Force Exerted by Wedge on Block in an Inclined Plane System

Introduction:
In this scenario, a block of mass m is placed on a smooth wedge inclined at an angle θ. The entire system is accelerated horizontally in such a way that the block does not slip on the wedge. We need to determine the magnitude of the force exerted by the wedge on the block.

Analysis:
To solve this problem, we will break down the forces acting on the block and analyze them separately. There are two main forces acting on the block:

1. Gravitational Force (Weight):
The weight of the block can be decomposed into two components: one parallel to the inclined plane (mg sin θ) and the other perpendicular to the inclined plane (mg cos θ).

2. Normal Force:
The normal force acts perpendicular to the inclined plane and opposes the component of the weight perpendicular to the plane. It prevents the block from sinking into the wedge. The normal force can be decomposed into two components: one parallel to the inclined plane and the other perpendicular to the inclined plane.

Force Analysis:
- The force exerted by the wedge on the block can be divided into two components: one perpendicular to the inclined plane (Fn ⊥) and one parallel to the inclined plane (Fn //).
- The perpendicular component of the force (Fn ⊥) is equal in magnitude and opposite in direction to the component of the weight perpendicular to the inclined plane, i.e., Fn ⊥ = mg cos θ.
- Since the block is not slipping on the wedge, the parallel component of the force (Fn //) must cancel out the component of the weight parallel to the inclined plane, i.e., Fn // = mg sin θ.

Result:
Therefore, the magnitude of the force exerted by the wedge on the block is equal to the component of the weight perpendicular to the inclined plane, which can be calculated as:

|Fn| = √(Fn ⊥² + Fn //²)
= √((mg cos θ)² + (mg sin θ)²)
= √(m²g² cos² θ + m²g² sin² θ)
= √(m²g² (cos² θ + sin² θ))
= √(m²g²)
= mg

Thus, the magnitude of the force exerted by the wedge on the block is equal to its weight, which is mg.