PhysicsCBSE Class 11 · Laws of Motion

Friction on Horizontal Surfaces and Inclines Class 11

Q: Why is it harder to start a body than to keep it moving?

Because limiting static friction is larger than kinetic friction.

Q: Is friction always a hindrance?

No. Walking, braking and the grip of tyres need friction.

Q: What does the normal reaction equal on a horizontal surface?

mg only if there is no vertical applied force.

Q: Why does the angle of repose not depend on mass?

Both the sliding component mg sin θ and the friction μmg cos θ are proportional to m, so m cancels.

Q: What is the direction of kinetic friction?

Opposite to the relative velocity of the surfaces.

Updated 17 October 2026 · Fact-checked

Static friction adjusts to match the applied force up to a limit fs(max) = μs N; once sliding starts, kinetic friction fk = μk N (with μk < μs) acts opposite to the motion. On an incline of angle θ the block just slips when tan θ = μs (angle of repose), moves up with deceleration g(sin θ + μ cos θ) and down with acceleration g(sin θ − μ cos θ).

At rest friction f equals mg sin θ; the block just slips when tan θ = μs, the angle of repose.

Illustration drawn for QuestionBuddy.

Understand Friction on Horizontal Surfaces and Inclines

Types of friction. Friction opposes relative motion (or its tendency) between surfaces in contact.

*Static friction* $f_{s}$ matches the applied force and can take any value up to the limiting friction $f_{s}^{ma x} = μ_{s} N$ .
*Kinetic friction* $f_{k} = μ_{k} N$ acts once sliding begins; $μ_{k} < μ_{s}$ for most pairs, and $f_{k}$ does not depend on the speed or the contact area (at this level).
*Rolling friction* is much smaller than sliding friction.

Here $N$ is the normal reaction, which is not always $m g$ : for a pull at angle $θ$ above the horizontal, $N = m g - F sin θ$ .

Angles. The angle of friction $λ$ satisfies $tan λ = μ_{s}$ (the angle between $N$ and the resultant of $N$ and $f_{s}^{ma x}$ ). The angle of repose $θ$ is the incline angle at which a block just starts to slide: $tan θ = μ_{s}$ , so $θ = λ$ .

**Incline of angle $θ$ .** Along the plane, $m g sin θ$ drives the block down; $N = m g cos θ$ .

Moving down with kinetic friction: $a = g (sin θ - μ_{k} cos θ)$ .
Moving up (friction acts down the plane): deceleration $= g (sin θ + μ_{k} cos θ)$ .
At rest on the incline while it is slowly raised, $f_{s} = m g sin θ$ increases until it reaches $μ_{s} m g cos θ$ ; then the block slips and friction drops to $μ_{k} m g cos θ$ .

Applied force at an angle. To drag a body at constant speed, $F cos θ = μ_{k} (m g - F sin θ)$ , so $F = \frac{μ _{k} m g}{cos θ + μ _{k} sin θ}$ . This is smallest when $tan θ = μ_{k}$ .

Block on block. If a force is applied to one block of a stack and no slipping is allowed, the friction between them supplies the acceleration of the other block; the largest allowed common acceleration is limited by $μ_{s}$ between them. Draw a free-body diagram for each block and equate accelerations.

Friction on Horizontal Surfaces and Inclines: formulas and key facts

Limiting friction: $f_{s}^{ma x} = μ_{s} N$; Static.
Kinetic friction: $f_{k} = μ_{k} N$; Constant during sliding.
Angle of repose: $tan θ = μ_{s}$; Same as the angle of friction.
Down a rough incline: $a = g (sin θ - μ_{k} cos θ)$; Sliding down.
Up a rough incline: $a = - g (sin θ + μ_{k} cos θ)$; Deceleration.
Pull at angle to the horizontal: $F = \frac{μ _{k} m g}{cos θ + μ _{k} sin θ}$; Constant velocity.

How to solve Friction on Horizontal Surfaces and Inclines questions

Always draw the free-body diagram, find N first, and decide whether the body is at rest (static friction) or sliding (kinetic friction).

1Draw all forces: weight, normal reaction, applied force and friction.
2Resolve along and perpendicular to the surface; the perpendicular equation gives N.
3Compare the applied force with μs N to decide whether the body moves.
4If it moves, use f = μk N and F_net = ma along the surface.
5On an incline, friction acts down the plane for upward motion and up the plane for downward motion.
6For stacked blocks, find the largest common acceleration allowed by friction, then the force.

Friction on Horizontal Surfaces and Inclines in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Pushing a wardrobe

Suppose a 50 kg wardrobe stands on a floor with μs = 0.5 and μk = 0.4 (g = 10 m/s²).

What push starts it moving, and what push keeps it moving steadily?

Try it first, then show the solutionSolution

N = 500 N; limiting friction = 0.5 × 500 = 250 N.
Kinetic friction = 0.4 × 500 = 200 N, so 200 N keeps it moving at constant speed.

Answer: 250 N to start, 200 N to keep it moving.

Memory hook: It is easier to keep something moving than to start it.

A parked scooter on a slope

Suppose a scooter parked with its brakes locked stays at rest on a ramp until the ramp angle exceeds 22° (tan 22° ≈ 0.40).

Estimate the coefficient of static friction between the tyres and the ramp.

Try it first, then show the solutionSolution

Slipping begins at the angle of repose.
μs = tan θ ≈ 0.40.

Answer: About 0.40.

Memory hook: Repose angle gives μs directly, no mass needed.

Tricks to remember Friction on Horizontal Surfaces and Inclines

Short memory aids. Say each one aloud once, then use it on a question the same day.

Start vs keep going
μs > μk, so starting needs more force than sustaining.
Use it when: Static versus kinetic questions.
Tan of repose
μs = tan θ; mass does not matter.
Use it when: Incline-angle questions.
Up adds, down subtracts
Up the incline: sin + μ cos. Down the incline: sin − μ cos.
Use it when: Choosing the sign.
Pull reduces N
Pulling upward at an angle reduces N; pushing downward increases it.
Use it when: Angled-force problems.

Friction on Horizontal Surfaces and Inclines: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Static versus kinetic friction · 3 marks · Short answerBoard level

A 20 kg box rests on a floor with μs = 0.4 and μk = 0.3 (g = 10 m/s²). Find (a) the limiting friction, (b) the friction when it is pushed with 70 N, (c) the acceleration when pushed with 100 N.

Try it first, then show the solutionSolution

(a) N = 200 N; f(max) = 0.4 × 200 = 80 N. (1)
(b) 70 N < 80 N: the box stays at rest and friction = 70 N. (1)
(c) Sliding: fk = 0.3 × 200 = 60 N; a = (100 − 60)/20 = 2 m/s². (1)

Answer: (a) 80 N; (b) 70 N, at rest; (c) 2 m/s².

Example 2: Angle of repose · 2 marks · Very short answerBoard level

A block just starts to slide when the incline is raised to 30°. Find μs and the force of friction on a 10 kg block at that instant (g = 10 m/s²).

Try it first, then show the solutionSolution

μs = tan 30° = 0.577 (1/√3). (1)
f = mg sin 30° = 10 × 10 × 0.5 = 50 N, which also equals μs mg cos 30°. (1)

Answer: μs = 0.58; f = 50 N.

Example 3: Up and down a rough incline · 4 marks · Long answerHarder board level

A 4 kg block is projected up a rough incline of angle 37° (sin 37° = 0.6, cos 37° = 0.8) with speed 10 m/s. The coefficient of kinetic friction is 0.5 (g = 10 m/s²). Find (a) how far up it goes, (b) its acceleration on the way down (assuming it slides back) and (c) its speed on returning to the start.

Try it first, then show the solutionSolution

(a) Deceleration = g(sin θ + μ cos θ) = 10(0.6 + 0.4) = 10 m/s². (1)
Distance = v²/2a = 100/20 = 5 m. (1)
(b) Down: a = g(sin θ − μ cos θ) = 10(0.6 − 0.4) = 2 m/s² (it slides back because tan 37° = 0.75 exceeds the coefficient of friction). (1)
(c) v² = 2 × 2 × 5 = 20, so v ≈ 4.5 m/s. (1)

Answer: (a) 5 m; (b) 2 m/s²; (c) about 4.5 m/s.

Example 4: Pulling at an angle · 3 marks · Short answerHarder board level

A 10 kg block on a horizontal floor (μk = 0.5) is pulled at constant velocity by a rope at 37° above the horizontal (sin 37° = 0.6, cos 37° = 0.8; g = 10 m/s²). Find the tension and compare it with the pull needed if the rope were horizontal.

Try it first, then show the solutionSolution

N = mg − T sin 37° = 100 − 0.6T. (1)
T cos 37° = μN: 0.8T = 0.5(100 − 0.6T), so 1.1T = 50 and T ≈ 45.5 N. (1)
A horizontal pull needs μmg = 50 N; the angled pull is smaller because it reduces N. (1)

Answer: About 45.5 N, less than 50 N.

Example 5: Block on block · 4 marks · Long answerHOTS / competency

A 2 kg block rests on a 4 kg block that lies on a frictionless floor. The coefficient of static friction between the blocks is 0.3 (g = 10 m/s²). Find the largest horizontal force that can be applied (a) to the lower block and (b) to the upper block without the upper block slipping.

Try it first, then show the solutionSolution

The friction on the upper block cannot exceed 0.3 × 2 × 10 = 6 N. (1)
(a) Force on the lower block: the upper block's maximum acceleration = 6/2 = 3 m/s², so F = (2 + 4) × 3 = 18 N. (1)
(b) Force on the upper block: the lower block's maximum acceleration = 6/4 = 1.5 m/s². (1)
F = (2 + 4) × 1.5 = 9 N. (1)

Answer: (a) 18 N; (b) 9 N.

Where students lose marks in Friction on Horizontal Surfaces and Inclines

The slips examiners see most often, and what to write instead.

Mistake: Taking N = mg when the force has a vertical component.
Not resolving perpendicular to the surface.
Do this: Write the perpendicular equation first and solve for N.
Usually costs: The whole numerical.
Mistake: Using μk for a block at rest.
Confusing static and kinetic friction.
Do this: If the block does not move, f equals the applied force (up to μsN).
Usually costs: 1 to 2 marks.
Mistake: Wrong direction of friction on an upward-moving block.
Drawing it up the plane.
Do this: Friction opposes motion: down the plane for upward motion.
Usually costs: Sign error in a.
Mistake: Writing the angle of repose as sin θ = μ.
Mixing up trigonometric ratios.
Do this: μs = tan θ.
Usually costs: Method mark.

Exam tips

Draw one free-body diagram per body and label N and friction directions.
State whether the body is static or sliding before choosing the friction formula.
For stacked blocks, write the condition for no slipping as a common acceleration.
Give the direction of friction in the answer for conceptual items.
A "just begins to slide" statement means limiting friction.

Practice questions on Friction on Horizontal Surfaces and Inclines

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQHarder board level
A block rests on an incline that is slowly tilted. Which statement about the friction on the block is correct as the angle increases from 0 until the block slips?
(a) It stays constant (b) It increases as mg sin θ up to the limit and then drops to the kinetic value (c) It decreases steadily (d) It is always μmg cos θ
Hint
Static friction balances mg sin θ.
Answer
(b).
Q2 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The force of static friction can have any value from zero up to μsN.
Reason (R): Static friction adjusts itself to the applied force until the limiting value is reached.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Self-adjusting force.
Answer
(a).
Q3 · 2 marks · Very short answerBoard level
A 5 kg block lies on a floor with μs = 0.5 (g = 10 m/s²). Find the minimum horizontal force that makes it just start to move.
Hint
Limiting friction.
Answer
f = 0.5 × 50 = 25 N.
Q4 · 3 marks · Short answerHarder board level
A block takes twice as long to slide down a rough 30° incline as it takes on a smooth incline of the same length. Find μk.
Hint
t ∝ 1/√a, so a_rough = a_smooth/4.
Answer
a_smooth = 5 m/s²; a_rough = 1.25 m/s²; 5 − 10 μ cos 30° = 1.25, so μ = 3.75/8.66 = 0.43.
Q5 · 4 marks · Long answerHOTS / competency
A box rests on the flat bed of a truck travelling at 72 km/h; μs = 0.4 and μk = 0.3 between box and bed (g = 10 m/s²). (a) Find the shortest distance in which the truck can be stopped without the box slipping. (b) If the truck instead stops with a uniform deceleration of 10 m/s², find how far the box slides relative to the bed before the truck stops.
Hint
Maximum deceleration without slipping = μs g.
Answer
(a) a(max) = 0.4 × 10 = 4 m/s²; d = 400/(2 × 4) = 50 m. (b) The truck stops in 2 s, moving 20 m. The box decelerates at μk g = 3 m/s² for 2 s, covering 20 × 2 − ½ × 3 × 4 = 34 m. Relative slide = 34 − 20 = 14 m (the box is still moving at 14 m/s).

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Limiting friction?Show answer
μsN.
2.Kinetic friction?Show answer
μkN.
3.Angle of repose?Show answer
tan θ = μs.
4.Acceleration down a rough incline?Show answer
g(sin θ − μ cos θ).
5.Which is larger, μs or μk?Show answer
μs.
6.Does friction depend on contact area?Show answer
No (at this level).

Friction on Horizontal Surfaces and Inclines: frequently asked questions

Why is it harder to start a body than to keep it moving?

Because limiting static friction is larger than kinetic friction.

Is friction always a hindrance?

No. Walking, braking and the grip of tyres need friction.

What does the normal reaction equal on a horizontal surface?

mg only if there is no vertical applied force.

Why does the angle of repose not depend on mass?

Both the sliding component mg sin θ and the friction μmg cos θ are proportional to m, so m cancels.

What is the direction of kinetic friction?

Opposite to the relative velocity of the surfaces.