Work problem - chain hanging from a ceiling

[marc321]

Greetings.

I've been working on this work (as in physics) problem, and I would like to see if my reasoning is correct.

Problem: A 10 foot chain weighing 25 lbs hangs from a ceiling. How much work is done in lifting the lower end of the chain to the ceiling so that it's level with the upper end?

Clearly only half of the chain is going to be affected, which tells me that integration will be on the interval [5,10]. Now, if I divide the chain below x=5 (picture the chain on a downward facing positive x-axis) into small intervals, each one will have to move up by an amount of about xi* - (10 - xi*) = 2xi* - 10, where xi* is a point in the ith interval. Since each foot weighs 2.5 lbs, the work done on each interval is 2.5(2xi* - 10)(delta-x).

So my integral to solve is 2.5(2x-10)dx on [5,10], correct? It gives a believable answer. The logic seems fairly sound from an algebraic perspective, but intuitively, it still seems a little weird to me. If the integral is thought of as a Riemann sum, it seems the chain is actually being pulled up from the middle (which may not matter, I don't know).

Thank you for taking the time to read this and help with or verify/criticize my solution.

 

[skeeter]

let the bottom of the chain be y = 0, then the top is y = 10 ft.

the part of the chain being lifted is from 0 to 5 ft, so the limits of integration are from 0 to 5.

let a small representative "piece" (dy) of chain in the interval from 0 to 5 have height y above the bottom. after it is lifted, it will be a distance y from the top ... therefore,
that small representative piece is lifted a vertical distance (10 - 2y).

last of all, the linear weight-density of the chain is 25 lbs/10 ft = 2.5 lbs/ft

summing all the work done on each representative piece from 0 to 5 ft gives you the following integral ...

\(\displaystyle \L W = \int_0^5 2.5(10-2y)dy = \frac{125}{2} ft-lbs\)

 

[marc321]

Why didn't I think of that? The picture in my head was unnecessarily complex and unintuitive. Use the positive y-axis... I'm an idiot.

At least my method results in the same solution, so I should give myself some credit.

Thanks for sharing that perspective. It's very helpful.