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Tricky problem here!

  1. Oct 12, 2012 #1
    1. The problem statement, all variables and given/known data

    A uniform chain has mass m = 7.51 kg and length L = 8.5 m. It is held on a frictionless table with L/12 hanging over the edge. Find the work needed to pull the hanging part back onto the table.

    2. Relevant equations

    * Must do the energy considerations!

    Let's see...

    →½mv²
    →mgh

    3. The attempt at a solution

    Attempt?

    ½ * 1/12 * m * v² = mgd?
     
  2. jcsd
  3. Oct 12, 2012 #2
    a twelfth of the chain is hanging over the edge of the table

    you want to figure out the work needed to raise that twelfth up onto the table.

    Figure out how much mass is in that twelfth and how high it needs to be raised in order to be brought back onto the table.
     
  4. Oct 13, 2012 #3

    CWatters

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    You only need use mgh not 0.5mv^2.
     
  5. Oct 13, 2012 #4
    Why do you need to use mgh?
     
  6. Oct 13, 2012 #5
    Then, is it W = 1/12 * mgh?
     
  7. Oct 13, 2012 #6

    OmCheeto

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    This is a tricky question! The bottom part of the chain would be the only part where mgh=(infinitesimal fraction of 7.51kg/12)*g*(8.5m/12) would hold true. The top part of the chain would require no work as it is already at table level. The bits in-between would require work proportional to their distances from the table top.

    I do believe that without knowing some universal rule, or knowing integral calculus, you'll have to figure out a clever trick to bring all of the mass to one point, without changing the energy of the system, and calculate from there.

    That's how I would do it, as I don't know that universal rule, and have forgotten all of my calculus. :redface:

    We don't know. You haven't told us what you're using for m, g, and h. If it is the m, g, and h from the original question, then the answer is no. That would only hold true if all of the mass was at the bottom of the chain.
     
  8. Oct 13, 2012 #7
    The center of mass of the hanging part of the chain is halfway up that part of the chain. For this kind of problem, you can treat the mass of the chain as concentrated at its center of mass, and use [itex]W=mgh[/itex] with that distance.

    If you can use integration, you would simply use [itex]\int^h_0\rho g h' dh'[/itex] where [itex]\rho=M/L[/itex].
     
  9. Oct 13, 2012 #8
    How is the center of mass relevant to this problem? I don't see it.
     
  10. Oct 13, 2012 #9
    Gravity acts on the center of mass. In most problems where the distribution doesn't change shape and that don't involve torques, you only have to consider forces acting on the center of mass. It's also true for work, since work is just force integrated (or summed) over distance.
     
  11. Oct 13, 2012 #10
    Without calculus, it's just...

    W = mgh?!

    What about the integral part? Is it...

    ∫^(h' = 0,h) ρgh' dh'
    = ∫^(h' = 0,h) mgh'/L dh'
    = mgh'²/(2L) | 0,h
    = mgh²/(2L)
    = mg(L/12)²/(2L)?

    Still don't get how to approach this problem. Please help.
     
    Last edited: Oct 13, 2012
  12. Oct 13, 2012 #11
    Without calculus, the h is the height of the center of mass, (L/12)/2.

    You did the integration correct. Both answers give you the same answer of mgL/32.

    I brought in center of mass because I wasn't sure if you were doing algebra or calculus based physics. You can just ignore it, since the integral method is more rigorous.
     
  13. Oct 13, 2012 #12
    Actually, I need to use energy considerations in any of the two methods - algebra or calculus.

    Then, it's just...

    W = mgL/32
    = 7.51 * 9.81 * 8.5 / 32 ≈ 19.6

    Am I on the right track?
     
    Last edited: Oct 13, 2012
  14. Oct 13, 2012 #13
    How is it that you get mgL/32?
     
  15. Oct 13, 2012 #14
    My mistake, I was thinking 16, not 12. It should be 24, not 32.
     
  16. Oct 13, 2012 #15
    What about this?

    ∫^(h' = 0,h) ρgh' dh'
    = ∫^(h' = 0,h) mgh'/L dh'
    = mgh'²/(2L) | 0,h
    = mgh²/(2L)
    = mg(L/12)²/(2L)
    = mgL²/144 / (2L)
    = mgL/288
     
  17. Oct 13, 2012 #16
    I just can't do math today. I also forgot to account for the fact that only 1/12 of the mass is hanging for the CoM version, so it should be (m/12)g(L/12)/2=mgL/288.
     
  18. Oct 13, 2012 #17
    Hah! I'm right about this! It's mgL/288!
     
  19. Oct 16, 2012 #18

    CWatters

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    1/12th of the mass is hanging at an average of 1/24th L below the table.

    1/12 * 1/24 = 1/288

    So mgl/288 seems right to me.
     
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