Physics › Mechanics › Scalars and vectors

Scalars and vectors

Half the quantities in mechanics carry a direction and half do not, and the maths only works once the two kinds are treated differently. Adding, resolving and closing the triangle are three moves that recur throughout mechanics.

Year 12AQA 3.4.1.1 CIE 1.4.1, 1.4.2, 1.4.3, 4.2.3 OCR A 2.3.1(a), 2.3.1(b), 2.3.1(c), 2.3.1(d), 3.2.3(f)EDEXCEL 12, 13, 14

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IN THIS TOPIC

Classify quantities as scalars or vectors, using the paired examples examiners keep returning to.
Add two perpendicular vectors by calculation, and any other pair by scale drawing.
Subtract one coplanar vector from another by reversing it and adding, and use that to find a change in a vector quantity.
Resolve a vector into two perpendicular components, including weight on an inclined plane.
State and use the equilibrium condition for two or three coplanar forces acting at a point.

COMMON MISCONCEPTION

Adding forces means adding the numbers.

Two kinds of quantity

A scalar has size only. A vector has size and direction, and that direction does real work in every equation it appears in. Learn the standard pairs cold. Distance is a scalar and displacement its vector partner, and speed pairs with velocity the same way. Mass is a scalar and weight a vector, but they are different quantities rather than two versions of one: weight is the gravitational force on the mass. 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 and subtracting 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, Pythagoras gives the magnitude and trigonometry gives the direction.

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 expected method is a scale drawing. Choose a scale, draw tip to tail carefully, and measure the resultant's length and angle with a ruler and a protractor. Write the scale on the diagram. It is part of the answer, and it is left off more often than it is put on.

WORKED EXAMPLE

Adding two perpendicular velocities

A plane flies at 60 m s⁻¹ due north while a crosswind blows it at 25 m s⁻¹ due east. Find the plane's resultant velocity.

Perpendicular vectors mean Pythagoras for the magnitude. The resultant is the square root of 60² + 25² = the square root of 4225 = 65 m s⁻¹.

The direction needs a line of its own. tan θ = 25/60, so θ = 23° east of north. A magnitude with no direction attached is half a vector, and half the marks.

Check it lands sensibly. The resultant beats either part on its own but falls short of their plain sum, which is exactly where a tip-to-tail triangle has to put it.

Subtraction needs no new rule, only a reversal. The negative of a vector, written −B, has the same magnitude as B and points the opposite way, so A − B means A + (−B). Draw A, turn B round, and join the two tip to tail exactly as you would for a sum.

One shortcut is worth knowing alongside it. Draw A and B from the same starting point, and A − B is the vector running from the tip of B to the tip of A. Both routes give the identical arrow, so use whichever the diagram in front of you already has.

FIG. 2Left, A and B drawn from one point, with A − B running from the tip of B to the tip of A. Right, the same answer assembled tip to tail as A followed by a reversed B. The two constructions are the same arrow, drawn two ways.

Every change in a vector quantity is a subtraction of this kind. A change in velocity is $\Delta v = v - u$ with both treated as vectors, which is why a ball rebounding off a wall at its original speed has not undergone a change of zero. Reverse u, add, and the change comes out at twice the speed, directed away from the wall. The same move gives the change in momentum in a collision, and the velocity of one object relative to another, which is one velocity minus the other.

Resolving into components

Addition run backwards is resolution. Any vector can be swapped for two components at right angles, and the pair behaves identically to the original in every calculation. A vector F at angle θ to a chosen axis splits into F cos θ along that axis and F sin θ perpendicular to it.

FIG. 3A 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 drives the block. W cos θ presses into the surface, and that is the component the normal contact force has to match.

FIG. 4Weight 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.

GUIDED PRACTICE

Resolve a launch velocity

A jet leaves the runway at 250 m s⁻¹, climbing at 60° above the horizontal. Decide which component takes the cosine, then find both components.

Show the working

The angle is measured from the horizontal, so the horizontal component takes the cosine. 250 cos 60° = 125 m s⁻¹.

Vertically, 250 sin 60° = 217 m s⁻¹. Adjacent takes the cosine. All the work is in spotting which direction the angle leans on.

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.

FIG. 5Three forces act on one point, left. To test for equilibrium, walk them head to tail: draw the first, start the second from its head, the third from that one's head. Here the walk arrives exactly back where it began, the triangle closes, and that closure is the statement that the forces balance. If the walk had fallen short or overshot, the gap would have been the resultant.

Zero resultant has a graphical signature. Drawn head to tail, forces in equilibrium form a closed triangle. Problems yield to either route, resolving in two perpendicular directions and setting each sum to zero, or drawing the closed triangle and doing trigonometry on it. Pick whichever makes the angles you were given easy.

INDEPENDENT PRACTICE

A hanging mass on two strings

A 3.0 kg lamp hangs from two identical strings, each at 40° above the horizontal. Find the tension in each string, resolving vertically. Work it through.

Show the working

The weight is 3.0 × 9.81 = 29.4 N. Each string supplies a vertical component T sin 40°, and there are two of them.

Equilibrium vertically gives 2T sin 40° = 29.4, so T = 29.4/(2 × 0.643) = 23 N.

Each tension exceeds half the weight, because only part of each string's pull acts upward. Flatter strings would need more still, and that is why nobody ever gets a washing line properly straight.

ASSESSMENT FOCUS

“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.
Subtraction is asked for as often as addition, and it is always the same instruction: reverse the second vector and add it. A “change in velocity” or “change in momentum” question is a vector subtraction, so a straight rebound gives twice the speed and not zero.
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. Test a slip against a limit. At θ = 0 the along-slope component W sin θ ought to 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

For the magnitude, $30^{2} + 40^{2} = 2500$ , and the square root gives 50 N.

For the direction, tan θ = 40/30, so θ = 53° to the 30 N force. A 3-4-5 triangle in disguise, and examiners are fond of these numbers for exactly that reason.

Vectors add tip to tail.

Only their components add as plain numbers.

Equilibrium closes the triangle.

WORKBOOK

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

20 questions on this topicAnswer them one at a time and mark yourself against the mark scheme.Practise this topic

Or read them with their mark schemes on the scalars and vectors questions page.

9 flashcards on this topicDefinitions, off-sheet equations and a spot-the-error card, scheduled by spaced repetition in your browser.Revise with flashcards

WHERE TO GO NEXT

Trigonometry and resolving vectors is the maths this lesson leans on, worked through from GCSE.

CHECK YOUR PROGRESS

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Classify quantities as scalars or vectors, using the paired examples examiners keep returning to.
Add two perpendicular vectors by calculation, and any other pair by scale drawing.
Subtract one coplanar vector from another by reversing it and adding, and use that to find a change in a vector quantity.
Resolve a vector into two perpendicular components, including weight on an inclined plane.
State and use the equilibrium condition for two or three coplanar forces acting at a point.

Open the full revision checklist to track your progress across the whole unit.