# Calculus Work Application

• Apr 19th 2011, 08:15 PM
SC313
Calculus Work Application
Hello,

The following question I have been having difficulties with:

"A cable is 20 m in length, and has a linear density of 3kg/m. It is hanging from a winch that is 30m above the ground. How much work is done in winding up:
a) Half of the cable? B) All of the cable?

I have attempted part a by

For the bottom portion which is all lifted 10m:

10*10*3*9.8

and for the top portion which varies:

The integrand of 3*9.8*y evaluated from 0-10. I am sorry but I can't get the symobls to work at all. :(

When this is all done and calculated, I do not get the correct answer. If someone could tell me what I am missing, that would be awesome! :D
• Apr 19th 2011, 08:43 PM
TaylorM0192
I don't understand the question -- it appears that the "winch" from which the rope hangs being 30m from the ground is immaterial to the question. All that matters is the length of the rope (i.e. the amount of displacement or work that must be done against the gravitational field). I assume that you are not expected to account for 1/r^2 deviations haha.

Clarify before I work out the problem please.
• Apr 19th 2011, 08:47 PM
SC313
Quote:

Originally Posted by TaylorM0192
I don't understand the question -- it appears that the "winch" from which the rope hangs being 30m from the ground is immaterial to the question. All that matters is the length of the rope (i.e. the amount of displacement or work that must be done against the gravitational field). I assume that you are not expected to account for 1/r^2 deviations haha.

Clarify before I work out the problem please.

I think that the fact it si 30m above the ground is, as you said, pointless to the quesetion. And no, thank goodness we are not expected to account for deviations! =D
• Apr 19th 2011, 09:15 PM
TaylorM0192
There is an easy and a hard way to do this problem. Introduce an xy-coordinate system. Place the cable on the y axis from 0 to 20 (m). Note that we are in KGS (real) units.

(a). Lifting the full length of the cable.

Method 1: Since the cable has uniform density (i.e. uniform mass distribution) you can find the center of mass without computing any integrals - it's clearly at the point y = 10m. From physics, we can know that we can "pretend" that all mass is concentrated at the center of mass when doing mechanics calculations. The uniform density is lamba = 3kg/m and therefore the total mass is 60kg. Furthermore, the center of mass moves 10m when the entire rope is hoisted. Therefore, the total work is given by: W = (M-total)(g)(d) = (60 kg)(10 m/s^2)(10 m) = 6000 J.

Method 2: This problem involes straight-line displacement, therefore work is simply force through linear displacement (no dot products). On our coordinate system, the rope occupies y1 = 0m to y2 = 20m. Cut up the rope into tiny pieces "dy" m. Each dy contribtes an element of work dW. The distance that such a dy must travel is (20 - y) against the gravitational field (What you are doing work against to raise the rope). The total work is then: Integral from y1 = 0m to y2 = 20m of: ((20-y) m) * (10 m/s^2) * (3 kg/m) * (dy) m = 6,000 J..

(b.) Lifting half the cable

You should be able to do this yourself. Follow Part A-ii and determine what the new limits should be. Note that your answer will not simply be HALF of part (a), since you do more work to lift the first half of the rope than you do the second half of the rope (do you know why?).

Good luck ~
• Apr 19th 2011, 09:18 PM
SC313
Thanks! =)
• Apr 19th 2011, 09:23 PM
TaylorM0192
NP!