Find Force to Lift Chain of Length L and Mass ρ Up

[Dick]

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.

 

[haruspex]

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.

 

[ehild]

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

 

[D H]

... 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 weight^1[/color]. 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 ρgx+ρu². 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.^1[/color]This 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.

 

[TSny]

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 W_net = 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.

 

[Dick]

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.

 

[D H]

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.

 

[Dick]

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.

 

[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]

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.

 

[haruspex]

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 weight^1[/color]. 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.

 

[sankalpmittal]

Any comments regarding my post #71 ?

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.

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

 

[D H]

Any comments regarding my post #71 ?

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.

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

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.

All that said, I think I see why it's such a divisive issue.

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 \rho gh 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
\rho g h + \frac{1}{2} \rho u^2
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]

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.

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]

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.

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]

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.

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:
\frac{\partial T}{\partial h} = \rho g
(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:
\Delta T = \rho g \Delta h
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:
F_{pull}= A(T+ \rho g h)
(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:
F_{pull} = T + \rho g h
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 \rho g h 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.