Friction for JEE Main

Friction questions look like Newton's laws questions with one extra force, and that extra force is where the marks go. Static friction does not have a fixed value. It adjusts itself, up to a limit, and the most common error in this chapter is writing μN for a friction force that never reached its limit.

This guide explains how friction really behaves, gives a check to run before every problem, and works through inclines, minimum-force problems and the block-on-block questions that JEE Main likes.

Static, limiting and kinetic friction

Static friction acts when surfaces are in contact but not sliding over each other. It takes whatever value is needed to prevent sliding, from zero up to a maximum:

f_s ≤ μₛN

The maximum value, μₛN, is called limiting friction.

Kinetic friction acts once sliding begins. It has a nearly constant value:

f_k = μₖN

For most surface pairs, μₖ is a little less than μₛ, which is why it is harder to start a heavy box moving than to keep it moving.

Friction (in this simple model) does not depend on the area of contact, and it acts along the surface, opposing relative motion or the tendency of relative motion between the two surfaces. That is not always opposite to the body's motion: when you walk, friction on your foot points forward.

Contact force. The surface exerts both N and f. Their resultant has magnitude √(N² + f²).

Angle of friction and angle of repose

• Angle of friction λ: the angle between the contact force and the normal when friction is limiting. tan λ = μₛ.
• Angle of repose: the steepest incline on which a block stays at rest. At that angle, mg sin θ = μₛ mg cos θ, so tan θ = μₛ.

The two angles are equal.

The "is there motion?" check

Before using μ anywhere, run this check:

1. Find the force that is trying to cause sliding (the applied force, or mg sin θ on a slope).
2. Find the limiting friction μₛN.
3. If the driving force ≤ μₛN, there is no sliding. Friction equals the driving force, not μₛN.
4. If the driving force > μₛN, the body slides and friction = μₖN.

Rough inclines

For a block on a rough incline of angle θ, N = mg cos θ.

The minimum pulling force

To drag a block on a rough floor, pulling at an angle φ above the horizontal lifts part of the weight off the floor, which lowers N and so lowers friction.

Horizontal balance: F cos φ = μ(mg − F sin φ), so F = μmg / (cos φ + μ sin φ).

The denominator is largest when tan φ = μ, giving F_min = μmg / √(1 + μ²).

Blocks on blocks

For two stacked blocks, ask first: do they move together?

1. Assume they move together and find the common acceleration.
2. Find the friction the top block needs for that acceleration.
3. Compare it with the limiting friction between the blocks. If needed ≤ limit, they move together. If not, they slip, and kinetic friction acts on each in opposite directions.

Worked problems

Problem 1: up and back down an incline. A block is sent up a rough 37° incline at 10 m/s. μ = 0.5 (take μₛ = μₖ), g = 10 m/s², sin 37° = 0.6, cos 37° = 0.8. How far up does it go, and what happens next?

Going up: a = 10(0.6 + 0.5 × 0.8) = 10 m/s², opposing the motion.
Distance = v²/2a = 100/20 = 5 m.
At the top, tan 37° = 0.75 > μ = 0.5, so it cannot stay at rest. It slides back with a = 10(0.6 − 0.4) = 2 m/s².

Problem 2: is there motion? A 2 kg block rests on a 30° incline with μₛ = 0.7 (g = 10 m/s²). Find the friction force.

Driving force = mg sin 30° = 20 × 0.5 = 10 N.
Limiting friction = 0.7 × 20 × cos 30° ≈ 0.7 × 17.3 ≈ 12.1 N.
10 N < 12.1 N, so the block stays at rest and friction = 10 N, up the slope.

Problem 3 (numerical answer): minimum pull. A 10 kg block lies on a floor with μ = 0.75 (g = 10 m/s²). Find the least force that can move it, and its direction.

tan φ = 0.75, so φ = 37° above the horizontal.
F_min = 0.75 × 100 / √(1 + 0.5625) = 75 / 1.25 = 60 N.
Check: F cos φ = 60 × 0.8 = 48 N; N = 100 − 60 × 0.6 = 64 N; μN = 0.75 × 64 = 48 N. It balances. (Pulling horizontally would need 75 N.)

Problem 4: block on block. A 1 kg block sits on a 4 kg block on a smooth floor. μ between the blocks is 0.4 (static and kinetic), g = 10 m/s². A horizontal force F acts on the lower block. Find the largest F for which they move together, and the accelerations when F = 30 N.

Top block's maximum acceleration = μg = 4 m/s². Moving together at that acceleration needs F = (1 + 4) × 4 = 20 N.
At F = 30 N they slip. Friction on each = 0.4 × 1 × 10 = 4 N.
Top block: a = 4/1 = 4 m/s² forward. Lower block: a = (30 − 4)/4 = 6.5 m/s².
(At F = 10 N, both move at 2 m/s², and friction on the top block is only 1 × 2 = 2 N.)

Practice set

1. μₛ = 1/√3. What is the angle of repose?
2. A 5 kg block on a floor with μₛ = 0.3 is pulled horizontally with 10 N (g = 10 m/s²). What is the friction force?
3. The same block is pulled with 20 N, and μₖ = 0.2. Find its acceleration.
4. A block sliding on a floor at 10 m/s comes to rest under friction with μₖ = 0.5 (g = 10 m/s²). How long does it take, and how far does it go?
5. What is the least horizontal force that holds a 2 kg block against a vertical wall if μₛ = 0.5 (g = 10 m/s²)?
6. A block slides down a 30° incline at constant velocity. What is μₖ?
7. A block takes twice as long to slide down a rough 45° incline as down a smooth one of the same length. Find μₖ.
8. A car at 20 m/s brakes hard on a road with μ = 0.5 (g = 10 m/s²). What is the shortest stopping distance without skidding?

Answers

1. tan θ = 1/√3, so 30°.
2. Limiting friction = 15 N > 10 N, so no motion; friction = 10 N.
3. 20 N > 15 N, so it slides. a = (20 − 0.2 × 50)/5 = 2 m/s².
4. Deceleration = μg = 5 m/s²; time = 2 s; distance = 100/10 = 10 m.
5. Friction must hold the weight: μF = mg, so F = 20/0.5 = 40 N.
6. μₖ = tan 30° = 1/√3 ≈ 0.58.
7. Time ∝ 1/√a. So a_smooth / a_rough = 4, giving sin 45° / (sin 45° − μ cos 45°) = 4, so 1/(1 − μ) = 4 and μ = 0.75.
8. Maximum deceleration = μg = 5 m/s²; distance = 400/10 = 40 m.

What to do next

• Run the "is there motion?" check in writing on every friction problem for the next week.
• Solve 20 incline problems, including ones where the block stays at rest.
• Do ten block-on-block problems using the μg shortcut, then check with full FBDs.
• Revise Newton's laws if FBDs still take long, and continue to circular motion, where friction supplies the centripetal force on a flat curve.