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Surface tension of a torus raindrop

  1. Nov 25, 2012 #1
    1. The problem statement, all variables and given/known data

    When calculating the difference in pressure inside a spherical raindrop, the force exerted by the surface tension is calculated to be 2pi*R*gama, where R is the radius of the drop and gamma is dE/dS (dyne/cm).
    When the shape of the raindrop is said to be that of a torus, the force exerted by the surface tension is calculated to be 2pi*R*gamma + 2pi(R+2r)*gamma (please see attachment).
    My question is simply why does the force in the case of the torus have two components, whereas in the case of a sphere it has only one?

    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Nov 25, 2012 #2

    mfb

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    In both cases, the expression is the length of the boundary (as seen in a 2D-projection) multiplied by gamma. A circle has one boundary, the torus has 2 (inner+outer).
     
  4. Nov 25, 2012 #3
    But isn't the circle's circumference considered a boundary? Or shouldn't it be? Is it not a thin layer of fluid verging on air?
     
  5. Nov 25, 2012 #4

    mfb

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    Of course. That should be the origin of 2pi*R*gamma.
    You get a layer of fluid/air in contact there.
     
  6. Nov 25, 2012 #5
    Okay. Thank you.
     
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