LearnYPH11WPH11 · Unit 1: Mechanics and MaterialsMechanicsScalars and vectors

Mechanics · Subtopic 8 of 8

Scalars and Vectors

Scalars have magnitude only; vectors have magnitude and direction. Vectors add tip-to-tail; for two at right angles use Pythagoras and trigonometry, and for other angles use a scale drawing or the cosine rule. Any vector can be resolved into two perpendicular components: $F\cos\theta$ along the reference direction and $F\sin\theta$ perpendicular to it.

Cambridge 9702 Paper 1Cambridge 9702 Paper 2Edexcel YPH11 WPH11AQA 7408 Paper 1≈ 30 minRevision 5 · updated 2026-10-27

Section 1 of 14

Learning objectives

Classify quantities as scalars or vectors.
Add and subtract coplanar vectors by drawing or calculation.
Resolve a vector into two perpendicular components.
Resolve forces on an inclined plane.

Pause and think

You walk $3$ km east and then $4$ km north. How far have you walked, and how far are you from where you started?

Section 2 of 14

Core idea

Start hereRowing across a river

Row straight across a river and you'll land downstream: your velocity and the current's combine. Splitting every vector into perpendicular components turns such problems into simple addition.

The big idea

Vectors add tip-to-tail. Resolve: $F\cos\theta$ and $F\sin\theta$ . Add components; recombine with Pythagoras and $\tan$ .

1. Components

Components of a 10 N force

θ (degrees) = 10

The two components together have exactly the same effect as the original force.

F_x=F\cos\theta,\quad F_y=F\sin\theta,\quad|F|=\sqrt{F_x^2+F_y^2},\quad\tan\theta=\frac{F_y}{F_x}

$\theta$: angle from the $x$ -direction.

Worked exampleCrossing a river

Boat 3.0 m s $^{-1}$ across; current 2.0 m s $^{-1}$ ; river 60 m wide.

Goal: Resultant velocity; crossing time.
Strategy: Perpendicular components.

$\sqrt{3^2+2^2}=3.6\text{ m s}^{-1}$
Pythagoras.
$\tan^{-1}\tfrac23=33.7^\circ$
Downstream of straight across.
$t=\frac{60}{3.0}=20\text{ s}$
Only the across component crosses the river.

Answer

3.6

m s

^{-1}

33.7^\circ

;

20

✓ CheckDrift

2\times20=40

m downstream;

40/60=\tan33.7^\circ

✓.

2. Non-perpendicular vectors

Non-perpendicular vectors: use a scale drawing or the cosine rule

R^2=6^2+4^2-2(6)(4)\cos120^\circ=76

, so

R=8.72

N. Or resolve both and add components.

Worked example6 N and 4 N at 60°

Resultant.

Goal: Size and direction.
Strategy: Take $x$ along the 6 N force.

$R_x=6+4\cos60^\circ=8$
Add $x$ -components.
$R_y=4\sin60^\circ=3.46$
Add $y$ -components.
$\sqrt{64+12}=8.72\text{ N at }23.4^\circ$
Recombine.

Answer

8.72

N at

23.4^\circ

to the 6 N force.

✓ CheckCosine rule:

36+16+2(24)(0.5)=76

✓.

Common misconceptionAdding magnitudes

6+4=10

only if the forces are parallel.

Pause and think

20 N at 30° above horizontal: components?

Summary

Resolve, add, recombine.

Resultant: Single equivalent vector.

HistoryStevin's wreath of spheres. Around

1586

, Simon Stevin imagined a loop of linked balls draped over a triangular wedge. Since it doesn't spin by itself forever, the forces along each slope must balance. From this he deduced how forces resolve along inclines: an early version of vector components.

Fun factPlanes fly sideways. In a strong crosswind, a landing plane points partly into the wind (a "crab angle") so that its resultant velocity lines up with the runway. Moments before touchdown, the pilot straightens the nose.

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