Question

the coefficient of friction between a hard rubber and normal street pavement is 0.8 What is the steepest incline on which a car can be safely parked​

Answer 1

38.7°

Step-by-step explanation:

Draw a free body diagram of the car. There are three forces:

Weight force mg pulling down,

Normal force N pushing up perpendicular to the incline,

and friction force Nμ pushing parallel up the incline.

Sum of forces in the perpendicular direction:

∑F = ma

N − mg cos θ = 0

N = mg cos θ

Sum of forces in the parallel direction:

∑F = ma

Nμ − mg sin θ = 0

Nμ = mg sin θ

μ = (mg sin θ) / N

μ = (mg sin θ) / (mg cos θ)

μ = tan θ

Given μ = 0.8:

θ = tan⁻¹(0.8)

θ = 38.7°