks
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courtrigrad
#2
Sep15-06, 02:41 AM
P: 1,237
any ideas?
Bob Jacobson
#3
Oct11-08, 10:37 PM
P: 1
Consider going back to basics. Force equals Pressure times Area. F = pgdA where pg equals 62.5 lb/ft^3.

But the Pressure (and therefore the Force) on the end wall varies with depth unlike the bottom of the tank.

You need to set up an integral to add up all of the strips of area along the side of the tank.

Imagine that each strip has a height of delta-y and a width of 2 feet. The depth is (3-y).

Thus each delta-F = 62.5*(3-y)*2*delta-y.

I picked the depth = (3-y) because the y-axis measures the height of the water. When y is 3, the depth is 0; when y is 0, the depth is 3.

Now integrate that from [3,0].

62.5*FINT((2)*(3-y),y, 3,0)
62.5*FINT((6-2y),y, 3,0)
62.5*(6y-y^2)|[3,0]
62.5(18-9)
562.5


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