Question

A block weighing 87.0 N rests on a plane inclined at 24.1° to the horizontal. The coefficient of the static and kinetic frictions are 0.25 and 0.13 respectively. What is the minimum magnitude of the force F, parallel to the plane, that will prevent the block from slipping?

Answer 1

Answer:

15.7 N

Explanation:

Draw a free body diagram.  The block has four forces acting on it.  Gravity pulling down, normal force pushing perpendicular to the plane, friction pointing up the plane, and applied force F pushing up the plane.

Sum of the forces normal to the plane:

∑F = ma

N − mg cos θ = 0

N = mg cos θ

Sum of the forces parallel to the plane:

∑F = ma

Nμ + F − mg sin θ = 0

F = mg sin θ − Nμ

Substituting:

F = mg sin θ − mgμ cos θ

F = mg (sin θ − μ cos θ)

Given mg = 87.0 N, θ = 24.1°, and μ = 0.25 (because the block is not moving):

F = 87.0 N (sin 24.1° − 0.25 cos 24.1°)

F = 15.7 N