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Impulsive force due to a falling chain

  1. Aug 15, 2012 #1
    A uniform chain of mass M and length L is held in vertically in such a way that its lower end just touches the floor. The chain is released from rest in this position. Any portion that strikes the floor comes to rest. Assuming that the chain does not form a heap on the floor, calculate the force exerted by it on the floor when a length x has reached the floor.


    I assumed chain that the chain is made up of many small mass elements and an impulsive force exerted by the floor brings each of these to rest.
    Velocity of the mass element at height of x when it reaches the ground is √2gx
    Impulse is change in momentum by time that is (m*v-m*0)/dt.
    m=M*dx/L
    So Impulse = Mv*dx/dt*(1/L)
    =Mv2/L
    v=√2gx
    so 2Mgx/L
    But the answer was given as 3Mgx/L
    Please show where i was wrong
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 15, 2012 #2
    You missed the normal force due to floor. :wink:
     
  4. Aug 15, 2012 #3
    Ok i got it thanks.
     
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