Distinguishing scalars and vectors, adding vectors by calculation and scale drawing, resolving a vector into perpendicular components, and the conditions for equilibrium of coplanar forces.

What this dot point is asking

AQA specification point 3.4.1.1 wants you to tell scalars from vectors, add vectors both by calculation (using Pythagoras and trigonometry) and by scale drawing, resolve a single vector into two perpendicular components, and apply the conditions for the equilibrium of coplanar forces.

Scalars and vectors

Knowing which is which matters because vectors must be combined with their directions taken into account, while scalars add arithmetically. Walking $3 \text{ m}$ east then $4 \text{ m}$ north gives a total distance (scalar) of $7 \text{ m}$, but a displacement (vector) of only $5 \text{ m}$ at an angle.

Adding vectors

To add two vectors, draw them tip to tail; the resultant runs from the start of the first to the tip of the last. For two perpendicular vectors $x$ and $y$ the resultant has magnitude $R = \sqrt{x^2 + y^2}$ and direction $\theta = \tan^{-1}\!\left(\dfrac{y}{x}\right)$ measured from the $x$ direction. For non-perpendicular vectors you can either use an accurate scale drawing (a ruler and protractor) or split each vector into components first and add the components separately.

Resolving a vector

Any single vector $F$ at angle $\theta$ to a chosen axis can be split into two perpendicular components.

Resolving turns an awkward angled problem into two independent straight-line problems, which is why it is the key skill for slopes, projectiles and forces in equilibrium. The component adjacent to the angle uses cosine; the component opposite the angle uses sine.

Equilibrium of coplanar forces

For an object on a slope, resolving the weight gives a component $W\sin\theta$ down the slope and $W\cos\theta$ into the surface, which must be balanced by friction and the normal contact force respectively.

Try this

Q1. State two examples each of a scalar and a vector quantity. [2 marks]

• Cue. Scalars: mass, energy, speed. Vectors: velocity, force, displacement.

Q2. A force of $20 \text{ N}$ acts at $60^{\circ}$ above the horizontal. Find its horizontal and vertical components. [2 marks]

• Cue. $F_x = 20\cos 60^{\circ} = 10 \text{ N}$; $F_y = 20\sin 60^{\circ} = 17.3 \text{ N}$.

Q3. State the condition for a point object to be in equilibrium. [1 mark]

• Cue. The resultant force on it is zero.