A rough inclined plane is inclined at 30° to the horizontal as shown in the figure. A uniform chain of length L is partly on the inclined plane and partly hanging from the top of the incline. If the coefficient offriction between chain and inclined plane is `mu`, the maximum length of the lianging part to prevent the chain from falling vertically is:

Explanation

`sigma xg = sigma (L-x)g sin 30^(@) + mu sigma(L-x) g cos 30^(@)`
`x=(L-x) 1/2 + mu(L-x) sqrt(3)/2, therefore x=((1+sqrt(3)mu)L)/(sqrt(3)(mu + sqrt(3)))`