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I Surface Tension and force required to pull a thin ring

  1. Dec 25, 2016 #1
    Hi everyone,

    While studying surface tension I came across a(numerical example) problem which asks to find the force required to pull a circular glass plate of some given radius from the surface of water(assume plane of plate on water surface). Value of surface tension of water is also given in problem. The solution was done by simply multiplying surface tension of water with length of circumference. And it ends.

    Now what I am thinking is that what would be the solution if instead of water the circular plate were in mercury which do not stick on glass. Since mercury does not stick(wet) on glass (as cohesive force of mercury molecules is much greater than adhesive force between mercury and glass) therefore no or very little force should be required to remove plate from mercury than that required for glass. The force required to remove glass plate from water would be high because molecules of water strongly holding the glass and will not leave it so force is required to break cohesive force of water that is surface tension of water. But in case of mercury, mercury molecules do not hold glass so tightly therefore while trying to remove plate from mercury, the mercury molecules leave the plate even when the applied force is less than that required to break the cohesive force between mercury molecules.

    This is what I suppose. Is my understanding true?

    Regards!
    Thanks a bunch.
     
  2. jcsd
  3. Dec 26, 2016 #2

    Simon Bridge

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    You seem to be wondering if the example problem was perhaps being a little simplistic - did they neglect the adhesion between water and the plate, considering only cohesive effects? In Mercury, the surface tension would be higher because of the cohesion is higher, but adhesion to the plate would be lower.

    You should look at the rest of the passage before the example to see what assumptions they used in the problem, and see if you can investigate how surface tension relates to adhesion.
     
  4. Dec 26, 2016 #3

    Andy Resnick

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    Without seeing the problem it's unclear where the misunderstanding is. For example, using a circular *ring* to measure interfacial energy is common (https://en.wikipedia.org/wiki/Du_Noüy_ring_method) as is using a rectangular plate (https://en.wikipedia.org/wiki/Wilhelmy_plate), neither of which seems to correspond to your problem.
     
  5. Dec 31, 2016 #4
    Exactly.

    The problem is, the complete passage of question is not mentioning any assumption but just solved ignoring how (less)adhesive force can affect the answer.

    Below is the example problem with its solution:

    Regards
     

    Attached Files:

  6. Dec 31, 2016 #5

    Andy Resnick

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    Thanks for posting the actual question: this is a poorly-posed problem that makes several badly-conceived assumptions. For example, as the disc is pulled up two things can happen. If the disc is non-wetting, the contact line must move across the face of the disc (dewetting) in order to separate the disc from fluid. If the disc is wetting, there will be some mass of fluid that remains attached to the disc (pendant drop). While drop necking/rupture process is mostly understood, contact line motion is not. Neither process can be ignored.

    Even if the question is restricted to an initial raising force, the calculation neglects the pressure jump that occurs across the curved air-fluid interface.- the pressure jump will be very high initially, resulting in a force much greater than your question indicates. For some reason I think this type of measurement has been performed and published, it's related to capillary pressure and applied to multiphase flow problems.
     
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