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

[Dick]

Suppose we have just 2 links each of mass m on a horizontal frictionless surface as shown. The right link is already in motion with speed u while the left link is still at rest. Left F Δt be the impulse that the right link gives to the left link to get the left link moving also at speed u. So, F Δt = mu. There will be a reaction impulse of equal magnitude acting to the left on the right link which we arrange to be exactly balanced by an external impulse (from the blue force) so that the right link maintains speed u. Then the work done by the blue force is F Δx = F(uΔt) =(F Δt) u = mu u = mu² = twice the increase in KE of the system. So, more work was done by the blue force than shows up as KE.

Now that I agree with. Picking up the links is an inelastic process.

 

[D H]

Now that I agree with. Picking up the links is an inelastic process.

Of course it is. The correct answer is that $F\ge \rho(gx+\frac 1 2 u^2)$, with equality resulting only if the process is reversible (i.e., picking up the chain is an elastic process).

There's not one thing in the F=dp/dt approach that suggests an irreversible process. In fact, this erroneous approach yields a reversible process. Simply apply this exact same concept but with the chain being lowered rather than raised.

 

[tms]

This is one of many cases where naively using F=dp/dt is flat out wrong. It's F=ma, not F=dp/dt.

What!? $F = dp/dt$ is the definition of force. $F = ma$ only applies when the mass is constant. In this problem, since the speed is constant, there is no acceleration, and, according to $F = ma$, there is no force.

 

[TSny]

We can replace the linked chain by an “elastic” system. Two masses (each m) are connected by a spring as shown. Initially the spring is unstretched with the mass on left at rest and the mass on right moving to the right with speed u. Also we have arranged a variable force F acting on the right mass that will always have the same magnitude as the spring force. Thus, the right mass always moves with the same speed u. The force F acts between t = 0 and t = t_f where t_f is chosen to be the instant when the velocity of the center of mass (cm) of the two-mass system is u.

The total momentum at t_f will be $(M_{system})V_{cm} = (2m)u = 2mu$. Since we started at time t = 0 with total momentum $mu$, the change in momentum of the system between t = 0 and t = t_f is $mu$. This change in momentum must come from the impulse of the applied force:

$\int{Fdt} = mu$.

The energy of the system at t = 0 is $E_o =mu^2/2$.

The energy of the system at t = t_f is $E_f = KE_{system} + PE_{spring}$.

We can always write the KE of the system as KE due to motion of the cm plus KE relative to cm. The KE due to motion of the cm is

$(1/2)(2m)V_{cm}^2 = mV_{cm}^2 = mu^2$ at t_f.

The KE relative to cm is just the “vibrational” kinetic energy associated with the rate of separation of the two masses. We combine the vibrational KE with the spring PE into a total “internal energy”. So, our energy at time t_f is $E_f = mu^2 +E_{int}$.

Hence, the change in energy from t = 0 to t = t_f is

$\Delta E = mu^2/2 + E_{int}$.

The work done by the applied force betwen t = 0 and t = t_f is

$W = \int{Fdx} = \int{F\frac{dx}{dt}dt} = \int{Fudt} = u\int{Fdt} = u(mu) = mu^2$

Suppose for some reason we didn’t know about the “internal energy”. (Maybe the spring is hard to see and it’s very stiff so that the vibrational amplitudes are two small for us to notice.) Then we would only be aware of the energy due to overall motion of the center of mass. Thus, we would claim that the change in energy of the system between t = 0 and t = t_f would be just the term $mu^2/2$ in the expression for $\Delta E$.

So, it would appear that the work done was twice the increase in energy of the system in apparent violation of the work-energy theorem. But we see that in fact the other half of the work went into “hard-to-see” internal energy (i.e., vibrational KE and PE of the spring).

 

[Dick]

Of course it is. The correct answer is that $F\ge \rho(gx+\frac 1 2 u^2)$, with equality resulting only if the process is reversible (i.e., picking up the chain is an elastic process).

There's not one thing in the F=dp/dt approach that suggests an irreversible process. In fact, this erroneous approach yields a reversible process. Simply apply this exact same concept but with the chain being lowered rather than raised.

When the chain is being lowered the force is not the same as when the chain is being raised. It's not reversible. This isn't even a uniquely confusing problem. Pouring sand onto a conveyor belt gives the same sort of result. Even the standard air table experiment with pucks and velcro shows it. As Tsny points out, you can replace the velcro with locking springs and get the same result. Then you can see where the kinetic energy went. It becomes internal to the two puck system instead of dissapated into friction. This is the same sort of problem.

 

[Saitama]

Thanks everyone for taking your time to explain the things so nicely. I am trying to understand each and every single post in this thread. This is probably going to take some time at my part. This thread is really interesting. Please continue your further discussions. I am following this thread, please don't think that I have left this thread. Its just that I am an average student and it takes me some (or a lot of) time to understand the things.

 

[Dick]

Thanks everyone for taking your time to explain the things so nicely. I am trying to understand each and every single post in this thread. This is probably going to take some time at my part. This thread is really interesting. Please continue your further discussions. I am following this thread, please don't think that I have left this thread. Its just that I am an average student and it takes me some (or a lot of) time to understand the things.

That's good news! If people with a lot of experience with stuff like this can disagree, no reason it shouldn't take you time to digest it. Feel free to chip in.

 

[Saitama]

One more question.

@D H: Then what should be the answer to the second part of the question mentioned in the pdf you posted on the first page. Please don't take me wrong but the pdf you posted is hosted under IIT Kanpur's (a reputed institution here) website and its very unlikely for them to do such mistakes (or maybe they did this time). Were you able to find the solution key of that problem sheet?

 

[haruspex]

Only just found this intriguing thread. Can't resist chipping in.
Yes indeed, TSny, it's the inelasticity. (It could be made even harder to comprehend by dealing with an inelastic string instead of a chain of links, but the same applies.) In time δt, a length uδt, mass ρuδt, accelerates from 0 to u. Note that this is an unbounded acceleration, so we're dealing with impulses; energy cannot be taken to be conserved. Momentum is conserved, so the force is (ignoring gravity - this would also work in free fall) ρu². The KE gained in δt = ρu³δt/2, while the work done, as the force moves distance uδt, is twice that.
With an elastic string, each uδt would take time to reach speed u, and would thereafter oscillate. Now that would make one helluva spring problem!

 

[Dick]

Only just found this intriguing thread. Can't resist chipping in.
Yes indeed, TSny, it's the inelasticity. (It could be made even harder to comprehend by dealing with an inelastic string instead of a chain of links, but the same applies.) In time δt, a length uδt, mass ρuδt, accelerates from 0 to u. Note that this is an unbounded acceleration, so we're dealing with impulses; energy cannot be taken to be conserved. Momentum is conserved, so the force is (ignoring gravity - this would also work in free fall) ρu². The KE gained in δt = ρu³δt/2, while the work done, as the force moves distance uδt, is twice that.
With an elastic string, each uδt would take time to reach speed u, and would thereafter oscillate. Now that would make one helluva spring problem!

Right. But energy is always conserved. What's not conserved is the center of mass kinetic energy of a composite system. It can hit something and absorb it into internal vibration. That was the wiggling chain I was referring to earlier.

 

[ehild]

I got an idea while sleeping . We assumed the situation on the left in the pic, that the chain stays at the same place while lifted, as if it was in a tube. But that involves that the parts on the floor have to move to that tube somehow, which involves some motion.

Imagine that the chain is straight on the floor and the person who lifts the chain moves his hand always above the end of the horizontal part. That means, the vertical part moves not only up but also sideways. The length of the chain is constant, so both components of the velocity are the same u. That means
KE=1/2 m (v²(vertical)+v²(horizontal))=1/2 (ρx)(2u²)
E = PE+KE=1/2 ρx²g+ρxu², dW=F(lift)dx=(ρxg+ρu²[/B])dx. F(lift)=(ρxg+ρu²) from work energy considerations
Anyway, it can not be assumed that the chain does not have any horizontal component of velocity. Either that part which is on the floor, or that part which is in the air or both. The KE has to include both terms. But the time derivative of the vertical component of momentum is equal to the vertical force. ehild

 

[D H]

OPlease don't take me wrong but the pdf you posted is hosted under IIT Kanpur's (a reputed institution here) website and its very unlikely for them to do such mistakes (or maybe they did this time). Were you able to find the solution key of that problem sheet?

No, I wasn't able to find the solution key.

Don't take something written by reputed institutions as sacrosant. MIT makes mistakes. So does Caltech. So do highly reputable publishers. That's why they all write erratum sheets. Those mistakes typically don't propagate in countries with strong copyright laws. There's a problem in India. India either has weak copyright law or has rather weak enforcement of them. Here's a google book search for the exact phrase "A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration" that illustrates what happens in India: http://www.google.com/search?q=A+pa...ius+r+such+that+its+centripetal+acceleration"

Aha! Pranav-Arora, it turns out you were involved in the thread that instigated this search: [thread]616190[/thread].

What!? $F = dp/dt$ is the definition of force. $F = ma$ only applies when the mass is constant.

Where did you get that idea? It's certainly not anywhere in Newton's Principia. (Then again, neither is F=ma. Newton's Principia is almost entirely calculus-free.)

There's a big problem with F defined via F=dp/dt: The $\dot m v$ term makes force a frame dependent quantity in the case of variable mass systems. On the other hand, defining force defined via F=ma means that force is the same quantity in all reference frames since both mass and acceleration are invariant quantities in Newtonian mechanics. This is why aerospace engineers almost inevitably choose F=ma rather than F=dp/dt.

In this problem, since the speed is constant, there is no acceleration, and, according to $F = ma$, there is no force.

Speed isn't constant. The links at rest on the platform change velocity by u as they are picked up. There is a problem is with F=dp/dt, however. The momentum of the rising part of the chain is identically zero from the perspective of a reference frame moving upwards at a constant speed u with respect to the platform. In this frame, F as calculated with dp/dt is zero if you are looking only at the momentum of the rising part of the chain.

There is a way around this mess, and that's not to use variable mass. Look at what happens to the chain as a whole. Now F=dp/dt and F=ma yield the same answer because $\dot m$ is zero. Unfortunately that will not solve the problem. The problem is that this is a kinematic view of force rather than a dynamic view of force. We need to attribute this kinematical net force to root causes to convert this kinematical POV to a dynamical POV. Naively attributing all of this force to the hoist, or whatever is lifting the chain, is incorrect.

To see that it must be incorrect, simply reverse the process: Lower the chain rather than raising it. To keep the chain moving downward at a constant velocity we'll let connect the upper end of the chain to some physical device and let the lowering chain do work on that device. How much work is performed? If we attribute all of the change in momentum to work done by that the device we get an over unity system. More work will be performed than is allowed by the conservation laws. That doesn't make sense.

One way out of this morass (and the way out of the raising morass as well) is to attribute some of the change in the momentum of the chain to the platform. For example, one could look at the transients as a link is picked up from the platform. Yech. That's getting into finite element analysis, something that is far beyond the scope of an introductory physics course.

The easiest way out of this morass is to look to the work energy theorem. This yields a nice simple answer (one that is not listed in the provided choices) if all forces are conservative. All bets are off if the system is not conservative. You aren't going to get a nice simple answer in the case of nonconservative forces. Nonconservative forces almost inevitably yield ugly solutions.

 

[sankalpmittal]

Pranav, from wherever you posted the question, what answer it gives ? I got (D) as my answer and then only can the work energy theorem be justified. Else , it appears magic.

If this is not the correct answer, please look at posts 20 and 21 and tell, where I went wrong.

@DH: As far as the other problem https://www.physicsforums.com/showthread.php?t=616190 is concerned that textbook answer which OP there posted is likely wrong. My textbook too has the question but its answer is way different...

The easiest way out of this morass is to look to the work energy theorem. This yields a nice simple answer (one that is not listed in the provided choices) if all forces are conservative. All bets are off if the system is not conservative. You aren't going to get a nice simple answer in the case of nonconservative forces. Nonconservative forces almost inevitably yield ugly solutions.

Well, I always use this equation if non conservative forces are involved.

W_nc=E_f-E_i

W_nc:Work done by non conservative forces.
E_f: Final Mechanical Energy.
E_i: Initial Mechanical energy.

 

[haruspex]

The easiest way out of this morass is to look to the work energy theorem. This yields a nice simple answer (one that is not listed in the provided choices)

... 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.

 

[ehild]

Any comment about my post#46?

ehild

 

[haruspex]

Any comment about my post#46?

ehild

I would have thought that the chain/string could be arbitrarily thin. The thinner it is, the less it will matter if it's all piled up.

 

[ehild]

It has constant length. It can not happen that the lifted part moves vertically upward and the horizontal part does not move at all. So the KE can not be calculated from u, the vertical component alone.

ehild

 

[haruspex]

It has constant length. It can not happen that the lifted part moves vertically upward and the horizontal part does not move at all. So the KE can not be calculated from u, the vertical component alone.

ehild

If you lay it out straight, sure, but in the OP it was piled up, and as you make it progressively thinner that matters less and less.

 

[Dick]

To see that it must be incorrect, simply reverse the process: Lower the chain rather than raising it. To keep the chain moving downward at a constant velocity we'll let connect the upper end of the chain to some physical device and let the lowering chain do work on that device. How much work is performed? If we attribute all of the change in momentum to work done by that the device we get an over unity system. More work will be performed than is allowed by the conservation laws. That doesn't make sense.

It's no more reversible than any other inelastic process. Hence there is no energy paradox. But I think I've said that before.

 

[tms]

... 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.

 

[haruspex]

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.

 

[tms]

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.

 

[Dick]

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.

 

[haruspex]

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.

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.

 

[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.

 

[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 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 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.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.

 

[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.