Acrate which has a mass of m1 = 125kg is sitting on an icy surface. a rope is attached to the crate and held at angle theta 28.0 above the horizontal. if the rope is pulled by a force p= 585 n and the crate resultantly accelerates at a rate of a =3.30m/s squared , find the coefficient of sliding friction between the crate and the icy surface.

μ = 0.109

Explanation:

Draw a free body diagram of the crate.  There are four forces:

Weight force mg pulling down.

Normal force N pushing up.

Applied force P pulling at θ above the horizontal.

Friction force Nμ pushing to the left.

Sum of the forces in the y direction:

∑F = ma

N + P sin θ − mg = 0

N = mg − P sin θ

Sum of the forces in the x direction:

∑F = ma

P cos θ − Nμ = ma

P cos θ − ma = Nμ

μ = (P cos θ − ma) / N

μ = (P cos θ − ma) / (mg − P sin θ)

Given:

P = 585 N

θ = 28.0°

m = 125 kg

a = 3.30 m/s²

μ = (585 cos 28.0° − 125 kg × 3.30 m/s²) / (125 kg × 9.8 m/s² − 585 sin 28.0°)

μ = 0.109

vapour or gas assuming atmospheriv pressure