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_1acostheta`
`m_1[g(sintheta-mu_1costheta]costheta`
Again applying NLM in vertical direction
`(m_1+m_2)-N_2=m_1asintheta`
`N_2=(m_1+m_2)g-m_1sintheta(gsintheta-mu_1gcostheta)`
For limiting condition
`f_2=mu_2N_2`
From (1) and (2)
`mu_2=(m_1costheta(gsintheta-mu_1gcostheta))/((m_2+m_2)g-m_1sintheta(gsintheta-mu_1gcostheta))`
Using values `mu_2=(1)/(8)`