I've been working on this problem for the last hour and can't quite seem to get it right.

A tank is full of water. Find the work required to pump the water out of the spout. Use the fact that water weighs 62.5 lb/ft3. (Assume a = 4 ft, b = 5 ft, and c = 6 ft.)


I've figured out that the volume is 4*5*6/2= 60 and the mass= 60*62.5 = 3750 I then integrated 3750x from 0 to 4. 1875x^2]=30,000

Have I done this problem correctly?

  • $\begingroup$ No idea - what steps did you take to get that answer? $\endgroup$ – Noah Schweber May 1 '16 at 0:40
  • $\begingroup$ Your are right that the mass is $60 ft^3\cdot 62.5 lb/ft^3=3750 lb$. But what is your question ? $\endgroup$ – callculus May 1 '16 at 0:49
  • $\begingroup$ How much work would it take to pump the water out of the spout? $\endgroup$ – TanBro May 1 '16 at 0:59
  • $\begingroup$ Just a guess, did you use integration to find the total work done by considering the infinitesimal work done by pumping a slice at a distance h from the top? $\endgroup$ – user247608 May 1 '16 at 1:14
  • $\begingroup$ math.stackexchange.com/questions/1185952/… $\endgroup$ – TanBro May 1 '16 at 1:27

Your first problem is not taking advantage of units. The volume $V=\frac12(4 ft)(5 ft)(6 ft)=60 ft^3$, Then the weight is $w=\gamma V=(62.5\,lb/ft^3)(60 ft^3)=3750\,lb$. Since work is force$\times$distance, you need a distance here, and the appropriate one is the distance you must raise the center of mass to get the $H_2O$ out of the spout. Since the center of mass of a uniform triangular area is $\frac13$ of the way from any side, the distance is $(4 ft)/3$ so the work is $(3750\,lb)(4 ft)/3=5000ft\cdot lb$. In calculus, the area of a horizontal section $y\,ft$ from the lowest point is $(5ft)\times(6ft)\times y/(4ft)=7.5\,y\, ft^2$. You have to lift that slice $4ft-y$, so work is $$w=\int_0^4(62.5)(4-y)(7.5y)dy=(62.5)\left[15y^2-2.5y^3\right]_0^4=5000ft\cdot lb$$


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