Answer:

a) θ = α

Explanation:

The forces acting on the mass m in both cases are

1. WEIGHT of the mass, mg which can be resolved into two components one along the plane(downward inclination) and other perpendicular to the plane.

2. Friction force, f along upward incline direction.

3. Normal reaction, N from the wedge perpendicular to the plane

The free body diagram of the mass in both cases is attached.

case 1

The mass will START to slide down just as the component of weight along the plane overcome the friction force.Hence

mgsin(θ) = f = μN

Also,

mgcos(θ) = N

Dividing the equations we get

sin(θ)/cos(θ) = μ

or tan(θ) = μ

or θ = tan⁻¹(μ) ........(1)

Case 2.

As the wedge is ACCELERATING upward with acceleration a, the mass will experience a pseudo force ma in the downward direction. This can be resolved into two components mgsin(α) and mgcos(α) similar to that of mg.

The mass will start to slide just as the components of weight and pseudo force along the plane balance the friction. Hence

mgsin(α) + masin(α) = f = μN

and

mgcos(α) + macos(α) = N

Dividing the equations, we get

sin(α)/cos(α) = μ

or tan(α) = μ

or α = tan⁻¹(μ) ...........(2)

From (1) and (2)

θ = α = tan⁻¹(μ)

This angle is known as angle of repose and is CONSTANT for a pair of surfaces.