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Scalars and vectors

Half the quantities in mechanics carry a direction and half do not, and the maths refuses to work until you treat the two kinds differently. Adding, resolving and closing the triangle: three moves that run the whole of mechanics.

Year 12AQA 3.4.1.1

IN THIS TOPIC

  • Classify quantities as scalars or vectors, using the standard paired examples.
  • Add two perpendicular vectors by calculation, and any vectors by scale drawing; resolve a vector into components at right angles, including on an inclined plane.
  • State and use the equilibrium condition for two or three coplanar forces acting at a point.

WHAT YOU PROBABLY THINK

Adding forces means adding the numbers.

Two kinds of quantity

A scalar has size only; a vector has size and direction, and the direction is not decoration. The standard pairs are worth knowing cold: distance is a scalar, displacement a vector; speed is a scalar, velocity a vector; mass is a scalar, weight a vector. Force and acceleration are vectors through and through.

The distinction bites the moment you combine quantities. Two 5 N forces can add to 10 N, to zero, or to anything between, depending entirely on their directions. Adding the numbers is only ever the special case of forces pointing the same way.

Adding vectors

Vectors add tip to tail: draw the first, start the second from its head, and the resultant runs from the very start to the very end. For two vectors at right angles, which is the only case AQA asks you to calculate, Pythagoras gives the magnitude and trigonometry the direction.

Two perpendicular forces added tip to tail: a 30 N and a 40 N force give a 50 N resultant30 N40 N50 Nθ
FIG. 1A 30 N and a 40 N force at right angles. The resultant is 50 N, and tan θ = 40/30 fixes its direction.

For vectors at other angles, the specification asks for a scale drawing: choose a scale, draw tip to tail carefully, and measure the resultant's length and angle with a ruler and protractor. State the scale on the diagram; it is part of the answer.

Resolving into components

Addition run backwards is resolution: any vector can be replaced by two components at right angles, and the pair behaves identically to the original. A vector F at angle θ to a chosen axis splits into F cos θ along it and F sin θ perpendicular to it.

Resolving a force into two components at right angles: F cos theta along, F sin theta perpendicularθFF cos θF sin θ
FIG. 2A force F at angle θ resolved into F cos θ along the reference direction and F sin θ perpendicular to it.

The exam's favourite setting is the inclined plane. Resolve the weight along and perpendicular to the slope, never horizontally and vertically: W sin θ acts down the slope, and it is what drives the block; W cos θ presses into the surface, and it is what the normal contact force must match.

Weight on an inclined plane resolved along the slope and perpendicular to itθWW sin θW cos θ
FIG. 3Weight on a 30 degree slope, resolved along the slope and into it. The angle between the weight and the perpendicular component equals the slope angle.

Equilibrium at a point

Two or three coplanar forces acting at a point are in equilibrium when their resultant is zero, and equilibrium means the object is at rest or moving at constant velocity. The second half of that sentence is tested as often as the first.

Three coplanar forces in equilibrium drawn head to tail close into a triangleF₁F₂F₃head to tail, and the triangle closes
FIG. 4Three forces in equilibrium drawn head to tail close into a triangle, ending exactly where they began.

Zero resultant has a graphical signature: drawn head to tail, forces in equilibrium form a closed triangle. Problems can be solved either way, by resolving in two perpendicular directions and setting each sum to zero, or by drawing the closed triangle and using trigonometry on it. Pick whichever makes the given angles easy.

THE EXAM BIT

  • “State whether X is a scalar or a vector, and explain the difference” is a routine opener: a vector has direction as well as magnitude. Two easy marks that go missing when rushed.
  • Calculations are limited to perpendicular pairs; anything else is a scale drawing. If angles other than 90° appear and no drawing is asked for, resolve first.
  • On a slope, resolve along and perpendicular to the slope. Horizontal and vertical components tangle the normal force with both directions and the algebra collapses.
  • The component adjacent to the angle takes cos, the opposite takes sin. Checking with a limit catches slips: at θ = 0 the along-slope component W sin θ should vanish, and it does.
  • “Constant velocity” in a question means equilibrium: resultant zero. Treat it exactly as you would “at rest”.

CHECK YOURSELF

Forces of 30 N and 40 N act on a point at right angles to each other. Find the magnitude of the resultant and the angle it makes with the 30 N force.

Show a hint

Tip to tail, then Pythagoras for the size and tangent for the angle.

Show the answer

Magnitude: 302 + 402 = 2500, and the square root gives 50 N.

Direction: tan θ = 40/30, so θ = 53° to the 30 N force. A 3-4-5 triangle, which is why examiners are so fond of these numbers.

Vectors add tip to tail.

Only their components add as numbers.

No animated video for this topic yet; these notes stand alone. InkPhysics on YouTube.