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Fluid Statics Problem

  1. May 10, 2007 #1
    1. The problem statement, all variables and given/known data
    The gate shown is 3m wide and for analysis can be considered massless. For what depth of water will this rectangular gate be in equilibrium as shown?

    [​IMG]


    2. Relevant equations

    These are the ones that I'm trying to use, my guide does it differently and gets the right answer. I'm trying to figure out where my mistake is..

    [tex]F_Ry'-F*L=0[/tex]
    [tex]F_Ry'=\int{pydA}[/tex]


    3. The attempt at a solution
    NB! The variable y in my equation is the êta in my drawing. Please ignore the y and h symbols in the drawing.

    [tex]d=ysin{\theta}[/tex]

    [tex]p=\rho gysin{\theta}[/tex]

    [tex]F_Ry'=sin{\theta}\rho gw \int_0^{sin{\theta}y}y^2dy[/tex]
    =
    [tex]\frac{3}{16}\rho gwy^3[/tex]

    I put this into the momentum equation and solve for d giving me 2,44m.
    The answer in the book is 2,66m.
    Does anyone see some major errors in my thinking, I'm sure there are some :)

    I hope my formulation is good enough..
     
    Last edited: May 10, 2007
  2. jcsd
  3. May 10, 2007 #2
    Hello,

    The pressure in the depth [tex]h[/tex] is [tex]\rho gh[/tex].
    In your formulation, the depth in your pressure [tex]p[/tex] should be [tex]d-y\sin\theta[/tex].
    The torque by the water is
    [tex]\int_{0}^{\frac{d}{\sin\theta}}\rho g(d-y\sin\theta)y(w\cdot dy)=\frac{1}{6}\frac{\rho gwd^3}{\sin^2\theta}[/tex]

    Hope it helpful.
     
  4. May 10, 2007 #3
    Yes! I get it, thanks alot. Variables are freaking me out...
     
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