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Friction is force that resists a moving body from slipping against each other by creating resistance. Frictional force between two surfaces is responsible for transforming kinetic energy or work into thermal energy or heat. Friction helps vehicles in acceleration, deceleration, and changing the direction. A vehicle without frictional force can lead to an accident due to losing control.
Read Also: Avogadro's law
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Key Terms: Friction, Frictional force, Static force, Normal force
Types of Friction
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There are four categories of Friction- static friction, sliding friction, rolling friction, and fluid friction.
Sliding Friction
In Sliding Friction, the weight of the sliding object calculates the amount of sliding friction present between the two objects. The sliding friction is supposed to be greater as the pressure exerted by the heavy object on the surface it slides over is comparably more.
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Rolling Friction
Friction between a circular object and the surface is called as Rolling Friction. It is required to overcome sliding friction is more than the force required to overcome the rolling friction.
Static Friction
Friction that keeps an object at rest without initiating any relative motion between the body and the surface is termed as Static Friction. For example, a parked car resting on the hill, a hanging towel on the rack. The maximum force of static friction is directly proportional to the normal force.
Fluid Friction
Layer of the fluids that are moving parallel to each other experience Fluid Friction. All the fluids provide some sort of internal resistance which makes them thick. Internal friction happens due to movement of molecules inside fluid and external friction happens as fluid interacts with other matters. Speed of the body, nature of the fluid, thickness of the fluid, shape of the body, and temperature are factors that affect fluid friction.
Friction Formula
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Friction Formula is expressed as:
Ff = µ\(\eta\)
Where,
- Ff = frictional force
- µ= coefficient of friction
- \(\eta\)= normal force
Normal Force formula is expressed as:
Fn (normal force)= mg
Where ,
- m = mass
- g= gravity
Rolling frictional force is expressed as:
F(rolling frictional force) = CrrN
Where,
- Crr = coefficient of rolling friction
- N = normal force
Things to Remember
- Movement of two bodies over one another is known as friction
- Moving the body by gradually increasing the value of \(\eta\) results in attaining the optimum value of \(\eta\) in case of static friction.
- Interaction of two items that create resistance is known as the frictional force
- Natural world consists of two kinds of opposing forces such as friction (solid) and viscous forces (liquid).
- Static friction does not exist without any applied force. The Law of static friction is fs≤ µs N.
- One can reduce friction by maintaining a thin layer of air between two solid surfaces in a relative motion.
- Force is required to keep a body in uniform motion.
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Sample Questions
Ques. A body of mass 50kg is resting on the floor. It does not move with a force of 5N. calculate the frictional force acting on the object. (2 Marks)
Ans. The static force of friction and the external force are equal and opposite when a body is resting without any motion.
fs = Fest
therefore, the statistic force of friction (fs) = 5N
Ques. A man begins to push his boat out of the mud to the water. the coefficient of friction between the boat and the mud is µ = 0.400. 50.0 kg is the mass of the boat. Calculate the frictional force acting on the boat. (2 Marks)
Ans. Ff = µ\(\eta\)
Ff = µ mg
Ff = (0.400)(50.0kg)(9.8m/s2)
Ff = 196kg.m/s2
Ff = 196N
Ques. Two boards are attached with two bolts. The force between the bolts is 500lbs. the complete strength of a single bolt is 5000lbs and the coefficient of static friction is µ = 0.5. Calculate the maximum force exerted to the boards to separate the two boards from one another. (2 Marks)
Ans. The frictional force along with the total strength of the bolts must be increased in order to pull the boards apart. The maximum force to pull apart the boards must be lower than the normal force required for the activity.
Fmax = 2 x 5000 + 500 x 0.5
= 10250lbs
Ques. Suppose a huge chunk of ice is being dragged across a frozen lake. The mass of the ice block is 250kg. The coefficient between the two surfaces (µk) is 0.05. Calculate the frictional force on the block of ice. (2 Marks)
Ans. Normal force (Ff) = µ\(\eta\)
Ff = µ mg
Fk = (0.05)(250kg)(9.8m/s2)
Fk = 122.5kg . m/s2
Fk = 122.5N
Ques. Calculate the force of friction between a girl of 40kg mass who is slipping on the frost and the coefficient of friction which is 0.45. (3 Marks)
Ans. mass (m) = 40kg
Coefficient of friction (µ) = 0.3,
Normal force (Fn) = mg
= 40kg x 9.8 m/s2
= 392N
Frictional force (Ff) = µFn
= 0.45 x 392N
= 176.4 N
Ques. A body of mass 2kg is resting on the floor. The coefficient of static friction between the floor and the object is µs= 0.8. calculate the amount of pressure to be exerted on the object to bring it to motion. (3 Marks)
Ans. N = mg
As the body is at rest, the normal force by the floor is equal to the gravitational force experienced by the body.
The maximum static frictional force fsmax
= µsN = µsmg
fsmax = 0.8 x 2 x 9.8 = 15.68N
The maximum static friction should be equal to the external force in order to bring the object to the motion.
Fext = 15.68N
Ques. A 10 kg box is being pulled by a child with a coefficient of friction µ being 0.3. Calculate the normal force as well as the frictional force. (3 Marks)
Ans. mass (m) = 10kg
Normal force (Fn) = mg
Fn= 10 kg x 9.8m/s2
= 98N
Frictional force (Ff) = µFn
= 0.3 x 98N
= 29.4N
Ques. Find the coefficient of the kinetic energy of a block of mass m which slides down the plane with an angle inclination of 60\(^\circ\). It has an acceleration of g/2. (4 Marks)
Ans. Kinetic friction is produced when the block is moving on the surface. The forces that are acting on the block are – normal force, downward gravitational force, and kinetic friction fk .
Along the x direction
mg sin \(\theta\) - fk = ma
but a is given as g/2
mg sin60\(^\circ\) - fk = mg/2
3/2 mg – fk = mg/2
fk = mg(3/2 – ½)
fk = (3-12 ) mg
mg cos \(\theta\) balances the normal force which results in no motion along the y- axis
mg cos \(\theta\) = N = mg/2
fk = µk N = µk mg/2
µk = (3-12)mg/ (mg/2)
µk = √3 – 1
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