Find minimum horizontal force required to keep the block and wedge stationary?

Question

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.

1. For this Q , total acc of $M+m$= $F/M+m. $

2. Considering inertial frame of reference , for mass M and m.FBD is as follows :

I haven’t marked N1 or N2 but can assume them according to FBD.

For mass M ,

1. μN1 on left means the friction from surface.

2. $N_m$ 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.

Answer 1

What you are learning here is the effect of the "inertia force" due to motion.

In order for both blocks to move together, we can simply write the equation, $F = (M + m)*a$, by the law of motion.

Then for the small block not to fall during the motion, there must have a normal force exerted from the smaller block on the larger block to produce friction force required to maintain the small block in place. In here, the friction force is $m*g$, and the inertia force is $m*a$, thus $N = m*g/\mu = m*a$, $a = g/\mu$. (The normal force $N$ is the effect of the "Inertia Force", which has the same intensity as the mass in motion and always in the reversed direction of the motion.)

Answer 2

mg does not equal 0. It's the weight of the block. The friction that the weight would have to exceed to start moving is [mu]ma. F is only moving M and m. Even if m is glued to M it will be (M+m)a.