An elevator weighing $3000$ lbs. is supported by a $12$ ft. cable that weights $14$ lbs./ft. Find the work a winch has to do by pulling the rope to lift the elevator $9$ feet.

I eventually figured out how to do this problem as follows: $$ 9 (3000+3\cdot 14) + \int_3^{12} 14 (12-y) \, \mathrm{d}y $$

where since the elevator only moving 9ft the bottom 3ft of the cable can be considered part of the elevator weigh so the combined weight that must be pulled a fixed distance of $9$ feet is $(3000 + 3 \cdot 14)$ lbs.

However, I also found that the following integral also gives the correct answer: $$ 9\cdot 3000 + \int_0^9 14 (12-y) \, \mathrm{d}y $$

However I cannot understand why this integral also works, as when you slice the rope into vertically stacked discs the above integral posits that the bottom most slice moves $12$ feet, which cannot be right.

Could someone please explain visually how the second integral calculates the work done?

  • $\begingroup$ How about $y_1=y+3,\ 12-y=15-y_1$, $$\int_0^9 14 (12-y) \, dy=\int_3^{12} 14 (15-y_1) \, dy_1 = \int_3^{12} \left(14\cdot 3+ 14(12-y_1)\right) \, dy_1 = 9\cdot 3\cdot 14+\int_3^{12} 14(12-y_1) \, dy_1\hbox{ ?}$$ I.e. these two are technically the same... $\endgroup$ Commented May 26, 2015 at 6:02
  • $\begingroup$ @AlexeyBurdin Ok, it makes sense numerically now but graphically what is going on. If you consider slices of the rope how much is each slice moving...? $\endgroup$
    – 1110101001
    Commented May 26, 2015 at 6:07
  • $\begingroup$ Is the "rope" the same as the "cable"? Should "the work a which has to do" be "the work a winch has to do"? $\endgroup$
    – hardmath
    Commented Jun 5, 2015 at 2:51

1 Answer 1


The bounds of the integral are being passed to (12-y), so smaller y results in larger areas of integration.

The first equation seems easier to "visualize" to me too, but I think the second one can be intuited as well. The first term is the work done for the elevator. The second term is the work done for the cable as it moves from position zero to position nine. At the start, there are 12-0=12 units of cable, and as expected there will be 12-9=3 units remaining at the end.

I'm not sure exactly what you mean by 'the bottom disk must move by 12.'

  • 1
    $\begingroup$ Oh I think I get it now. In the first integral, I was considering the rope as comprised of thin, vertically stacked circles and the distance that each circle moved was $12 - y$, starting from the circle 3 ft. from the bottom. Whereas in the second one, instead of considering the work done in moving slices a variable distance ($12 - y$) with a incremental force $14 dy$, you are considering the work done in moving the rope itself, of variable weight $14(12 - y)$ a distance of $dy$. $\endgroup$
    – 1110101001
    Commented May 26, 2015 at 6:13

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