Comparing the rate at which two chains fall in different scenarios

Jimbo07 said:
ETA: jj, the amount of chain over the edge should only make a difference where friction is concerned.
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Not quite right. Go back to Newton and Force = Mass * Accel. When the first link is over the edge (applying an arbitrary 1 gram per link) a force of F = (0.001)(9.8) = 0.0098 Newtons is accelerating the entire 1000 m chain. If there are 100 links per meter, a=F/m=0.0098/(100)(1000)(0.001)=0.000098 m/s^2.

When the second link goes over, F=(0.002)(9.8)=0.0196 N, so the acceleration of the entire chain at this moment is a=0.000196 m/s^2.
 
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ceptimus said:
The horizontal velocity imparted to the chain still on the level doesn't transfer into vertical velocity unless there is some kind of pipe guiding the chain around the edge.
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Ha. That's very good. I hadn't thought of that at all.

I imagine we're supposed to pretend there is such a pipe.
 
69dodge said:
Ha. That's very good. I hadn't thought of that at all.

I imagine we're supposed to pretend there is such a pipe.
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How can the horiztonal rate of the chain across the ground on top the precipice be any different from the rate of the fall?

It's a chain.
 
ImaginalDisc said:
How can the horiztonal rate of the chain across the ground on top the precipice be any different from the rate of the fall?

It's a chain.
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Yeah, but the part that's resting on the ledge is moving sideways, not down. As it slips over the edge, it's not yet falling at all.

If the chain were in a guiding pipe, however, as soon as it left the edge it would already be falling at the same speed that it was previously moving sideways, so gravity would start speeding it up from that already high speed instead of from zero downward speed.
 
In my mind I see a gravitational force on each link which is counteracted by resting on the ground as it moves off the edge then there is no counterforce to gravity causing it to fall. As it crosses the edge it starts to accelerate vertically while it has been accelerating horizontally all along. If it is moving very fast it will still fall just as fast unless there is a force to fight against gravity. Assuming no friction and the ground is level then there is no force. I can see the whole no friction thing confuses me. In my mind I think the lower drop chain will make several zig zags on the ground and the higher one will just be one big zig with the end link whipping out like the end of a whip. The lack of friction is confusing because I don't see that in my life at all (friction everywhere). I guess when I really think about it the whole mass is accelerating horizontally at a constant rate because there is no friction which should result in the same speed. The lower height chain is losing mass affected by gravity as it hits the ground however since there is no friction to overcome it matters not.
 
TPrime said:
Not quite right. Go back to Newton and Force = Mass * Accel. When the first link is over the edge (applying an arbitrary 1 gram per link) a force of F = (0.001)(9.8) = 0.0098 Newtons is accelerating the entire 1000 m chain. If there are 100 links per meter, a=F/m=0.0098/(100)(1000)(0.001)=0.000098 m/s^2.

When the second link goes over, F=(0.002)(9.8)=0.0196 N, so the acceleration of the entire chain at this moment is a=0.000196 m/s^2.
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So there I was, on my hour and a half drive home, going over what I had posted at work... I slowly started to get the heebie jeebies! It has been well over a year and a half since I drew a free body diagram, and I should have drawn one in the first place.

69dodge said:
You just said, correctly, that some of the downward gravitational force on the chain is cancelled out by the upward force of the ledge on the chain. So why do you still think that the acceleration of the chain will be g? It will be less, because there's less force acting on it than on a freely falling chain.
Google for "Atwood machine".
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I'll have to get back to it... I have to build a bench tonight. I hope I use tools better than... well, anyway...

:D
 
Assuming that we overlook the problem of horizontal motion, the longer chain has a final velocity (i.e velocity at the point where the first link hits the ground and the last link slips over the edge) slightly over three times greater than that of the short chain.
 
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Why wouldn't it be the same as dropping 2 chains past a point? One which fell to a height 10 meter below the point and one that fell to a point 100 meters below. It seems without friction the lasts link would travel at the same speed. Why would it be different going from horizontal to vertical?
 
Jimbo07 said:
ETA: jj, the amount of chain over the edge should only make a difference where friction is concerned. A large amount of chain would need to be over for the force of gravity to overcome the force of friction holding it back. In a no-friction case, the downward force of the moon's gravity would get it going. Of course, your constant of acceleration would be different.
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No, Jimbo, no. Friction is irrelevant.

The amount of chain that's going down (i.e. over the edge but not at the ground yet), compared to the amount of chain yet to go over the edge, sets the how much of 'g' the chain accilerates by.

Friction is not being considered.
 
To be precise, I just did the math, and the "final velocity" of the chain is equal to

square root(g * length of chain)

Taking g = 9.8, the final velocities are 9.9 m/s for the short chain and 31.3 m/s for the longer chain.
 
TPrime said:
Not quite right. Go back to Newton and Force = Mass * Accel. When the first link is over the edge (applying an arbitrary 1 gram per link) a force of F = (0.001)(9.8) = 0.0098 Newtons is accelerating the entire 1000 m chain. If there are 100 links per meter, a=F/m=0.0098/(100)(1000)(0.001)=0.000098 m/s^2.

When the second link goes over, F=(0.002)(9.8)=0.0196 N, so the acceleration of the entire chain at this moment is a=0.000196 m/s^2.
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And the maxium amount of chain "over the edge" that's still falling, in one case, will be 10 meters worth. In the other case it will be 1000 meters worth. Factor of 100 difference in mass.
 
Could this be imagined as a horizontal plane dividing a zone of no acceleration due to gravity from a normal G zone, two vertical chains held by the weight of one link each which protrudes below the plane, and restrained at the top, simultaneously released with one having only ten meters below the plane to fall and the other having it's full 1000 meter length to bottom out?

It would get rid of the right angle turn, and simplify things I think.

Cheers,
Dave

Edit: Corrected to meters from feet. DUH.
 
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I don't see why any great mathematical ability is required. The only difference between the two chains is that the 10 meter one will, after a while, experience an additional upwards force, and will therefore not fall as quickly.

If you take it to an extreme, obviously a drop of 0 meters would pull the chain much more slowly.

Dr Adequate said:
To be precise, I just did the math, and the "final velocity" of the chain is equal to

square root(g * length of chain)

Taking g = 9.8, the final velocities are 9.9 m/s for the short chain and 31.3 m/s for the longer chain.
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I think that you should check your math.
 
Well, I only have 8 year old high school physics to go on, but I think the following is right:

f=ma
a=f/m

In this scenario f=g on the part of the chain that is already falling.

So, when the 100m of the 1000m chain has slipped over the edge of the cliff, f = 9.81*100(X) (X = the mass of 1m of chain).
At this same time f = 9.81*10(X) for the other chain.

So for the first chain's acceleration we get a = 981(X)/1000(X)
For the second chain's acceleration we get a= 98.1(X)/1000(X)
The X's cancel, and we are left with:
(1) = 0.981 and
(2) = 0.0981

Clearly the one that experiences more acceleration at any point in the fall will be the one that is moving faster at the end of the fall.
 
Dilb said:
Two dimensional physics is hard, so I made a simulation. Enjoy
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Thanks Dilb. Assuming your simulation is accurate (and it looks good to me) then it confirms my intuition about the wave propogating down the falling chain. :)
 
Art Vandelay said:
I don't see why any great mathematical ability is required. The only difference between the two chains is that the 10 meter one will, after a while, experience an additional upwards force, and will therefore not fall as quickly.
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Okay, you're a buffoon. Do you hear me? A buffoon. You tell me to check my maths. Check your physics! The force acting on the chain is proportional to the mass of the chain which is over the edge. The acceleration of the chain is proportional to this divided by the total mass of the chain.

You have to be ... ooh ... really adequate at maths to derive the beautiful result I just gave. You have to be a physics idiot to suppose that the two chains would move the same until the 10 meter chain hits bottom.

Your ignorance is matched only by your impertinence. Do not, please, attempt to correct your betters.

Argue the post, not the poster.
Replying to this modbox in thread will be off topic  Posted By: Lisa Simpson
 
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Dr Adequate said:
Okay, you're a buffoon. Do you hear me? A buffoon. You tell me to check my maths. Check your physics! The force acting on the chain is proportional to the mass of the chain which is over the edge. The acceleration of the chain is proportional to this divided by the total mass of the chain.

You have to be ... ooh ... really adequate at maths to derive the beautiful result I just gave. You have to be a physics idiot to suppose that the two chains would move the same until the 10 meter chain hits bottom.

Your ignorance is matched only by your impertinence. Do not, please, attempt to correct your betters.
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Beautiful. Unimpeachable. Correct. And in complete agreement with my own independent analysis of the situation via calculus, which is, I think, reasonable if one assumes that the chain links are tiny compared to the chain's length.

Calculus you must learn. Difficult it will be.

The final chain velocities, i.e. when the last piece of chain just crosses the edge, assuming g = 9.8 m/s^2 come out as 99.0 m/s for the 1,000 m drop, and 30.1 m/s for the 10 m drop.

The formulae are:
v = Sqrt(g*L) for the full drop
v = Sqrt(g*a*(a/L + 2*Ln(L/a))) for a reduced drop of "a" metres
where v is the final velocity, L is the length of the chain, g is the acceleration due to gravity, and a is the height of the reduced drop. The equations are valid provided that the full drop height is greater or equal to L, and the reduced drop height a is less than or equal to L.

Thus, the reduced drop accelerates at a lower overall rate than the full drop.

'Luthon64
 
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