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[Hanging/Sliding Chain Problem] Confusion in Normal Force

  1. May 8, 2013 #1
    This is a very basic problem which you all must have seen somewhere. But recently an itching doubt has come up to me regarding this problem.

    A chain lies on a table with some part of it hanging over the edge. Now if the coefficient of friction is μ, then find out the maximum part of the chain that can hang without sliding.


    The Usual Solution
    See the solution to the problem here: http://cnx.org/content/m14806/latest/

    Now, in solving the problem, they considered the part of the chain on the table as "m1" and the hanging part as "m2".

    Then, they say that:
    μ*NormalForce = m2*g​
    [itex]\Rightarrow[/itex] μ*m1*g = m2*g​

    Now, my confusion is that why did they take normal force to be = m1*g??

    Consider an FBD of the whole chain. Now the only force in the vertically above direction is the normal force. And it should balance all of the weight i.e. (m1+m2)*g.

    So according to me, normal force = (m1+m2)*g.

    However, all the solutions that I've seen take normal force = m1*g. But I just don't get it.

    Thanks in advance.
  2. jcsd
  3. May 8, 2013 #2


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    hi champu123! :smile:
    i think you're right …

    the usual question has zero friction, and asks you for the acceleration of the chain,

    and even with no friction, that's less than g because of the increased normal force
    Last edited: May 8, 2013
  4. May 8, 2013 #3

    Meir Achuz

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    What you say seems reasonable, but the result mu=m2/(m1+m2) is not reasonable.
    That would say that as m1--> 0, the string would still remain at rest if mu were equal to at least 1.
    Look at a free body diagram of the horizontal part of the string up to just short of the edge
    The normal force on it is indeed m1g, since the only other vertical force is m1g down.
    The answer is that there is a vertical upward force m2g on the hanging part of the string by the edge of the table. This is more clearly seen if you consider a pulley at the end of the table. Then the upward force by the pulley is seen from a FBD of the pulley.
  5. May 8, 2013 #4


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    Hi Meir! :smile:
    Yes, there has to be a vertical upward force m2g from the table somewhere

    it depends on the structure of the chain where it is …

    but unless the edge of the table is smooth, doesn't that still give the same maximum friction of µ(m1 + m2)g ? :wink:
  6. May 8, 2013 #5
    The only vertical force on it might not be just m1g. Depending on the structure of the chain, the hanging chain will also transmit some tangential force (tangential to the cross section of the chain) apart from the axial force. And this tangential force will add to the vertical force on m1.


    I think the structure/design of the edge plays a crucial role.

    • If there is pulley, then it's all right. The solution is correct in that case. All the weight of the hanging part will be taken up by the pulley. And as the pulley rotates freely, it will also offer no friction. So, in this case the net friction will only be μm1g.

    • However, if we introduce a rough edge. Then, whatever normal force it takes, it also apply friction accordingly. In this case, the net friction should be μ(m1+m2)g. Shouldn't it??

    Correct me if I am wrong. But I think this is a huge blunder all the textbooks have been making. In a real case, the answer should be somewhere between the two extremes. It will be a complex encounter of the structure of chain/edge and the forces acting.

    And only one thing can clear it for once and for all: Experiment. If someone could perform some experiment using a simple chain and see how far it can hang without falling. The value of "μ" could be obtained from a different experiment. And then see what the results match with. This should be fairly easy to perform in a school physics lab (I'm not in my school right now, vacations!!).
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