Work done lifting cable against gravity

[physicsnnewbie]

Homework Statement

A cable weighing 2 pounds per foot of length (and so having a mass of 2 pounds per foot) is suspended from the top of a well 200 feet deep and extends to the bottom. Find the work done in raising the cable to the surface. Over the distance of 200 feet you may use 32m as the constant weight of a mass 'm'.

Homework Equations

3. attempt at a solution
dW/dr = F
dW/dr = 32m
dW/dr = 32(2r)
dW/dr = 64r
W = 32r^2 + C = 32(200)^2 = 1280000

The fact that 'r' is measuring cable length (instead of distance between two objects which was used to derive the original equation) is screwing with me. I need someone to help me get my head around it.

 

[lswtech]

find the CM and you are done :)

 

[HallsofIvy]

For a problem like this, where the weight density is constant, lswtech's idea is easiest- the center of mass is exactly at the middle of the cable and lifting the cable requires the same work as if the total mass were concentrated at the center of mass. Just a couple of simple arithmetic operations.

More generally (and a way of showing that the previous statement is true), focus on some small section of chain, of length \Delta x, at height x below the surface. If the weight density is \gamma(x), then the weight of that little section is (approximately) \gamma(x)\Delta x and it has to be lifted a distance x. It takes (approximately) \gamma(x) x\Delta x work to lift it to the surface. Dividing the entire cable into pieces and adding the work required to lift each piece, the total work would be (approximately) \sum \gamma(x) x \Delta x.

The reason for the repeated "approximately"s is that if the density is NOT constant, and \Delta x is not "infinitesmal", the weight is not simply a product. But that is a "Riemann" sum, as is used to define the definite integral. If take the limit as \Delta x gets smaller and smaller, becoming "infinitesmal", it becomes exact: the work is given by
\int \gamma(x) x dx

If, in particular, \gamma is a constant, we can take it out of the integral:
\gamma \int_0^L x dx
(I have put in 0 and L as the distances of the two ends of the cable- it extends down from 0 to its length L.)

That is easily seen to be
\left[\gamma \frac{x^2}{2}\right]_0^L= \gamma \frac{L^2}{2}= \left(\gamma L\right)\left(\frac{L}{2}\right)
where at the end I have factored to show it is "\gamma L" times "L/2", the weight of the cable times the distance it midpoint is lifted.

 

[physicsnnewbie]

Thanks, that was very helpful!