1. Analysis of the System

We have a wedge of mass $M$ and a block of mass $m$. The block is released on the frictionless slant face. Due to the normal reaction from the block, the wedge will experience a force to the right and will accelerate. Since there is friction between the floor and the wedge, this acceleration is opposed by a frictional force.

Let:

• $a$ = Acceleration of the wedge (towards the right).
• $N$ = Normal reaction between the block and the wedge.
• $N_g$ = Normal reaction from the ground on the wedge.
• $f$ = Frictional force between the ground and the wedge.

2. Equations of Motion for the Wedge

The forces acting on the wedge in the horizontal and vertical directions are:

• Horizontal: The horizontal component of normal reaction ($N\sin\theta$) pushes it right. Friction ($f$) acts left.
• Vertical: The ground normal ($N_g$) acts up. Weight ($Mg$) and the vertical component of block’s normal ($N\cos\theta$) act down.

Vertical equilibrium of the wedge:

$$N_g = Mg + N\cos\theta$$

Horizontal equation (Newton’s 2nd Law):

$$N\sin\theta – f = Ma$$

Assuming the wedge moves, friction is kinetic (limiting case for calculating N):

$$f = \mu N_g = \mu(Mg + N\cos\theta)$$

Substituting $f$ into the horizontal equation:

$$N\sin\theta – \mu(Mg + N\cos\theta) = Ma$$

Solving for acceleration $a$:

$$a = \frac{N(\sin\theta – \mu\cos\theta) – \mu Mg}{M} \quad \dots(1)$$

3. Equations of Motion for the Block

We analyze the block in the frame of the accelerating wedge. We introduce a pseudo force $ma$ acting to the left. Resolving forces perpendicular to the incline:

• Component of Weight: $mg\cos\theta$ (into the incline).
• Component of Pseudo Force: The pseudo force is horizontal (left). Its component perpendicular to the incline is $ma\sin\theta$ (out of the incline).

Since the block does not leave the surface or penetrate it, forces perpendicular to the incline balance:

$$N + ma\sin\theta = mg\cos\theta$$ $$N = mg\cos\theta – ma\sin\theta \quad \dots(2)$$

4. Solving for Normal Reaction (N)

Substitute the expression for $a$ from (1) into (2):

$$N = mg\cos\theta – m\sin\theta \left[ \frac{N(\sin\theta – \mu\cos\theta) – \mu Mg}{M} \right]$$

Multiply by $M$ to clear the denominator:

$$MN = Mmg\cos\theta – m\sin\theta [ N(\sin\theta – \mu\cos\theta) – \mu Mg ]$$

Expand the terms:

$$MN = Mmg\cos\theta – N m\sin\theta(\sin\theta – \mu\cos\theta) + \mu mMg\sin\theta$$

Group terms containing $N$ on the left:

$$N [ M + m\sin\theta(\sin\theta – \mu\cos\theta) ] = Mmg\cos\theta + \mu mMg\sin\theta$$ $$N = \frac{mMg(\cos\theta + \mu\sin\theta)}{M + m\sin\theta(\sin\theta – \mu\cos\theta)}$$