A wedge `B` of mass 2 m is placed on a rough horizontal suface. The coefficient of friction between wedge and the horizontal surface is `mu_(1)`. A block of mass m is placed on wedge as shown in the figure. The coefficient of friction between block and wedge is `mu_(2)`. The block and wedge are released from rest.
Q. Suppose the inclined surface of the wedge is at `theta=37^(@)` from angle from horizontal and `mu_(2)=0` are wedge will remain in equilibrium if

Options

A. will remain in equilibrium if `mu_(1)=0.5`
B. will accelerate towards left if `mu_(1)=0`
C. will acceleration toward left if `mu_(1)=0.25`
D. will remain in equilibrium if `mu_(1)=0.3`

Answer

A, D

Explanation

Since `mu_(2)gttantheta` so block will remain at rest, so net content force onblock will be vertically upward. Net contact force on wedge due to block will be vertically downward. There is no tendency of sliding of wedge so wedge will remain in equlibrium for any value of `mu_(1)`