MHB How Much Work to Pull Up a Second Rope on a Building?

[MarkFL]

Here is this week's POTW:

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Suppose you are at the top of a building of height $h$, and you have one rope of mass $m_1$ hanging over the side tied to a second rope of mass $m_2$ on the ground. Both ropes are of length $h$. Show that the work required to haul the second rope up to you, such that it is now hanging over the side of the building is the product of the height of the building and the average of the weights of the two ropes.

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[MarkFL]

Congratulations to the following member for the correct solution:

• lfdahl

His solution, which was very similar to mine, is as follows:

Spoiler

I assume, that the mass is uniformly distributed on both ropes. Now, suppose the second rope is hauled to height $x$ and $0 \le x \le h$. The first rope has length $h-x$.

At this point the free hanging weight of rope $1$ is: $\frac{m_1g}{h}(h-x)$

The free hanging weight of rope $2$ is: $\frac{m_2g}{h}x$

The work to haul the ropes a short distance, $\delta x$, is: $\delta W = \left (\frac{m_1g}{h}(h-x)+ \frac{m_2g}{h}x \right )\delta x = \frac{1}{h} \left (w_1(h-x)+ w_2 x \right )\delta x $

- where $w_1$ and $w_2$ are the respective total weights of the ropes.

The total work is obtained by integration:

\[ W = \frac{1}{h} \int_{0}^{h}\left ( w_1(h-x)+w_2x \right )dx = \frac{1}{h}\left [hw_1x-\frac{1}{2}w_1x^2+\frac{1}{2}w_2x^2 \right ]_0^h = \left ( \frac{w_1+w_2}{2} \right )h.\]

I now turn the mic back over to anemone. (Yes)