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Work Done (integral)2

  1. Dec 22, 2011 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations
    [itex]W = \int F dx[/itex]

    For density of water i used
    D = 62.4lb/ft^3
    g = 32.2 ft/s^2

    [itex]g(D = \frac{m}{V})[/itex]

    [itex]gDV = mg[/itex]

    [itex]gDV = F[/itex]

    3. The attempt at a solution
    i use limit from 0 to 6
    [itex]W = \int_{0}^{6} F dx[/itex]

    [itex]W = \int_{0}^{6} gDV dx[/itex]

    [itex]W = \int_{0}^{6} (32.2)(62.4)(\pi{6}^{2}{15}) dx[/itex]

    [itex]W = 20451979.29 [/itex]

    help check this pls
  2. jcsd
  3. Dec 22, 2011 #2


    Staff: Mentor

    This is incorrect. You are assuming that all of the water has to be lifted 15 feet, and this is not the case. Mentally divide the water into cylindrical slices, each of which is [itex]\pi[/itex]*36*[itex]\Delta y[/itex] in volume. From the volume, you can get to the weight of each slice, but the distance a slice of water has to be pumped depends on where the slice is in the tank. The layers (slices) at the top require almost no effort to pump out, since they don't need to be lifted very far. The ones at the bottom require a lot more work, since the have to be lifted much farther.
  4. Dec 23, 2011 #3
    found the answer thx for the advice

    slice of cylindrical element anywhere inside cylinder
    [itex]dV = 36{\pi}dy[/itex]

    [itex]dF = 32.2(62.4(\pi{36})dy)[/itex]

    work done from this differential element to the top: then (distance = 15 - y)
    [itex]dW = 32.2(62.4(\pi{36})dy)(15-y) [/itex]

    [itex]W = \int_{0}^{15} dW[/itex]

    [itex]W = 25564974 [/itex]lb-ft

    [itex]W = 793943 [/itex]slugs-ft
  5. Dec 23, 2011 #4


    Staff: Mentor

    I didn't verify your numbers, but what you have looks to be set up correctly.
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