# Pressure on a Dam (go figure)

1. Nov 4, 2007

### Saladsamurai

Water stands at a depth D=35 m behind the vertical upstream face of a dam of width W=314 m.

Find the net horizontal force on the dam from the gauge pressure of the water:

I am quite lost here. I thought it would just be F=Pressure*Area
=(rho*g*D)(D*W). But that is horribly wrong. What am I missing?

Casey

2. Nov 4, 2007

### Staff: Mentor

Well =(rho*g*D)(D*W) assumes constant pressure along D, which is not the case.

The pressure increases with depth rho * g * x, and one can then use dA = W dx.

The pressure at the surface of the water is pg = 14.7 psi, or 0.101325 MPa (i.e. 1 atm).

The pressure as a function of depth is P(x) = pg + rho*g*x

Try an integral F = Int (P(x) dA) = Int (P(x) W dx) with limits of 0 to D = 35 m.

3. Nov 4, 2007

### Saladsamurai

Hmm. I have never used integration in physics (yet)...I was just picking random problems to practice. this will be a nice first.

So dA=W dx means a differential change in Area corresponds to one in depth. That is, W is constant and here you used "x" for depth.

Now it says use the "gauge pressure" ...so I don't think I need the leading term of 1atm since gauge pressure is just the change due to depth.

So F=Int[P(x)*W dx]
=Int[rho*g*x*W*DX]
=(rho*g*W)Int[ x dx] from x=0-->35

WOW! That worked great! 1.88 *10^9 N. (correct)

Thanks Astronuc!

Casey

not to push my luck, but if you have a moment, maybe you could clear up my misunderstandings Here.

Edit:nevermind, I got it using an integral! whoooaahhhh Integrals!

Last edited: Nov 4, 2007
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