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Forces and Newton second law

  1. Sep 12, 2016 #1
    1. The problem statement, all variables and given/known data
    A chain of mass M and length L is suspended vertically with its lowest end touching a scale. The chain is released and falls onto the scale. What is the reading of the scale when a length of chain, x , has fallen? (Neglect the size of individual links.)

    2. Relevant equations
    Weight = mg, Resultant force = ma, Newton third law.

    3. The attempt at a solution
    When the chain is suspended, by drawing the free body diagram of the chain, it should experienced three forces, upward tension, normal reaction by the scale, and its own weight.
    When the chain falls onto the scale, it experience 3 forces too, upward tension, normal reaction and the weight of the the fallen chain.The weight of the fallen chain, W1 = (x/L)Mg. The upward tension has a reaction pair too, which is pointing downwards and has a magnitude of the weight of the chain which is still suspended. Therefore N2 = W2 = (L-x/L)Mg.
    There are a total of 2 forces acting downward on the scale, the reaction of the tension, and the weight of the fallen chain. Therefore, the scale should read N2+W1=W1+W2 = W, the weight of the chain itself.
    I do not have the answer for this question. Can someone please help me to check if my solution is correct because I am not confident with this answer. Thank you so much !
     
  2. jcsd
  3. Sep 12, 2016 #2

    PeroK

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    I suspect that you are supposed to assume that the tension in the chain disappears almost instantaneously and the entire chain immediately goes into free fall.

    If you do not assume this, then you would need to some additional assumptions about how quickly the tension is lost. If the chain were a "slinky", for example, this could happen:

     
  4. Sep 12, 2016 #3
    Yeah, there shouldnt be any tension left when it is released, the string stays on top due to inertia. Therefore I should be calculating the downward resultant force, which is the chain that is still falling and the weight of the fallen chain. Is that right?
     
  5. Sep 12, 2016 #4

    PeroK

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    The scale is doing two things:

    1) Supporting the weight of the chain that has already fallen. And:

    2) What else?
     
  6. Sep 12, 2016 #5
    Supporting the weight of the chain that is still falling? Since the chain that is still falling is in contact with the scale, there must be a normal contact force right?
     
  7. Sep 12, 2016 #6

    PeroK

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    Not quite, the chain is falling and then it isn't falling any more. Something must have applied a force to slow it down!
     
  8. Sep 12, 2016 #7
    Is it the tension of the chain that stop it from falling down?
     
  9. Sep 12, 2016 #8

    PeroK

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    If someone dropped a chain on your head, would it be the tension in the chain or your head that stopped it falling? Why would your head hurt?
     
  10. Sep 12, 2016 #9
    So the scale is the one who exerted a force on the chain?
     
  11. Sep 12, 2016 #10

    PeroK

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    Exactly. The scale is not only supporting the weight of the chain has has fallen and come to rest, but is exerting a force on the chain that is currently hitting it and coming to rest.

    To give you a hint: force equals rate of change of momentum. That will be key in solving this.
     
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