Static and Kinetic Friction on an incline?

Question

A 10 kg crate is placed at rest on an incline plane (mus =0.6, muk =0.4), whose angle can be adjusted

a) The maximum angle of the incline for which the crate will stay at rest is:___degrees

b) if the angle of incline is 15 degrees, what is the static friction force? ___N

what is the component of weight acting along incline?___N

Net force on crate?___N

acceleration of crate?___m/s2

c) if the angle of incline is 50 degrees, what is the static friction force? ___N

what is the component of weight acting along incline?___N

Net force on crate?___N

acceleration of crate?___m/s2

d) For the 50o angle, find how long it would take the crate to slide down 5 m along the incline? ___s

Answer 1

a) Normal force = Fn = W * cos θ

max Friction force = μs * Fn = 0.6 * W * cos θ

component of gravity acting along incline = W * sin θ

Setting those equal:

W * sin θ = 0.6 * W * cos θ

(sin θ) / (cos θ) = 0.6

tan θ = 0.6

θ = 30.96°

b) θ = 15°

Fs(max) = μs * Fn = μs * W * cos θ = 0.6 * 98 * cos 15°

Fs(max) = 56.8 N

Fg along incline = W * sin θ = 98 sin 15° = 25.4 N, which is less than Fs(max), so actual force of static friction is 25.4 N as well.

Net force on crate = 0 N

acceleration = 0 m/s²

c) θ = 50°

Fs(max) = μs * Fn = μs * W * cos θ = 0.6 * 98 * cos 50°

Fs(max) = 37.8 N

Fg along incline = W * sin θ = 98 sin 50° = 75.1 N, which is greater than Fs(max), so the block overcomes static friction, and kinetic friction takes over.

Fk = μk * Fn = μk * W * cos θ = 0.4 * 98 * cos 50°

Fk = 25.2 N

Fnet = Fg along incline - Fk = 75.1 N - 25.2 N = 49.9 N, along the incline.

a = F/m = 49.9 / 10 = 4.99 m/s²

d) (assuming starting from rest)

d(t) = at²/2

t² = 2d/a = 2 * 5 / 4.99 = 2.0

t = √2.0 = 1.4 s