I was thinking about this situation with a friend of mine, and we had opposing views:

Suppose you have a frictionless table with mass M resting on top. You attach a chain to the mass M, a chain with another mass already attached to the other side. You let this other mass hang from one end of the frictionless table, while making sure mass M is stationary, holding it down with your hands. As soon as you stop exerting a force on mass M, letting it fall under the influence of the hanging mass, the hanging mass accelerates downwards with an acceleration of g.

The chain is very heavy, and its weight is not negligible.

How would the mass M accelerate?

1) It makes sense to me that the mass M would accelerate at a constant acceleration of g - although the weight of the chain is not negligible, the mass of an object doesn't play a role in its acceleration. As the hanging mass would be in free fall, the mass M, because rested upon a frictionless surface, would also be, in a sense, in free fall.

2) My friend thinks that as the hanging mass falls, the force downwards (the gravitational force) would increase: F = ma, where although a is constant, because the mass of the chain is not negligible, the total mass is increasing as more of the chain is pulled by the hanging mass in free fall, and therefore the mass M would be pulled with a greater force, leading to an increasing acceleration against time.

Which seems to be more accurate, and why?

  • $\begingroup$ why would you think the hanging mass is in free fall?(with an acceleration equal to g)? $\endgroup$ Oct 6, 2015 at 13:07
  • $\begingroup$ In my mind, I can visualise how a block on a frictionless surface, say a block on a big ice cube, would fall with the same acceleration as the block its attached to, where the block its attached to is in free fall. Does this only seem to make sense if the weight of the chain is negligible? Is that what I'm misunderstanding? $\endgroup$ Oct 6, 2015 at 13:10
  • $\begingroup$ if you draw the free body diagram of the mass ahngign from the end of the chain/rope, there is a tension acting vertically upwards.. so, in effect its downward acceleration is [mg - T(tension)]/m ,which is obviuosly less than g,thus the hanging mass itself is not in free fall.. $\endgroup$ Oct 6, 2015 at 13:14
  • $\begingroup$ Ahh yes...As a follow-up question, is it right to say that on a frictionless inclined plane of angle theta, the acceleration of any object would be g? The normal force would have no influence? $\endgroup$ Oct 6, 2015 at 13:24
  • $\begingroup$ acceleration down the plane would be g*sin(theta) $\endgroup$ Oct 6, 2015 at 13:30

1 Answer 1


Your friend is right - the acceleration increases. To see why, simplify your system and get rid of the masses so you just have a chain of significant mass with one end hanging off the table.

In such a system you have a small part (say 10%) of the chain that is free to fall (the bit hanging off) and another part that is not being acted on by gravity (g is normal to the table so it can't move and since there's no friction it doesn't increase that either).

Initially, therefore, you effectively have the force due to 10% of the chain acting on 100% of the chain, so the system accelerates at 0.1g. As the chain goes off the edge, more and more of it is acted on by gravity so the force increases. But the total mass remains constant, so the acceleration increases ($A = F/M$). Finally, when all the chain is off and being acted on by gravity, the acceleration is g.

This is a pretty easy experiment to do - any chain and a polished table will do. The increasing rate of clicks as the links go over the edge provides a clearly audible indication of what's going on.


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