Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
    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


    User Avatar
    Science Advisor
    Homework Helper

    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.

  5. Feb 4, 2007 #4


    User Avatar
    Science Advisor
    Homework Helper

    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook