In the figure, a ladder of mass m is shown leaning against a wall. It is in static equilibrium making an angle `theta` with the horizontal floor. The coefficient of friction between the wall and the ladder is `mu_1` and that between the floor and the ladder is `mu_2.` the normal reaction of the wall on the ladder is `N_1` and that of the floor is `N_2.` if the ladder is about to slip. than

Options

A. `mu_(1) = 0, mu_(2) ne 0` and `N_(2), tan theta =mg//2`
B. `mu_(1) = 0, mu_(2)` and `N_(1) tan theta =mg //2`
C. `mu_(1) ne 0 mu_(2) ne 0` and `N_(2) = (mg)/(1+mu_(1)mu_(2))`
D. `mu_(1) = 0 mu_(2) ne 0` and `N_(1)tan theta = mg//2`

Answer

C, D

Explanation

Consider of translational equilibrium,
`N_(1) = mu_(2)N_(2)`………..(i)

`N_(2) + mu_(1)N_(1) = Mg`……..(ii)
Solving `N_(2) = (mg)/(1+ mu_(1) mu_(2)), N_(2) = (mu_(1) mg)/(1+ mu_(1)mu_(2))`
Appling torque equation about corner (left) point on the florr
`mg l/2 cos theta = N_(1) l sin theta + mu_(1) N_(1) l cos theta`
Solving `tan theta = (1-mu_(1)mu_(2))/(2 mu_(2))`