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Chain falling in a scale

  1. Jan 14, 2015 #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
    Momentum, conservation of energy

    3. The attempt at a solution

    Hello, I have checked threads who dealt with this problem but did not understand everything !

    After reading different threads, I understand that the problem is to find the normal force from the scale on the chain.

    Firstly, the net force on the scale is equal to the time derivative of momentum:

    ## \frac{dP_x}{dt} = N_x - W_x = N_x - \frac{M}{l}xg##

    The change of momentum is

    ## \left\{
    \begin{array}{}
    P_x(t) = \frac{M}{l}(l-x(t)) \times (-\dot x(t) )\\
    P_x(t+\triangle t) = \frac{M}{l}(l-x(t+\triangle t)) \times (-\dot x(t+\triangle t))
    \end{array}
    \right. ##

    So

    ##\frac{dP_x}{dt} = -M \ddot x + \frac{M}{l}({\dot x}^2+ x \ddot x ) = \frac{M}{l}g(x-l) +\frac{M}{l}{\dot x}^2
    ##

    because ##\ddot x = g ##.
    In order to find ##\dot x^2##, I use conservation of energy on the point mass located at the top of the chain, which is accelerated by gravity:

    ## \left.
    \begin{array}{}
    K_0 = 0 ,\ U_0 = \frac{M}{l} g l = Mg \\
    K_x = \frac{1}{2} \frac{M}{l}{\dot x}^2, \ U_x = \frac{M}{l} g (l-x) \\
    K_0 + U_0 = K_x + U_x
    \end{array}
    \right\}
    \Rightarrow {\dot x}^2 = 2gx \Rightarrow \frac{dP_x}{dt} = 3\frac{M}{l}gx - Mg ##

    So

    ## N_x = \frac{M}{l}gx + \frac{dP_x}{dt} =4 \frac{M}{l}gx - Mg ##

    Is that correct?
     
    Last edited: Jan 14, 2015
  2. jcsd
  3. Jan 14, 2015 #2

    BiGyElLoWhAt

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    What is N_x in your force equation (dp/dt)? I think there may be an inconsistency in that statement.
     
  4. Jan 14, 2015 #3
    Hello, ##N_x## is the normal force from the scale on the chain
     
  5. Jan 14, 2015 #4

    BiGyElLoWhAt

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    Ok, so then what's W_x?

    **Oops read it backwards.
     
  6. Jan 14, 2015 #5
    It is the weight of the chain on the scale after length x has fallen
     
  7. Jan 14, 2015 #6
    No obviously I'm wrong, the net external force in the system chain-scale is ## N_x - Mg ## and not ## N_x - \frac {M}{l}gx##, so ## N_x =3 \frac {M}{l}gx##,
     
  8. Jan 14, 2015 #7

    BiGyElLoWhAt

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    Haha, yea sorry about that, I forgot how a scale worked for a minute. I just went through and double check you, it looks good.
     
  9. Jan 14, 2015 #8

    BiGyElLoWhAt

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    The chain that's still in the air doesn't exert a weight force on the scale.
     
  10. Jan 14, 2015 #9
    yes, but it must be taken into account because it is an external force in the system chain-scale
     
  11. Jan 14, 2015 #10

    BiGyElLoWhAt

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    Gravity accelerates the chain at a constant velocity. If you drop two balls, one on top of the other (neglecting drag) does on exert a force on the other?
     
  12. Jan 14, 2015 #11

    BiGyElLoWhAt

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    Also, if you replace your mass/unit length with the entire mass, wouldn't the scale always read more than the weight of the chain?
     
  13. Jan 14, 2015 #12
    I think that in my first attempt I have cheated with the relationship between net external force and momentum.
    I am using momentum of the chain-scale system, but don't take into account the weight of the falling chain, but it is an external force.
    So without cheating,
    ## N_x = 3 \frac{M}{l}gx## and not ## N_x = 4 \frac{M}{l}gx - Mg ##
     
  14. Jan 14, 2015 #13

    BiGyElLoWhAt

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    Doesn't the falling chain have mass (m/l)(l-x)? The chain that's already on the scale has 0 momentum. What do you mean you didn't take into account the weight of the falling chain? Maybe I'm missing something here.
     
  15. Jan 14, 2015 #14
    The definition is net external force = dP/dt.
    I used the momentum of the chain-scale system, there is a part that is on the scale at rest, and a part falling.
    But I only took into account the weight of the chain on the scale, which is not correct in this system, because the falling part contributes to total external force.
     
  16. Jan 14, 2015 #15
    A way to be 100% sure that It was wrong at first attempt is that ##N_x < 0## for ## 0 \le x \le \frac{l}{4}##, which is not admissible.
     
  17. Jan 14, 2015 #16

    haruspex

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    You can get there by looking at the whole system, but it's unnecessarily complicated. All the weigh pan cares about is the speed and density of the chain hitting it.
    At time t after release, with what velocity is the chain hitting the pan? What is the rate of change of momentum on those links?

    If you want to look at the whole system, that will include gravity acting on the falling part of the chain, and the rate of change of momentum of that part of the chain. These two will balance, which is why you can ignore them.
     
  18. Jan 14, 2015 #17

    BiGyElLoWhAt

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    From where I'm sitting, I see the scale can read the part that's sitting on it, at rest, definitely. Also, there will be some portion of the chain that has recently come into contact with the scale, and the scale must exert a force greater than the weight force of that segment to slow it down. That's newtons second law. But I for some reason am not seeing how the top most point of the chain, or the point under that, or... will exert a force on the scale or vice versa.
     
  19. Jan 14, 2015 #18

    I'm trying, but I don't understand your reply. Can you restate it? Do you agree with the solution ?
     
  20. Jan 14, 2015 #19

    haruspex

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    Yes, sorry, I was concentrating on the impact part. I should have written:
    All the weigh pan cares about is the speed and density of the chain hitting it, plus the weight of what is already in the pan
    Which solution? It seems to have evolved, and I'm not sure what your current proposal is. At one point you had ##3 \frac{M}{l}gx##, which is what I get.
     
  21. Jan 14, 2015 #20
    I get that too, thank you!
     
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