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Feb27-12, 01:34 PM
PF Gold
boneh3ad's Avatar
P: 1,503
You have the total weight right, but you still aren't getting the lateral force right. At a given depth, the pressure on the wall at that point is:
[tex]P = \rho g h[/tex]

[itex]P[/itex] is pressure (force per area)
[itex]\rho[/itex] is density (mass per volume)
[itex]g[/itex] is the acceleration due to gravity (length per time per time)
[itex]h[/itex] is depth (length)

You have to make sure that since you insist on using Imperial units, that your units remain consistent. If you are taking your density in pounds per cubic foot, that is essentially the [itex]\rho g[/itex] term.

Now, that gives you the pressure at a given depth. Force is pressure times area, but you know that the pressure varies with depth, so you have to integrate through the depth to get the total force.

[tex]dF = P\;dA = P\;dx\;dh[/tex]

[itex]dA[/itex] is a differential area
[itex]dx[/itex] and [itex]dh[/itex] are differential lengths that describe the area [itex]dA[/itex]

Now let's say the x-direction just the other coordinate for a given wall. You need to integrate the pressure over the area of the wall to get the force.

[tex]F = \int\limits_0^D \int\limits_0^W \rho g h\;dx\;dh[/tex]

[itex]F[/itex] is force
[itex]D[/itex] is the total depth
[itex]W[/itex] is the width of the side
and the rest have already been defined.

For vertical, rectangular walls, this simplifies to:
[tex]F = \rho g W \int\limits_0^D h\;dh[/tex]
[tex]F = \rho g W \left[\frac{h^2}{2}\right]_0^D[/tex]

So, that leaves you with:
[tex]F = \frac{\rho g W D^2}{2}[/tex]

Just make sure your units are consistent.


Looks like Jasso beat me to it. Just see whose description makes more sense to you, haha.