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I Surface Tension of a soap film

  1. Jul 7, 2017 #1

    K41

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    I don't understand, when calculating the force to change the surface area of the film, how the length obtained in the diagram below is "l" (lowercase L).
    Figure_12_08_04a.jpg


    I understand that there are two surfaces to overcome, hence the factor of 2, but why is the length perpedicular to the force used and not the length parallel to the force, since surface tension is defined as the force parallel to a surface per unit length of that surface??
     
    Last edited by a moderator: Jul 7, 2017
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  3. Jul 7, 2017 #2

    Orodruin

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    It is the force parallel to the surface and orthogonal to the section.

    There are three dimensions so in order to specify a direction it is not sufficient to say that it is orthogonal to one direction, which would leave an ambiguity.
     
  4. Jul 7, 2017 #3

    K41

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    I do not understand. Infact I am also now unsure why the factor isn't 4, because I see this as effectively a flattened bubble, which is then being stretched in one direction.
     
  5. Jul 7, 2017 #4
    Surface tension acts like a stretched sheet of rubber. The sheet is attached to the bar that you are pulling to the right. The sheet is pulling back on the bar of length l to the left.
     
  6. Jul 7, 2017 #5

    K41

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    I sketched how I view the problem and attach it. The force due to surface tension acts parallel to a given surface. Going around the path of the sheet, shows that the lengths that contribute to counteracting the force are also parallel (assuming that the sheet has no curvature etc) ??
     

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  7. Jul 7, 2017 #6

    Orodruin

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    As I said, that is wrong. The surface is two dimensional snd what you have drawn is the direction of the section. The force is orthogonal to the section.
     
  8. Jul 7, 2017 #7

    Orodruin

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    What you need to do is a free body diagram of the bar ... If you integrate around the film you get zero, which is the total force on the film.
     
  9. Jul 7, 2017 #8
    i agree with orodruin. The surface tension acts perpendicular to contact surfaces, not parallel. It's like pressure.
     
  10. Jul 7, 2017 #9

    K41

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  11. Jul 7, 2017 #10

    K41

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    Well clearly the applied force acts on the bar! There is also an inter-molecular adhesion force from the liquid onto the bar but I don't know what direction this is in, I assume its perpendicular to the bar. I don't know what other forces act on the bar, provided we also neglect its weight and forces due to motion of air etc.
     
  12. Jul 7, 2017 #11
  13. Jul 7, 2017 #12

    Orodruin

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    To be honest, at this level you really should be reading textbooks, not papers.
     
  14. Jul 7, 2017 #13

    K41

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    I have read Cengel's Fluid Mechanics book Chapter 2 and Streeter's fluid mechanics book (Chapter 1) and neither go into surface tension in any significant detail and that is why I came here. I had already read those papers previously for different reasons. I haven't checked White's book specifically, but for other items, it seemed similar to Streeter's book.

    Any further suggestions?
     
    Last edited: Jul 7, 2017
  15. Jul 7, 2017 #14
    Get a physical chemistry textbook.
     
  16. Jul 7, 2017 #15

    K41

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    Are there any specific ones you recommend and specific chapters?
     
  17. Jul 7, 2017 #16

    Nidum

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  18. Jul 7, 2017 #17

    K41

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    I will leave the explanation due to energy arguments here:

    The work done in keeping the film at its current position is F x d where d is the horizontal length of the film (from the left to the right). The surface energy is equivalent to the surface tension, multiplied by the surface area of the liquid film, which in this case, is the horizontal length, multiplied by the vertical length.

    If it is assumed all the applied work is acting to create the surface (in reality some energy is lost to friction on the frames), then:

    Fd = 2σ dA = 2σdL

    The factor of two accounts for the fact that there are two surfaces. Thus, the d's are cancelled and the subsequent equation is σ = F / 2L. This allows you to calculate the surface tension of a liquid with a known applied force, and known dimensions of a frame, if it is assumed frictional effects are negligible.

    Sources:
    Cengel and Cimbala - Fluid Mechanics Fundamentals and Applications (Mechanical Engineering) (3rd edition)

    Section 3.2.1 Surface Tension of Pure Liquids
    Wakeham, W., et al., Material Properties: Measurement and Data, in Springer Handbook of Experimental Fluid Mechanics, C. Tropea, A.L. Yarin, and J.F. Foss, Editors. 2007, Springer Berlin Heidelberg. p. 85-177.

    I also have several sources that say the surface tension acts parallel to a surface:
    http://aapt.scitation.org/doi/abs/10.1119/1.3619866?journalCode=ajp
    http://iopscience.iop.org/article/10.1088/0031-9120/6/2/001/meta
    Physics vol 2 - Sunil Batra Section 19.9.2 (free preview is available on google books)
    An Introduction to Fluid Dynamics - Batchelor (pg 61) (The passage here will be understood using the diagram in Sunil's book) (free preview is available on google books).

    I would also refer people to the discussions here:
    https://physics.stackexchange.com/questions/150836/why-is-surface-tension-parallel-to-the-interface

    I think Orodruin's explanation, though it was not clear at the time, becomes clearer by imagining the liquid film is quite literally on the plane of the webpage (or book). Another way to look at it, is to imagine we are looking directly on top of the liquid film. So imagine you have a glass of water, and you are literally looking into the glass of water onto the surface. That is the viewpoint.

    Thus the applied force is acting tangential to the surface of the film (remembering we are looking from the top, down onto the film). The surface tension force opposes this action by generating an equal but opposite force, tangential to the surface and also within the plane of the paper, thus the system is in a state of equilibrium. The value of the surface tension is constant, but the opposing force generated is larger because the area has increased and the surface tension force is defined as surface tension multiplied by length.

    If the applied force is removed, the film moves to the left, whereby the film uses the least amount of energy to maintain its state of equilibrium.

    PS: The liquid film remains attached to the sliding rod because of adhesion forces.
     
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