The coefficient of friction between the block and the wedge is `mu`. Friction between the wedge and the table is to be neglected.

Answer

`Q=(mMg cos alpha cos varphi)/(M cos varphi +m sin alpha sin (alpha-varphi))`

Explanation

The equation of motion will assume the form (see fig.)
`-Q sin alpha + T cos alpha = Mb_(x)`
`Q sin alpha- T cos alpha = ma_(x), " " - mg + Q cos alpha + T sin alpha = ma_(y)`
`a_(y)= (-a_(x) +b_(x)) tan alpha, T=muQ`
Hence `Q=(mMg cos alpha)/(M+m sin alpha (sin alpha-mu cos alpha))`
`a_(x)= (Mg cos alpha(sin alpha-mu cos alpha))/(M+m sin alpha(sin alpha - mu cos alpha))`
`b_(x)= -(mg cos alpha(sin alpha- mu cos alpha))/(M+m sin alpha(sin alpha - mu cos alpha))`
`a_(y) = -((M+m)g sin alpha (sin alpha- mu cos alpha))/(M+m sin alpha (sin alpha- mu cos alpha))`
The block can either slide down the wedge, or remain stationary. It can slide down if `sin alpha gt mu cos alpha`, i.e. if `tan alpha gt mu`. In this case

`a_(x) gt 0, a_(y) lt 0 and b_(x) lt 0`. If, on the other hand, `tan alpha le mu`, the block and the wedge will remain stationary.