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7.3MechanicsStretch

Friction and Static Particles

Combining friction with equilibrium lets you analyse a particle on the point of sliding — often on a rough inclined plane. The limiting friction F = μR closes the system of equations.

26 min Applications of Forces
Edexcel A Level Maths: 7.3 Friction and Static ParticlesCurated by Kepler, not produced by us. Watch it before the notes below.
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What you'll be able to do

  • Combine friction with equilibrium
  • Analyse a particle on a rough slope
  • Use F = μR at the point of slipping
  • Find the limiting angle or force
1

On a rough slope

For a particle in limiting equilibrium on a rough slope: resolve perpendicular to find R=mgcos⁡θR = mg\cos\theta, then up the slope friction balances mgsin⁡θmg\sin\theta, with F=μRF = \mu R at the point of slipping.

mgsin⁡θ=μ mgcos⁡θ ⇒ μ=tan⁡θmg\sin\theta = \mu\, mg\cos\theta \ \Rightarrow\ \mu = \tan\theta
At the point of sliding on a slope.
Example 1A particle is about to slide on a rough slope at angle θ. Show μ = tan θ.
1Perpendicular: R=mgcos⁡θR = mg\cos\theta.
2Along: mgsin⁡θ=μR=μmgcos⁡θmg\sin\theta = \mu R = \mu mg\cos\theta.
3Divide: μ=tan⁡θ\mu = \tan\theta.
Answer
μ=tan⁡θ\mu = \tan\theta

Tip — On the point of sliding down a slope, μ = tan θ where θ is the slope angle.

Formula recap

R=mgcos⁡θR = mg\cos\theta
Normal reaction on a slope.
μ=tan⁡θ\mu = \tan\theta
At the point of sliding on a slope.

Common mistakes to avoid

Using F = μR before the particle is at the point of slipping.
F = μR only in limiting equilibrium; otherwise F ≤ μR.
Using mg for the normal reaction on a slope.
R = mg cos θ on an inclined plane.

Key takeaways

  • Combine friction (F = μR) with equilibrium.
  • On a rough slope R = mg cos θ.
  • At the point of sliding, μ = tan θ.

Test yourself

Answer a few questions, then review anything you missed.

Prefer paper? Open the written self-check

Answer each question before opening its answer. This self-check does not record a quiz score.

  1. At the point of slipping, friction equals:

    • μR\mu R
    • RR
    • 00
    • mgmg
    Show answer and explanation

    Answer: μR\mu R

    Limiting friction.

  2. On a slope the normal reaction is:

    • mgcos⁡θmg\cos\theta
    • mgmg
    • mgsin⁡θmg\sin\theta
    • μmg\mu mg
    Show answer and explanation

    Answer: mgcos⁡θmg\cos\theta

    Perpendicular component.

  3. On the point of sliding down a slope at θ:

    • μ=tan⁡θ\mu = \tan\theta
    • μ=sin⁡θ\mu = \sin\theta
    • μ=cos⁡θ\mu = \cos\theta
    • μ=1\mu = 1
    Show answer and explanation

    Answer: μ=tan⁡θ\mu = \tan\theta

    sin/cos = tan.

  4. The down-slope component of weight is:

    • mgsin⁡θmg\sin\theta
    • mgcos⁡θmg\cos\theta
    • mgmg
    • μR\mu R
    Show answer and explanation

    Answer: mgsin⁡θmg\sin\theta

    Along the slope.

  5. Before slipping, friction satisfies:

    • F≤μRF \le \mu R
    • F=μRF = \mu R
    • F>μRF > \mu R
    • F=0F = 0
    Show answer and explanation

    Answer: F≤μRF \le \mu R

    Adjusts to hold equilibrium.

Next step

Check your understanding

Work through questions on Friction and Static Particles. Use the feedback to decide what needs another look.

2 past-paper questions · 3 marks

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