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
`sumF_(x)=0 , N_(1)=mu_(2)N_(2)sumF_(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 subsutituting the values
`N_(1)=(2w)/(9)& N_(2)=(8w)/(9):. N_(2) gtN_(1)`