Friction: Static, Kinetic, and the Angle of Repose

Friction is the most misunderstood force in JEE mechanics: static friction is adjustable, kinetic friction is (almost) constant, and the angle of repose decides whether a block slides. Learn the rules that make every friction problem mechanical.

Friction1 min · Free lecture

In this lesson

  • Distinguish static, kinetic and rolling friction and their laws
  • Apply f_s ≤ μ_s N and f_k = μ_k N correctly
  • Solve incline problems using the angle of repose

Friction opposes relative motion (or the tendency of it) between surfaces in contact. Static friction fsf_s adjusts itself to match the applied force up to a maximum fs,max⁡=μsNf_{s,\max} = \mu_s N; once motion starts, kinetic friction takes over at the roughly constant value fk=μkNf_k = \mu_k N, with μk<μs\mu_k < \mu_s.

fs≤μsN,fk=μkN,tan⁡θ0=μsf_s \le \mu_s N, \qquad f_k = \mu_k N, \qquad \tan\theta_0 = \mu_s

The three friction laws: static limit, kinetic value, angle of repose

  • f_s — static friction (adjustable) (N)
  • \mu_s — coefficient of static friction
  • \mu_k — coefficient of kinetic friction
  • N — normal reaction (N)
  • \theta_0 — angle of repose
Static friction is NOT always μ_s N. It is whatever it needs to be, up to that ceiling. Writing f_s = μ_s N for a block that is not about to slip is the #1 friction error in JEE papers.

On an incline of angle θ\theta, a block starts sliding exactly when the component of weight down the slope exceeds the static ceiling: mgsin⁡θ>μsmgcos⁡θmg\sin\theta > \mu_s mg\cos\theta, i.e. tan⁡θ>μs\tan\theta > \mu_s. The critical angle θ0=tan⁡−1μs\theta_0 = \tan^{-1}\mu_s is the angle of repose. Once sliding, the acceleration is a=g(sin⁡θ−μkcos⁡θ)a = g(\sin\theta - \mu_k \cos\theta).

a=g (sin⁡θ−μkcos⁡θ)a = g\,(\sin\theta - \mu_k \cos\theta)

Acceleration of a block sliding down an incline

  • a — acceleration down the incline (m/s^2)
  • \theta — incline angle (rad)

Worked example

A 5 kg block rests on an incline at 30° with μ_s = 0.6, μ_k = 0.4 (g = 10 m/s²). (a) Does it slide? (b) If the incline is raised to 40°, find its acceleration.

  1. (a) tan 30° ≈ 0.577 < μ_s = 0.6 → static friction can hold it: it does NOT slide; f_s = mg sin 30° = 25 N
  2. (b) tan 40° ≈ 0.839 > 0.6 → it slides: a = g(sin 40° − μ_k cos 40°) = 10(0.643 − 0.4 × 0.766) ≈ 3.37 m/s²

Answer: (a) no, f_s = 25 N; (b) a ≈ 3.37 m/s²

Transcript (1 min)
WEBVTT

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Friction: Static, Kinetic, Angle of Repose
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f_s ≤ μ_s N · f_k = μ_k N
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f_s = μ_s N only at the slipping threshold
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tan θ₀ = μ_s — the sliding threshold
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a = g(sin θ − μ_k cos θ)
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Example: 30° → holds (f_s = 25 N); 40° → a ≈ 3.37 m/s²
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Recap: f_s ≤ μ_s N · f_k = μ_k N · tan θ₀ = μ_s
Printable notes (free)

Practice

Free · 4 questions with full solutions
  1. Q1 · Numerical · difficulty 1/5

    A block of mass 20 kg rests on a horizontal surface with coefficient of static friction 0.7. Taking g = 9.8 m/s², find the maximum horizontal force that can be applied without moving it.

  2. Q2 · Numerical · difficulty 2/5

    A block slides down an incline at 50° with coefficient of kinetic friction 0.25. Taking g = 10 m/s², find its acceleration down the incline.

  3. Q3 · Numerical · difficulty 2/5

    A block rests on an incline with coefficient of static friction 0.6. Find the angle of repose in degrees.

  4. Q4 · MCQ · difficulty 2/5

    A block is at rest on a horizontal surface with μ_s = 0.5. You push it with a horizontal force of 10 N and it does NOT move. The static friction acting on it is:

    • 10 N
    • greater than 10 N
    • 0 N
    • 5 N