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Homework Help: Find the hydrostatic force exerted on a sphere

  1. Dec 24, 2009 #1
    1. The problem statement, all variables and given/known data

    a spherical vessel is made by fitting together identical hemispherical halves across a horizontal plane.It is filled completely with water of weight W.What is the the resultant hydrostatic pressure force on the bottom half of the vessel?

    anyone please help me how to work this problem..

    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 28, 2009 #2
    Sounds like a trick question to me. If I say any more I will give it away... but ok. hint: How much of the weight of the water W is being held up by the lower half of the sphere?
     
  4. Dec 29, 2009 #3
    After my post yesterday I realized that there is probably more to it since pressure is the force normal to surface. suda, if you still care about this post, send me a PM and I will help further.
     
  5. Dec 30, 2009 #4
    i m unable to work that.i can't even get a start.please help.
     
  6. Jan 1, 2010 #5
    Ok. I'm going to take a stab at this.

    Essentially what we will do is calculate the weight of an infinitesimally thin column of water, then find the component of that force normal to the surface of the sphere, then integrate. It will be a little difficult to see without a diagram but here goes.

    The volume of a sphere in cylindrical coordinates is

    [tex]\int_{0}^{2\pi}{\int_0^R{\int_{-\sqrt{R^2-\rho^2}}^{+\sqrt{R^2-\rho^2}}{\rho dz d\rho d\phi}}}[/tex]

    If we drop the rho and phi integration and perform just the z integration we get

    [tex]2\rho \sqrt{R^2-\rho^2} d\rho d\phi[/tex]

    This is the volume of a column that lies at angle phi and runs from [tex]z=-\sqrt{R^2-\rho^2}[/tex] to [tex]z=+\sqrt{R^2-\rho^2}[/tex]

    (phi doesn't appear in the expression because of the symmetry)

    Multiplying by [tex]3W/4\pi R^3[/tex], the weight-density of the water, we get

    [tex]\frac{3W}{2\pi R^3}\rho \sqrt{R^2-\rho^2} d\rho d\phi[/tex]

    This is the weight of that column of water. If we performed the integration over rho and phi of this quantity as is, we would get W, the weight of the water in the sphere. The force of this infinitesimal column weight is in the vertical direction. What we want to do is find the component normal to the surface of the sphere and integrate that over rho and phi. I'll let you take it from here. I got [tex]\frac{3}{8}W[/tex]. Do you know if that is the answer?
     
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