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Hydrostatic force

  1. Feb 27, 2013 #1
    1. The problem statement, all variables and given/known data
    A triangular plate with height 6 ft and a base of 8 ft is submerged vertically in water so that the top is 3 ft below the surface. Express the hydrostatic force against one side of the plate as an integral and evaluate it. (Recall that the weight density of water is 62.5 lb/ft3.)


    2. Relevant equations

    hydrostatic pressure is [itex] p = \rho g h[/itex]
    3. The attempt at a solution
    The hydrostatic force is the pressure times the area of the surface that is submerged

    the width of the triangle is solved using the following ratio:
    [tex] \frac{w}{8}=\frac{9-x}{6} [/tex]

    [tex] w = \frac{8}{6} (9-x) [/tex]

    The force is then given by:

    [tex] \int_{3}^{9} \rho g \frac{8}{6} (9-x) [/tex]

    Are the upper and lower bounds of the integral correct?
    In this question the weight density is given in lb/ft^3 - does this value replace the value of rho in the above equation or does it represent rho times g?

    Thanks for your help!
     
  2. jcsd
  3. Feb 27, 2013 #2

    SteamKing

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    The weight density of fresh water is 62.5 lb force / ft^3. This value represents rho*g.

    You don't explicitly define x. Your integral has an expression for the width of the triangle (which I would check again), but it lacks dx.

    I would check to see if your relation for the width gives zero width for depths from 0 to 3 feet. If this relation is in error, I suggest a change in the limits of integration to match the depths of the base and apex of the triangle.
     
  4. Feb 28, 2013 #3

    SteamKing

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    Your integral is also missing h to convert rho*g to a pressure.
     
  5. Feb 28, 2013 #4
    Thank you for your help. Based on what you've told, I correct the integral to this:

    [tex] F = \int_{3}^{9} \rho g \frac{8}{6}\left( 9 - x \right) x dx [/tex]

    How does this look?
     
  6. Feb 28, 2013 #5

    SteamKing

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    Since the apex of the triangle is submerged, there can be no additional area from a depth of 3 feet to 0 feet. Your expression for the width of the triangle needs some adjustment to account for this.
     
  7. Mar 1, 2013 #6
    Does this mean that the 9 - x should be actually 6 - x?

    From the question, the triangle is inverted and submerged 3 ft under water... wouldn't the force be from 0 depth to 9 ddepth??
     
    Last edited: Mar 1, 2013
  8. Mar 1, 2013 #7

    SteamKing

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    I think you need to take a moment and make a sketch of the problem.

    The bottom of the triangle is 9 feet below the surface, and the top is 3 feet below.
    From 3 feet below to the surface, there is no triangle, so the width is zero.

    If x is going to be your overall depth measured from the surface, then the vertical position relative to the base of the triangle must be the depth minus 3 feet. The depth is required to determine hydrostatic pressure, but the depth coordinate must be modified when determining position between the base and the top of the triangle.
     
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