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Force exerted by fluid

  1. Nov 29, 2014 #1
    1. The problem statement, all variables and given/known data
    A solid cube, mass m, side length l, is placed in a liquid of uniform density, ρ(rho), at a depth h0 below the surface of the liquid, which is open to the air.
    The upper and lower faces of the cube are horizontal.

    Find the magnitude of force, Fs, exerted on each vertical face of the cube and express it in terms of ρ (rho), l, h, atmospheric pressure p0 and gravity g.

    2. Relevant equations
    Pressure, p, at any point in the liquid = (p0 + ρg(h0+x))


    3. The attempt at a solution
    surface integral
    Fs = -n ∫ (p0 + ρg(h0 +x))dA
    where n is the unit vector normal to the surface

    Area integral of a rectangle with side lengths a & b written as 2 single integrals
    Fs = -n ∫ [ ∫ (p0 + ρg(h0 +x))dy]dx . . . . . . . . . . (first integral limits 0 & a, second limits 0 & b)
    Fs = -n ∫ (p0 + ρg(h0 +x)) b dx
    Fs = -b (p0 + ρgh0) a + ½*ρ0ga2) n
    F
    s = -ab (p0 + ρ0g(h0 + ½a)) n
    remembering that ab = l*l = area of cube side = l2
    so
    Fs = -l2 (p0 + ρ0g (h0 +½a)) n

    is that OK?
    Is the integral look right.
    I worry as I'm doing this at home by myself.
     
  2. jcsd
  3. Nov 29, 2014 #2

    SteamKing

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    Staff Emeritus
    Science Advisor
    Homework Helper

    You've still got an 'a' in the expression for Fs. That's not one of the variables requested by the OP.

    You can check your answer w/o using an integral.
     
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