Lifting up a chain

sankalpmittal

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.

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.

sankalpmittal

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.

haruspex

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.
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.
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.

sankalpmittal

Any comments regarding my post #71 ?

BTW, I am also of the opinion that force from the platform on chain can be ignored.

D H

Look at the problem in parts. The lifter is bearing the weight of the chain that is already lifted off the platform and is applying a force to lift even more of the chain off the platform. Both of those forces point upwards when the chain is being lifted upwards. Answer D makes no sense.

Why? What justifies that opinion?

The work energy theorem, done right, yields an unambiguous answer, one that isn't listed. You'll get the same answer if you look at the problem from a Lagrangian or Hamiltonian perspective.



I think it's just a bad problem. You get a consistent answer, one not listed, if you use work energy concepts, a Lagrangian formulation, or a Hamiltonian formulation. You get a different answer if you ignore that the platform is somehow of assistance.

The real answer? You'd have to have a very detailed model of the elasticity/inelasticity of the chain, detailed models of the transients as parts of the chain are lifted off the platform, etc. A freshman physics problem should not require finite element analysis to get the right answer.

BruceW

The answer depends on the assumptions made. If we assume the chain is fed through to us (i.e. we don't need to provide its KE), then the pulling force only needs to be [itex]\rho gh[/itex] In other words, we only need to provide the GPE of the chain.

If we assume that we must provide the KE and GPE of the chain, then the pulling force is
[tex]\rho g h + \frac{1}{2} \rho u^2 [/tex]
This does agree with using dP/dt As long as we assume that the tension in the chain near the ground is non-zero.

Finally, we could assume that some of the force we provide gets wasted in friction or whatever, as the chain is being accelerated on the ground. In this case, we can just say the pulling force must be greater, and that the tension in the chain near the ground is also greater. In this case, the pulling force depends on the kind of model we use for energy dissipation, so I don't think this is the answer to this question.

Anyway, neither of the simplest answers seem to be in the possible answers given.

Dick

That hits the nail on the head. If you look back at my bogus energy argument in post 66 and ignore the stupid part about offhandedly saying the lifting force is unchanged, it's a way to get the chain to the right final state with the right amount of input energy. But now look at momentum balance. The final momentum of the chain is Lρu. The F=ρgx lifting force as DH correctly points out, just supports the chain. It doesn't provide any momentum input to the chain. It's all coming from the machine on the platform that's putting upward momentum into the chain by putting downward momentum into the platform. If it's not there then I would say that the source of all upward momentum must be the hoist. I would say that the ONLY THING that the platform should be allowed to do is provide normal forces to the weight on it.

Now go back to having the hoist alone lifting the chain and use the proposed ρgx+ρu^2/2 (coming from energy conservation). Since the ρgx provides no momentum input then all the the momentum input must be coming from the ρu^2/2 part of the force. If I compute that boldly using Δp=FΔt (which is pretty close to F=dp/dt which is really getting a bad rap here, but there must some aspect of momentum conservation to believe in) putting Δt=L/u I get Lρu/2. Comparing that with Lρu I see the hoist provided ONLY ONE HALF of the final momentum.

You really have to attribute the rest to some contribution from the platform. It has to somehow push the chain up with more than normal forces. I think this is what BruceW is talking about when he is saying there is a nonzero tension at the bottom of the chain. I don't think that's a very reasonable thing.

haruspex

Agreed. As I keep suggesting, we can get rid of this 'contribution from the platform' furphy by casting it in a free fall context. We can also get rid of any orthogonal movement by taking the chain/string/plates to be exceedingly thin. E.g. it could be a tiny coil of string around the axis of movement. The hard part, for the imagination, is to understand that:
1. Pulling on the free end of the string is not going to make the coil as a whole start to move. This is because the string must be taken as completely limp - it can sustain a tension, but no bending moment.
2. The newly-pulled part of the string will not 'overshoot', i.e. will not at any time exceed the speed u. Any real string would have some elasticity. This would mean that at first it travels more slowly than u, so when at speed u there is still some extension, and it will continue to accelerate. This overshooting would then assist in drawing off the next part of the string.
What might clinch the argument, if anyone can manage this, is to develop the full (wave) equation for a given elasticity, then observe what happens as the elasticity tends to 0.

BruceW

The point I was making is that simple mechanics tells us the behaviour of the chain above the ground level, even if the behaviour at ground level is more complicated. To show this, here is the equation for balance of momentum in the suspended part of the chain:
[tex]\frac{\partial T}{\partial h} = \rho g [/tex]
(Where h is height, T is tension and g is the absolute value of gravitational acceleration)
The equation assumes that the suspended part of the chain is moving upward at a steady rate. And since the density doesn't change, we get:
[tex]\Delta T = \rho g \Delta h [/tex]
Now we can use this equation to relate tension at the bottom of the suspended part of the chain to the tension at the top. This is useful because the tension at the top (times by cross-section area of chain), is equal to the pulling force. Explicitly:
[tex]F_{pull}= A(T+ \rho g h) [/tex]
(Here, A is the cross-section area, T is the tension at the lowest part of the suspended section of chain and h is the height of the chain above the ground level)
The formula can be used in the limit of thin chain, by assuming that as A gets very small, rho and T get very large, so that in the limit, none of the terms explode, and we get:
[tex]F_{pull} = T + \rho g h [/tex]
Where T is now interpreted as the tension times area at the lowest part of the suspended section of chain, and rho is now mass per length. The equation is now in a form which is useful to us. It is useful because the pulling force is what we want to find out, and T essentially represents our assumption on what happens to the chain at ground level.

The form of the equation above tells us that the pulling force is split into two parts. The term [itex]\rho g h[/itex] negates the effect of gravity on the suspended section of chain. The other term (T) is the force transmitted through tension at the bottom of the suspended section, to the ground-level section of the chain. The term T is the one we don't know.