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For example, a half-full cylindrical tank that holds a liquid that is 12 pounds per cubic foot(4ft radius by 8 ft tall). How do we find the work done in pumping the fluid out of the tank from the top outlet?

I tried using Work = Force * distance:

  • determined the force is the volume and weight of the liquid, $\pi*(4)^2 * 8/2 * 12 = 768\pi$
  • Because work is the integral of force: Work = $\int_{4}^{8}768 \pi dx = 3072\pi$ , which is incorrect (Ans: 4608ft-lb)

Where have I gone wrong here, did I get the value of the force wrong, where is my knowledge gap?

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  • $\begingroup$ What does $x$ signify in your integral? $\endgroup$
    – David K
    Commented Sep 30, 2021 at 12:48
  • $\begingroup$ Change in distance? $\endgroup$
    – nvs0000
    Commented Oct 1, 2021 at 7:19
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    $\begingroup$ A change in distance could work, but you have to be much, much more precise than that in how you think about it. First, what object undergoes a change in distance? Second, how is this distance measured? Distance is from point A to point B; what is point A, what is point B? Is it straight line distance, distance along a curve, only the horizontal component of distance, only the vertical component? $\endgroup$
    – David K
    Commented Oct 1, 2021 at 11:10

1 Answer 1

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The work done to pump out liquid from the top outlet is different for the liquid at different depths.

Volume of infinitely thin layer of liquid ($dx$) at depth $x$ from top is,

$dV = \pi \cdot 4^2 \cdot dx = 16 \pi ~ dx ~ $ and this liquid needs to be taken $x$ distance against gravity.

So work done in pumping out the liquid in unit ft-lb is,

$ \displaystyle \int_4^8 \rho~ x ~ dV $

where $\rho = 12 $ lb / ft$^3$

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  • $\begingroup$ Incorrect though, the correct answer was 4608 ft-lb. $\displaystyle \int_4^8 \rho~ x ~ dV \implies \displaystyle \int_4^8 12 x ~ dV= 288$ ? $\endgroup$
    – nvs0000
    Commented Oct 1, 2021 at 7:24
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    $\begingroup$ $dV = 16 \pi dx$, right? $\endgroup$
    – Math Lover
    Commented Oct 1, 2021 at 7:31
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    $\begingroup$ So it is $ \int_4^8 (12 \cdot 16 \pi) x ~ dx = 4608 \pi $ $\endgroup$
    – Math Lover
    Commented Oct 1, 2021 at 7:34

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