FRICTION FORCE AT INCLINED PLANE (SURFACE)

Consider the diagram below, a mass of body sliding down the incline plane

At constant speed a = 0m/s²

F_k = mgsinθ

R = mgcosθ

But:

Fk = μ_kR – make μ_k subject

μ_k = F_k/R

μ_k = mgsinθ/mgcosθ

μ_k = sinθ/cosθ

μ_k = Tanθ

Therefore, at constant speed coefficient of kinetic friction is given by the formula above

At rest a = 0m/s²

Fs = mgsinθ

R = mgcosθ

But:

Fs = μ_sR – make μ_s subject

μ_s = Fs/R

μ_s = mgsinθ/mgcosθ

μ_s = sinθ/cosθ

μ_s = Tanθ

Therefore, at rest coefficient of static friction is given by the formula above

Example 1

A mass is placed on an inclined plane such that it can move at constant speed, when slightly tapped. If the angle of the plane makes with the horizontal plane is 30⁰. Find the coefficient of kinetic friction.

Solution

Data given:

Angle of the plane, θ = 30⁰

Coefficient of dynamic friction, μ =?

From:

μ_s = Tanθ

μ_s = Tan30⁰

μ = 0.56

The coefficient of kinetic friction = 0.56

Example 2

A block of wood of mass 5kg is placed on a rough plane inclined at 60^0. Calculate its acceleration down the plane if coefficient of friction between the block and the plane

Solution

Data given:

Angle of the plane, θ = 60⁰

Mass of the wood, m = 5kg

Acceleration due to gravity, g = 10m/s²

Friction force, Fr =?

Diagram:

First Find coefficient of dynamic friction, μ = Tan60⁰

μ_k = Tanθ

μ_k = Tan60⁰

     = 0.3

Net force, F =?

F = mgsin60^o - Fr

F = 50sin60^o - Fr

But:

Fr = μ_{k x} mgcosθ

Fr = μ_k x 50cosθ

Fr = 0.3 x 50cos60^o

Now:

F = 50sin60^o – (0.3 x 50cos60^o)

F = 35.8N

But:

F = ma – make a subject

a = F/m

   = 35.8/5

   = 7.1m/s²

Its acceleration down the plane = 7.1m/s²