Topic 2 of 8 · Form 5
Forces, equilibrium and friction
Resolve forces into perpendicular components (usually horizontal and vertical, or parallel and perpendicular to a slope). For equilibrium, the components in each direction sum to zero. Lami's theorem: for three forces in equilibrium, P/sin α = Q/sin β = R/sin γ (angles opposite each force).
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Resolving forces
Resolve forces into perpendicular components (usually horizontal and vertical, or parallel and perpendicular to a slope). For equilibrium, the components in each direction sum to zero. Lami's theorem: for three forces in equilibrium, P/sin α = Q/sin β = R/sin γ (angles opposite each force).
Weight, normal reaction and tension
Weight W = mg acts downwards; the normal reaction R acts perpendicular to the surface; tension acts along strings.
Friction
Limiting friction F = μR when the body is on the point of moving (or moving); otherwise F ≤ μR.
Worked example: A 5 kg block rests on a rough slope inclined at 30°. It is on the point of slipping. Find μ (g = 9.8 m s⁻²).
Perpendicular: R = 5g cos 30° = 42.4 N
Parallel: F = 5g sin 30° = 24.5 N
μ = F/R = tan 30° = 0.577
Angle of friction
The angle λ with tan λ = μ; a body on a slope slips when the slope angle exceeds λ.
Worked example: A horizontal force P keeps a 10 kg box in equilibrium on a smooth slope at 20°. Find P.
Resolve along the slope: P cos 20° = 10g sin 20° → P = 10g tan 20° = 35.7 N
Check yourself
- State the condition for limiting friction.
- Three forces of 3 N, 4 N and 5 N are in equilibrium. What angle is between the 3 N and 4 N forces?
- A 2 kg mass rests on a horizontal rough table with μ = 0.4. Find the maximum friction force (g = 9.8).
Show answers
- Friction equals μR when the body is on the point of moving.
- 90° (3–4–5 triangle of forces)
- 0.4 × 2 × 9.8 = 7.84 N
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