Resultants of forces · Statics: forces, equilibrium & moments

A force is a vector: it has a size (in newtons, N) and a direction. So forces add like the vectors in vectors. The single force that has the same effect as several forces together is their resultant.

Two forces at an angle

Draw the two forces $P$ and $Q$ from the same point, with the angle $\theta$ between them. Complete the parallelogram: its diagonal is the resultant $R$ .

The parallelogram of forcesR² = P² + Q² + 2PQ cos θ

R^2 = P^2 + Q^2 + 2PQ\cos\theta

The angle $\alpha$ between $R$ and $P$ comes from $\tan\alpha = \dfrac{Q\sin\theta}{P + Q\cos\theta}$ , or from the sine rule once you know $R$ .

The resultant of two forcesSet P, Q and the angle

11.85 Nresultant R18.9°angle between R and P

P (N) = 8Q (N) = 5θ = 50°

R² = 8² + 5² + 2(8)(5) cos 50° = 140.42, so R = 11.85 N. sin α = 5 sin 50° ÷ 11.85, so α = 18.9°. Below 90°, cos θ is positive, so R is more than √(P² + Q²).

Worked example · NECO 2023

NECO 2023 · Paper 1 · Q38

Find the magnitude of the resultant of two forces of 5 N and 4 N acting at a point if the angle between them is $60^\circ$ .

Use the formula
- ${R^2 = 5^2 + 4^2 + 2(5)(4)\cos 60^\circ}$ .
- ${= 25 + 16 + 20 = 61}$ .
Think first. P = 5, Q = 4 and θ = 60°.
Take the square root
- ${R = \sqrt{61} \approx 7.8}$ N: option D.

More: two forces

Several forces: resolve and add

For three or more forces, resolve each one into components, add the components, then put the total back together. With bearings, the east part is $F\sin\theta$ and the north part is $F\cos\theta$ :

Resolving a force on a bearing(F, θ) = F sin θ east + F cos θ north

Worked example · NECO 2023

NECO 2023 · Paper 2 · Q7

Three men pushed a bus with forces of 204 N in the direction $030^\circ$ , 300 N in the direction $090^\circ$ and 225 N in the direction $120^\circ$ . Find the magnitude of the resultant force, correct to the nearest newton.

East parts
- ${204\sin 30^\circ = 102}$ , ${300\sin 90^\circ = 300}$ and ${225\sin 120^\circ = 194.86}$ .
- Total east: ${102 + 300 + 194.86 = 596.86}$ .
Think first. F sin θ for each force.
North parts
- ${204\cos 30^\circ = 176.67}$ , ${300\cos 90^\circ = 0}$ and ${225\cos 120^\circ = -112.5}$ .
- Total north: ${176.67 + 0 - 112.5 = 64.17}$ .
Think first. F cos θ for each force.
The resultant
- ${R = \sqrt{596.86^2 + 64.17^2}}$ .
- ${= \sqrt{360\,360} \approx 600}$ N.

Forces along the sides of a shape are resolved the same way, along two directions at right angles.

Worked example · WAEC 2014

WAEC 2014 · Paper 2 · Q15 (a)

$PQRS$ is a square. Calculate the magnitude and direction of the resultant of the forces $5\text{ N}$ acting along $PQ$ , $3\sqrt2\text{ N}$ acting along $PR$ and $3\text{ N}$ acting along $PS$ .

Two directions
- Along $PQ$ : ${5 + 3\sqrt2\cos 45^\circ = 5 + 3 = 8}$ N.
- Along $PS$ : ${3 + 3\sqrt2\sin 45^\circ = 3 + 3 = 6}$ N.
Think first. Take PQ and PS as the two directions. PR is the diagonal, at 45° to both.
Magnitude and direction
- ${R = \sqrt{8^2 + 6^2} = 10}$ N.
- ${\tan^{-1}\frac68 \approx 36.9^\circ}$ to $PQ$ .

More: several forces

Your turn

WAEC 2018 · Paper 2 · Q14 (a)

(a)
Three forces $6\text{ N}$ , $4.5\text{ N}$ and $8\text{ N}$ act on a body $A$ of mass $1.2\text{ kg}$ as shown in the diagram: the $6\text{ N}$ force acts due north, the $4.5\text{ N}$ force at $120^\circ$ clockwise from it, and the $8\text{ N}$ force at $150^\circ$ anticlockwise from it. Calculate the magnitude of the: (i) resultant force; (ii) acceleration of the body $A$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)(i)

The forces are on bearings $000^\circ$ (6 N), $120^\circ$ (4.5 N) and $210^\circ$ (8 N).
East: $0 + 4.5\sin120^\circ + 8\sin210^\circ = 3.897 - 4$
$= -0.103$ .
North: $6 + 4.5\cos120^\circ + 8\cos210^\circ = 6 - 2.25 - 6.928$
$= -3.178$ .
$|\mathbf R| = \sqrt{0.103^2 + 3.178^2}$
$\approx 3.18\text{ N}$ .

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