A-Level — mathsStandard2 min read

Friction

Apply friction force models at A-Level. Use F ≤ μR and solve problems on rough surfaces.

Tutor AI Team
Tutor AI Team

Friction opposes motion (or tendency to move). A-Level uses the model F≤μRF \leq \mu R where μ\mu is the coefficient of friction and RR is the normal reaction.

Core Concepts

Friction Law

Fmax⁡=μRF_{\max} = \mu R

  • Static: F≤μRF \leq \mu R (at rest or about to move).
  • Kinetic: F=μRF = \mu R (in motion, limiting friction).

On a Horizontal Surface

R=mgR = mg. Fmax⁡=μmgF_{\max} = \mu mg.

On an Inclined Plane

R=mgcos⁡θR = mg\cos\theta. Fmax⁡=μmgcos⁡θF_{\max} = \mu mg\cos\theta.

Component of weight down slope: mgsin⁡θmg\sin\theta.

About to slide: mgsin⁡θ=μmgcos⁡θmg\sin\theta = \mu mg\cos\theta → μ=tan⁡θ\mu = \tan\theta.

Worked Example: Example 1

Problem

3 kg on rough horizontal surface, μ=0.4\mu = 0.4. Force 20 N applied horizontally.

R=3g=29.4R = 3g = 29.4 N. F=0.4×29.4=11.76F = 0.4 \times 29.4 = 11.76 N.

Resultant = 20−11.76=8.2420 - 11.76 = 8.24 N. a=8.243=2.75a = \frac{8.24}{3} = 2.75 m/s².

Solution

Worked Example: Example 2

Problem

Box on rough slope at 25°, μ=0.3\mu = 0.3. Acceleration down slope:

Force down = mgsin⁡25°−μmgcos⁡25°mg\sin 25° - \mu mg\cos 25°. a=g(sin⁡25°−0.3cos⁡25°)=9.8(0.423−0.272)=1.48a = g(\sin 25° - 0.3\cos 25°) = 9.8(0.423 - 0.272) = 1.48 m/s².

Solution

Practice Problems

    1. 5 kg block on rough surface (μ=0.25\mu = 0.25). Minimum force to start moving.
    1. At what angle does a block start to slide if μ=0.5\mu = 0.5?

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Key Takeaways

  • ✓

    Fmax⁡=μRF_{\max} = \mu R.

  • ✓

    Friction opposes the direction of (potential) motion.

  • ✓

    On a slope: μ=tan⁡θ\mu = \tan\theta at the angle of friction.

  • ✓

    Find RR first (perpendicular equilibrium), then use friction law.

Tags:
#a-level#maths#mechanics#friction

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