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Chain falling through a hole

  1. May 19, 2014 #1
    1. The problem statement, all variables and given/known data

    Pls help me with the (d) option of the question asked in the link
    https://www.physicsforums.com/showthread.php?t=724332&page=1

    Correct expression for tension is ρgx/6 (as given in the answer sheet)

    2. Relevant equations



    3. The attempt at a solution

    Velocity as a function of x is ##\sqrt{\frac{2gx}{3}}## (acc. to the solution of previous post)
    [itex]\Rightarrow[/itex] a = [itex]dv/dt[/itex] = [itex]g/3[/itex]
    applying Newton's law
    ρg[itex]x/2[/itex] - T = ρ[itex]g/3[/itex]*[itex]x/2[/itex]
    [itex]\Rightarrow[/itex] T = ρg[itex]x/3[/itex]

    Pls help me to find the correct answer.
     
    Last edited: May 19, 2014
  2. jcsd
  3. May 19, 2014 #2

    For tension in any point only the part of the chain UNDER the point contributes, while the whole chain is falling.
     
  4. May 19, 2014 #3
    I am a bit confused here !!!
    bcoz when i am considering the lower half of the chain as a system and applying Newton's law on this part i must have to use the mass of the lower half only rather than considering the mass of whole part that is hanging.
    but this method is leading to wrong answer while using the mass of whole part leads to the right one.
    Pls further explain the system that you are taking and its free body diagram.
     
  5. May 19, 2014 #4
    Actually, I apologize for my first post. You can't write Newton's law. Tension is an internal force and has nothing to do with Newtons law.

    By definition ##F(t)=ma(t)##, but you already found out that ##a## is constant and equal to ##g/3##, therefore mass of the chain has to be a function of time, which it is. So ##T(t)=m(t)a=\rho x(t)a##

    That's how I would do it. I'm not sure if that's correct hopefully some others, smarter physicists will comment this too.
     
  6. May 19, 2014 #5
    You cant apply Newton's 2nd directly to the mass of the bottom half of the chain since this "object's" mass is constantly changing. Instead, apply the second to each link of length dy individually and integrate to find the tension at height y = x/2.
     
  7. May 19, 2014 #6

    haruspex

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    I don't see the flaw.
    Consider the piece of chain which is the lower half at some instant. Its downward acceleration is g/3, and its mass is ρx/2, so the net force on it must be ρgx/6, and that must equal ρgx/2-T. Hence T = ρgx/3.
    I didn't notice at the time, but the referenced thread never discussed whether (D) was correct.
    If you look at my post #22 there you will see I showed that the tension at the top of the chain (just below the hole) must be ρv2 = 2ρgx/3. Since the tension elsewhere must be proportional to distance from the bottom, the tension at the halfway point is ρgx/3.
    So I conclude the answer sheet is wrong here. All four statements are correct.
     
  8. May 19, 2014 #7
    You're right. I calculated by my method and also got T=ρgx/3.
     
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