How do I decide whether to use sine or cosine?

Sketch the right-angled triangle with the vector as the hypotenuse. The component next to the angle uses; the one across from it uses. Never memorise it as "cos is horizontal".

Do axes have to be horizontal and vertical?

No, and choosing otherwise often saves work. On a slope, axes along and perpendicular to the surface mean only the weight needs resolving.

Is a scale drawing acceptable in the exam?

Where the question allows it, yes — but it must be drawn accurately with a stated scale. Calculation is faster and not subject to drawing error, so prefer it unless a drawing is specifically asked for.

Can a resultant be zero?

Yes. If all forces on an object cancel, the resultant is zero and the object is in equilibrium — at rest or moving at constant velocity, by Newton’s first law.

Back to course

2.1MechanicsFoundation

Scalars, Vectors and Resolving

Forces rarely act along convenient axes. A rope pulls at an angle, a slope tilts gravity, a wind blows across a flight path. $Resolving$ splits any such vector into perpendicular components that can be handled independently — and that independence is what turns an awkward two-dimensional problem into two simple one-dimensional ones.

30 min Topic 2 · Mechanics

What you'll be able to do

Classify quantities as scalar or vector and explain the difference
Add vectors by scale drawing and by calculation
Resolve a vector into perpendicular components
Choose axes that simplify a problem, including on an inclined plane

Which quantities carry a direction

A $scalar$ has magnitude only: mass, time, energy, temperature, speed, distance, power, charge. A $vector$ has magnitude and direction: displacement, velocity, acceleration, force, momentum, and every field strength you meet later.

A common exam question simply asks you to sort a list. The pairs that catch people out are distance/displacement, speed/velocity, and mass/weight — mass is a scalar in kilograms, weight is a force in newtons and therefore a vector.

Adding vectors

Two vectors add head-to-tail: draw the second starting where the first ends, and the $resultant$ runs from the start of the first to the end of the second. For perpendicular vectors the resultant follows from Pythagoras and trigonometry directly.

For vectors that are not perpendicular, either resolve both into components and add componentwise, or use a scale drawing. Edexcel accepts a careful scale drawing where the question permits it, but calculation is quicker and avoids drawing error.

R = F_{1}^{2} + F_{2}^{2}

Resultant of two perpendicular vectors.

tan θ = \frac{F _{2}}{F _{1}}

Its direction from the

F_{1}

axis.

1They are perpendicular, so

R = 6. 0^{2} + 8. 0^{2} = 100

2Direction:

tan θ = \frac{8.0}{6.0}

, so

θ = 5 3^{\circ}

north of east.

Answer

10 N

5 3^{\circ}

north of east

Tip — Magnitudes never simply add unless the vectors are parallel. $6 N$ and $8 N$ give $10 N$ perpendicular, $14 N$ parallel, and $2 N$ opposed.

Resolving into components

Resolving reverses addition: it replaces one vector with two perpendicular ones that together have the same effect. For a vector of magnitude $F$ at angle $θ$ to an axis, the component $along$ that axis is $F cos θ$ and the component $perpendicular$ to it is $F sin θ$ .

The reliable way to remember it is not "cos for horizontal" — that fails as soon as the angle is measured from the vertical. The rule that always holds is that $cos$ goes with the component $adjacent$ to the angle and $sin$ with the component $opposite$ it. Sketch the triangle and read it off.

F_{∥} = F cos θ

Component adjacent to the angle.

F_{⊥} = F sin θ

Component opposite the angle.

1The angle is measured from the horizontal, so the horizontal component is adjacent:

250 cos 3 0^{\circ}

2The vertical component is opposite:

250 sin 3 0^{\circ}

AnswerHorizontal

217 N

; vertical

125 N

Tip — Check the size of your components: each must be smaller than the original vector, and they must satisfy Pythagoras back to it. If a component comes out larger, sine and cosine have been swapped.

Choosing sensible axes

Axes do not have to be horizontal and vertical. Choosing them to line up with the motion usually removes most of the work.

On an inclined plane, take one axis $along$ the slope and the other perpendicular to it. Then the normal contact force lies entirely on one axis, the friction entirely on the other, and only the weight needs resolving — into $m g sin θ$ down the slope and $m g cos θ$ into it.

That choice is what makes slope problems routine. With horizontal and vertical axes instead, three of the forces need resolving rather than one.

W_{∥} = m g sin θ

Weight component down the slope.

W_{⊥} = m g cos θ

Weight component into the slope.

1Weight:

W = m g = 5.0 \times 9.81 = 49.05 N

2The angle between the weight and the perpendicular to the slope equals the slope angle, so the down-slope component is

W sin 2 0^{\circ}

49.05 \times sin 2 0^{\circ}

Answer

16.8 N

down the slope

Equation recap

R = F_{1}^{2} + F_{2}^{2}

Perpendicular resultant.

F_{∥} = F cos θ

Adjacent component.

F_{⊥} = F sin θ

Opposite component.

W_{∥} = m g sin θ

Down a slope.

W_{⊥} = m g cos θ

Into a slope.

Common mistakes to avoid

Always using

cos

for the horizontal component.

cos

goes with the side

adjacent

to the angle. If the angle is from the vertical, the horizontal component uses

sin

Adding force magnitudes without regard to direction.

Only parallel vectors add arithmetically. Otherwise resolve and add componentwise.

Using

m g cos θ

for the force down a slope.

Down the slope is

m g sin θ

;

m g cos θ

presses into the surface.

Calling weight a scalar because mass is.

Weight is a force in newtons and therefore a vector; mass is a scalar in kilograms.

Key takeaways

Scalars have magnitude only; vectors have magnitude and direction.
Vectors add head-to-tail; perpendicular ones combine by Pythagoras.
Resolving splits a vector into perpendicular components that act independently.
$cos$ pairs with the component adjacent to the angle, $sin$ with the opposite one.
On a slope, use axes along and perpendicular to it: $m g sin θ$ and $m g cos θ$ .

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Scalars, Vectors and Resolving

Which quantities carry a direction

Adding vectors

Resolving into components

Choosing sensible axes

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Scalars, Vectors and Resolving

Which quantities carry a direction

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