Scalars, vectors and equilibriumAQA A-Level Physics: Revision notes

Section 1

Scalars and vectors

A scalar has magnitude only; a vector has magnitude and direction.

scalars: speed, distance, mass
vectors: velocity, displacement, force (including weight), acceleration

Weight is a vector (a force towards the centre of the Earth) whereas mass is a scalar. Two vectors are equal only if both their magnitudes and their directions are equal.

Key termsscalarvector

Common mistake

Speed and velocity are not interchangeable. A hiker walking 14 km and ending 10 km from the start has an average speed based on 14 km but an average velocity based on 10 km.

Section 2

Adding vectors

Vectors are added head to tail; the resultant goes from the start of the first to the end of the last.

By calculation (two vectors at right angles): $R = \sqrt{a^2 + b^2}$ with $\tan\theta = b/a$ for the direction.
By scale drawing (any angle): choose a scale, draw the vectors head to tail with a ruler and protractor, then measure the resultant and its angle.

Worked example. 6.0 km east then 8.0 km north gives a displacement $\sqrt{6.0^2 + 8.0^2} = 10$ km at $\tan^{-1}(6.0/8.0) = 36.9°$ east of north.

Key termsresultantscale drawing

Exam tip

State a reference direction with every angle, for example '36.9° east of north'. An angle on its own is not a direction.

Section 3

Resolving vectors

A vector can be split into two components at right angles. For a force $F$ at angle $\theta$ to the horizontal:

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

The two components together have the same effect as the original force. Components at right angles are independent: a component does not affect motion at right angles to it. Choose axes to make the problem easy, often along and perpendicular to a surface.

Key termscomponentresolving

Exam tip

The side next to the angle uses cos, the side opposite uses sin. Check by testing 0° and 90°: at 0° the horizontal component should be the whole force.

Section 4

Forces on an inclined plane

For an object of weight $W$ on a slope at angle $\theta$ to the horizontal, resolve the weight along and perpendicular to the slope:

component down the slope: $W\sin\theta$
component perpendicular to the slope: $W\cos\theta$

The normal contact force balances the perpendicular component. If the object is at rest, friction or a rope balances the component down the slope.

Worked example. $W = 60$ N, $\theta = 30°$ : down the slope $60 \sin 30° = 30$ N, perpendicular $60 \cos 30° = 52$ N.

Key termsnormal contact forceinclined plane

Section 5

Equilibrium of coplanar forces

An object is in equilibrium when the resultant force on it is zero. It is then either at rest or moving with constant velocity.

For coplanar forces acting at a point there are two methods:

Resolving: the sum of the components in any direction is zero (for example, up equals down and left equals right).
Closed triangle: for three forces, the vectors drawn head to tail form a closed triangle, because their resultant is zero.

Worked example. A 30 N lamp hangs from two strings, each at 25° to the horizontal. Vertically, $2T\sin 25° = 30$ , so $T = 35.5$ N.

Key termsequilibriumcoplanar forcesclosed triangle

Common mistake

Equilibrium does not mean at rest. A car moving at constant velocity in a straight line is in equilibrium.

Must Know

Scalars have magnitude only; vectors have magnitude and direction
Add vectors head to tail: $\sqrt{a^2+b^2}$ at right angles, scale drawing otherwise
Components: $F\cos\theta$ and $F\sin\theta$ at right angles
On a slope: $W\sin\theta$ down the slope, $W\cos\theta$ perpendicular to it
Equilibrium: resultant force zero; at rest or constant velocity
Three forces in equilibrium form a closed triangle

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Scalars, vectors and equilibrium

A hiker walks 6.0 km due east and then 8.0 km due north, taking 2.0 hours in total, and stops.
Calculate the direction of the hiker's displacement, giving your answer as an angle east of north.2 marks
A box of weight 60 N is at rest on a rough ramp that is inclined at 30° to the horizontal.
The ramp is raised so that it is inclined at 40° to the horizontal and the box is still at rest. Calculate the frictional force acting on the box.2 marks
A lamp of weight 30 N hangs at rest from the middle of two identical light strings. Each string is attached to a ceiling hook and makes an angle of 25° with the horizontal. The lamp and strings are in equilibrium.
Calculate the tension in each string.3 marks

See the full worksheet

Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).