In the adjoining figure, the coefficient of friction between wedge (of mass M) and block (of mass m) is μ. Find the minimum horizontal force F required to keep the block stationary with respect to wedge.
For this Q , total acc of $M+m$= $F/M+m. $
I haven’t marked N1 or N2 but can assume them according to FBD.
For mass M ,
μN1 on left means the friction from surface.
$N_m$ means normal force by the mass m which is equal to $N_M$.
For mass m ,
1 )μN2 on left is because frictional force is always in direction opposite to motion. Since the N2=0(no surface from ground for mass m) (as per Q), So $μN2=0$.
Q 1 Why did we not take $N_M$ as the normal force in μN2 ?
2 )$mg=0 $
Total F= μN1 + $N_m$ where $N_m$ = μ*N = m * acc of mass m. Since ,
μN=0. Therefore , either m or a = 0 it has to be .
So , total $F= μN$ only.
But correct answer is $(M+m)*a$ where $a=g/μ$
So , I just wish to know where am I wrong in my calculation.