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Homework Help: Hydrostatic force on water tank problem

  1. Feb 4, 2007 #1
    1. The problem statement, all variables and given/known data
    The end of a tank containing water is vertical and has the indicated shape (in attached picture). Explain how to approximate the hydrostatic force against the end of the tank by a Riemann sum. Then express the force as an integral and evaluate it.

    2. Relevant equations
    P=1000gx
    F=P*A (pressure * area)

    3. The attempt at a solution
    I think my main problem is finding the area of the ith strip. The pressure is relatively easy to calculate, and if I know the pressure and area I can integrate to find the hydrostatic force.
     

    Attached Files:

  2. jcsd
  3. Feb 4, 2007 #2

    AlephZero

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    Does the fact that the picture says "10 m (radius)" mean the end of the tank is a semicircle?

    If so, you can use the equation of the circle, x^2 + y^2 = r^2, to find the area of each strip.

    If it is some other shape, the question should tell you what the shape is.
     
  4. Feb 4, 2007 #3
    Yes, the end of the tank is a semicircle. Looking at a similar example in the book, is the length of each strip [tex]2 \sqrt{100-y_i^2}[/tex]? Because in the book's example, the end of the tank is a full circle with radius 3, and they got the length of each strip to be [tex]2 \sqrt{9-y_i^2}[/tex]. I'm not sure how to get this though from the equation of the circle.

    Thanks
     
  5. Feb 4, 2007 #4

    AlephZero

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    The equation of the circle is [itex]x^2 + y^2 = 10^2[/itex]

    So [itex]x = \pm\sqrt{100-y^2}[/itex]

    The strip at height [itex]y_i[/itex] goes from [itex]x = -\sqrt{100-y_i^2}[/itex] to [itex]x = +\sqrt{100-y_i^2}[/itex] so its length is [itex]2\sqrt{100-y_i^2}[/itex]
     
  6. Feb 4, 2007 #5
    Thanks. Can you quickly check if I did this right:

    [tex]\displaystyle{\int_{0}^{5} 62.5y(2 \sqrt{100-y}) \;dy}[/tex]
    [tex]\displaystyle{125 \int_{0}^{5} y \sqrt{100-y} \;dy}[/tex]

    Substituting u=100-y, du=-dy:

    [tex]\displaystyle{-125 \int_{100}^{95} (u+100) \sqrt{u} \;du}[/tex]

    I get the answer to be 1218880 lb for hydrostatic force. This seems really big, so did I do something wrong above?
     
    Last edited: Feb 4, 2007
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