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

Rod is in equilibrium (but just about to slip)
`rArr sum F=0 rArr N_(1) = mg` and `N_(2) = mu_(S)N_(1) rArr tau_("about any point") =0`
`tau_("about bottom") = mg l/2 cos theta - N_(2) l sin theta =0`
`tan theta = (mg)/(2 N_(2)) = (mg)/(2mu_(2)mg) rArr tantheta =1/(2mu_(S))`