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# Difference between Static Friction and Dynamic Friction

Friction is defined as the force that opposes the motion between any surfaces, fluid layers, and any other material in contact with each other. Friction is produced due to the irregularities between the surfaces. It produces heat. It depends on the nature of the surfaces. Friction always acts between any two contact surfaces of bodies. Friction is a contact force.

For Example, rubbing our hands generates heat. So friction is produced while rubbing our hands. While driving a car, if we use the brake to stop or slow down the car, friction is produced between the brakes and wheels, which makes the car slow down or stop.

Mathematically, friction is given as,

F = μη

where,

• F is the Frictional Force,
• μ is the coefficient of friction,
• η is the Normal force (= mg),
• m is the mass of the object, and
• g is the acceleration due to gravity (= 9.8m/s2)

### Unit Of Friction

Since friction is considered to be a force therefore the SI unit of Friction is Newton (N).

According to the position of the objects or materials in contact, friction can be categorized into two,

• Static Friction
• Dynamic Friction Let’s now discuss the difference between the two, starting with understanding their basic concepts as,

## What is Static friction?

Static friction is defined as the frictional force which acts between the two surfaces when they are in the rest position with respect to each other.

Some Examples of Static friction are,

• Pushing heavy objects such as trains, buses, etc.
• Clothes hanging on the hanger.
• Car parked on a slope.
• Man standing on the mountain.

Mathematically, static friction is defined as,

Fs = μsη

where,

• Fs is the Static Frictional Force,
• μs is the coefficient of static friction,
• η is the Normal force (= mg),
• m is the mass of the object, and
• g is the acceleration due to gravity (= 9.8m/s2).

## What is Dynamic Friction?

Dynamic friction is defined as the frictional force which is created between any two surfaces when they are in a moving position. It is also called Kinetic Friction

Some Examples of Dynamic friction are,

• While walking the force experience on the foot is dynamic friction.
• Force experience by the wheels of the moving bicycle.
• Force experienced by the blades of the skates while ice skating.
• Resistive force experienced by the hull of the boat while it sails.

Mathematically, kinetic friction is defined as,

Fk = μkη

where,

• Fk is the kinetic Frictional Force,
• μk is the coefficient of kinetic friction,
• η is the Normal force (= mg),
• m is the mass of the object, and
• g is the acceleration due to gravity (= 9.8m/s2).

## Solved Examples

Example 1: A man is moving a box with a mass of 20 kg. Find out the frictional force if it has a coefficient of friction of 0.4.

Solution:

Given that,

mass of the box m = 20 kg.

coefficient of friction μ = 0.4

Formula of friction

F = μη

η = mg = 20 × 9.8 = 196N

F = 0.4 × 196 = 78.5 N

Therefore, the frictional force required is 78.5 N.

Example 2: A normal force of 600N is exerted on a table of 12kg on the floor. It has a coefficient of friction of 0.6. Find the static friction exerted on it.

Solution:

Given that,

mass of the table m = 12kgs

Normal force η = 600N

coefficient of static friction μs = 0.6

Formula of Static Friction Fs = μsη

Fs = 0.6 × 600 = 360N.

Therefore, the static frictional force required is 360 N

Example 3: A boy playing with you who has a mass of 2 kgs and is at rest on the floor. If it has the force of 80 N of static frictional force. Find its coefficient of friction.

Solution:

Given

mass of toy m = 2 kgs

Static force Fs = 80 N

Formula Fs = μsη

normal force η = mg = 2 × 9.8

η  = 19.6N

Fs= μsη

80 = μs × 19.6

μs =  80/19.6

μs =  4.08

Therefore, the coefficient of static friction required is 4.08.

Example 4: A girl is playing with a football and then she kicked it with 250 N. calculate the kinetic friction if it has a coefficient of kinetic friction of 0.4.

Solution:

Given

The normal force η= 250 N

coefficient of kinetic friction μk = 0.4

Formula Fk = μkη

Fk= 0.4 × 250

= 100N

Therefore, the Kinetic frictional force required is 100 N.

Example 5: A train is moving at a uniform speed with a force of 500 N. If the kinetic friction is applied on the train is 1000 N. Calculate the coefficient of kinetic friction.

Solution:

Given

normal force η = 500N

Kinetic force Fk = 100 N

Formula Fk = μkη

1000 = μk × 500

μk = 1000/500

μk = 2.

Therefore, the coefficient of kinetic friction required is 2.

Example 6: An object of mass 2 kgs is moving on a table. If kinetic friction is applied to the object is 120N and it has a coefficient of friction of 0.6. Calculate the normal force applied to it.

Solution:

Given

The mass of object m = 2 kgs

Kinetic friction Fk = 120 N

coefficient of kinetic friction μk = 0.6

Formula Fk= μkη

120 = 0.6 × η

η = 120/0.6

= 20 N

Therefore, the normal force applied on the object is 20 N.