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Problem on Work

  1. Dec 30, 2006 #1
    1. The problem statement, all variables and given/known data
    A tank full of water has the shape of a parabloid of revolution with shape obtained by rotating a parabola about a vertical axis.
    a) If its height is 4 ft and the radius at the top is 4 ft, find the work required to pump the water out of the tank.
    b) After 4000 ft-lb of work has been done, what is the depth of the water remaining in the tank?

    2. Relevant equations
    m = density*volume

    3. The attempt at a solution
    I don't know how to do part (b). This is what I have for (a):
    I labeled the radius of cross section as Ri (ith subinterval)
    Ri/(4-Xi) = 4/4
    Ri = 4-Xi
    Volume of ith layer of water = pi(4-Xi)^2 dx
    Mass of ith layer of water = 62.5pi(4-Xi)^2 dx
    Force to raise ith layer = (9.8 m/s^2)(62.5pi(4-Xi)^2 dx
    W to raise ith layer = 612.5pi*x*(4-x)^2 dx
    Total work = Integral of 612.5pi*x*(4-x)^2 dx on [0, 4]

    The answer is not right, so can anybody tell me what I did wrong and how to fix it? Also, how would you do part (b)?

    Thanks :smile:
     
  2. jcsd
  3. Dec 30, 2006 #2

    AlephZero

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    You seem to have got the wrong shape for the tank. The equation of any parabola is y = ax^2 (with y vertical and x horizontal) and the problem says y = 4 when x = 4 so you can find the value of a.

    Your "Ri = 4-Xi" seems wrong - that would be a cone, not a parabola.

    In your "W to raise ith layer" you are not using consistent units - you used g in m/s^2.

    For part (b), just find the work to pump out the water to depth D (a similar integral to the first part).
     
    Last edited: Dec 30, 2006
  4. Dec 30, 2006 #3
    I see. What do you do with the 'a' if I plug in 4 for both x and y? I get a=1/4, but not sure where to go with that.

    I thought of it as a cone, so that was wrong. How would you find the radius then for a parabola?
     
  5. Dec 30, 2006 #4

    AlephZero

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    You know y = 1/4 x^2, so rearranging that, x = 2 sqrt(y)

    Or in words: at height y above the base of the tank, the radius is 2 sqrt(y).
     
  6. Dec 30, 2006 #5
    thanks. But instead of using m/s^2 for acceleration of gravity, what should the units be? The problem uses feet, so is it ft/s^2? ft/h^2? I'm not sure.
     
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