Lifting up a chain

by Pranav-Arora
Tags: chain, lifting
P: 616
 Quote by haruspex ... and which is wrong. The correct answer is achievable by consideration of conservation of momentum and can be done without involving the dp/dt formulation.
Why is momentum conserved? That implies no net force.
 Conservation of momentum and conservation of work energy (as opposed to thermal energy) lead to different answers - one must be flawed.
Or both.
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P: 9,849
 Quote by tms Why is momentum conserved? That implies no net force.
No, it only requires that all forces have been taken into account. Using the work energy theorem requires all losses have been taken into account, and there's a good reason to suspect there are losses: an elastic string would have been accelerated on take up and overshot equilibrium, leading to oscillations. The energy that would have gone into these must have been dissipated somehow.
P: 616
 Quote by haruspex No, it only requires that all forces have been taken into account. Using the work energy theorem requires all losses have been taken into account, and there's a good reason to suspect there are losses: an elastic string would have been accelerated on take up and overshot equilibrium, leading to oscillations. The energy that would have gone into these must have been dissipated somehow.
But the problem involves idealized object, the way most intro physics problems do. There is no information about possible sources of energy loss. Since a supposed loss is calculated without such information, it implies that every situation involves the same loss. I find it very hard to believe that pulling up an anchor chain (with links 2 feet across) and pulling up a shoestring both create exactly the same loss.

Using F = dp/dt, on the other hand, presupposes only that you know the change in momentum in order to get the net force, and I don't see how that could be wrong.
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P: 25,228
 Quote by tms But the problem involves idealized object, the way most intro physics problems do. There is no information about possible sources of energy loss. Since a supposed loss is calculated without such information, it implies that every situation involves the same loss. I find it very hard to believe that pulling up an anchor chain (with links 2 feet across) and pulling up a shoestring both create exactly the same loss. Using F = dp/dt, on the other hand, presupposes only that you know the change in momentum in order to get the net force, and I don't see how that could be wrong.
I think haruspex would agree that F=dp/dt is the right thing to do, and would also agree the loss doesn't depend on the detailed dynamics of how the energy was lost. Was just saying you can infer by thinking about possible ways the loss can occur that you can conclude it must happen. You don't disagree. DH is the one who disagrees.
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P: 9,849
 Quote by Dick I think haruspex would agree that F=dp/dt is the right thing to do, and would also agree the loss doesn't depend on the detailed dynamics of how the energy was lost. Was just saying you can infer by thinking about possible ways the loss can occur that you can conclude it must happen. You don't disagree. DH is the one who disagrees.
Thanks Dick, well put.
 Quote by tms I find it very hard to believe that pulling up an anchor chain (with links 2 feet across) and pulling up a shoestring both create exactly the same loss.
The mechanism of the loss may be rather different; I picture the anchor chain links as swinging side to side briefly as they leave the ground, rather than bouncing up and down. But it would appear that about half the energy is lost either way.
 P: 616 I just think we're missing something---I don't know what. It's as if we were doing a typical block-on-frictionless-surface type problem, and the answer had a frictional loss in it---a frictional loss that was the same for every block and surface.
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P: 25,228
 Quote by tms I just think we're missing something---I don't know what. It's as if we were doing a typical block-on-frictionless-surface type problem, and the answer had a frictional loss in it---a frictional loss that was the same for every block and surface.
If you picture two blocks colliding and sticking together then the amount of energy loss is exactly predictable using conservation of momentum or force balance. This is the same thing. Of course it's not the same for every every collision. Depends on the momenta and masses. Tsny had a lucid picture of the links of the chain being joined. We aren't missing anything. DH is in a state of denial.
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P: 9,849
 Quote by tms I just think we're missing something---I don't know what. It's as if we were doing a typical block-on-frictionless-surface type problem, and the answer had a frictional loss in it---a frictional loss that was the same for every block and surface.
I don't see any difficulty. Try this analogy. A chain of buckets (ideally, flat platforms) moves up past a fixed platform at (on average) constant speed u. An operator slides masses m from the fixed platform onto the moving ones as they go past. (These masses stand for bits of string/chain.) Clearly, each such move is an impact. If the total mass of the chain and its existing masses is M then each impulse costs it speed um/M, and the energy needed to restore its speed is M(u2-(u-um/M)2)/2 = mu2(1 - m/M). As m/M→0, that's mu2, double the kinetic energy imparted to the mass m.
 HW Helper Thanks P: 10,623 When we say that the KE of the whole system is 1/2 (ρx)u2 we totally neglect the motion of the piece still on the ground. It has to move somehow, as the length of the chain is constant. And we do not know anything about that motion, but it contributes to KE. I have shown an arrangement in #46 when the chain does not move on the table. In that case, the work-energy method gives the same result as the momentum method. The problem says that the chain is piled up, but it cannot be treated as a single point as it has length. As we do not know the details of the internal motion of the parts,the best thing is treating the chain as whole. See TSny's post #23. The acceleration of the CM is equal to the net force, including the lifting force, weight of the whole chain and the normal force from the table. ( the question: is the normal force really equal to the weight of the chain still on the ground?) The CM of the whole chain does accelerate while the lifted part moves with constant vertical speed. ehild
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P: 15,167
 Quote by haruspex ... and which is wrong. The correct answer is achievable by consideration of conservation of momentum and can be done without involving the dp/dt formulation. Conservation of momentum and conservation of work energy (as opposed to thermal energy) lead to different answers - one must be flawed.
Do it right and they don't disagree.

First let's look at my magical chain in post #17 where links magically appear out of nowhere and attach themselves to the chain. In that case I would agree with the answer implied by the text. All of the force needed to accelerate a newly created link to a velocity of u is supplied by the lifter. There's a problem here. Physics and magic don't mix well.

Next let's look at what happens during the process of lifting a single link of the chain off the platform. Between the point in time when the link is lying flat on the platform and the point in time when the link is vertical and lifts off the platform, the lifter only applies a force equal to half of the link's weight. The platform bears the other half of the link's weight1. It's only until the link is vertical and lifted off the platform that the lifter bears the entire weight of the link. By this time, the link has already been accelerated to an upward velocity of u. That the lifter bears only half of the weight during the process of lifting a link resolves the apparent discrepancy between a conservation of momentum and conservation of energy approach.

Next, let's look at a number of different approaches to solving this problem. Use the work energy theorem and you'll find that the lifter exerts a force equal to ρgx+½ρu2. Use Lagrangian mechanics and and you'll get a force equal to ρgx+½ρu2. Use Hamiltonian mechanics and you'll get a force equal to ρgx+½ρu2. Use conservation of momentum with the platform bearing half the weight during the process of lifting a single link and you'll get a force equal to ρgx+½ρu2. They all agree.

Finally, let's look at what happens if we use F=dp/dt. In a reference frame in which the platform is stationary you'll get a force equal to ρgxu2. In a reference frame whose origin is moving upward at a constant velocity u with respect to the platform you'll get a force equal to ρgx. The cost of picking up the chain is free in this frame per F=dp/dt. Use some other reference frame and you'll get yet another contradictory answer.

1This assumes the link being lifted is directly below the lifter; i.e., that the suspended part of chain is vertical. The force needed to start lifting a link is a bit more than half the link's weight if the link is not directly below the lifter. That this is never truly the case does introduce a tiny bit of lossiness. There's lots of other little bits of lossiness that appear here and there. They don't add up to the erroneous answer given by F=dp/dt.
 HW Helper Thanks P: 4,842 Suppose vertical slabs are placed on a frictionless horizontal surface as shown. All the slabs have a small hole through the center except the first slab on the left. A string is attached to the center of the first slab and the string is threaded through the holes in all the other slabs. In the first scenario there is initially a small space between each slab. The string is then pulled to the right so that the first slab maintains a constant speed u. In this case, each slab attains a speed u before slamming into the next slab. The second scenario has all the slabs in contact before the string is pulled. In each case there will be a certain amount of work done to get all the slabs moving at speed u. The work required for the first scenario is approximately Mu2 while for the second scenario it is Mu2/2, where M is the total mass of all the slabs. The reason for saying “approximately” Mu2 in the first case is that the work required to get the first slab moving is just mu2/2 while the work for each succeeding slab is mu2, where m is the mass of each slab. For a large number of slabs, the total work will be approximately Mu2. The work is greater for the first scenario because internal energy is generated in the slabs due to the completely inelastic collisions. The work-energy theorem stated in the form Wnet = ΔK is only applicable to a single point particle or to a system that can be treated as a single point particle. It is not generally true for a deformable system where parts of the system move relative to one another. Attached Thumbnails
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P: 25,228
 Quote by D H Do it right and they don't disagree.
One more time. Here's a work-energy picture. Suppose the length of the chain is L. Just so I don't have to worry about inelastic stuff, I'll put a machine on the platform that accelerates each link of the chain to velocity u before it gently attaches it to the rising chain. So when the chain clears the deck it has PE (1/2)L^2ρg and KE (1/2)Lρu^2 and the force on the chain was just as you say. But I also had to feed the machine on the platform energy (1/2)Lρu^2. Everything is all happy and conserved. Now you seem to be telling me that I can unplug the machine on the platform because I'll get to the same final state (maybe with some small losses) without it. THAT is violating energy conservation.
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P: 15,167
 Quote by Dick One more time. Here's a work-energy picture. Suppose the length of the chain is L. Just so I don't have to worry about inelastic stuff, I'll put a machine on the platform that accelerates each link of the chain to velocity u before it gently attaches it to the rising chain. So when the chain clears the deck it has PE (1/2)L^2ρg and KE (1/2)Lρu^2 and the force on the chain was just as you say.
That's wrong. If links magically attach themselves to the chain with the same velocity as the chain itself, the lifter only needs to support the weight of the chain, so F=ρgx. There is no u term with this style of magic.

It's best not to do physics in a world where magic occurs, but if you insist on doing so, you at least should get the mathematics right.
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P: 25,228
 Quote by D H That's wrong. If links magically attach themselves to the chain with the same velocity as the chain itself, the lifter only needs to support the weight of the chain, so F=ρgx. There is no u term with this style of magic. It's best not to do physics in a world where magic occurs, but if you insist on doing so, you at least should get the mathematics right.
Yes, you are right about that. I made it reversible and elastic. So the KE in the chain is coming from the machine on the platform. Not the same. Ooops. Sorry.
 P: 758 Oh please !! Why is everybody ignoring my previous posts in this thread. Please see my posts:20,21 and 48. I am just curious as to where I went wrong/or not. I (being a student) also did lot of such questions, but this one seem to bewilder me as well.
 Mentor P: 15,167 sankalpmittal, the correct answer is not (D), F=ρ(u^2-gx). It makes no sense. Force decreases with increasing height? The force obviously must increase as x increases.
P: 758
 Quote by D H sankalpmittal, the correct answer is not (D), F=ρ(u^2-gx). It makes no sense. Force decreases with increasing height? The force obviously must increase as x increases.
Thanks for your reply. My thinking in that was: When chain was pulled to a height x by a variable force "F" , then only the lower part receive normal reaction. Still most of its part was in air (consider vacuum). Thus the net force according to me should be F-ρgx and F can be evaluated. Using the answer (D), I could justify work energy theorem (post 21). How? I am confused.
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P: 9,849
 Quote by D H Next let's look at what happens during the process of lifting a single link of the chain off the platform. Between the point in time when the link is lying flat on the platform and the point in time when the link is vertical and lifts off the platform, the lifter only applies a force equal to half of the link's weight. The platform bears the other half of the link's weight1. It's only until the link is vertical and lifted off the platform that the lifter bears the entire weight of the link. By this time, the link has already been accelerated to an upward velocity of u.
In your model, where you lift each link up by one end until vertical, the acquired velocity of the link, at the point where it reaches vertical, has become horizontal.
 Use conservation of momentum with the platform bearing half the weight during the process of lifting a single link and you'll get a force equal to ρgx+½ρu2.
Eh? So if I hesitate with a link part lifted I get extra momentum from the platform?! None of this discussion needs to involve gravity. The same problem can be posed in free fall. Hence we can safely ignore any force from the platform.
 Finally, let's look at what happens if we use F=dp/dt.
If you mean in the sense of matter being created with zero velocity in the reference frame, I have not used that in any of my posts. I agree it's wrong.

All that said, I think I see why it's such a divisive issue. What will happen in practice depends on details of the system. Some of the models I've offered, in particular the one with the rising chain of 'buckets', accentuate the coalescence aspect, so maximising the energy loss. This may well be appropriate for, for example, a fine gold chain. (There is something distinctive about how it feels when you lift a fine gold chain from one end. It feels heavier while lifting it than when held steady.)
All of the models discussed will be lossy in some way, but some more than others. E.g. with your heavy chain lift, with large links, losses could be kept to a minimum by varying the lifting rate. As each link tilts, reduce the force, allowing the angular momentum of the link to assist in completing getting it to the vertical. By the time it becomes vertical, it is almost at rest. Now increase the force, initiating the lifting of the next link. I believe you can get arbitrarily close to lossless that way. But, as the link size tends to zero, the start-stop technique remains incompatible with the notion of lifting at steady speed, so this is not applicable to the OP.

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