Question

A body of mass 8kg is in limiting equilibrium overt an inclined plane of inclination 30°. If the inclination is made 60°, the minimum force required to prevent the body from sliding down is​

Answer 1

Coefficient of friction = 1/√3

∵ tan θ = μ

where,  θ is the angle of repose and μ is the coefficient of friction.

∴ tan  θ = tan 30

⇒ θ = 30

This means angle of repose is equal to the angle of the Inclined Plane.

It also means that the body will stay at rest.

To keep the body in motion, the Force must me equal to the sum of friction and horizontal component of Normal reaction.

Refer to the attachment for the answer.

From there,

Force = mg sinθ + Friction.

⇒ Force = mg sin θ + μN

⇒ Force = 10 × 10 × Sin 30 + 1/√3 × mg cosθ

⇒  Force = 50 + 150

⇒ Force = 200 N

Hope it Helps.

Answer 2

Step-by-step explanation:

Refer to the attachment...