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?
1 Answer | Add Yours
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).
`m*a = m*g*(x/l)` and `a =g*(x/l)`
We’ve answered 317,379 questions. We can answer yours, too.Ask a question