A uniform rod of length L rests against a smooth roller as shown in figure. Find the friction coefficeint between the ground and the lower end if the minimum angle that rod can make with the horizontal is `theta`.

Explanation

Forces acting on the rod are as shown in the figure.

Roller is smooth so only normal force `N_(2)` acts perpendicular to the rod. `N_(1)` and f are normal reaction and friction from the floor. Mg is weight of rod acting at a distance L/2 from the bottom end O.
The is a case of equilibrium.
Vertical equilibrium `impliesN_(1)+N_(2)costheta=Mg" "......(i)`
Horizontal equilibrium `impliesf=N_(2)sintheta" "......(ii)`
Rotational equilibrium about point O
`impliesN_(2)(h)/(sintheta)=Mg""(L)/(2)costheta`
`impliesN_(2)=(MgLsinthetacostheta)/(2h)" "......(iii)`
Substituting equation (iii) in equation (i) we get the following:
`impliesN_(1)+((MgLsinthetacostheta)/(2h))costheta=Mg`
`impliesN_(1)=Mg=(MgLsinthetacos^(2)theta)/(2h)" ".....(iv)`
Rod is on the verge of slipping.
`f=muN_(1)`
Using equation (ii) we can rewrite the above relation as following:
`N_(2)sintheta=muN_(1)`
Substituting from equations (iii) and (iv) we get the following:
`(MgLsinthetacostheta)/(2h)sintheta=mu(Mg-(MgLsinthetacos^(2)theta)/(2h))`
`impliesmu=(Lcosthetasin^(2)theta)/(2h-Lcos^(2)thetasintheta)`