A uniform rod is made to lean between a rough vertical wall and the ground. The coefficient of friction between the rod and the ground is `mu_(1)` and between the rod and the wall is `mu_(2)`. Find the angle at which the rod can be leaned without slipping.

Explanation

Let the weight of ladder be `W` and length `2a`.

Since the ladder is in equilibrium
`SigmaF_(x)=0implies N_(2)=mu_(1)N_(1)` .....(i)
`SigmaF_(y)=0implies N_(1)+mu_(2)N_(2)=W` ......(ii)
`Sigma tau_A=0`
`implies Wxxacostheta=N_(2)xx2asintheta+mu_(2)N_(2)xx2acostheta` .......(iii)
From (`i`) and (`ii`), we have
`N_(1)=(W)/(1+mu_(1)mu_(2)),N_(2)=(mu_(1)W)/(1+mu_(1)mu_(2))`in (`iii`)
`Wcostheta=(2mu_(1)Wsintheta)/(1+mu_(1)mu_(2))+(2mu_(2)mu_(1)Wcostheta)/(1+mu_(1)mu_(2))`
`(2mu_(1)sintheta)/(1+mu_(1)mu_(2))=(1-(2mu_(1)mu_(2))/(1+mu_(1)mu_(2)))costheta`
`tantheta=(1-mu_(1)mu_(2))/(2mu_(1))`
`theta=tan^(-1)(1-(mu_(1)mu_(2))/(2mu_(1)))`