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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.enter image description here

  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 :enter image description here

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.

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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.)

enter image description here

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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.

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