0
$\begingroup$

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.)

Diagram

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?

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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 $$w=\int_0^4(62.5)(4-y)(7.5y)dy=(62.5)\left[15y^2-2.5y^3\right]_0^4=5000ft\cdot lb$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.