The block of mass `m_(1)` is placed on a wedge of an angle `theta`, as shown. The block is moving over the inclined surface of the wedge. Friction coefficient between the block and the wedge is `mu_(1)`, whereas it is `mu_(2)` between the wedge and the horizontal surface. if `mu_(1)=1/2, theta=45^(@), m_(1)=4 kg, m_(2)=5 kg ` and `g=10 m//s^(2)`, find minimum value of `mu_(2)` so that the wedge remains stationary on the surface. express your answer in multiple of `10^(-3)`

Explanation

Taking block `+` wedge as system and applying `NLM` in horizontal direction
`f_(2) =m_(1)acos theta` ltbr. `=m_(1)[g(sintheta-mu,costheta)]cos theta` .... (1)
Again applying `NLM` in vertical direction
`(m_(1)+m_(2))g-N_(2)=m_(1) a sin theta`
`N_(2) =(m_(1)+m_(2))g-m_(1)sin theta(g sin theta-mu_(1)g costheta)`
For limiting condition `f_(2)=mu_(2)N_(2)`.....(2)
`mu_(2)=(m_(1)costheta(g sintheta=mu_(1)gcostheta))/((m_(2)+m_(2))g-m_(1)sintheta(gsntheta-mu_(1)gcostheta))`
Using values
`mu_(2)=(1)/(8)=125xx10^(-3)`
`=125`