Force components and vector addition

Key ideas

Force is a vector — it has size and direction. Two skills cover everything the exam asks:

1. Adding vectors (finding a resultant). Draw the forces head-to-tail; the resultant runs from the start of the first to the head of the last. For two perpendicular forces, use Pythagoras for the size and trigonometry for the angle.

2. Splitting a force into components. A force $F$ at angle $\theta$ to the horizontal is equivalent to:

$F_x = F\cos\theta \qquad F_y = F\sin\theta$

Bring a ruler and protractor — the exam expects tidy scale/vector diagrams, and marks reasoning shown in diagrams.

Worked ExampleResolving a force into components

A child pulls a sled with a $100$ N force at $30°$ above the horizontal. Find the horizontal and vertical components.

The horizontal (adjacent) component uses cosine; the vertical (opposite) component uses sine:

$F_x = 100\cos 30° = 86.6 \approx 87 \text{ N}$

$F_y = 100\sin 30° = 50 \text{ N}$

Only the $87$ N horizontal component drags the sled forward; the $50$ N vertical component just lifts a little of its weight off the snow.

Worked ExampleAdding perpendicular forces

Two ropes pull a boat: $30$ N east and $40$ N north. Find the resultant force.

Step 1 — Size, by Pythagoras

The forces are at right angles, so:

$\sqrt{30^2 + 40^2} = \sqrt{2500} = 50 \text{ N}$

Step 2 — Direction, by trig

$\tan\theta = \frac{40}{30} \;\Rightarrow\; \theta = 53° \text{ north of east}$

Practice question

A lawnmower is pushed with $80$ N along the handle, which slopes at $60°$ to the horizontal. What is the horizontal component pushing the mower forward?

Worked solution: $F_x = 80\cos 60° = 40$ N. (Note the steeper the handle, the smaller the useful component.)

Test yourself

Practice by grade

One question each at Achieved, Merit and Excellence. Have a go, then compare with the model answer.

Achieved

A $60$ N force is applied to a box at an angle of $25°$ above the horizontal.

Calculate the horizontal component of the force.

Merit

A child pulls a $15$ kg sled across smooth ice with a $40$ N force directed at $30°$ above the horizontal.

Calculate the horizontal component of the force, and hence the acceleration of the sled.

Excellence

Using the $15$ kg sled on smooth ice above, explain with supporting calculations why pulling with $40$ N at $30°$ produces a smaller acceleration than pulling with the same $40$ N horizontally.