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Intro Fluid Dynamics homework question

  1. Dec 17, 2009 #1
    How much force is necessary to lift a ring, diameter 20mm, made of fine wire, and placed on the surface of water at 20 degrees celsius?

    All the equations from the chapter:
    Shearing stress = F/A = Viscosity*(velocity parallel/h)
    Kinematic Viscosity = viscosity/density
    Change in pressure = 4(surface tension/diameter)
    Capillarity height = 4Tcos(theta)/(g*diameter*density)
    Bulk modulus = -V*(change in pressure/change in V)
    compressibility = 1/Bulk modulus

    I'm learning fluid dynamics on my own for possible research so I don't have any professor or guidance, just a book. This makes solving problems much more difficult. With this problem specifically, I'm lost because I feel like the mass of the ring must be known as part of the problem, but no other information is given. This leads me to believe it has something to do with surface tension, compressibility, or shearing stress (this equation contains force).
    My initial reaction was to solve the first equation for force and substitute the second equality in for shear stress. This leaves:
    F=A*(viscosity*velocity parallel/h)
    but there is no velocity parallel or required height.
    The other equations don't seem to be of any help.
    The answer the book gives is 9.15*10^-3 N
    Thanks for any help,
  2. jcsd
  3. Dec 17, 2009 #2


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    Sounds interesting, what book are you reading?

    I'm quite lost myself when it comes to fluid dynamics, but I'd relate this problem to surface tension. More specifically, the Du Noüy ring: http://en.wikipedia.org/wiki/Du_Noüy_ring_method.

    I got the right answer, but using a bit ad hoc methods (well, maybe not, but had you not supplied the answer, I would probably not be writing this reply). I thought of the thing as creating two surfaces (inner, outer) when you begin to lift the ring. I wrote an equation for the energy, and thus obtained the force.

    EDIT: And yeah, I ignored the weight of the ring thing, as I suppose if you do these things experimentally, you add a counterbalance to get rid of it + I suppose it's supposed to be very, very light.
    Last edited: Dec 17, 2009
  4. Dec 17, 2009 #3
    The book is "Introduction to Fluid Mechanics" by Y. Nakayama. so are you saying you don't use any of the equations the book gives? If possible could you please list out the energy equations, my physics is better but i don't see a solution using energy either. Gravitational is mgh, energy from surface tension would be... ?
  5. Dec 17, 2009 #4


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    Well, that's how I would do it, but again, having little experience with fluid dynamics, there might be other, better, solutions.

    Energy from surface tension would be the tension times the area, i.e. gamma * 2 * 2*pi*r*h. Differentiate w.r.t. h, and you've got the force, I suppose.
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