# Chain falling out from a tube

Saitama

## Homework Statement

A chain AB of length ##l## is located in a smooth horizontal tube so that its fraction of length ##h## hangs freely and touches the surface of the table with its end B. At a certain moment, the end A of the chain is set free. With what velocity will this end of the chain slip out of the tube.

(Ans: ##v=\sqrt{2gh\ln(l/h)}## )

## The Attempt at a Solution

I don't see how should I start with this problem. I tried to see the forces acting on the hanging part of the chain but that didn't help me.

The forces acting on the hanging part of the chain are weight, tension (T) due to the part of chain in the tube and the normal reaction from ground(N). Let the length of chain remaining in tube be x and mass per unit of chain be ##\lambda##. Applying Newton's second law on hanging part of chain,
$$\lambda hg-T-N=\lambda ha$$
Newton's second law for part in the tube,
$$T=\lambda xa$$
I am not sure if this helps and I don't know how to calculate N here.

Any help is appreciated. Thanks!

#### Attachments

• chainintube.png
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voko
The reaction from the table, even though it does act on the end of the chain, cannot be transferred to the rest of the chain due to the nature of the chain.

Saitama
The reaction from the table, even though it does act on the end of the chain, cannot be transferred to the rest of the chain due to the nature of the chain.

Do you mean ##N=0##? Nature of chain? What it has got to do with N?

voko
A chain can only transfer tension. It cannot transfer compression. Reaction here acts as a compressive force.

LeonhardEu
A good point to start is that the chain could not be extended. So all points of the chain have the same speed and acceleration.

Saitama
A chain can only transfer tension. It cannot transfer compression. Reaction here acts as a compressive force.

If we had a string of some finite mass instead of chain, would N=0 still hold?

LeonhardEu
Voko did't say that N = 0. The reaction from the ground acts upon the end of chain just to oppose to the weight of the part of the chain that already has fallen and to make zero the momentum of the point of chain that is arriving. But it doesn't act to the rest of the chain.

voko
Yes, a massive string would be identical with a chain in this problem.

Saitama
Yes, a massive string would be identical with a chain in this problem.

I tried to set N=0 in the equations I wrote above.
From the two equations,
$$\lambda hg-\lambda xa=\lambda ha$$
$$\Rightarrow a=\frac{gh}{h+x}$$
I use ##a=vdv/dx## here. The limits of integration for x are from l-h to 0 and for v are 0 to v.
Solving gives,
$$v=\sqrt{2gh\ln(h/l)}$$
which is incorrect.

voko
What direction is positive for ## a ##, ## v ##, ## x ##? Are you consistent in your equations?

1 person
Saitama
What direction is positive for ## a ##, ## v ##, ## x ##? Are you consistent in your equations?

Oh yes, sorry. Very silly of me. Thanks a lot voko! :)

Homework Helper
Gold Member
There's a serious flaw in the way the question is posed. It ought to say that the tube turns down at the end, directing the chain vertically. Otherwise the horizontal momentum of the chain as it leaves the tube will carry it into an arc, interacting with the suspended section in a complex way.

voko
There's a serious flaw in the way the question is posed. It ought to say that the tube turns down at the end, directing the chain vertically. Otherwise the horizontal momentum of the chain as it leaves the tube will carry it into an arc, interacting with the suspended section in a complex way.

I had the same concern initially. But then I looked at the picture :)

Saitama
I had the same concern initially. But then I looked at the picture :)

I am not sure what you both talk about but is it related to the bent in the tube at the end?

voko
I am not sure what you both talk about but is it related to the bent in the tube at the end?

It is. Without it, the chain would have a horizontal velocity component at the end, complicating analysis enormously.

Saitama
It is. Without it, the chain would have a horizontal velocity component at the end, complicating analysis enormously.

Does this mean that the equations I have written above won't apply to this case? Even if I consider them valid, I would get the same last equation (post #9) to integrate and it looks to me that the solution would be horizontal component of velocity.

voko
Without the tube ensuring 1D motion for each segment, the hanging part will be curved. Because it is curved, there will be more of it, meaning more weight, so your basic equation will not be correct.

Saitama
Without the tube ensuring 1D motion for each segment, the hanging part will be curved. Because it is curved, there will be more of it, meaning more weight, so your basic equation will not be correct.

Ok, thanks for the info. I guess I shouldn't try this as you say its complicated.