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

`N_(1)=m"g"`
`N_(2)=f=muN_(1)`
` N_(2)=muN_(1)`
`N_(2)=mu"mg"`
torque about point 'O'
`"Mg"(1/2costheta)=N_(2)lsintheta`
`"mg"1/2costheta=mu"mg "lsintheta`
`tantheta=(1)/(2mu)`
`theta=tan^(-1)((1)/(2mu)) `
`vec(tau)"anticlockwisc"^(=)vec(tau)"clockwisc"`