Question

A rubber boot is at rest on an inclined plane. When the angle of inclination of the plane is increased to 36.5°, the boot begins to slide down the incline. Then, when the angle is decreased to 30.5°, the speed of the boot is constant. What are the coefficients of static and kinetic friction between the boot and the incline?

Answer 1

To solve this problem we will use the equations of kinetic and mechanical energy.

We start for the first angle where the given angle is 36.5 ° and the acceleration is zero.

The force equations would be given by:

`F= mgsin\theta\\N= mgcos\theta\\F_r = \mu_s N`

Where
`\mu` is coefficient of friction (static)

If we make a sum of forces in the direction in which the block moves we will have the SinΘ Component and that of the friction force, so

`\sum F = 0\\F-F_r=0\\F=F_r \\mg sin\theta = \mu N\\mg sin\theta= \mu mgcos\theta`

solving for the coefficient,

`\mu = (sin\theta)/(cos\theta)=tan\theta \\\mu = tan (36.5) \\\mu = 0.740`

We apply the same consideration for the angle of 30.5°, then we know that the coefficient (dinamic) is
`\mu_k = tan\theta \\\mu_k = tan (30.5) \\\mu_k = 0.589`