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Force on a body due to pressure of a fluid

  1. Jul 12, 2015 #1
    • If I placed a anti-symmetric object in water, and if it floats,is it possible for it to move left or right due to a force imbalance in calculating [tex] \int p * \vec{dA} [/tex].
    • I know that if the object is a closed surface, we can apply the divergence theorem and because we also know how pressure changes with position ([itex]\rho g z [/itex]), we can prove that the resultant force can only be in the z direction.
     
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  3. Jul 13, 2015 #2

    Svein

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    Have you tried it in practice?
     
  4. Jul 13, 2015 #3
    I don't think I have something that it very anti-symmetric and floats.

    Ohh, and correction to my second statement, "I know I know that if the object is a closed surface, and if it is completely submerged, we can apply the....."
     
  5. Jul 13, 2015 #4
    So which pressure are you talking here? Are we applying any pressure from above, equivalent to the formula mentioned?? Or is it the pressure inside?? It's that I'm not clear about the procedure here.
     
  6. Jul 13, 2015 #5

    Orodruin

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    You can still apply the divergence theorem when the object floats. The only difference is that the pressure gradient changes at the water surface.
     
  7. Jul 13, 2015 #6

    A.T.

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    Don't you think, that after thousands of years of building boats, we would have found that self-propelling hull shape by now?
     
  8. Jul 13, 2015 #7
    @GlenMedina Essentially, I am talking about the buoyant force.
    @Orodruin Wait, so divergence theorem still holds even if the vector field is discontinuous in a sense?
    @A.T. : I guess not, but I can only put this to rest with a mathematical or very thorough physical explanation . :)
     
  9. Jul 13, 2015 #8

    Orodruin

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    Yes, you can show it for distributions in general. As long as you are careful with what you do with the discontinuities you will be fine. In this case the discontinuity is in the derivative of the pressure field and the integral is well defined without any further assumptions. It is worse for point charges and other types of singularities where the discontinuities essentially give you delta functions.
     
  10. Jul 13, 2015 #9
    Ok, thanks! Also, can you link me to somewhere/some book that explains how to deal with discontinuities? I suspect it would just be in any regular multi book, but I just remember them saying when stuff isn't "nice", you can't apply the theorem.
     
  11. Jul 13, 2015 #10

    A.T.

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    How about considering the pressure forces from air and water separately? The two partial object surfaces aren't closed, but the missing parts are planar and horizontal, so they cannot introduce any horizontal force disbalance.
     
  12. Jul 13, 2015 #11

    Orodruin

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    This is true for the theorem as it is usually presented. If you allow for distributions where things such as ##\Delta \phi(\vec x) = - \delta^{(3)}(\vec x)## are well defined, you will essentially be safe also in these cases.

    Edit: The way this would be handled with the "usual" theorem would be to use the fact that the ##\delta## function is zero (and therefore nice) everywhere except at ##\vec x = 0## to rewrite integrals as an integral over a small surface around ##\vec x = 0##.
     
  13. Jul 13, 2015 #12

    SteamKing

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    Floating objects tend to try to find a point of stability, such that the buoyant force and the weight of the object line up to produce zero net moment. Once stability is achieved, the object is in equilibrium and remains motionless until some external force disturbs its equilibrium.
     
  14. Jul 13, 2015 #13
    Do you mean the surface that intersects the horizontal plane at the surface? So if I had a sphere where half of it is submerged, the missing part would be the great circle around the equator?
     
  15. Jul 15, 2015 #14
    If you could have a boat that just accelerated due to equal pressure of the water on two asymmetric sides, then that would violate laws of thermodynamics.
     
  16. Jul 15, 2015 #15
    How so?
     
  17. Jul 15, 2015 #16
    Well, you would get work for free.
     
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