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Homework Help: Free falling chain problem

  1. Oct 11, 2005 #1
    A chain of mass M and length L is suspended vertically with its lowest end touching the scale. The chain is released and falls ono the scale. What is the reading of the scale after a length x of the chain has fallen?
    So what I really have to find is the force applied by the string onto the scale for any x. This force is composed of the gravitational force of the part that is already lying on the scale, plus some other force, which i suspect is coming from the fact that the chain is falling with a certain velocity. The problem is I don't know what exactly this force is.
    I assumed that it would be
    [tex]F = \frac{dp} {dt}[/tex]
    which doesn't make a lot of sense to me.

    Any hints as to how I could at least model this problem would be helpful.
    Last edited: Oct 11, 2005
  2. jcsd
  3. Oct 11, 2005 #2
    This is how far I've been able to get since my last post.

    Considering a length x of the chain, the force it will apply onto the scale is F=mg+dp/dt, where m is the mass of this length x and p is its momentum. There seems to me that this is too simplistic. Isn't this like considering the length x as a single particle with momentum p? Am I allowed to do this, even though its momentum is not constant (depends on x)?

    In any case, a particle falling from a height x has an initial potential energy [itex] x \Delta m g [/itex], which translates, once the particle has fallen, into kinetic energy
    [tex] \Delta K = \frac{1}{2} \Delta m {v_f}^2 [/tex]
    So from these two,
    [tex] v_f = \sqrt{2gx} [/tex]

    Now that I have the final v, I found that
    [tex] \frac{dp}{dt} = \frac{dp}{dx} \frac{dx}{dt} = 2mg [/tex]. Correct?

    So in this case the total force applied on the scale is 3mg, where m=x*(density).

    There seems to be something fishy in all this, though I can't figure out what.

    I'd appreciate any feedback on this.
    Last edited: Oct 11, 2005
  4. Oct 12, 2005 #3
    Little correction

    Well, it seems this is not correct after all. Due to a technical error in my derivations :redface: , that is in fact equal to mg.

    Well, in fact this seems to be completely wrong, since I assumed m was a constant, which is not (silly me).

    Thanks for your invaluable help anyways.
  5. Oct 12, 2005 #4


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    Hmmm, this is an interesting one. I'm not entirely sure how to solve it, but you might get somewhere by considering your original equation:

    [tex]F = \frac{dp} {dt}[/tex]

    which I believe indeed does apply. But, remember that p = mv, both of which are functions of t. You need to remember how to differentiate a product of 2 functions. Like I said, I'm not sure this will go anywhere, but it may be a start in the right direction.
    Last edited: Oct 12, 2005
  6. Oct 12, 2005 #5

    Doc Al

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    Here are a few hints. At any given time, the scale must:
    (1) support the weight of the chain segment already on it
    (2) stop the momentum of the moving chain hitting it​
    #2 is where dp/dt comes in. Consider an element of chain mass (dm) hitting the scale in time dt. You'll need to figure out how fast the chain is falling as a function of time (and thus length). Then figure out how much mass hits in time dt (what length of chain hits in that time). Then you can figure out dp/dt.
  7. Oct 12, 2005 #6


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    This is somewhat less accurate than it needs to be, since it ignores the possible effects of induced stress in the chain.
    Here's how to do it slightly more accurate:
    For each time interval [itex]\delta{t}[/itex], we consider the partition [itex]P^{(\delta{t})}[/itex] of the chain into 3 parts, each that has constant mass in the said time interval between instants t and [itex]t+\delta{t}[/itex]:
    a) [itex]C_{U}^{(\delta{t})}(t)[/itex], the upper part of the chain that remains strictly above the scale in the whole interval.
    We assign to this part the velocities [itex]v_{U}^{(\delta{t})}(t),v_{U}^{(\delta{t})}(t+\delta{t})[/itex] where [itex]v_{U}^{(\delta{t})}(t)[/itex] is the velocity at time t and [itex]v_{U}^{(\delta{t})}(t+\delta{t})[/itex] is the velocity at time [itex]t+\delta{t}[/itex].
    We require that the velocity function is continuous.
    The part has mass [itex]m^{(\delta{t})}(t)[/itex] throughout our interval.
    b) [itex]C_{L}^{(\delta{t})}(t)[/itex], the lower part of the chain that was strictly lying on the scale at time t, and still is so ever after.
    This is assumed to be at rest throught the time interval; its mass is [itex]m_{U}^{(\delta{t})}(t)[/itex]
    c)[itex]C_{E}^{(\delta)}(t)[/itex], the connecting element that at time t has velocity [itex]v_{U}^{(\delta{t})}(t)[/itex] and has fallen to rest at time [itex]t+\delta{t}[/itex]. Its mass is [itex]\delta{m}^{(\delta{t}}[/itex].
    Now, we split up our chain in two sub-systems, [itex]S_{U+E}^{(\delta{t})}[/itex], that consists of a)+c), and [itex]S_{L}^{(\delta{t})}[/itex], that consists of b).
    We may now state how Newton's laws are approximately given for the two subsystems:
    where [itex]F_{U+E}^{(\delta{t})}[/tex] express the forces from either the lower chain or from the scale at an instant within the time interval were looking at.
    We may now, of course, sum together these expressions to get rid of any stresses within the chain and proceed, but I choose to keep them in the following in order to illustrate something later on.
    Now is the time for a rather convoluted limiting process!
    By going to the limit [itex]\delta{t}\to{0}[/itex], what we are actually doing, is to go through DISTINCT PARTITIONS [itex]P^{(\delta{t})}[/itex].
    So, by going through these different chain-partitions, we get to the material system for which only an infinitesemal part changes its velocity abruptly.
    We define: [tex]a_{U}(t)=\lim_{\delta{t}\to{0}}\frac{v_{U}^{(\delta{t})}(t+\delta{t})-v_{U}^{(\delta{t})})}{\delta{t}}[/tex]
    and we have [tex]\lim_{\delta{t}\to{0}}\frac{\delta{m}^{(\delta{t})}}{\delta{t}}=\frac{M}{L}|v_{u}(t)|[/tex] where M is the total chain's mass, L the total length of the chain, and [tex]v_{U}(t)=\lim_{\delta{t}\to{0}}v_{U}^{(\delta{t})}(t)[/tex]
    With x(t) being the length on the table, we therefore gain Newton's second laws:
    Summing together removes the possible stress within the chain, and remembering that [itex]v_{U}(t)\leq0[/tex] we gain the expression for the normal force:
    It can be seen that the only resulting difference between between the suggested approach and my own is that my method allows for [itex]a_{U}\neq-g[/itex]
    Finally, a word about the possible stress in the chain:
    Let us set [itex]F_{U+E}=F_{U,S}+F_{C}[/itex] with (U,S) and C subscripts for scale force on the system and stress force from the rest of the chain.
    Similarly, we write [itex]F_{L}=F_{L,S}-F_{C}[/itex] using Newton's 3.law.
    Since we have [itex]F_{U,S}=0[/itex] when the whole chain lies on the scale, we get from (1), [itex]F_{C}=0[/tex], that is, once the chain is lying wholly on the scale, any stress that may have been has vanished..:wink:
    Last edited: Oct 13, 2005
  8. Oct 13, 2005 #7

    Wow, arildno, thanks a lot for the beautiful solution. I can't guarantee that I understood all of it, but still apreciate a lot. I don't think we're expected to consider the strain in the chain though (only intro to mechanics, you know:biggrin: )

    I see what you mean. I thought the same in the beggining, but it is not so. Indeed, the mass of the chain that's already on the scale, mg, is somehow already "built-in" dp/dt. Indeed
    [tex]F= \frac{d}{dt}mv=\frac{dm}{dt}v+m\frac{dv}{dt} [/tex]
    If [itex]\frac{dm}{dt}=0[/itex], then F is just mg. So if the mass was falling in one piece, the force would still be equal to dp/dt, not dp/dt+mg.

    In any case, thanks a lot. I ended up solving it all right using the above equation
    [tex]F= \frac{d}{dt}mv=\frac{dm}{dt}v+m\frac{dv}{dt}
    F=\lambda v^2+mg
    F=3mg [/tex]
  9. Oct 13, 2005 #8

    Doc Al

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    While it's true that the net force on the chain will equal the rate of change of its momentum, the way that I suggested is perfectly OK as well. (They are, of course, equivalent.)

    Using my method, I'm only interested in the rate of change of the piece of chain that hits the table in time dt:
    [tex]v \frac{dm}{dt} = \frac{M}{L} v \frac{dx}{dt}[/tex]

    The chain above the table is in free fall, so after x length has hit the table the speed is given by:
    [tex]v^2 = 2 g x[/tex]
    [tex]v \frac{dx}{dt} = 2 g x[/tex]

    Thus "dp/dt" for that small piece of chain is:
    [tex]\frac{M}{L} v \frac{dx}{dt} = \frac{M}{L} 2 g x[/tex]

    And the total force (adding the weight already on the scale) is:
    [tex]F = \frac{M}{L} 2 g x + \frac{M}{L} g x = 3 \frac{M}{L} g x[/tex]
    Not quite sure what you did here.
    Last edited: Oct 13, 2005
  10. Oct 13, 2005 #9


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    For OP:
    As can be seen from this expression, this cannot be correct for ALL times, because when the chain is fully at rest on the weight (x=L), we must have F=Mg, not thrice its own weight.

    Again, this is consistent with Doc Al's assumption that we may model the upper part of the chain experiencing FREE FALL:
    It seems reasonable, does it not, that this assumption holds only for small values of the parameter x/L, i.e, when most of the chain is still in the air?
    But that would be the time when it won't be much induced stress in the chain, either..

    To understand my solution better, just write it as follows:
    But this is just Newton's 2.law "f=ma"stated for the WHOLE chain:
    External forces: Scale force F and weight Mg.

    The two terms on the right hand side is "ma" for the upper part of the chain and the element that collides with the chain. "ma" doesn't appear for the part lying on the scale, since its acceleration is zero.
    Last edited: Oct 13, 2005
  11. Oct 13, 2005 #10

    Doc Al

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    You are certainly correct that I make the simplifying assumption that the part of the chain not yet on the scale is in free fall (and I ignore any induced stress on the falling chain). But I don't think that that assumption only applies for small values of x/L. I think it entirely reasonable that at the moment when x = L the scale force will equal 3mg (not merely mg!), since that's when the chain is hitting with maximum speed. (The scale force will quickly become mg once the chain is completely on the scale, of course.)
  12. Oct 13, 2005 #11


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    What I basically wanted to point out, is that if you just consider Newton's 2.law for PARTS of the chain (say the colliding bit and the bit lying on the scale), then any force acting upon these parts from the upper part (i.e, stress, for example) is strictly speaking an EXTERNAL force that should be included.
    Alternatively, we may say from the start that any such stress is negligible;

    but it is only when we set up Newton's 2.law for the WHOLE chain that such stress would be a strictly internal force..

    For what it's worth, I don't disagree with you a lot that x/L almost 1 might well be modelled by your approach, but for "safety's sake", we might limit ourselves to state that the approach would definitely work for tiny x/L's.
    Last edited: Oct 13, 2005
  13. Oct 13, 2005 #12
    Yeah, the strain in the chain seems to be a fairly important factor in the end, but I think the momentum approach to this problem provides a reasonable model. But if I do take the strain into account, wouldn't I need some extra information in order to completely solve for the force? Because, not only am I missing an equation, but I suppose the strain in the chain would somehow be related to the internal structure of that chain (density, elasticity, etc).

    In any case, I now understand the problem as well as I need to for my purposes and a bit more (thanks arildno)

    Thanks a lot guys
  14. Oct 14, 2005 #13


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    It's not that you take the stress into the account of my approach; rather, Doc Al's approach is explicitly based upon that such stress is negligible.
    That is quite a different statement, and I would like to emphasize that for "practical purposes", Doc Al's approach is probably good enough.
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