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/s^{2}).
Since friction is a force only 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 and 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,
 Lamp resting on a table, and
 While rubbing hands, static friction is produced.
Mathematically, the static friction is defined as:
F_{s} = μ_{s}η
where,
 F_{s} is the Static Frictional Force,
 μ_{s} 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/s^{2}).
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,
 Wheels of the moving object,
 Rolling of objects on the ground,
 Swimming, and
 Skating.
Mathematically, the kinetic friction is defined as:
F_{k }= μ_{k}η
where,
 F_{k} is the kinetic Frictional Force,
 μ_{k} 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/s^{2}).
Difference between Static Friction and Dynamic Friction
Static Friction  Dynamic 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.  Dynamic Friction is defined as the friction which is created between any two surfaces when they are in a moving position. 
Common examples are a Lamp resting on the table and While rubbing hands, static friction is produced  Common examples are Wheels of the moving vehicles, Rolling of objects on the ground, and Skating. 
Mathematically, the static friction is defined as: F_{s} = μ_{s}η where,
 Mathematically, the kinetic friction is defined as: F_{k }= μ_{k}η where,

The magnitude is greater than dynamic friction because it has a greater value of the coefficient.  The magnitude is less than static friction because it has a lesser value of the coefficient. 
It has maximum magnitude as long as the applied force does not exceed its maximum and the surfaces will not move.  It has constant magnitude regardless of the speed at which the two objects are moving 
It depends on the magnitude of the applied force.  It is independent of the magnitude of the applied force. 
The coefficient of static friction depends on the nature of materials that are in contact.  The coefficient of dynamic friction depends on the nature of the material and the temperature of the material. 
Value of Static Friction can be zero.  Value of Kinetic Friction can never be zero. 
Sample Questions
Question 1: A man is moving a box with a mass of 20 kgs. Find out the frictional force if it has a coefficient of friction of 0.4.
Answer:
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.
Question 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.
Answer:
Given that,
mass of the table m = 12kgs
Normal force η = 600N
coefficient of static friction μ_{s} = 0.6
Formula of Static Friction F_{s} = μ_{s}η
F_{s} = 0.6 × 600 = 360N.
Therefore, the static frictional force required is 360 N.
Question 3: A boy is 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?
Answer:
Given
mass of toy m = 2 kgs
Static force F_{s} = 80 N
Formula F_{s} = μ_{s}η
normal force η = mg = 2 × 9.8
η = 19.6N
F_{s}= μ_{s}η
80 = μ_{s} × 19.6
μ_{s} = 80/19.6
μ_{s }= 4.08
Therefore, the coefficient of static friction required is 4.08.
Question 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.
Answer:
Given
The normal force η= 250 N
coefficient of kinetic friction μ_{k} = 0.4
Formula F_{k }= μ_{k}η
F_{k}= 0.4 × 250
= 100N
Therefore, the Kinetic frictional force required is 100 N.
Question 5: A train is moving with a uniform speed with the force of 500 N. If the kinetic friction is applied on the train is 1000 N. Calculate the coefficient of kinetic friction?
Answer:
Given
normal force η = 500N
Kinetic force F_{k} = 100 N
Formula F_{k} = μ_{k}η
1000 = μ_{k} × 500
μ_{k }= 1000/500
μ_{k }= 2.
Therefore, the coefficient of kinetic friction required is 2.
Question 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.
Answer:
Given
The mass of object m = 2 kgs
Kinetic friction F_{k} = 120 N
coefficient of kinetic friction μ_{k} = 0.6
Formula F_{k}= μ_{k}η
120 = 0.6 × η
η = 120/0.6
= 20 N
Therefore, the normal force applied on the object is 20 N.