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Homework Help: Momentum Problem on a chain

  1. Jan 20, 2009 #1
    1. The problem statement, all variables and given/known data
    A chain of length L and total mass M is released from rest with its lower end just touching the top of a table. Find the force exerted by the table on the chain after the chain has fallen through a distance x. (Assume each link comes to rest the instant it reaches the table.)


    2. Relevant equations
    p= mv
    ek= (1/2)mv^2


    3. The attempt at a solution
    First, I tried to find the force by taking the derivative of kinetic energy:
    ek= (1/2)mv^2
    m= M(L-x)/L, v^2= 2xg
    ek= Mxg(L-x)/L
    d/dt(ek)= Mgv-2Mgxv/L

    F= [d/dt(ek)]/v= Mg-2Mgx/L
    But this turns out to be the wrong answer. Any advice? Thanks.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 20, 2009 #2
    Assuming that the mass is distributed equally along the chain, what proportion of the mass M is on the table when a length x out of a total length L is on the table?
     
  4. Jan 20, 2009 #3
    mass= M(x/L)
    But I don't think its as simple as the normal force of the mass on the table: Mg(x/L).
     
  5. Jan 21, 2009 #4
    Well the answer does seem a bit simple and obvious. But I think it is right.
    Let's ask ourselves what force, if any, does the table exert on the part of the chain which is falling?
     
  6. Jan 21, 2009 #5
    In addition to the normal force, I think there is also a force somehow related to the motion of the part of the chain that is in the air. Mg(x/L) is not the right answer according to the book.
     
  7. Jan 22, 2009 #6
    OK a force is characterised by a change in momentum.

    At the beginning the momentum of the chain is zero.

    After the chain has fallen X the part of the chain in motion has a certain momentum.

    Maybe they want you to express the instantaneous change in momentum.
     
  8. Jan 25, 2009 #7
    Thanks for the help.
    So d/dt[ (M(L-X)/L) * sqrt(sgX) ] = ... Mg - 3Mgx/L
    -> Force= 3Mgx/L - Mg
    This is fairly close to the answer = 3Mgx/L. Why is there the Mg?
     
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