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Math Help - [SOLVED] Derivation of Equilibrum for a Fluid

  1. #1
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    [SOLVED] Derivation of Equilibrum for a Fluid

    Okay, we have the volume forces: \int \rho \overset{\rightharpoonup }{F}dV

    and the surface forces: -\int p\overset{\rightharpoonup }{n}dA=-\int \nabla p \, dV

    If the fluid is in equilibrum: \int \rho \overset{\rightharpoonup }{F}-\nabla pdV=0

    My question is how from the above you come to \rho \overset{\rightharpoonup }{F}-\nabla p=0
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  2. #2
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    Quote Originally Posted by fobos3 View Post
    Okay, we have the volume forces: \int \rho \overset{\rightharpoonup }{F}dV

    and the surface forces: -\int p\overset{\rightharpoonup }{n}dA=-\int \nabla p \, dV

    If the fluid is in equilibrum: \int \rho \overset{\rightharpoonup }{F}-\nabla pdV=0

    My question is how from the above you come to \rho \overset{\rightharpoonup }{F}-\nabla p=0
    The crucial point is that the formula, \int \rho \vec{F}-\nabla pdV=0, is not just true for integration over a fixed region but is true over any region of the fluid. Further, the functions involved here are continous. If \rho \vec{F}-\nabla p=0 were NOT true at some point then an integral over some very small region around that point could not be 0; if the value were positive there would have to be some small region around the point in which the function was positive and the integral over that region would be positive, not zero. Similarly if it were negative at some point.
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    Quote Originally Posted by fobos3 View Post
    Okay, we have the volume forces: \int \rho \overset{\rightharpoonup }{F}dV

    and the surface forces: -\int p\overset{\rightharpoonup }{n}dA=-\int \nabla p \, dV

    If the fluid is in equilibrum: \int \rho \overset{\rightharpoonup }{F}-\nabla pdV=0

    My question is how from the above you come to \rho \overset{\rightharpoonup }{F}-\nabla p=0
    The crucial point is that the integral, : \int \rho \overset{\rightharpoonup }{F}-\nabla pdV=0 is 0 not just over a fixed region but over any region of fluid. Further, the functions involved are continuous. If \rho \overset{\rightharpoonup }{F}-\nabla p=0 were, say, positive at some point, then there would be a small region in which it were positive and integration over that region would be positive, not negative. Similarly if there were a point at which it was negative.
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