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Old Apr 5th 2016, 06:49 PM   #1
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Chain on table

This is from "Fundamentals of Physics" 6th Edition. It seems like a simple problem but I'm racking my brain.
A chain is held on a frictionless table with one-fourth of its length hanging over the edge. If the chain has length L and mass m, how much work is required to pull the hanging part back onto the table?

The answer in the back of the book is mgL/32.

My approach is the following. The mass hanging off the side is m/4 (assuming uniform density). The distance lifted is L/4. So the work is mgL/16. Now I realize that the mass hanging off changes with the length that is hanging off, so I suspect there is some calculus involved in the correct solution, but I'm having trouble setting it up. Help?
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Old Apr 6th 2016, 02:29 AM   #2
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You don't need calculus for this one.
It is basically the same as the work to stretch a spring, where the same factor of one half appears.
Well done for recognising that the mass and therefore the lifting force varies.

The work is the area under the force x distance graph.

As you note this is a straight line through the origin.

So the work is simple a calculation as shown in the attachment.

Calculus will give you the same answer, but if you want to know how to use calculus then do the following.
Note that this time you must work with the mass per unit length m/L, not the total mass of the hanging chain.

1) Divide the hanging length into small pieces dl, each a distance l down from the table top.
2) The mass of each segment dl is (m/L)dl and you need to calculate the work as above ie W = (m/L)gldl
3) Then you need to sum the work for all the segments dl from l = 0 to l = L/4
4) That is total work = integral ((m/L)gldl) from l = 0 to l = L/4

It is important to distinguish between the constant length of the chain and the distance variable l.
Attached Thumbnails
Chain on table-chain1.jpg   Chain on table-chain2.jpg  

Last edited by studiot; Apr 6th 2016 at 05:29 AM.
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Old Apr 6th 2016, 06:36 AM   #3
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One way to think of this is to consider the weight hanging off the table as being a point load at the center of mass. The center of mass is 1/2 of the length that is hanging over the edge, or L/8. The work to lift this force times distance, where force = mg/4, and the distance it needs to be raised is L/8, hence work done = mgL/32.
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