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Lifting up a chain

  1. Feb 4, 2013 #1
    1. The problem statement, all variables and given/known data
    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)

    2. Relevant equations



    3. 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!
     
    Last edited: Feb 5, 2013
  2. jcsd
  3. Feb 4, 2013 #2

    ehild

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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
     
  4. Feb 4, 2013 #3

    tms

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    The mass of the chain being pulled up by the variable force is also variable.
     
  5. Feb 4, 2013 #4
    I understand it now but I am unable to form an equation to find F.
     
  6. Feb 4, 2013 #5

    tms

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    Start from first principles:
    [tex]\begin{align}
    F &= \frac{dp}{dt} \\
    &= v\, \frac{dm}{dt} + m\, \frac{dv}{dt}.
    \end{align}
    [/tex]
     
  7. Feb 5, 2013 #6
    [tex]F=u \cdot \frac{d(ρx)}{dt}+ρx \frac{dv}{dt}[/tex]
    I can't figure out what should replace dv/dt here.

    I have formulated an expression for the work done but not sure how to proceed further. Here's the expression I have reached:
    [tex]W=-ρgxdx-\frac{ρu^2dx}{2}[/tex]
     
  8. Feb 5, 2013 #7

    ehild

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    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
     
    Last edited: Feb 5, 2013
  9. Feb 5, 2013 #8
    So the term dv/dt equals to zero.
    Do you mean I need to add an extra term in the LHS? I think the first equation should be this:
    [tex]F-ρgx=ρu^2[/tex]
    [tex]F=ρgx+ρu^2[/tex]
    Ahah, I reached the answer, thanks a lot! :smile:

    Yes, you are right, I need to use dW and I guess I made a sign error. The infinitesimal work should be
    [tex]dW=ρgxdx+\frac{ρu^2dx}{2}[/tex]
    Substituting dW=Fdx doesn't give me the right answer.
     
  10. Feb 5, 2013 #9
    ehild, can you please explain why the energy method doesn't work?
     
  11. Feb 5, 2013 #10

    ehild

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    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
     
  12. Feb 5, 2013 #11

    D H

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    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 [Broken], 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.
     
    Last edited by a moderator: May 6, 2017
  13. Feb 5, 2013 #12

    tms

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    The force in the equation is the net force, of course. It is composed of the upwards force and gravity.
     
  14. Feb 5, 2013 #13

    ehild

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

    ehild
     
  15. Feb 5, 2013 #14

    tms

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    I did read it (after I made my own post). What other force is there?
     
  16. Feb 5, 2013 #15

    ehild

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    Normal force from the ground.

    ehild
     
  17. Feb 5, 2013 #16

    tms

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    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.
     
  18. Feb 5, 2013 #17

    D H

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    OK, let's do that.

    Suppose that at some point in time [itex]t[/itex] the length of the chain being held off the platform is [itex]x(t)[/itex]. This length of chain is moving upwards at a constant velocity [itex]u[/itex] with respect to the platform. Given that the mass per unit length of the chain is [itex]\rho[/itex], this means the momentum of the chain with respect to the platform is [itex]p(t)=\rho x(t) u[/itex], directed upward. Some very short time [itex]\Delta t[/itex] later, the length of the chain moving upward is [itex]x(t)+u\Delta t[/itex], making the momentum [itex]p(t+\Delta t)=\rho (x(t) + u\Delta t)u[/itex]. The change in momentum is [itex]\Delta p = \rho u^2 \Delta t[/itex]. Applying [itex]F_{\text{net}}=\lim_{\Delta t \to 0} \Delta p / \Delta t[/itex] yields [itex]F_{\text{net}}=\rho u^2[/itex]. Adding the weight of the chain yields the total force needed to keep the chain moving at a constant velocity, [itex]F_{\text{tot}} = \rho gx + \rho u^2[/itex]. 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 [itex]F_{\text{net}}=\rho u^2[/itex], a constant. Calculating the work performed by this constant net force yields [itex]W=\int_0^x F\,dl = \rho x u^2[/itex]. 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 [itex]dm/dt = \rho u[/itex] 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.
     
  19. Feb 5, 2013 #18

    tms

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    It's late, and I'm not thinking all that clearly, but I think the problem is with the integration. [itex]\int F\,dx[/itex] 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.
     
  20. Feb 5, 2013 #19
    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?
     
  21. Feb 5, 2013 #20
    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.
     
    Last edited: Feb 5, 2013
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