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Immersing an object in a fluid

  1. Nov 27, 2009 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    Consider a liquid whose density [tex]\rho[/tex] varies such that [tex]\rho (x)=Kx[/tex] where [tex]K[/tex] is a constant and [tex]x[/tex] is measured from the surface of the liquid. The liquid is contained into a cylinder, and the liquid height is [tex]L=\frac{2m}{AeK}[/tex].
    We introduce a parallelepiped horizontally where its upper and bottom surfaces are worth [tex]A[/tex], its mass is [tex]m[/tex] and its height is [tex]2e[/tex].
    Depreciate the viscosity of the liquid.

    1)Determine the equilibrium position of the mass.
    2)Determine the motion equation of it, if we apart it a [tex]\Delta x[/tex] from its equilibrium position.
    2. Relevant equations
    None given.


    3. The attempt at a solution

    I don't think I can apply Archimedes' principle because the fluid hasn't a constant density.
    What I did was to calculate the pressure difference between the upper and bottom surfaces of the mass.
    if X is the distance between the surface of the liquid and the center of mass of the parallelepiped, I get that the force acting on the upper surface is [tex]AK\int _0^{X-e}xdx=-\frac{AK}{2}(X-e)^2[/tex].
    Similarly I get the force acting on the bottom surface :[tex]\frac{AK}{2}(X+e)^2[/tex].
    I sum them up to get the total force due to the pressure's difference : [tex]2AKeX[/tex].
    This force acts upward. However to calculate the net force on the mass, I have to add the last force : its weight : [tex]mg[/tex].
    When the mass is at equilibrium, [tex]mg=2AKeX \Rightarrow X=\frac{mg}{2AKe}[/tex].

    Am I right? If so, I'll try to continue alone.
     
  2. jcsd
  3. Nov 27, 2009 #2

    kuruman

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    You are correct here.
    It is better to calculate the pressure as a function of depth from the free surface then calculate the difference between the upward and downward force. You will see that the upward (buoyant) force has two terms not just one.
     
  4. Nov 27, 2009 #3

    fluidistic

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    Thanks for helping!
    Ok, glad to know this.

    Ok so I went wrong somewhere.
    [tex]P=\frac{F}{A}=\rho (x)gx=Kx^2g[/tex]... or I can't even use the formula [tex]P=\rho g h[/tex]?
     
  5. Nov 27, 2009 #4

    kuruman

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    No. Consider a rectangular element of fluid of area A and thickness dx such that its upper face is at the free surface. What is the pressure at the bottom of the element? To find the pressure at depth x, integrate from zero to x.
     
  6. Nov 27, 2009 #5

    fluidistic

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    [tex]P_0[/tex] : atmospheric pressure.
    So the pressure at the bottom of the element is [tex]P_0+\rho gdx[/tex]. Integrating, [tex]P_0+g\int _0^h \rho (x)dx=P_0+\frac{Kgh^2}{2}=P(h)[/tex].
     
  7. Nov 27, 2009 #6

    kuruman

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    Correct. Now proceed to find the buoyant force.
     
  8. Nov 27, 2009 #7

    fluidistic

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    Force=PA thus at the bottom of the mass, [tex]F=-\frac{AKg(X+e)^2}{2}[/tex] while at the top of it, [tex]F=\frac{AKg(X+e)^2}{2}[/tex], adding them up, the buoyant force is [tex]-2AKgeX[/tex], which negative sign seems good since I took the positive sense to be from the surface of the water to the bottom. Ok so in my first post I forgot the g factor and the negative sign disappeared.
    Did I go wrong? I'd get [tex]mg=-2AKgeX\Rightarrow X=-\frac{m}{2AKe}[/tex], hmm yes I went wrong. I shouldn't have this negative sign at least.
     
  9. Nov 27, 2009 #8

    kuruman

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    I understand that e is half the height of the parallelepiped. How have you defined X?
     
  10. Nov 27, 2009 #9

    fluidistic

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    Oh sorry, X is the depth of the center of mass of the parallelepiped.
     
  11. Nov 27, 2009 #10

    kuruman

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    Then at the bottom of the mass, the force is up and equal to
    [tex]
    F_{Bot}=+\frac{AKg(X+e)^2}{2}
    [/tex]
    and at the top the force is
    [tex]
    F_{Top}=-\frac{AKg(X-e)^2}{2}
    [/tex]
    What is the sum of the two which should be the buoyant force?
     
  12. Nov 27, 2009 #11

    fluidistic

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    I get my answer but with a positive sign... how is that possible? A positive force is a force with the same direction than the x axis, namely a force pointing at the bottom of the tank.
     
  13. Nov 27, 2009 #12

    kuruman

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    My convention is that "up" is positive and "down" is negative. That's why the force acting at the bottom of the box is Positive and the force acting at the top is negative.
     
  14. Nov 27, 2009 #13

    fluidistic

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    Ah ok, so we've the same result. Then I just have to equate this with mg, right?
     
  15. Nov 28, 2009 #14

    kuruman

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    Right, on the assumption that the entire mass is immersed in the fluid and sinks until it reaches equilibrium at some depth denoted by X.
     
  16. Nov 28, 2009 #15

    fluidistic

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    Ok, thanks a lot for your help.:smile:
     
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