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Homework Help: Pressure on a Dam (go figure)

  1. Nov 4, 2007 #1
    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. jcsd
  3. Nov 4, 2007 #2

    Astronuc

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    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.
     
  4. Nov 4, 2007 #3
    Hmm. I have never used integration in physics (yet:redface:)...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.:wink:

    Edit:nevermind, I got it using an integral! whoooaahhhh Integrals!
     
    Last edited: Nov 4, 2007
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