agaricus5
Level: Smitemaster
Rank Points: 1838
Registered: 02-04-2003
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Hey, DRODers!

My friend and I have been stumped by a deceptively easy-looking mechanics question involving a chain and a table, and as there are quite a few physicists and mathematicians around here, I was wondering if someone could give us some help, preferably in the next three hours or so.

The problem is this:

"A uniform chain of length 'l', total mass 'm' and containing many links, is held at one end over a table with the other end just touching the table top.

1) The chain is released and falls freely. What is the speed of the falling section of the chain at time 't' after release? What is the force between the links?

2) Work out the increment of mass 'delta m', which hits the table in the increment of time 'delta t'. Find the corresponding change in momentum and hence the instantaneous force on the table.

3) What is the total force acting on the table as a function of time? Show that the maximum value of the total force is three times the total weight of the chain."

We'd had a good try at it, but got all tangled up in a set of delta things and infinitesimals. If anyone is able to help us solve the problem, we'd be very grateful for it, and thanks in advance.

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Oneiromancer
Level: Legendary Smitemaster

Rank Points: 2936
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Ugh. I always sucked at these problems. I'm afraid this is one physicist that won't be able to help with this. Ask me about the stuff that really matters (i.e. what they don't teach you in class) instead.

Game on,

____________________________
"He who is certain he knows the ending of things when he is only beginning them is either extremely wise or extremely foolish; no matter which is true, he is certainly an unhappy man, for he has put a knife in the heart of wonder." -- Tad Williams

TripleM
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Registered: 02-05-2003
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Well, its been a while since I've done physics, but I've just been tutoring someone in it the last week or two, and I've just worked through this together with a friend online just now.
So:
1) Acceleration is just gravity, so v = gt. I guess the force between the links is 0, since everything is falling at the same speed.

2) Using the speed above, a total length of gt * delta t of the chain hits the table, so delta m is (m/l)*gt*delta t. Momentum is mass*velocity, which (m/l)*gt*delta t*gt, or (mg^2t^2*delta t)/l. Instantaneous force is that over delta t, which is mg^2t^2/l.

3) Total force is weight of chain on table + the above. The length of the rope on the table (integrating velocity) is 0.5gt^2, so, like before, the mass on the table is 0.5gt^2*(m/l), and the weight is 0.5gt^2*(m/l)*g = 0.5mg^2t^2/l.
So total weight is 1.5mg^2t^2/l.
Now, the maximum value of t is when all of the chain is on the table, ie 0.5gt^2 = l. So t^2 = 2*l/g, and substituting that in the above gives 3mg, which is 3 times the total weight of the chain.

Whee!

TripleM
Level: Smitemaster
Rank Points: 1373
Registered: 02-05-2003
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Ah, and you could probably simplify that last step. The weight of the chain on the table is 0.5mg^2t^2/l = mg at the end, so the instantaneous force is mg^2t^2/l = 2mg at the end, so (2mg+mg) = 3mg.

agaricus5
Level: Smitemaster
Rank Points: 1838
Registered: 02-04-2003
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Thanks for the quick and very helpful response, TripleM; we were able to follow your explanation and solve the problem with your help.

It turns out we'd actually managed to get about three-quarters of the solution ourselves correctly. The bit we messed up was bringing back all those delta terms when we shouldn't have, thereby over-complicating the problem.

Again, thanks; my friend really appreciated your help.



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Resident Medic/Mycologist