LAWS OF FRICTION

The following are the laws of friction:

1.  Frictional force is directly proportional to the normal force between the two surfaces in contact.

Fr α R – remove proportionality constant

Fr = KR

K = is the coefficient of frictional force, μ

Now:

Fr = μR

Fr = μR – make μ subject

μ = Fr/R

Therefore:

Coefficient of friction is the ratio friction force to the normal reaction. Since it involves ratio of the same unit it has no SI unit

2.   Friction depends on the nature of surface in contacts.

3.   Friction does not depend on the surface areas in contact.

4.   The ration of limiting friction over normal reaction is constant for two surfaces in contact.

FRICTION FORCE AT HORIZONTAL PLANE (SURFACE)

From the laws of friction above,

Fr = μR

But R=mg

Fr = μmg

Example 1

A block of mass 270kg is pulled along a horizontal surface. If the coefficient of kinetic friction between the block and the surface is 0.4, what is the friction force acting on the block as it slides?

Solution:

Data given:

Mass of block, m = 270kg

Acceleration due to gravity, g = 10 m/s²

Coefficient of kinetic friction, μ = 0.4

Kinetic Friction force, Fr =?

From:

Fr = μmg

Fr = 0.4 x 270x10

Fr = 1, 080N

The friction force acting on the block as it slides = 1,080 N

Example 2

A box of mass 2kg rest on a horizontal surface, a force of 4.4 N is required to just start the box moving. What is the coefficient of static friction between the block and the surface?

Solution:

Data given:

Mass of block, m = 2kg

Acceleration due to gravity, g = 10 m/s²

Static Friction force, Fr = 4.4 N

Coefficient of kinetic friction, μ =?

From:

Fr = μmg – make μ subject

μ = Fr/mg

μ = 4.4/(2x10)

μ = 0.22

The coefficient of static friction between the block and the surface = 0.22

Example 3

An aluminium block of mass 2.1kg rests on a steel platform. A horizontal force of 15N is applied to the block.

a)       Given that coefficient of limiting friction 0.6, will the block move?

b)      If will moves, what will be its acceleration where Given that coefficient of kinetic friction is 0.47

Solution

Data given:

Mass of block, m = 2.1kg

Acceleration due to gravity, g = 10 m/s²

Force applied, F = 15 N

Coefficient of limiting friction, μ_s = 0.6

Coefficient of kinetic friction, μ_k = 0.47

a)        

Find the static friction force, Fs =?

From:

F = μmg

Then:

F_s = μ_s x mg

F_s = 0.6 x 2.1 x 10

F_s = 12.81N

Since: F > F_s, hence the car will move

b)      Acceleration acquired, a =?

First find the kinetic friction force, F_k =?

From:

F = μR

Then:

F_k = μ_k x mg

F_k = 0.47 x 2.1 x 10

F_k = 9.87N

F_k it opposes the direction of F so the force which causes motion (net force) is given;

F_net = F - F_k

F_net = 15 – 9.87

F_net = 5.13 N

From: Newton’s second law;

F = ma

Then:

F_net = ma,       make a subject

a = F_net/m

   = 5.13/2.1

   = 2.44 m/s²

Acceleration = 2.44 m/s²

FRICTION FORCE AT INCLINED PLANE (SURFACE)