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Homework Help: Basic Question about Force for Fluid Mechanics

  1. Nov 1, 2013 #1
    1. The problem statement, all variables and given/known data

    The block of ice (temperature 0°C) shown in Figure P9.57 is drawn over a level surface lubricated by a layer of water 0.10 mm thick. Determine the magnitude of the force needed to pull the block with a constant speed of 0.50 m/s. At 0°C, the viscosity of water has the value η = 1.79 × 10^–3 N ∙ s / m^2.

    I have attached the image. The surface is 0.80 x 1.20m, with force pointing to the right.

    2. Relevant equations

    Force = Pressure x Area
    m = pV, p = 1.00 x 10^3 kg/m^3

    not sure how these are helpful.

    3. The attempt at a solution

    Can I just multiply the viscosity by the speed by the area of the surface divided by the thickness of the water to get the force?

    So F= ηv x A / thickness = 1.79 x 10^-3 Ns/m^2 x 0.5 m/s x 0.8 x 1.2 m /0.10 = 8.6 x 10^-2 N?

    The units cancel out.. not sure if this is the right approach though. I don't really know where to begin. Can anyone please help / guide me?

    EDIT: Sorry... it accidentally posted when I couldn't complete the subject title. I would want it entitled "Basic Question about Force for Fluid Mechanics

    Attached Files:

  2. jcsd
  3. Nov 1, 2013 #2
    I've just checked dynamic viscosity from Wikipedia. http://en.wikipedia.org/wiki/Viscosity

    There's the equation ##F=\mu A \frac{u}{y}##. Maybe that can be helpful to check your answer. ##u## is the speed, ##A## is the area, and ##y## the separation.

    Make sure though that you verify that the equation can be applied to your problem.
  4. Nov 1, 2013 #3
    Hmm so if my u = 1.79 × 10–3 N ∙ s / m2, A = 0.80m x 1.20m, v = 0.50 m/s, and y = .10 mm I should get...

    F = uAv/y = 1.79x10-3 Ns/m2 * (0.80m x 1.20m) x 0.50 m/s / (0.10 mm x 10^-3 m/1mm) = 8.6 N

    Can someone please check my math/logic and that my variables are correct? Is "0.10 mm" the right value to use for the separation?

    Thanks for the formula.
  5. Nov 1, 2013 #4


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    Staff: Mentor

  6. Nov 1, 2013 #5
    Thanks for fixing the title :)
  7. Nov 2, 2013 #6
    The way this works is: the tangential force F exerted by the layer of water on the block (and by the ground on the layer of water) is equal to the shear stress τ times the contact area A. According to Newton's law of viscosity, the shear stress in the liquid layer is equal to the velocity gradient dv/dy within the fluid times the viscosity of the fluid μ. In this example, the velocity of the block is V, the velocity of the ground is zero, and thickness of the gap (0.1mm) is h, so the velocity gradient is V/h. So the shear stress on the block is τ = μ (V/h). So the tangential force is F = μ (V/h)A.
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