A uniform rod is made to lean between rough vertical wall and the ground. Friction coefficient between rod and walls is `mu_(1)=(1)/(2)` and between the rod and the ground is `mu_(2)=(1)/(4)`. The rod is about to slip at contact surfaces. The correct options are:

Options

A. the normal reaction between rod and wall is `(mu_(2)W)/(1 + mu_(1) mu_(2))`
B. normal reaction between rod and ground is `(W)/(1 + mu_(1) mu_(2))`
C. `N_(2) gt N_(1)`
D. `N_(1) gt N_(2)`

Answer

A, B, C

Explanation

For equilibrium of rod
`sumFx = 0 , N_(1) = mu_(2) N_(2) sum F_(y) = 0 , N_(1) = (mu_(2)W)/(1 + mu_(1) mu_(2))`
So , choice (a) is correct and `N_(2) = (W)/(1 + mu_(1) mu_(2))`
So , choice (b) is correct . By substituting the values
`N_(1) = (2W)/(9)` and `N_(2) = (8W)/(9)`
`therefore " " N_(2) gt N_(1)`