The difference between scalars and vectors, adding vectors by scale diagram and calculation, and resolving a vector into perpendicular components.

What this dot point is asking

CCEA wants you to distinguish scalar and vector quantities, add two or more vectors by scale diagram and by calculation, and resolve a single vector into two perpendicular components. This underpins every mechanics topic in AS 1, from equilibrium to projectile motion.

Scalars versus vectors

The practical consequence is that vectors must be combined with regard to their directions. Two forces of $5\ \text{N}$ acting in the same line add to $10\ \text{N}$; the same two forces acting at right angles give a resultant of only about $7.1\ \text{N}$, and acting in opposite directions they cancel to zero.

Adding vectors

Vectors are added tip to tail: the tail of the second vector is placed at the tip of the first, and the single vector drawn from the start of the first to the end of the last is the resultant. For two perpendicular vectors of magnitudes $A$ and $B$, the resultant magnitude is

and its direction is $\theta = \tan^{-1}\!\left(\frac{B}{A}\right)$ measured from $A$. For non-perpendicular vectors, draw a labelled scale diagram (for example $1\ \text{cm} \equiv 10\ \text{N}$) and measure the resultant's length and bearing with a ruler and protractor.

Resolving a vector

This is the key to inclined-plane and projectile problems, where weight is resolved along and perpendicular to a slope, or velocity is split into horizontal and vertical parts that obey separate equations of motion.

Examples in context

A free-kick in football is struck at $25\ \text{m s}^{-1}$ at $40^{\circ}$ above the ground. Its horizontal velocity is $25\cos 40^{\circ} = 19.2\ \text{m s}^{-1}$ and its vertical velocity is $25\sin 40^{\circ} = 16.1\ \text{m s}^{-1}$; the horizontal part stays roughly constant while gravity slows and reverses the vertical part. A plane flying north at $200\ \text{m s}^{-1}$ in a crosswind of $30\ \text{m s}^{-1}$ from the west has a resultant ground velocity of $\sqrt{200^2 + 30^2} = 202\ \text{m s}^{-1}$ at $8.5^{\circ}$ east of north, which the pilot must correct for to stay on track.

Try this

Q1. A force of $20\ \text{N}$ acts at $30^{\circ}$ above the horizontal. Find its horizontal and vertical components. [2 marks]

• Cue. Horizontal $= 20\cos 30^{\circ} \approx 17.3\ \text{N}$; vertical $= 20\sin 30^{\circ} = 10\ \text{N}$.

Q2. Two perpendicular displacements of $3\ \text{m}$ east and $4\ \text{m}$ north are combined. State the magnitude of the resultant. [1 mark]

• Cue. $\sqrt{3^2 + 4^2} = 5\ \text{m}$.

Q3. Explain why pulling a heavy case along on a strap at an angle reduces the upward force needed but does not give the largest horizontal pull. [2 marks]

• Cue. The vertical component $F\sin\theta$ supports part of the weight, but the useful horizontal component $F\cos\theta$ shrinks as $\theta$ grows; a flat pull gives the largest forward force.