A ladder of length `l` and mass `m` is placed against a smooth vertical wall, but the ground is not smooth. Coefficient of friction between the ground and the ladder is `mu`. The angle `theta` at which the ladder will stay in equilibrium is

Options

A. `theta=tan^(-1)(mu)`
B. `theta=tan^(-1)(2mu)`
C. `theta=tan^(-1)(mu/2)`
D. none of these

Answer

D

Explanation

`mg=N_(1)`
`muN_(1)=N_(2)` or `N_(2)=mumg`
Taking moment about `O` we get
`mumgl sintheta=mg1/2costheta`
`tantheta=1/(2mu)` or `theta=tan^(-1)(1/(2mu))`