All surfaces are smooth in the adjoining figure. Find F such that block remains stationary with respect to wedge.

Explanation

Acceleration of (block +wedge) is `a=(F)/((M+m))`
et us slove the problem by using both frames From inertial frame of reference (Ground) F.B.D. of block w.r.t ground (Apply real forces) :

with respect to ground block is moving with an acceleration 'a'
`:.SigmaF_(y)=0rArrNcostheta=mg . . . . . . . . (i)`
and `SigmaF_(x)=marArrNsintheta=ma . . . . . . . .(ii)`
From Eqs. (i) and (ii)
`:.F=(M+m)a" "=(M+m)g tan theta`
From non-inertial frame of reference (Wedge) :
F.B.D. of block w.r.t wedge (rael forces + pseudo force)

w.r.t wedge, block is stationary
`:.SigmaF_(y)=0rArrsintheta=ma . . . . . . .(iii)`
`SigmaF_(x)=0rArrsintheta=ma . . . . . . . .(iv)`
From Eqs. (iii) and (iv), we will get the same result i.e. `F=(M+m)g tan theta`.