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

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SUMMARY

The discussion centers on calculating the force required to lift a chain of length L and mass per unit length ρ at a constant velocity u. The correct expression for the force F as a function of the height x of the chain above the platform is F = ρ(gx + u^2), which accounts for both gravitational force and the force needed to accelerate the chain. Participants emphasized the importance of understanding variable mass systems and the application of the work-energy theorem in this context. Misapplications of fundamental equations like F = dp/dt can lead to incorrect conclusions, highlighting the necessity of careful analysis in physics problems involving variable mass.

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  • Understanding of Newton's laws of motion, particularly F = dp/dt.
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Homework Statement


The end of a chain of length L and mass per unit length ρ, which is piled up on a horizontal platform is lifted vertically with a constant velocity u by a variable force F. Find F as a function of height x of the end above platform.
A)ρ(gx+2u^2)
B)ρ(gx+u^2)
C)ρ(2gx+ρu^2)
D)ρ(u^2-gx)

Homework Equations


The Attempt at a Solution


The chain goes up with a constant speed so at any instant the net force should be zero i.e. F=mg. If the height of the end is x, then m=ρx therefore F=ρgx. But there is no such option. :confused:

Any help is appreciated. Thanks!
 
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Think: The part of the chain still on the ground has zero speed. The force has to accelerate new and new pieces to speed u.

ehild
 
The mass of the chain being pulled up by the variable force is also variable.
 
ehild said:
Think: The part of the chain still on the ground has zero speed. The force has to accelerate new and new pieces to speed u.

ehild

I understand it now but I am unable to form an equation to find F.
 
Start from first principles:
\begin{align}<br /> F &amp;= \frac{dp}{dt} \\<br /> &amp;= v\, \frac{dm}{dt} + m\, \frac{dv}{dt}.<br /> \end{align}<br />
 
tms said:
Start from first principles:
\begin{align}<br /> F &amp;= \frac{dp}{dt} \\<br /> &amp;= v\, \frac{dm}{dt} + m\, \frac{dv}{dt}.<br /> \end{align}<br />

F=u \cdot \frac{d(ρx)}{dt}+ρx \frac{dv}{dt}
I can't figure out what should replace dv/dt here.

ehild said:
Or use energy. The work done when increasing the length by dx is the difference of the initial and final energies.
I have formulated an expression for the work done but not sure how to proceed further. Here's the expression I have reached:
W=-ρgxdx-\frac{ρu^2dx}{2}
 
The velocity is constant, isn't it? And think: what force you have to use in the first equation.
As for the method concerning change of energy, you have to use dW instead of W and it is the infinitesimal work Fdx when moving the end of the rope up by dx. Fdx=d(PE+KE). And mind the signs. But something is wrong, I do not see what. The two methods give different results.

ehild
 
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ehild said:
The velocity is constant, isn't it?
So the term dv/dt equals to zero.
ehild said:
And think: what force you have to use in the first equation.
Do you mean I need to add an extra term in the LHS? I think the first equation should be this:
F-ρgx=ρu^2
F=ρgx+ρu^2
Ahah, I reached the answer, thanks a lot! :smile:

ehild said:
As for the method concerning change of energy, something is wrong, I do not see what. You have to use dW instead of W and it is the infinitesimal work Fdx when lifting dx length.
Yes, you are right, I need to use dW and I guess I made a sign error. The infinitesimal work should be
dW=ρgxdx+\frac{ρu^2dx}{2}
Substituting dW=Fdx doesn't give me the right answer.
 
ehild, can you please explain why the energy method doesn't work?
 
  • #10
I do not know. But I am inclined to think that the other method can be wrong. There is force also between the two parts of the chain. I asked the other HH-s, no reply yet.

ehild
 
  • #11
The work-energy theorem is correct. A mis-application of F=dp/dt yields ρ(gx+u^2), which apparently is the supposedly correct answer. It's not. One has to be extremely careful in applying either F=ma or F=dp/dt to variable mass systems. Do it wrong (which is remarkably easy to do for variable mass systems) and you'll get the wrong answer. As a clue that this approach is wrong, I found an old version of the problem here, http://www.iitk.ac.in/phy/oldfiles/phy102N/Problem_Sheet_6.pdf , question #8. From that page, emphasis mine:
One end of an open-link chain of length L and mass ρ per unit length, piled on a platform, is lifted vertically with a constant velocity v by a variable force F. Find F as a function of the height x of the end above the platform. Also find the energy lost during the lifting of the chain.​
So, let's use that answer of F=ρ(gx+u^2) to calculate the work performed. The net force is that force less gravity, or Fnet=ρu^2. Integrating this to find the work performed while lifting the chain to a height x at a constant velocity u yields W=ρxu^2, which is twice the change in kinetic energy. Is the difference between this calculated amount of work performed and the change in the kinetic energy the amount of "energy lost during the lifting of the chain"? No. It's the amount by which the author of the problem messed up.
 
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  • #12
ehild said:
The velocity is constant, isn't it? And think: what force you have to use in the first equation.
The force in the equation is the net force, of course. It is composed of the upwards force and gravity.
 
  • #13
tms said:
The force in the equation is the net force, of course. It is composed of the upwards force and gravity.

Read DH's post. The upward force is not the lifting force alone.

ehild
 
  • #14
ehild said:
Read DH's post. The upward force is not the lifting force alone.
I did read it (after I made my own post). What other force is there?
 
  • #15
Normal force from the ground.

ehild
 
  • #16
That only acts on the part of the chain still on the ground, not in motion. Also, according to DH, the force calculated from F = dp/dt is too large, so if there is another force it must be downwards.
 
  • #17
tms said:
Start from first principles:
F = \frac{dp}{dt}
<br /> OK, let&#039;s do that.<br /> <br /> Suppose that at some point in time t the length of the chain being held off the platform is x(t). This length of chain is moving upwards at a constant velocity u with respect to the platform. Given that the mass per unit length of the chain is \rho, this means the momentum of the chain with respect to the platform is p(t)=\rho x(t) u, directed upward. Some very short time \Delta t later, the length of the chain moving upward is x(t)+u\Delta t, making the momentum p(t+\Delta t)=\rho (x(t) + u\Delta t)u. The change in momentum is \Delta p = \rho u^2 \Delta t. Applying F_{\text{net}}=\lim_{\Delta t \to 0} \Delta p / \Delta t yields F_{\text{net}}=\rho u^2. Adding the weight of the chain yields the total force needed to keep the chain moving at a constant velocity, F_{\text{tot}} = \rho gx + \rho u^2. That&#039;s answer (B). Done!<br /> <br /> Or maybe we&#039;re not done. It&#039;s always good to do a sanity check.<br /> <br /> Let&#039;s see how much work is done by this force and compare this to the change in kinetic energy. The net force is F_{\text{net}}=\rho u^2, a constant. Calculating the work performed by this constant net force yields W=\int_0^x F\,dl = \rho x u^2. The change in kinetic energy is half this amount. At this point we can do one of two things:<br /> (a) Attribute this discrepancy to energy that is somehow lost.<br /> (b) Figure out where we went wrong.<br /> <br /> The right option is (b), figure out where we went wrong. Energy is not somehow lost. It&#039;s a conserved quantity. Where we went wrong was in attributing all of this <i>F<sub>net</sub></i> to the hoist that is lifting the chain. We weren&#039;t solving the chain-lifted-off-a-platform problem. We were instead solving this problem:<div style="margin-left: 20px"><i>The end of a chain of length x and mass per unit length ρ is lifted vertically with a constant velocity u by a variable force F. At any point in time, mass magically appears out of nowhere at a rate dm/dt = \rho u with zero velocity with respect to the ground and attaches itself to the end of the chain. Find F as a function of height x of the end above platform.</i>&#8203;</div>It&#039;s best not to solve problems in universes where magic occurs.
 
  • #18
D H said:
So, let's use that answer of F=ρ(gx+u^2) to calculate the work performed. The net force is that force less gravity, or Fnet=ρu^2. Integrating this to find the work performed while lifting the chain to a height x at a constant velocity u yields W=ρxu^2, which is twice the change in kinetic energy.
It's late, and I'm not thinking all that clearly, but I think the problem is with the integration. \int F\,dx takes care of a force that varies with position acting on each bit of mass, but in the problem each bit of mass moves a different distance, so the varying force acts on each bit over a varying distance.
 
  • #19
D H said:
OK, let's do that.

Suppose that at some point in time t the length of the chain being held off the platform is x(t). This length of chain is moving upwards at a constant velocity u with respect to the platform. Given that the mass per unit length of the chain is \rho, this means the momentum of the chain with respect to the platform is p(t)=\rho x(t) u, directed upward. Some very short time \Delta t later, the length of the chain moving upward is x(t)+u\Delta t, making the momentum p(t+\Delta t)=\rho (x(t) + u\Delta t)u. The change in momentum is \Delta p = \rho u^2 \Delta t. Applying F_{\text{net}}=\lim_{\Delta t \to 0} \Delta p / \Delta t yields F_{\text{net}}=\rho u^2. Adding the weight of the chain yields the total force needed to keep the chain moving at a constant velocity, F_{\text{tot}} = \rho gx + \rho u^2. That's answer (B). Done!

Or maybe we're not done. It's always good to do a sanity check.

Let's see how much work is done by this force and compare this to the change in kinetic energy. The net force is F_{\text{net}}=\rho u^2, a constant. Calculating the work performed by this constant net force yields W=\int_0^x F\,dl = \rho x u^2. The change in kinetic energy is half this amount. At this point we can do one of two things:
(a) Attribute this discrepancy to energy that is somehow lost.
(b) Figure out where we went wrong.

The right option is (b), figure out where we went wrong. Energy is not somehow lost. It's a conserved quantity. Where we went wrong was in attributing all of this Fnet to the hoist that is lifting the chain. We weren't solving the chain-lifted-off-a-platform problem. We were instead solving this problem:
The end of a chain of length x and mass per unit length ρ is lifted vertically with a constant velocity u by a variable force F. At any point in time, mass magically appears out of nowhere at a rate dm/dt = \rho u with zero velocity with respect to the ground and attaches itself to the end of the chain. Find F as a function of height x of the end above platform.​
It's best not to solve problems in universes where magic occurs.

You explained it nicely. Thanks! :smile:

But do you mean that the question given is wrong? If I was asked to calculate the energy lost, how can I find it?

Can you give links to some good resources where I can learn more about these variable mass systems?
 
  • #20
Pranav-Arora said:
You explained it nicely. Thanks! :smile:

But do you mean that the question given is wrong? If I was asked to calculate the energy lost, how can I find it?

Can you give links to some good resources where I can learn more about these variable mass systems?

This question did surprise me too. I completed H.C. Verma work and energy and there was 57th question in which we had a chain just touching the ground and it was released from rest. It was asked that out of chain of length L , x length of it strike the floor. No heap formed. We had to calculate force exerted by chain on floor as a function of displacement x. I tried and got the wrong answer until I realized that weight of chain also acts in addition to change in momentum. In this case also , normal reaction acts. By using this logic , I get the same answer as DH. Mechanical energy is not conserved as I suspect, may be some non conservative forces be acting in chain+earth system.

But again, work energy theorem,i.e. Wnet=ΔK.E. still applies. Did you try applying it ? If work energy theorem fails, then I am sure that this question is somewhat a magic. Work energy theorem applies to every inertial Earth system, as per H.C. Verma.
 
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  • #21
D H said:
OK, let's do that.

Suppose that at some point in time t the length of the chain being held off the platform is x(t). This length of chain is moving upwards at a constant velocity u with respect to the platform. Given that the mass per unit length of the chain is \rho, this means the momentum of the chain with respect to the platform is p(t)=\rho x(t) u, directed upward. Some very short time \Delta t later, the length of the chain moving upward is x(t)+u\Delta t, making the momentum p(t+\Delta t)=\rho (x(t) + u\Delta t)u. The change in momentum is \Delta p = \rho u^2 \Delta t. Applying F_{\text{net}}=\lim_{\Delta t \to 0} \Delta p / \Delta t yields F_{\text{net}}=\rho u^2. Adding the weight of the chain yields the total force needed to keep the chain moving at a constant velocity, F_{\text{tot}} = \rho gx + \rho u^2. That's answer (B). Done!

Or maybe we're not done. It's always good to do a sanity check.

Let's see how much work is done by this force and compare this to the change in kinetic energy. The net force is F_{\text{net}}=\rho u^2, a constant. Calculating the work performed by this constant net force yields W=\int_0^x F\,dl = \rho x u^2. The change in kinetic energy is half this amount. At this point we can do one of two things:
(a) Attribute this discrepancy to energy that is somehow lost.
(b) Figure out where we went wrong.

The right option is (b), figure out where we went wrong. Energy is not somehow lost. It's a conserved quantity. Where we went wrong was in attributing all of this Fnet to the hoist that is lifting the chain. We weren't solving the chain-lifted-off-a-platform problem. We were instead solving this problem:
The end of a chain of length x and mass per unit length ρ is lifted vertically with a constant velocity u by a variable force F. At any point in time, mass magically appears out of nowhere at a rate dm/dt = \rho u with zero velocity with respect to the ground and attaches itself to the end of the chain. Find F as a function of height x of the end above platform.​
It's best not to solve problems in universes where magic occurs.

Wait , I may be stupid , but let's see. Wnet=ΔKE

Fnet = ρu2-ρgx

Now as x=u2/2g

Fnet = ρu2-ρu2/2
F net =ρu2/2

Integrating net F with respect to dx from 0 to x,

W=ρu2x/2

So W=ρu2x/2 which is equal to ΔKE. As length x is in air , its incorrect to consider normal reaction. I think this is where you blundered.

Edit: As per me the correct answer is D. Then only this question is justified by work energy theorem.
 
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  • #22
Pranav-Arora said:
But do you mean that the question given is wrong?
Yep. In a multiple choice type question, the question itself is wrong when the correct answer is not one of listed answers. That's the case here.

This is not the first time this has happened. Textbooks can be erroneous. Multiple choice: The problem apparently is particularly bad
(a) In far too many online automated physics homework problem systems.
(b) In far too many Indian physics texts.
(c) Both of the above are true.

Apparently India has rather weak enforcement of its copyright laws, which leads Indian textbook authors to copy problems from one another. It's much easier to copy some other author's already-worked problems rather than to create new ones. That's fine (other than copyright issues) if the problems are well written and the solutions are correct. It's not so fine if the question is poorly written, if the correct answer isn't present in the multiple choice list, or if the solution in the answer guide / worked example is incorrect. We've chased down a number of problems with Indian physics texts at this site, occasionally seeing the exact same bad question replicated in dozens of different texts.

If I was asked to calculate the energy lost, how can I find it?
The correct answer is that energy is not lost, at least not ideally. Work is being done against gravity in this problem Gravity is a conservative force. There are no losses in working against gravity.

In reality, there will be some energy loss due to entropy. (Note: the energy isn't really "lost". It's just converted to unusable energy: Heat.) A cable or chain may heat up a bit due to non-conservative interactions as the cable/chain grows taut, links shift, etc. We can ignore all that messiness by assuming an ideal chain. Even a terribly constructed, non-ideal chain will not heat up to the extent implied by the wrong answer to this problem.

Can you give links to some good resources where I can learn more about these variable mass systems?
The best thing to do in an introductory physics class is to try to pose the question without worrying about variable mass. It's messy, and easy to mess up. IMO, it's not a subject introductory physics classes should delve around in much (other than perhaps rockets as an exemplar). This particular problem can easily be answered without looking at variable mass. Simply look at the chain of fixed length L as a whole.
 
  • #23
I’ll add my 2 cents to the confusion! For any system ##F_{net}^{ext} = M a_{cm}## where M is the total mass of the system.

Let the system be the entire chain alone. The net external force is ##F_{net}^{ext} = F_{lift} – Mg + F_N## where ##F_{lift}## is the lifting force and ##F_N## is the normal force acting on the part of the chain still resting on the platform.

When amount ##x## of chain is off the table, the center of mass of the section that is off the table is at ##x/2## above the platform and has mass ##\rho x##. So, relative to the platform, the location of the center of mass of the whole chain is

##x_{cm} = \frac{(\rho x)(x/2)}{M} = \frac{\rho}{2M}x^2##.

Taking the second time derivative of this yields ##a_{cm} = \frac{\rho}{M}(\dot{x}^2 + x\ddot{x}) = \frac{\rho}{M}u^2## since ##\ddot{x} = 0##

So, we have ##F_{lift} – Mg + F_N = \rho u^2## or

##F_{lift}= \rho u^2 + Mg - F_N##

If we take the normal force to equal the weight of the part of the chain resting on the table, then ##F_N = (M-\rho x)g## and we get

##F_{lift}= \rho u^2 + \rho g x = \rho (u^2+g x)##
 
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  • #24
TSny said:
I’ll add my 2 cents to the confusion!
##F_{lift}= \rho u^2 + \rho g x = \rho (u^2+g x)##

It is very convincing 2 cents :) But what is the problem with the energy then?


ehild
 
  • #25
ehild said:
It is very convincing 2 cents :) But what is the problem with the energy then?
I think the problem is with the integration W = \int F\,dx. For any particular bit of the chain that is correct, but each bit of the chain goes a different distance so the force acts on each bit for a different distance, thus doing a different amount of work. I haven't yet figured out the correct integral, though. I know I'm not being abundantly clear, but I think the answer is in this direction.

At any rate, if the energy is right, what is wrong with the F = dp/dt solution? Since that is basically the definition of force, it, too, must be right. Picking the energy to be right because dm/dt is "magic" is not really good enough.
 
  • #26
F acts at the lifted end of the rod and its work is ∫Fdx. We do that work on the whole system, and that work increases the energy. Unless something else non-conservative force does also work. :confused:

ehild
 
  • #27
Regarding energy, I suspect that when each link is jerked off the platform, it's like an inelastic collision- causing internal energy to increase. The chain will get warmer!
 
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  • #28
TSny said:
Regarding energy, I suspect that when each link in jerked off the platform, it's like an inelastic collision- causing internal energy to increase. The chain will get warmer!

But why?

ehild
 
  • #29
ehild said:
F acts at the lifted end of the rod and its work is ∫Fdx. We do that work on the whole system, and that work increases the energy. Unless something else non-conservative force does also work. :confused:
I'm confused, too. I still don't see anything wrong with the F = dp/dt solution.
 
  • #30
TSny said:
Regarding energy, I suspect that when each link in jerked off the platform, it's like an inelastic collision- causing internal energy to increase. The chain will get warmer!
But there is nothing in the problem or solution(s) about the internal working of the "chain". The solution(s) apply just the same to the anchor chain of an aircraft carrier or to a piece of string.
 

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