PhysicsISC Class 11 · Laws of Motion

Friction and Motion on an Incline: Static, Kinetic and Connected Bodies

Updated 17 October 2026 · Fact-checked

Friction opposes relative motion between surfaces. Static friction adjusts up to a limit fₛ ≤ μₛN, and kinetic friction is fₖ = μₖN. On a rough incline of angle θ, a block slides down with a = g(sinθ − μ cosθ); the angle of repose α satisfies tanα = μₛ.

Split mg into mg sinθ along the slope and mg cosθ into it; N balances mg cosθ, and friction μN opposes the slide.

Illustration drawn for QuestionBuddy.

Understand Friction and Motion on an Incline

Static and kinetic friction. Static friction is a self-adjusting force equal to the applied force until the limiting value fₛ(max) = μₛN is reached. Once sliding begins, kinetic friction fₖ = μₖN acts, and μₖ is usually a little less than μₛ. Friction does not depend on the area of contact (for the ISC level), but depends on the nature of the surfaces and on the normal reaction N.

Always find N first. On a level floor N = mg only if there is no vertical force. If a rope pulls at an upward angle φ, N = mg − T sinφ, which lowers the friction. If it pushes downwards, N = mg + P sinφ.

Angles. The angle of friction λ satisfies tanλ = μ; it is the angle between the normal reaction and the resultant of N and friction. The angle of repose α is the inclination at which a body just begins to slide: tanα = μₛ.

Block on an incline. Along the incline: down the plane the pull is mg sinθ; the normal reaction is N = mg cosθ. Sliding down: a = g(sinθ − μₖ cosθ). Pushed or pulled up the plane at constant speed: P = mg(sinθ + μₖ cosθ). If the block is at rest, friction equals mg sinθ and is not μN.

Connected bodies. Draw a free-body diagram for each body. For a body of mass m on a rough table joined over a smooth pulley to a hanging mass M: Mg − T = Ma and T − μmg = ma. Adding gives a = (M − μm)g/(M + m).

Friction and Motion on an Incline: formulas and key facts

Friction laws: $f_{s} \leq μ_{s} N ∣ f_{k} = μ_{k} N$; N is the normal reaction.
Angles: $t anλ = μ ∣ t an α = μ_{s}$; λ = angle of friction, α = angle of repose.
Block on incline: $a = g (s in θ - μ cos θ) ∣ P = m g (s in θ + μ cos θ)$; Sliding down, and pushed up at constant speed.
Table and hanging mass: $a = (M - μ m) g / (M + m)$; Smooth pulley, light string.

Friction and Motion on an Incline 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.

Pulling a trunk with a rope

Suppose a 20 kg trunk lies on a floor with μ = 0.5 and a student pulls it with a rope at 37° above the horizontal (take sin37° = 0.6, cos37° = 0.8, g = 10 m s⁻²).

Find the smallest pull that just moves the trunk, and compare it with a horizontal pull.

Try it first, then show the solutionSolution

N = mg − T sin37° = 200 − 0.6T.
T cos37° = μN → 0.8T = 0.5(200 − 0.6T) = 100 − 0.3T.
1.1T = 100 → T = 90.9 N. A horizontal pull needs 0.5 × 200 = 100 N.

Answer: About 91 N, less than the 100 N for a horizontal pull.

Memory hook: An upward component of the pull reduces N and hence the friction.

Tricks to remember Friction and Motion on an Incline

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

N first, f second
Write the normal reaction from the vertical balance before using μ.
Use it when: Pulled, pushed and inclined blocks.
At rest, friction ≠ μN
If the body is not about to slide, friction equals the force it balances.
Use it when: Questions asking for friction on a block that stays at rest.

Friction and Motion on an Incline: solved examples

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

Example 1: Limiting friction · 1 mark · MCQBoard level

A 5 kg block rests on a horizontal surface with μₛ = 0.4 (g = 10 m s⁻²). The maximum static friction is:
(a) 2 N
(b) 12.5 N
(c) 20 N
(d) 50 N

Try it first, then show the solutionSolution

fₛ(max) = μₛmg = 0.4 × 5 × 10 = 20 N.

Answer: (c) 20 N

Example 2: Pushing a block · 3 marks · Short answerBoard level

A 10 kg block on a horizontal floor has μₛ = 0.25 and μₖ = 0.2. A horizontal force of 30 N is applied (g = 10 m s⁻²). Does it move? Find the acceleration and the speed after 4 s.

Try it first, then show the solutionSolution

Limiting friction = 0.25 × 100 = 25 N < 30 N, so the block moves. (1 mark)
fₖ = 0.2 × 100 = 20 N; a = (30 − 20)/10 = 1 m s⁻². (1 mark)
v = at = 1 × 4 = 4 m s⁻¹. (1 mark)

Answer: It moves; a = 1 m s⁻²; v = 4 m s⁻¹.

Example 3: Sliding down a rough incline · 4 marks · Short answerHarder board level

A block slides from rest down a 30° incline with μₖ = 0.2 (g = 10 m s⁻²). Find its acceleration and the time to slide 4 m.

Try it first, then show the solutionSolution

a = g(sinθ − μ cosθ) = 10(0.5 − 0.2 × 0.866) = 10 × 0.3268 = 3.27 m s⁻². (2 marks)
s = ½at² → t² = 2 × 4/3.268 = 2.448. (1 mark)
t = 1.56 s. (1 mark)

Answer: a = 3.27 m s⁻², t = 1.56 s.

Example 4: Pulling with a rope · 4 marks · Short answerHarder board level

A 20 kg block on a floor with μ = 0.5 is pulled by a rope at 37° above the horizontal. Find the least tension that moves the block, and the acceleration if T = 120 N (sin37° = 0.6, cos37° = 0.8, g = 10 m s⁻²; take μₖ = μₛ).

Try it first, then show the solutionSolution

N = 200 − 0.6T. Just moving: 0.8T = 0.5(200 − 0.6T) → 1.1T = 100 → T = 90.9 N. (2 marks)
For T = 120 N: N = 200 − 72 = 128 N; friction = 64 N. (1 mark)
Net force = 120 × 0.8 − 64 = 32 N; a = 32/20 = 1.6 m s⁻². (1 mark)

Answer: T = 90.9 N; a = 1.6 m s⁻².

Example 5: Block on a table with a hanging mass · 5 marks · Long answerHOTS / competency

A 4 kg block on a rough table (μ = 0.25) is joined by a string over a smooth light pulley to a 2 kg mass hanging freely (g = 10 m s⁻²). Find the acceleration, the tension, and the smallest hanging mass that would start the motion.

Try it first, then show the solutionSolution

Hanging mass: 20 − T = 2a. Table mass: T − 0.25 × 40 = 4a, that is T − 10 = 4a. (1 mark)
Adding: 10 = 6a → a = 5/3 = 1.67 m s⁻². (1 mark)
T = 10 + 4 × 5/3 = 16.7 N; check: 20 − 16.7 = 3.3 = 2 × 1.67. (1 mark)
Motion just starts when Mg = μmg: M = 0.25 × 4 = 1 kg. (2 marks)

Answer: a = 1.67 m s⁻², T = 16.7 N, minimum hanging mass 1 kg.

Where students lose marks in Friction and Motion on an Incline

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

Mistake: Using N = mg on an incline
Writing friction as μmg for a block on a slope.
Do this: N = mg cosθ on an incline; draw the force components.
Usually costs: 1 to 2 marks.
Mistake: Ignoring the vertical component of a pull
Using N = mg while the rope is inclined upwards.
Do this: N = mg − T sinφ for an upward-inclined pull.
Usually costs: 2 marks.
Mistake: Taking friction as μN for a block at rest
Using μN when the block is not about to slide.
Do this: Friction at rest equals the force it balances; μN is only the limit.
Usually costs: 1 to 2 marks.

Exam tips

Draw a free-body diagram and mark mg, N, friction and the applied force.
Resolve forces along and perpendicular to the plane before writing equations.
State whether you take μₛ = μₖ when the problem gives only one value.
For connected bodies, write one equation per body and add them.

Practice questions on Friction and Motion on an Incline

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 · MCQBoard level
The coefficient of friction between a block and an incline is 1/√3. The angle of repose is:
(a) 30°
(b) 45°
(c) 60°
(d) 90°
Hint
tanα = μ.
Answer
(a) 30°
Q2 · 2 marks · Short answerBoard level
A force of 50 N keeps a 20 kg box moving at constant speed on a horizontal floor (g = 10 m s⁻²). Find μₖ.
Hint
At constant speed, applied force = kinetic friction.
Answer
μₖ = 50/(20 × 10) = 0.25.
Q3 · 3 marks · Short answerHarder board level
A 5 kg block rests on a 30° incline with μₛ = 0.7 (g = 10 m s⁻²). Find the friction acting on it. Is it μₛN?
Hint
The block is at rest; compare mg sinθ with the limiting value.
Answer
mg sin30° = 25 N. N = 50 cos30° = 43.3 N, limiting friction = 30.3 N > 25 N, so the block rests and friction = 25 N, not μₛN.
Q4 · 3 marks · Short answerHarder board level
A car travelling at 20 m s⁻¹ on a level road (μ = 0.6 between tyres and road) is braked to a stop with the maximum retardation friction can give, without skidding. Find the shortest stopping distance (g = 10 m s⁻²).
Hint
Maximum retardation = μg.
Answer
a = 6 m s⁻²; s = u²/2a = 400/12 = 33.3 m.
Q5 · 3 marks · Short answerHarder board level
A 2 kg block is pushed up a 37° incline by a force parallel to the incline at constant speed; μₖ = 0.5 (sin37° = 0.6, cos37° = 0.8, g = 10 m s⁻²). Find the force.
Hint
P = mg(sinθ + μ cosθ).
Answer
P = 20(0.6 + 0.5 × 0.8) = 20 × 1.0 = 20 N.
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a 10 kg box rests on a truck bed (μₛ = 0.4, g = 10 m s⁻²).
(i) Find the greatest acceleration of the truck for which the box does not slip. (1 mark)
(ii) Find the friction on the box when the truck accelerates at 3 m s⁻². (1 mark)
(iii) The truck, travelling at 20 m s⁻¹, brakes with 5 m s⁻²; take μₖ = 0.4. Find the box's retardation. (1 mark)
(iv) How far does the box slide forward relative to the truck? (1 mark)
Hint
Friction is the only horizontal force on the box.
Answer
(i) μg = 4 m s⁻². (ii) 10 × 3 = 30 N. (iii) 4 m s⁻² (friction cannot supply 5). (iv) The truck stops in 20²/10 = 40 m; the box in 20²/8 = 50 m; relative slide = 10 m.

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.Define the angle of repose.Show answer
The inclination at which a body just starts sliding down; tanα = μₛ.
2.Does friction depend on contact area?Show answer
Not at this level; it depends on N and the surfaces.
3.Acceleration down a rough incline?Show answer
g(sinθ − μ cosθ).
4.Static or kinetic friction is larger (limiting)?Show answer
Limiting static friction is usually larger.

Friction and Motion on an Incline: frequently asked questions

Why is kinetic friction less than limiting static friction?

Once surfaces slide, the microscopic bonds between them have less time to form, so the resisting force is a little less. This is why it is harder to start a heavy box moving than to keep it moving.

Can the friction on a body be forward?

Yes. For a person walking, static friction on the foot is forward; friction on a box on an accelerating truck is also forward. Friction opposes relative sliding, not motion as such.

Is the angle of repose related to the angle of friction?

Yes, for limiting friction they are equal: tanα = μₛ = tanλ.