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Homework Help: Chain falling out from a tube

  1. Aug 29, 2013 #1
    1. The problem statement, all variables and given/known data
    A chain AB of length ##l## is located in a smooth horizontal tube so that its fraction of length ##h## hangs freely and touches the surface of the table with its end B. At a certain moment, the end A of the chain is set free. With what velocity will this end of the chain slip out of the tube.

    (Ans: ##v=\sqrt{2gh\ln(l/h)}## )

    2. Relevant equations

    3. The attempt at a solution
    I don't see how should I start with this problem. I tried to see the forces acting on the hanging part of the chain but that didn't help me.

    The forces acting on the hanging part of the chain are weight, tension (T) due to the part of chain in the tube and the normal reaction from ground(N). Let the length of chain remaining in tube be x and mass per unit of chain be ##\lambda##. Applying Newton's second law on hanging part of chain,
    $$\lambda hg-T-N=\lambda ha$$
    Newton's second law for part in the tube,
    $$T=\lambda xa$$
    I am not sure if this helps and I don't know how to calculate N here.

    Any help is appreciated. Thanks!

    Attached Files:

    Last edited: Aug 29, 2013
  2. jcsd
  3. Aug 29, 2013 #2
    The reaction from the table, even though it does act on the end of the chain, cannot be transferred to the rest of the chain due to the nature of the chain.
  4. Aug 29, 2013 #3
    Do you mean ##N=0##? Nature of chain? What it has got to do with N? :confused:
  5. Aug 29, 2013 #4
    A chain can only transfer tension. It cannot transfer compression. Reaction here acts as a compressive force.
  6. Aug 29, 2013 #5
    A good point to start is that the chain could not be extended. So all points of the chain have the same speed and acceleration.
  7. Aug 29, 2013 #6
    If we had a string of some finite mass instead of chain, would N=0 still hold?
  8. Aug 29, 2013 #7
    Voko did't say that N = 0. The reaction from the ground acts upon the end of chain just to oppose to the weight of the part of the chain that already has fallen and to make zero the momentum of the point of chain that is arriving. But it doesn't act to the rest of the chain.
  9. Aug 29, 2013 #8
    Yes, a massive string would be identical with a chain in this problem.
  10. Aug 29, 2013 #9
    I tried to set N=0 in the equations I wrote above.
    From the two equations,
    $$\lambda hg-\lambda xa=\lambda ha$$
    $$\Rightarrow a=\frac{gh}{h+x}$$
    I use ##a=vdv/dx## here. The limits of integration for x are from l-h to 0 and for v are 0 to v.
    Solving gives,
    which is incorrect. :confused:
  11. Aug 29, 2013 #10
    What direction is positive for ## a ##, ## v ##, ## x ##? Are you consistent in your equations?
  12. Aug 29, 2013 #11
    Oh yes, sorry. Very silly of me. Thanks a lot voko! :)
  13. Aug 29, 2013 #12


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    There's a serious flaw in the way the question is posed. It ought to say that the tube turns down at the end, directing the chain vertically. Otherwise the horizontal momentum of the chain as it leaves the tube will carry it into an arc, interacting with the suspended section in a complex way.
  14. Aug 29, 2013 #13
    I had the same concern initially. But then I looked at the picture :)
  15. Aug 30, 2013 #14
    I am not sure what you both talk about but is it related to the bent in the tube at the end?
  16. Aug 30, 2013 #15
    It is. Without it, the chain would have a horizontal velocity component at the end, complicating analysis enormously.
  17. Aug 30, 2013 #16
    Does this mean that the equations I have written above won't apply to this case? Even if I consider them valid, I would get the same last equation (post #9) to integrate and it looks to me that the solution would be horizontal component of velocity.
  18. Aug 30, 2013 #17
    Without the tube ensuring 1D motion for each segment, the hanging part will be curved. Because it is curved, there will be more of it, meaning more weight, so your basic equation will not be correct.
  19. Aug 30, 2013 #18
    Ok, thanks for the info. I guess I shouldn't try this as you say its complicated.
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