Consider a tube that lies in the vertical plane, with its shape being an arbitrary function $y=f(x)$. Its ends are on the same level, and a uniform chain lies in the tube.enter image description here

Now, it is a fairly simple exercise to show that the chain doesn't accelerate out of the tube, by integrating over the forces on each mass element. However, That is not what I am looking for. I have a feeling that a simple physical explanation exists to explain no acceleration, though I am unable to find something concrete.

It is something like a chain placed symmetrically on a sphere(i.e. its COM lies on the top of the sphere); the chain doesn't accelerate because it is being 'pulled' equally in both directions( in a hand-wavy way), as the mass on either side is the same. (To be more precise, we can say that the element of the chain on the very top is stationary, since it is pulled by equal tensions on either side of it. If that is stationary, the whole chain is).

The problem is, in the general case we have considered, it is not necessary that the mass on either side is the same, as the tube(hence chain) length may not be the same about the maxima of the function. In that case, it is not clear to me how can the 'pull' on either side be same.enter image description here

I suspect that it may have something to do with the fact the number of $y$ values on either side of the maxima is the same(they end up at the same level, and the function goes through the same points on either side of maxima); but I really cannot think of a definite solution.A simple physical explanation would be appreciated.

  • $\begingroup$ Turn your tube upside down and fill it with water $\endgroup$ Jan 30, 2018 at 17:15
  • $\begingroup$ Or ignore friction and think at a triangle shaped tube. . $\endgroup$
    – Alchimista
    Jan 30, 2018 at 19:21
  • $\begingroup$ @JohnRennie that's a very clever analogy; but I still don't see how this explains my problem. In the case you mentioned, the water stays in equilibrium because the pressure on both sides is equal, as it depends only on the height( which is same). I do not see how to extend this to the chain. I see that the push of the atmosphere is replaced by the pull of the gravity in my case, but I still think I'm missing something. Sorry, I think I'm missing something really stupid here :) $\endgroup$
    – GRrocks
    Jan 31, 2018 at 13:15
  • $\begingroup$ See Understanding Feynman's argument about Stevinus $\endgroup$ Nov 17, 2018 at 23:07


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