# Homework Help: Chain sliding on a pulley

1. Mar 24, 2017

### Eugen

1. The problem statement, all variables and given/known data
On a pulley with a very small radius and negligible inertia, that rotates without friction around its fixed horizontal axis, there is a chain of mass m and length l. The chain starts sliding from its equilibrium position. Let x be the distance between the ends of the chain. Express as a function of x: a) the acceleration of the chain. b) the force exerted by the chain on the pulley.

2. Relevant equations
F = ma

3. The attempt at a solution
a) The chain is composed of two parts: the part that starts sliding and keeps getting longer, of mass m1 and the part that keeps getting shorter, of mass m2. I think that on m1 act these forces:
- m1g, its weight, positive
- m2g, negative
So the acceleration of m1 should be g(m1 - m2)/m1. This should also be the acceleration of m2. But I don't know how to express this acceleration in terms of x and l.

As for the force acting on m2, it appears to me that it should also be m1g - m2g, which must be wrong, since the two bodies have the same acceleration and different masses.

2. Mar 24, 2017

### haruspex

No. That overlooks the additional force needed to accelerate m2.
Consider the tension at the top of the chain, each side of the pulley. Consider all the forces acting on each portion of chain. Write out the ΣF=ma equation for each of the two portions of chain.

3. Mar 24, 2017

### Eugen

I think the forces acting on the two parts of the chain are weight and tension:

T - m2g = m2a
m1g - T = m1a
a = g(m1 - m2)/m

It can be proven that (m1 - m2)/m = x/l, so that a = gx/l.
As for point b), I'm still clueless.

Last edited: Mar 24, 2017
4. Mar 24, 2017

### haruspex

What forces act on the pulley?

5. Mar 24, 2017

### Eugen

I think the tension acts on both left and right. The total force should be 2T. Not quite sure, though.

6. Mar 24, 2017

### haruspex

yes.

7. Mar 25, 2017

### Eugen

Well then. The force acting on the pulley is F.

F = m2a + m2g + m1g - m1a
F = g(m2 + m1) + a(m2 - m1)
F = gm + a(m2 - m1)

From the equation m2a + m2g = m1g - m1a
we find that m1 - m2 = xm/l, so m2 - m1 = - xm/l
Substituting, we find that
F = gm(1 - x2/l2)

Thank you, haruspex.