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Force needed to unstick a semi sphere on the ground

  1. Aug 15, 2009 #1

    fluidistic

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    0. The problem statement, all variables and given/known data

    What is the force required to unstick a semi sphere from the ground if the pressure inside the semi sphere is lesser than the one outside it?
    Assume the semi sphere has a radius of [tex]r[/tex].


    1. The attempt at a solution

    [tex]\Delta P = \frac{F_{\text {net}}}{A}[/tex] where [tex]A[/tex] is the surface area of the semi sphere, that is [tex]A=2\pi r^2[/tex] and P is the pressure.
    So I get that the net force (pointing upward) is worth [tex]2\pi r^2 \Delta P[/tex].
    I wonder if I'm correct.
     
  2. jcsd
  3. Aug 15, 2009 #2

    kuruman

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    Not so fast. I assume that one pulls in a direction perpendicular to the flat face. Now suppose that instead of pulling on the hemisphere, you pulled on the flat face. Then the area would be just πr2 giving half as much force. This is the correct answer because the force has the same magnitude and direction over the flat face so it is fair to say that "force is pressure times area" in this case. For the hemisphere, you have to do an integral in which case you will get the same answer. Area elements dA near the "equator" generate little force to oppose the direction of the pull along the "pole."
     
    Last edited: Aug 15, 2009
  4. Aug 15, 2009 #3

    fluidistic

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    Ok thanks kuruman. I over simplified the problem, now I know how it is.
     
  5. Aug 15, 2009 #4

    fluidistic

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    This problem is haunting my brain.
    I'm trying to do it via the integration method.

    I can use the fact that the surface area of the hemisphere is [tex]2\pi r^2[/tex] and it remains to find the force acting on a differential of area ([tex]dA[/tex]) and integrate it. But I don't success in it.
    I realize that only the vertical component of the force matters, that is [tex]F\sin \theta[/tex] but I have to replace [tex]F[/tex] by [tex]\Delta PA[/tex]. I have some difficulties when it comes to find the [tex]A[/tex] (more precisely the [tex]dA[/tex]). So I'd have [tex]dF=\Delta P dA[/tex] because the pressure is constant. Hence [tex]F=\Delta P \int dA[/tex].

    As you can see I'm confused.

    Can someone help me a bit more?
     
  6. Aug 16, 2009 #5

    kuruman

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    Take the axis of symmetry of the hemisphere to be z and assume that you pull along the z-axis. Use standard spherical angles θ and φ. Note that θ is measured from the z-axis. An area element on the surface of the sphere is

    dA = r2sinθ dθ dφ

    The force on this element is dF = p dA in a direction radially in and perpendicular to the surface. The z- component of this is

    (dF)z = r2sinθ cosθ dθ dφ

    Integrate. It is best to double the force over half a hemisphere because of the unconventional limits of integration for θ.
     
  7. Aug 16, 2009 #6

    fluidistic

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    Thanks infinitely. Although I did calculus III (didn't ace it) I feel that was a bit over my head. I'll think about it.
     
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