Question

A body of mass 10 kg is placed on an inclined surface of angle 30 degree if the coefficient of limiting friction is 1 by root 3 find the force required to just push the body of the inclination surface the force is being applied two parallel to inclined surface

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

Answer:

Explanation: hi