1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

(Mechanics) Falling chain

  1. Sep 19, 2010 #1
    1. The problem statement, all variables and given/known data

    This is from Serway's book Prob 9.71...(busying preparing for GRE)

    A chain of length L and total mass M is released from rest with its lower end just touching the top of a table, as in figure (a). Find the force exerted by the table on the chain after the chain has fallen through a distance x, as in figure (b).
    Assume each link comes to rest the instant it reaches the table.
    [PLAIN]http://img820.imageshack.us/img820/9648/971d.jpg [Broken]

    2. Relevant equations

    The solution given by the book is 3mgx/L, but I couldn't get it.

    3. The attempt at a solution
    My attempt is to consider the centre of mass of the chain.
    The centre of mass is calculated to be
    Then differentiate twice to get an expression of [tex]a_{CM}[/tex]
    and find the net force on the centre of mass, less the gravity Mg, should be the force to stop the chain.
    Finally, the normal force acting on the chain already on the table is Mg(x/L)
    Then add together.
    But the problem is, I could not figure out the derivatives of [tex]x_{CM}[/tex]:
    what should be dx/dt?

    Anyone could give me some hints? Or my approach is not correct?
    I know this is a challenging problem as it is a level 3 in the book. Thanks!!
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Sep 19, 2010 #2
    The table has to hold the weight only of the length of chain that has already fallen and come to a stop.

    At the same time, a force is exerted on it to bring to a stop every length of chain that comes crashing down.

    The center of mass and its velocity are not relevant here. What you should be looking at instead is the velocity of the length of chain that drops to the table at every point in time, that's the key to the answer.
  4. Sep 19, 2010 #3
    Sometimes these kinds of good-old-classical problems really make me start at such depth of physics, even if it's a subject long studied for hundreds of years as classical mechanics. I don't have the exact answer for the problem, but this thread may be helpful:

    P.S.: You may also find this paper interesting: http://arxiv.org/abs/1005.2887 (click on one of the three options on the right to download)
    Last edited: Sep 19, 2010
  5. Sep 19, 2010 #4
    A good simplifying assumption is that there is no up-transmission of the force as the chain falls, that is, that the velocity of all parts of the chain still in the air is uniform, and is given by the simple kinematic relation [tex]v=gt[/tex], that, incidentally, is also the rate at which the length of chain adds up on the table. With this assumption the solution is quite simple.

    The key to solving the problem at the level it is presented, is to make this assumption.

    The other key point is to look at the total force as two distinct components, one is the weight of the chain resting on the table, and the other is the force needed to bring the falling chain to a stop.
  6. Sep 19, 2010 #5
    I think it's more of a "practical" simplification. Whether it's a good one to physics or not, I have no answer.
  7. Sep 19, 2010 #6
    I think the trickiest part of this is the fact it says "assume the part that hits the table stops instantly", because if sounds like: if you're then going to try and use F = dp/dt on the little piece that hits the table, you'll get an infinite answer due to dt = 0. Well anyway, that is how I thought :) due to Royalcat's first post, I was encouraged to try it anyway, and the clue is that you're focussing on an infinitesimal piece so the dp turns zero too; taking the limit for [tex]\mathrm{d}t \to 0[/tex], you'll see the ratio stays finite :)

    ADDENDUM: I couldn't find the flaw in the OP's original method as I thought it was very insightful and it had to be right. Having calculated it his way, I also find the right answer :) So both methods are correct! As for his original question: dx/dt is the rate of the increase of x, and you can associate this with the velocity of the particle at the very top of the rope. Let v be the instantaneous speed of the top particle, then:
    [tex]a_{CM} = \frac{\mathrm d \left( \frac{x-L}{L} v \right)}{\mathrm d t}
    = \frac{x-L}{L} \frac{\mathrm d v}{\mathrm d t} + \frac{v^2}{L} = -g + \frac{gx}{L} + \frac{2gx}{L}[/tex] because dv/dt = g (this is true for every particle on the rope still in the air) and I also used the kinematic equation v² = 2gx.

    This leads right away to [tex]a_{CM} = \frac{3gx}{L} - g[/tex]

    This has made me quite happy: I remember not being able to understand this problem when I saw it in Serway this year (first year undergrad) (I had read about treating it like a system that loses mass so F = ma is not true anymore, but that was extremely uninsightful and I couldn't agree with it) And now I know two ways :) Thanks
  8. Sep 19, 2010 #7
    As dt -> 0, the element dm that hits the ground also goes to 0, and intuitively the increment in velocity dv also goes to 0. Therefore, dp approaches 0.
    I think that assumption (stopping instantly) is an impressive & beautiful thought of physicists: to model a chain, which looks so rough and so hard to analyze, as a continuum soft thin "rope". To me, it looks more like a stream of water.
    Yet, sometimes, this beautiful thought is not really applicable.

    There is no flaw in the OP's post. It's just that the OP didn't thought of a simplification: dx/dt = gt and dv/dt = g.
  9. Sep 20, 2010 #8
    Thanks all the discussions. I now understand how v=sqrt(2gx) fits the assumption.
    After some thoughts, I believe my approach, i.e. the solution by mr. vodka, should be correct, except for the following statements in my originial post, should be deleted: (because we should have already considered all the forces acting on the chain)
    Another approach is from the link to the paper posted by hikaru1221, which should be the solution as RoyalCat mentioned.

    However, in that paper, it mentions that
    How the mass varies in this question?
  10. Sep 20, 2010 #9
    It is completely applicable, I don't see what the problem is with an infinitesimal quantity going to 0. That's obvious.
    The important thing is that the interesting quantity, that is, [tex]\frac{dp}{dt}[/tex] is finite.

    If you're more comfortable working with finite quantities throughout, instead of infinitesimal ones, you could have just as well solved the problem using a length of chain [tex]\Delta \ell[/tex] which hits the table and comes to a stop during a period [tex]\Delta t[/tex] which is small enough so that you can assume the entire length [tex]\Delta \ell[/tex] was traveling at approximately the same velocity before the fall, and that the velocity of the bits of rope yet to come to a stop during the period [tex]\Delta t[/tex] did not change appreciably over that period.

    In the limit where [tex]\Delta t \rarrow 0[/tex] this approximation becomes exact, but the fact a quantity is infinitesimal, does not in any way invalidate your assumptions about the system.

    The only thing you're neglecting here is whatever stress develops in the chain as parts of it hit the table, which is a terrific approximation, considering how it isn't all that spring.

    The mass-variation we have in this question is that the table supports a different mass as time goes by. Similar questions would include a rope starting on a table with a hole, and gradually falling through, or a spool of rope being held stationary, and its end is pulled at a constant velocity, slowly unwrapping the rest of the rope, etc.

    The second problem I quoted is actually quite interesting, as it demonstrates that mechanical energy can never be conserved for that system, and that half of it will always go to heat. This happens because the way to accelerate additional rope as time goes by is solely by the tension in the rope that is already in motion, and this produces internal friction as additional rope is accelerated to the required velocity.
    Last edited: Sep 20, 2010
  11. Sep 20, 2010 #10
    When it comes to F=dp/dt instead of F=ma, peope usually think of mass-varied situations like rocket burning fuel. In this problem, it depends on the system you consider: if that's the whole chain, mass is conserved; if it's just the part on the air of the chain, mass is not conserved.

    I'm talking about how we model the chain and whether the model is applicable or not. And I would say that, it's sometimes not applicable. For the OP's problem, the paper I cited in post #3 also refutes the model people usually use to solve the problem.

    Here is a discussion on a similar problem, from "Introduction to Classical Mechanics" by David Morin (Harvard) (see page 170-173), regarding the same issue:
    http://books.google.com/books?id=Ni...ottom; see Fig. 5.17. Then the system&f=false
    And of course, dp/dt is finite, but how we calculate it is important.
    Last edited: Sep 20, 2010
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Threads - Mechanics Falling chain Date
Photo of rotating scale and falling coins Dec 31, 2017
Projection at an angle - throwing stones Sep 26, 2017
Deflection of a Falling Mass Jul 23, 2017
Mechanics Falling Chain Problem Nov 20, 2011