Friction Important Questions, Formulas and Solutions for JEE by PW

Friction questions often depend on identifying the tendency of relative motion before writing the equations of motion. You need to distinguish static friction from kinetic friction, calculate the correct normal reaction and check whether the friction required for a particular motion exceeds its limiting value. These steps become especially important in inclined-plane problems, accelerating frames and systems containing two or more blocks.

Physics Wallah’s important questions break friction into focused applications, from walking and limiting friction to rough inclines, chains and connected blocks. Use each solution to understand the force directions, revise the relevant formula and practise applying the correct friction condition to numerical problems.

Friction: Important Questions and Answers

Q1. What Is Friction, and How Do You Determine Its Direction?

Friction is a contact force that opposes the tendency of relative slipping or actual relative motion between two surfaces. It acts parallel to the contact surface, while the normal reaction acts perpendicular to it.

Friction does not always oppose the motion of an entire body. Its direction depends on the relative slipping tendency at the contact.

Examples:

• Walking: The foot pushes the ground backwards. Static friction from the ground acts forward on the foot, helping the person walk.

• Skidding: When a shoe slides forward relative to the ground, kinetic friction acts backwards on the shoe.

Key distinction:

• Static friction acts when the contacting surfaces do not slip relative to each other.

• Kinetic friction acts when the contacting surfaces slide relative to each other.

Memory tip: Determine which way the surfaces tend to slip relative to one another before assigning the direction of friction.

Q2. What Are Static Friction and Limiting Friction?

Static friction acts between surfaces that are in contact but do not slide relative to each other. It adjusts its magnitude to oppose the tendency of slipping.

The range of static friction is:

0 ≤ fₛ ≤ μₛN

The maximum value is called limiting friction:

fₛ(max) = μₛN

Here, μₛ is the coefficient of static friction and N is the normal reaction.

Solved example: A block rests on a horizontal surface. A horizontal force of 5 N is applied, and the limiting friction is 12 N. Find the actual static friction.

Since the applied force is below the limiting value, the block remains at rest.

Therefore, static friction balances the applied force:

fₛ = 5 N

It is not automatically equal to μₛN. Static friction reaches μₛN only when the block is just about to slip.

Important points:

• Static friction is self-adjusting.

• Limiting friction is the maximum possible static friction.

• A force slightly greater than the limiting friction can initiate slipping, assuming the force conditions remain unchanged.

The limiting force is both the maximum friction that can maintain rest and the minimum applied force needed to initiate motion in the corresponding setup.

Q3. What Is Kinetic Friction, and How Does It Differ from Static Friction?

Kinetic friction acts when two surfaces slide relative to each other.

Its magnitude is commonly modelled as:

fₖ = μₖN

Here, μₖ is the coefficient of kinetic friction.

For most ordinary surfaces:

μₛ > μₖ

Therefore, the force required to start sliding is generally greater than the force required to maintain sliding.

The coefficient of friction is dimensionless and depends on the pair of surfaces in contact. If a question gives only one coefficient, use it as instructed by the problem; in standard simplified questions, it may be used for both static and kinetic friction.

Memory tip: Static friction prevents slipping; kinetic friction acts during slipping.

Q4. Why Does Friction Arise Between Apparently Smooth Surfaces?

Even surfaces that appear smooth contain microscopic irregularities. When two surfaces touch, their actual contact occurs at a limited number of microscopic points.

The small real area of contact can produce high local pressure. Microscopic bonds may form between the contacting regions, and these bonds resist relative movement.

When an external force attempts to slide one surface over the other, the resistance at these contact points contributes to friction.

The normal reaction and friction are perpendicular contact-force components. Their resultant contact force has magnitude:

R = √(N² + f²)

Here, N is the normal reaction and f is the frictional force.

Memory tip: Microscopic irregularities → Small real contact area → High local pressure → Interfacial bonds → Resistance to slipping.

Q5. How Do You Calculate Friction and Acceleration on an Inclined Plane?

Consider a block of mass m on an inclined plane at angle θ to the horizontal.

The weight mg has two components:

• Perpendicular to the plane: mg cos θ

• Parallel to the plane: mg sin θ

If no other force acts perpendicular to the plane, the normal reaction is:

N = mg cos θ

When the block slides down, kinetic friction acts up the plane:

fₖ = μₖmg cos θ

Applying Newton’s second law along the plane:

mg sin θ − μₖmg cos θ = ma

Therefore:

a = g(sin θ − μₖ cos θ)

This expression applies when the block is sliding down the incline and the stated forces are the only relevant forces along the plane.

Solved example: A block slides down a 30° incline with μₖ = 0.2. Take g = 10 m/s². Find its acceleration.

Using sin 30° = 0.5 and cos 30° ≈ 0.866:

a = 10(0.5 − 0.2 × 0.866)

a = 10(0.3268)

Answer: a ≈ 3.27 m/s² down the plane

If the block moves with constant velocity, its acceleration is zero. The net force along the plane must then be zero.

Q6. What Is the Angle of Repose, and How Is It Related to the Coefficient of Friction?

The angle of repose is the inclination of a plane at which a body is just about to slide down under gravity.

At the limiting condition, static friction reaches its maximum value:

mg sin θᵣ = μₛmg cos θᵣ

Cancelling mg and rearranging:

tan θᵣ = μₛ

Therefore:

θᵣ = tan⁻¹(μₛ)

Solved example: If the coefficient of static friction is 1/√3, find the angle of repose.

tan θᵣ = 1/√3

Since tan 30° = 1/√3:

Answer: θᵣ = 30°

For an ideal block initially at rest on an incline:

• If θ < θᵣ, static friction can prevent sliding.

• If θ = θᵣ, the block is just about to slip.

• If θ > θᵣ, the block cannot remain at rest under the same conditions and begins sliding down.

Once the block slides, kinetic friction is used to calculate its acceleration.

Q7. What Force Is Required to Move a Block Up or Prevent It from Sliding Down an Incline?

For a block of mass m on an incline of angle θ, the component of gravity along the plane is mg sin θ.

Case 1: Moving the block up the incline at constant speed

Friction acts down the plane because it opposes the upward sliding.

The applied force required parallel to the plane is:

F = mg sin θ + μN

If N = mg cos θ:

F = mg(sin θ + μ cos θ)

Case 2: Preventing a block from sliding down

For a block at rest, the required force depends on the direction in which it is applied and whether static friction alone can hold the block.

If an applied force F acts up the plane and the block is just about to slide down, friction also acts up the plane at its limiting value. The force balance is:

mg sin θ = F + μₛmg cos θ

Therefore:

F = mg(sin θ − μₛ cos θ)

This expression gives the required upward applied force when the right-hand side is positive and the block is at the limiting condition for downward slipping. If static friction alone can hold the block, no additional force is required.

Important: Always determine the tendency of relative motion before assigning friction's direction. The formula changes when the direction of impending or actual motion changes.

Q8. What Is the Minimum Force Required to Pull a Block on a Rough Horizontal Surface?

Consider a block of mass m on a rough horizontal surface. A force F is applied at an angle θ above the horizontal.

The vertical force balance gives:

N = mg − F sin θ

At the limiting condition for impending motion:

F cos θ = μN

Substituting the normal reaction:

F cos θ = μ(mg − F sin θ)

Rearranging:

F(cos θ + μ sin θ) = μmg

Therefore:

F = μmg / (cos θ + μ sin θ)

The minimum value occurs when:

tan θ = μ

The minimum force is:

F(min) = μmg / √(1 + μ²)

Here, μ is the coefficient used for the limiting-friction condition.

Pulling at an upward angle reduces the normal reaction and therefore reduces the limiting friction. This is why pulling can be easier than pushing downward at the same angle magnitude.

Important: The result assumes the force is applied above the horizontal and the block is just about to move.

Q9. How Does Friction Act in an Accelerating Frame?

When analysing motion from a non-inertial frame accelerating with acceleration a, a pseudo-force acts opposite to the frame's acceleration.

Its magnitude is:

F(pseudo) = ma

Consider a block resting inside a truck that accelerates forward. Relative to the truck, the block tends to slip backwards because of its inertia. Static friction acts forward to prevent this relative slipping.

In an inertial frame, if static friction is the only horizontal force accelerating the block:

fₛ = ma

For the block to remain at rest relative to the truck, the required friction must not exceed the maximum static friction:

ma ≤ μₛmg

Cancelling m:

a ≤ μₛg

Therefore, the maximum truck acceleration for which the block can remain at rest relative to the truck is:

a(max) = μₛg

Key point: Actual static friction is ma in this situation, not automatically μₛmg. It reaches μₛmg only at the limiting condition.

Q10. How Is Friction Calculated for a Chain Partly on a Rough Surface?

Consider a uniform chain of length l and mass m. Its linear mass density is:

λ = m/l

Suppose a length x lies on a smooth surface and the remaining length (l − x) lies on a rough horizontal surface. If the coefficient of kinetic friction is μₖ, the rough portion has mass λ(l − x).

The normal reaction on that portion is λ(l − x)g. Therefore, the frictional force is:

fₖ = μₖλ(l − x)g

As more of the chain reaches the smooth surface, the length resting on the rough surface decreases, so the frictional force decreases.

For a constant pulling force F, the net force is:

F − μₖλ(l − x)g = λl a

Since the acceleration changes with position, use:

a = v(dv/dx)

If the chain starts from rest and is pulled until its entire length reaches the smooth surface, the idealised work-energy result is:

F l − μₖλg l²/2 = (1/2)λl v²

Therefore:

v = √(2F/λ − μₖgl)

This result assumes the chain is initially at rest, the pulling force remains constant, and the stated contact arrangement applies throughout the motion. The expression under the square root must be non-negative for the stated final state to be reached from rest under these assumptions.

Q11. How Do You Solve Friction Problems Involving Two Blocks?

Two-block systems often involve a force applied to one block while friction acts at the interface between the blocks. The main task is to determine whether the blocks move together or slip relative to each other.

Method:

1. Draw a separate free-body diagram for each block.

2. Calculate the normal reaction at each contact.

3. Determine the maximum available static friction.

4. Initially assume that the blocks move together with a common acceleration.

5. Calculate the friction required to produce that acceleration.

6. Compare the required friction with the maximum static friction.

7. If the required friction exceeds the limit, analyse the blocks separately using kinetic friction where applicable.

When the blocks move together:

a₁ = a₂ = a

Solved example: A 4 kg lower block rests on a smooth horizontal table and carries a 2 kg upper block. A horizontal force F pulls the lower block. The maximum static friction between the blocks is 6 N. Find the force at which slipping begins.

The total mass is:

M = 4 + 2 = 6 kg

Assuming both blocks move together:

a = F/6

The upper block is accelerated by friction alone. Therefore, the friction required is:

f = 2a

Substituting a = F/6:

f = 2(F/6) = F/3

At the limiting condition:

F/3 = 6

Answer: F = 18 N

Below 18 N, the blocks can move together under the stated conditions. At 18 N, the required friction reaches its limiting value. For a greater force, the common-acceleration assumption is no longer valid, and the motion must be analysed separately.

Q12. How Do You Analyse Friction in a Three-Block System?

For three blocks in contact, friction may act at more than one interface. A common mistake is to assume that all blocks necessarily move together.

Use the following method:

1. Draw separate free-body diagrams for all three blocks.

2. Assume a common acceleration and calculate it using the external forces on the system.

3. Determine the friction required at each interface to maintain the common motion.

4. Calculate the maximum static friction available at each interface.

5. Compare the required and maximum friction values separately.

6. If the required friction exceeds the limit at an interface, separate the motion of the blocks across that interface and solve the revised equations.

Important: Slipping begins at an interface when the friction required for common motion exceeds the maximum static friction available there. One interface may reach its limit before another.

Q13. How Can Relative Motion Be Used to Calculate the Separation of Two Blocks?

Relative motion is useful when two blocks start with the same velocity but have different accelerations.

If their initial relative velocity is zero:

u(rel) = 0

Their relative acceleration is:

a(rel) = a₁ − a₂

Use the signed accelerations consistently. If the question asks for the magnitude of separation, use the magnitude of the relative acceleration when it remains constant.

For constant relative acceleration and separation distance L:

L = u(rel)t + (1/2)a(rel)t²

Since u(rel) = 0:

L = (1/2)a(rel)t²

Therefore, for a(rel) > 0 in the chosen separation direction:

t = √[2L/a(rel)]

This method is useful for blocks accelerating differently on rough surfaces and for pulley systems where friction changes the acceleration of connected bodies.

Q14. What Are the Most Important Friction Formulas for JEE Physics?

Static friction

• Range: 0 ≤ fₛ ≤ μₛN

• Limiting friction: fₛ(max) = μₛN

Kinetic friction

• fₖ = μₖN

Resultant contact force

• R = √(N² + f²)

Inclined plane

• Normal reaction: N = mg cos θ

• Component of weight down the plane: mg sin θ

• Acceleration while sliding down: a = g(sin θ − μₖ cos θ)

Angle of repose

• tan θᵣ = μₛ

Minimum pulling force at an angle

• F = μmg / (cos θ + μ sin θ)

• tan θ(min) = μ

• F(min) = μmg / √(1 + μ²)

Accelerating frame

• Pseudo-force magnitude: F(pseudo) = ma

• Maximum acceleration without slipping: a(max) = μₛg

Uniform chain

• Linear mass density: λ = m/l

• Friction on rough portion: fₖ = μₖλ(l − x)g

Relative motion

• a(rel) = a₁ − a₂

• From rest relative to one another: L = (1/2)a(rel)t²

To solve friction questions accurately, identify the contact surfaces, draw the free-body diagrams and determine the tendency of relative slipping before choosing a formula. Physics Wallah’s questions will help you practise limiting-friction conditions, inclined-plane motion and multi-block systems while improving your ability to recognise when static friction is sufficient and when sliding begins.

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