# A uniform chain of length `l` has half of its length overhanging the edge of a smooth table. The chain is initially at rest. What will be the acceleration of the chain when a length `x` is...

A uniform chain of length `l` has half of its length overhanging the edge of a smooth table. The chain is initially at rest.

What will be the acceleration of the chain when a length `x` is overhanging?

*print*Print*list*Cite

### 1 Answer

The linear mass density of the chain is `lambda = m/l` . When a length `x` of the chain is overhanging the mass of this overhung part of the chain is

`m_x =x*lambda = m*(x/l)`

The corresponding weight of overhung mass `m_x` is

`G = m_x*g = m*g*(x/l)`

Now the entire chain system can be regarded as a single body with two forces acting on it: `G` (which pulls down the chain) and its inertia `m*a` (which is opposing the movement down of the chain) (here m is the total mass of the chain).

Therefore

`m*a = m*g*(x/l)` and `a =g*(x/l)`