Coefficient of Static Friction Formula

Friction is a kind of force that opposes the motion of two objects towards each other. It is a contact force when two objects are under contact they experience friction.

Friction is defined as the opposition offered by the surfaces that are in contact when they were in relative motion.

There are different kinds of frictions on the basis of type of objects, to oppose the movement of sliding objects or to resist the motion of two objects when they are under a relative motion to each other. Types of friction are as,

1. Static friction
2. Sliding friction
3. Rolling friction
4. Kinetic friction
5. Fluid friction.

Static Friction

Static friction as the name suggests static means at rest so static friction acts on objects when they are in a rest position. When both objects are at a rest or one of the objects is a little tilted or inclined then the friction that opposes the motion of the object is known as Static friction.

Formula of Coefficient of Static Friction

The coefficient of static friction can be calculated with the formula:-

μs = F /N

Where

F = Static force of friction

μs = coefficient of static friction

N = Normal force

Examples of Static friction

• A car parked on the slope the static friction of wheels with the road holds the car at its position.
• Spectacles over the nose are another common example of static friction.
• Wearing a watch, hanging a shirt over a hanger, ring in our hands, belts that hold our pants are some common life examples of static friction.
• Water droplet on a windowpane.
• Tiles that are stick over the roof due to some strong forces of cement, here the cause of friction is chemical.

Static friction force

Laws of Static Friction

1. In static friction, the maximum force is independent of the area of contact.
2. The normal force is comparative to the maximum force of static friction it means that, if the normal force increases, the maximum external force that the object can possess without moving, also increases.

Sample Problems

Question 1: A force of 400 N is exerted on a box of 10 kg still on the floor. If the coefficient of friction is 0.3, what is the value of static friction?

Solution:

Given,

Force F = 400 N,

Coefficient of friction, μ_s = 0.3

Static friction is given by F_s = μ_s F_n

= 0.3 × 400 N

F_s = 120 N.

Question 2: State the laws of Static friction.

Answer:

1. In static friction, the maximum force is  independent on the area of contact.
2. The normal force is comparative to the maximum force of static friction it means that, if the normal force increases, the maximum external force that the object can posses without moving, also increases.

Question 3: A box kept on the floor experience a force of 90N with a coefficient of static friction of 0.4. Find the friction force. 

Solution:

Given,

Force applied or normal force N = 90 N

Coefficient of friction = 0.4

Static friction can be calculated as: F = μ_s × N.

F = 0.4 ×  90

= 36 N

Question 4: A box of 40Kg kept on the floor experience a force of 60N horizontally and its coefficient of static friction of 0.2. Find the frictional force. Will the box move from its position.

Solution:

Given, 

Force applied = 60 N

Coefficient of friction = 0.2

Normal force F_n = mg = 40 × 10 = 400 N.

Static friction can be calculated as: F = μ_s × N.

F = 0.2 ×  400 

= 80 N

We can see that static frictional force i.e. 80N, is greater than applied force 60 N which means that box will remain in its position.

Question 5: A Force of 50 N is exerted on the box kept on the floor with the coefficient of static friction of 0.2. Find the friction force.

Solution:

Given,

Force applied or normal force N = 50 N

Coefficient of friction = 0.2

Static friction can be calculated as: F = μ_s × N.

F = 0.2 ×  50

= 10 N

Question 6: Force of 30N is exerted horizontally on the box of 20Kg kept on the floor with the coefficient of static friction of 0.3. Find the friction force. Determine the will box move or not?

Solution:

Given,

Force applied = 150 N

Coefficient of friction = 0.3

Normal force F_n = mg = 30 × 10 = 300 N.

Static friction can be calculated as: F = μ_s × N.

F = 0.3 ×  300

= 90 N

We can see that static frictional force i.e. 90N, is greater than applied force 30 N which means that box will not move from its position.

Question 7: An box having a mass of 20 kg is placed on a smooth surface. Static friction between these two surfaces is given as the 30 N. Find the coefficient of static friction?

Solution:

Given,

Mass of box, m = 20 kg

Friction between them, F = 30 N

Coefficient of static friction μ_s = ?

We know that,

Normal force, N = mg

So, N = 20 × 9.81 = 196.2 N (g = 9.81)

For coefficient of static friction is,

μ_s = F/N

μ_s = 30/196.2

μ_s = 0.153