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Digging a Well (Work Problem)

  1. Nov 20, 2011 #1
    1. The problem statement, all variables and given/known data

    A well is dug in the shape of a rectangular prism. It is 30ft deep and has a base with area of 40ft^2. Assuming that the soil weighs 150 lbs/ft^3, calculate the work W required to raise the soil to ground level.


    2. Relevant equations

    W = ∫dW
    dW = ρ(area)(distance)dx


    3. The attempt at a solution

    I had dW = ρ40(30-x)dx

    Then I said W = 150∫[from 0 to 30] 1200-40x dx

    Solved to get 2700000ft/lb, which is wrong. I don't really understand what I did wrong here. Isn't this just like pumping water out of a tank or something?
     
  2. jcsd
  3. Nov 20, 2011 #2

    LCKurtz

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    I get the same answer. I guess we are both assuming "rectangular prism" means the same thing as "rectangular parallelepiped" or "rectangular box".
     
  4. Nov 21, 2011 #3

    Ray Vickson

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    I think your answer is too large by a factor of 30. Go back to your dW formula to see why.

    RGV
     
  5. Nov 21, 2011 #4
    I don't understand why it would just be x for the distance rather than 30-x. Could someone explain this to me?
     
  6. Nov 21, 2011 #5

    LCKurtz

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    It doesn't matter whether you use x or 30 - x. I think you are correct and I am puzzled what RGV sees that we don't, since his answers are usually spot on.
     
  7. Nov 21, 2011 #6
    But x and 30-x give you totally different answers...
     
  8. Nov 21, 2011 #7

    LCKurtz

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    No they don't.
     
  9. Nov 21, 2011 #8

    Ray Vickson

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    The cross-sectional area at depth x is 40 at x = 0, but the formula given by the OP gives 40*30. I think the correct formula should be dW = ρ*40*(30-x)/30 dx.

    RGV
     
  10. Nov 21, 2011 #9

    LCKurtz

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    But the OP has just chosen his coordinates measuring depth from the other end; that's the 30-x, which varies from 30 to 0. You could measure from the other end and use x as I mentioned earlier. The cross section is a constant 40.

    We are both thinking of a hole shaped by a rectangular box, right?
     
  11. Nov 21, 2011 #10
    Oh, I'm sorry. I see they are the same now, I just don't understand how.
     
  12. Nov 21, 2011 #11

    LCKurtz

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    It is just a matter of which end you put your coordinate system and whether x is measured
    positive up or down.
     
  13. Nov 21, 2011 #12

    Ray Vickson

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    No, I was thinking of a wedge-shaped excavation (because of the word "prism" in the original post). In that case I would be wrong anyway, because (my) dW is weight, not work. For a wedge-shaped hole the answer would be
    Work = int rho * 40 * x*(1 - x/30) dx = 900000 ft-lb.

    RGV
     
    Last edited: Nov 21, 2011
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