All subjects · Mathematics · WME01 (M1)

Resolving vectors and finding resultant forces

Any vector can be split into two perpendicular components. Resolving is the most used skill in M1: you will do it in almost every force question.

Resolving

A force F at angle θ to a direction has:

The component next to the angle uses cos. Example: 20 N at 30° above the horizontal has horizontal part 20 cos 30° and vertical part 20 sin 30° = 10 N.

Resultant by components

Resolve every force horizontally and vertically (or along i and j), add each set, then use Pythagoras for the magnitude and tan for the direction.

Resultant by the cosine rule

For two forces P and Q with angle θ between them, the resultant R satisfies R2 = P2 + Q2 + 2PQ cos θ. (This is the cosine rule on the triangle of forces, where the angle in the triangle is 180° - θ.) Use the sine rule to find its direction.

Worked example

Two forces of 5 N and 7 N act at a point with an angle of 60° between them. Find the magnitude of their resultant.

  1. R2 = 25 + 49 + 2(5)(7)cos 60°
  2. R2 = 74 + 35 = 109

Answer: R = 10.4 N

Key idea: Along the direction: F cos θ; perpendicular: F sin θ. Then add components or use the triangle of forces.

Check yourself

  1. A force of 20 N acts at 30° above the horizontal. Find its horizontal component, to 3 significant figures.
    Show answer

    17.3 N. 20 cos 30° = 17.3 N.

  2. For the same force (20 N at 30° above the horizontal), find the vertical component.
    Show answer

    10 N. 20 sin 30° = 10 N.

  3. Forces of 8 N (east) and 6 N (north) act on a particle. Find the magnitude of the resultant.
    Show answer

    10 N. sqrt(64 + 36) = 10 N.

  4. Forces of 9 N and 4 N act at a point with an angle of 90° between them. Find the angle the resultant makes with the 9 N force, to 1 decimal place.
    Show answer

    24.0 °. tan θ = 4/9, so θ = 24.0°.

  5. Two horizontal forces of 10 N and 6 N act on a particle with an angle of 150° between them. Find the magnitude of the resultant, to 3 significant figures.
    Show answer

    5.66 N. R2 = 100 + 36 + 2(10)(6)cos 150° = 32.08, so R = 5.66 N.

  6. For the 10 N and 6 N forces at 150° to each other, find the angle between the resultant and the 10 N force, to 1 decimal place.
    Show answer

    32.0 °. Along the 10 N force: 10 + 6 cos 150° = 4.80; perpendicular: 6 sin 150° = 3. tan θ = 3/4.80, so θ = 32.0°.

  7. A force F makes an angle θ with a slope. Which expression is its component perpendicular to the slope?
    • F sin θ
    • F cos θ
    • F tan θ
    • F/cos θ
    Show answer

    F sin θ. The component along the slope (next to θ) is F cos θ, so the perpendicular one is F sin θ.

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Last updated 7 November 2026. Spotted a mistake? Tell us.