Summary — What This Chapter Covers
- Identify every force on a particle with a clear force diagram: weight, normal reaction, tension, friction and any applied force.
- Forces are vectors: a force F at angle θ to a direction has component F cos θ along it and F sin θ perpendicular to it.
- A particle is in equilibrium when the sum of the components of the forces in any direction is zero.
- A contact force has a normal component R and a frictional component F, with F ≤ μR; in limiting equilibrium F = μR.
- Newton's third law: the force A exerts on B is equal and opposite to the force B exerts on A.
- Use g = 10 m s−2, so the weight of a mass m kg is 10m N.
1. Resolving Forces and Resultants
Components
Turning forces into components
The examiners want calculations, not scale drawings. Choose two perpendicular directions (usually horizontal and vertical, or parallel and perpendicular to a slope) and resolve every force into those directions. A force of magnitude F making an angle θ with your chosen direction contributes F cos θ along it and F sin θ at right angles to it.
The resultant is the single force with the same effect as all of them. Add the components in each direction to get X and Y; then the magnitude is √(X2 + Y2) and the direction is given by tan α = Y/X. Always say which direction your angle is measured from.
Resultant from components
X = ΣF cos θ, Y = ΣF sin θ R = √(X2 + Y2), tan α = Y/X
Trig results you are expected to know
sin(90° − θ) = cos θ, cos(90° − θ) = sin θ, tan θ = sin θ / cos θ, sin2 θ + cos2 θ = 1
Worked Example
Two horizontal forces act on a particle: 20 N in the direction of the positive x-axis and 15 N at 60° to the positive x-axis. Find the magnitude and direction of their resultant.
Practice Question 1
Forces of 12 N due east and 5 N due north act on a particle. Find the resultant.
2. Equilibrium of a Particle
Equilibrium
Resolve twice, set each to zero
If a particle is in equilibrium, the resultant is zero, so the components in every direction sum to zero. Two perpendicular directions give two equations, enough for two unknowns. Lami's theorem and the triangle of forces are acceptable alternatives, but resolving is what the mark scheme expects and works every time.
On a slope, resolve parallel and perpendicular to the plane: the weight mg then has components mg sin α down the slope and mg cos α into the slope. A smooth contact exerts only a normal reaction — it is a model that ignores friction, and it is a limitation when the real surfaces are rough.
Weight on a slope inclined at α
Down the slope: mg sin α Into the slope: mg cos α
Worked Example
A particle of weight 30 N hangs in equilibrium from two light strings. One string makes 30° with the horizontal, the other 60° with the horizontal, on opposite sides. Find the tensions T1 (30° string) and T2 (60° string).
3. Friction
Friction
F ≤ μR, with equality only at the point of slipping