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Force due to pressure over a surface

  1. Feb 2, 2013 #1
    1. The problem statement, all variables and given/known data
    I'm trying to find the Force per unit length due to a pressure function across a cylinder immersed in a fluid.


    2. Relevant equations
    Bernouilli's equation,
    $$\mathbf{F}_s = -\mathbf{n} \int_s p dA$$


    3. The attempt at a solution

    I've got an expression for p which takes the form $$p = p(\theta)$$.
    My expression for the unit vector is $$\mathbf{n}=\cos \theta \mathbf{i} + \sin \theta \mathbf{j}$$
    I then get an equation:
    $$\mathbf{F}_s/length = -(\cos \theta \mathbf{i} + \sin \theta \mathbf{j}) \int_s p(\theta) r d\theta$$

    Now it's clear from the solution that I need to have the trig functions inside the integral...but is this mathematically justified? I know that the values of the unit vector change as we move around the surface, but does that mean I can just include them into the integral? Where the i-component for example would become:
    $$\mathbf{F}_i / length = \int_s -p(\theta)r \cos \theta d\theta$$
    Is this right?

    Thanks in advance
     
  2. jcsd
  3. Feb 2, 2013 #2

    jhae2.718

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    The general form of the equation is:
    $${\boldsymbol F}_\mathrm{s} = -\!\!\int\limits_s \!\!p\hat{\boldsymbol n}\mathrm{d}S$$
    If the normal is independent of the variables of integration, then you can write
    $${\boldsymbol F}_\mathrm{s} = -\hat{\boldsymbol n}\!\!\int\limits_s\!\! p\mathrm{d}S$$
    You can see this if you look at the force on a differential fluid element:
    $$\mathrm{d}{\boldsymbol F}_\mathrm{s} = -p\mathrm{d}{\boldsymbol S} = -p\hat{\boldsymbol n}\mathrm{d}S$$
    which is integrated over the surface to find the total pressure force.
     
  4. Feb 2, 2013 #3
    Thanks, that makes perfect sense.
     
  5. Feb 2, 2013 #4

    haruspex

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    Yes, but that's only going to be true for a flat surface. For a curved surface the unit vector will change according to position on the surface.
    tomwilliam2, I feel that you've missed some important facts in the statement of the problem. Please post it word for word.
     
  6. Feb 3, 2013 #5
    It's part of a much longer problem...my mistake was applying the formula which applies only to a flat surface (where the unit vector can be taken outside the integral) instead of applying the general formula, with the unit vector containing elements dependent on the variable of integration.
    All sorted now, and it works out as I was expecting.
    Thanks again
     
  7. Feb 3, 2013 #6
    Where is that general equation from? It makes sense, but I haven't seen it in that form before.
     
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