Resolving Vectors

Spec mapping: OCR H556 Module 2.3 — Nature of quantities, specifically the resolution of a vector into perpendicular components and the component method of vector addition. The technique developed here underpins every problem in Modules 3 (forces and motion), 5 (Newtonian world) and 6 (fields) that involves an angle. (Refer to the official OCR H556 specification document for exact wording.)

The previous lesson taught you how to add vectors with the tip-to-tail triangle method. In this final lesson we learn the inverse — and arguably more powerful — operation: resolving a single vector into two perpendicular components. Once a vector is split into its horizontal and vertical parts, the problem reduces to arithmetic on ordinary scalars, and almost every question in A-Level mechanics becomes a matter of substituting into one or two equations.

This technique underpins everything in mechanics, electricity, waves and fields. Master it here and you will save enormous effort across the rest of the course.

Key definition. Resolving a vector means decomposing it into two (or more) perpendicular components along chosen coordinate axes, such that the original vector equals the vector sum of those components.

The Idea: Any Vector Equals the Sum of Its Components

Given any vector $\mathbf{F}$ making angle $\theta$ with a chosen $x$ -axis, we can draw a right-angled triangle with $\mathbf{F}$ as its hypotenuse, one side along the $x$ -axis, and the other perpendicular to it.

Trigonometry on this right-angled triangle gives the four core relationships you must know absolutely:

$F_x = F\cos\theta \quad \text{(horizontal component)},$

$F_y = F\sin\theta \quad \text{(vertical component)},$

$F^{2} = F_x^{2} + F_y^{2} \quad \text{(magnitude reconstruction)},$

$\tan\theta = \frac{F_y}{F_x} \quad \text{(direction reconstruction)}.$

Where do cos and sin come from?

In the right-angled triangle the angle $\theta$ is between the $x$ -axis and the hypotenuse (the vector $\mathbf{F}$ ). The adjacent side to $\theta$ is the horizontal $F\cos\theta$ ; the opposite side is the vertical $F\sin\theta$ . SOHCAHTOA:

$\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{F_x}{F} \Longrightarrow F_x = F\cos\theta.$

$\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{F_y}{F} \Longrightarrow F_y = F\sin\theta.$

If $\theta$ is measured from the vertical rather than the horizontal, the cos and sin swap. Always check which axis the angle is measured from before substituting.

Sign conventions

Components carry signs that tell you which way along each axis they point. The usual convention is:

$+x$ = right (east).
$+y$ = up (north).

A force at $135^{\circ}$ from the positive $x$ -axis (that is, $45^{\circ}$ above the negative $x$ -axis) has

$F_x = F\cos 135^{\circ} = -\frac{F}{\sqrt{2}}\,(\text{negative}), \qquad F_y = F\sin 135^{\circ} = +\frac{F}{\sqrt{2}}\,(\text{positive}).$

You do not need to memorise the cos and sin of obtuse angles; sketch the vector and read the signs from where it points.

Worked Example 1 — Basic Resolution

A $50\,\text{N}$ force acts at $30^{\circ}$ above the horizontal. Find its horizontal and vertical components.

Solution.

$F_x = 50\cos 30^{\circ} = 50 \times 0.8660 = 43.3\,\text{N}.$

$F_y = 50\sin 30^{\circ} = 50 \times 0.5000 = 25.0\,\text{N}.$

Check by reconstruction: $\sqrt{43.3^{2} + 25.0^{2}} = \sqrt{2499.9} \approx 50.0\,\text{N}$ . ✓

Worked Example 2 — A Projectile's Initial Velocity

A ball is launched at $20\,\text{m s}^{-1}$ at $40^{\circ}$ above the horizontal. Find its initial horizontal and vertical velocities.

Solution.

$v_x = 20\cos 40^{\circ} = 20 \times 0.7660 = 15.3\,\text{m s}^{-1}.$

$v_y = 20\sin 40^{\circ} = 20 \times 0.6428 = 12.9\,\text{m s}^{-1}.$

These two components evolve independently under gravity: $v_x$ remains constant (no air resistance) while $v_y$ decreases as $v_y - gt$ . This is the foundation of all projectile-motion analysis in the mechanics module — and it works precisely because resolving lets us treat the two perpendicular directions independently.

Worked Example 3 — Resolving Weight on an Inclined Plane

A $5.0\,\text{kg}$ block rests on a frictionless slope inclined at $25^{\circ}$ to the horizontal. Calculate the components of its weight parallel to and perpendicular to the slope.

Solution. Weight magnitude:

$W = mg = 5.0 \times 9.81 = 49.05\,\text{N}\,(\text{vertically downward}).$

The key move on an inclined plane is to rotate the coordinate system so that the axes lie parallel and perpendicular to the slope (rather than horizontal and vertical). The angle of the slope $\theta$ then equals the angle the weight vector makes with the normal to the slope.

The crucial geometry: the slope is tilted by $\theta$ from horizontal, so the vertical weight vector now makes angle $\theta$ with the normal to the slope (not with the slope itself). Resolving:

$W_{\parallel} = W\sin\theta = 49.05 \times \sin 25^{\circ} = 49.05 \times 0.4226 = 20.7\,\text{N}\,(\text{down the slope}).$

$W_{\perp} = W\cos\theta = 49.05 \times \cos 25^{\circ} = 49.05 \times 0.9063 = 44.5\,\text{N}\,(\text{into the slope}).$

Newton's third law gives the normal reaction $N = W_{\perp} = 44.5\,\text{N}$ (slope on block, upward perpendicular). Newton's second law along the slope gives

$a = \frac{W_{\parallel}}{m} = \frac{20.7}{5.0} = 4.14\,\text{m s}^{-2}\,(\text{down the slope}).$

The hardest thing for candidates to remember is that $\sin\theta$ goes with the parallel component and $\cos\theta$ goes with the perpendicular component — the opposite of the naïve expectation. The reason is the geometric subtlety above: the angle $\theta$ between the slope and horizontal is also the angle between the (vertical) weight and the (perpendicular to slope) normal. Sketch it once and the swap becomes obvious.

Worked Example 4 — Non-Perpendicular Addition via Components

Revisit the previous lesson's two forces of $10\,\text{N}$ and $15\,\text{N}$ with $60^{\circ}$ between them. Solve by components.

Solution. Place the $15\,\text{N}$ force along the $+x$ axis.

$\mathbf{F}_1$ ( $15\,\text{N}$ along $+x$ ): $F_{1x} = 15$ , $F_{1y} = 0$ .
$\mathbf{F}_2$ ( $10\,\text{N}$ at $60^{\circ}$ above $+x$ ): $F_{2x} = 10\cos 60^{\circ} = 5.0$ , $F_{2y} = 10\sin 60^{\circ} = 8.66$ .

Sum:

$R_x = 15 + 5.0 = 20.0, \qquad R_y = 0 + 8.66 = 8.66.$

Magnitude:

$|\mathbf{R}| = \sqrt{20.0^{2} + 8.66^{2}} = \sqrt{475} \approx 21.8\,\text{N}.$

Direction:

$\theta = \arctan\frac{8.66}{20.0} = \arctan(0.433) = 23.4^{\circ}\,\text{above the 15 N force}.$

This matches the cosine-rule answer from the previous lesson exactly — but with much simpler arithmetic. The component method is the universal tool for vector addition at A-Level and beyond.

Worked Example 5 — Three Forces in Equilibrium

A mass hangs from two strings, one at $30^{\circ}$ to the vertical, the other at $45^{\circ}$ to the vertical, on opposite sides. Weight $= 100\,\text{N}$ . Find the two tensions $T_1$ (on the $30^{\circ}$ string) and $T_2$ (on the $45^{\circ}$ string).

Solution. Take $+x$ to the right, $+y$ upward. The weight is $(0, -100)$ . $T_1$ pulls up and to the left of the mass (since the $30^{\circ}$ string is on the right of the mass when the mass hangs, the tension pulls the mass back towards the support — up-left by convention). $T_2$ pulls up-right.

$T_1$ components: $(-T_1\sin 30^{\circ}, +T_1\cos 30^{\circ})$ .
$T_2$ components: $(+T_2\sin 45^{\circ}, +T_2\cos 45^{\circ})$ .

For equilibrium the $x$ and $y$ components separately sum to zero.

$x$ -direction:

$-T_1\sin 30^{\circ} + T_2\sin 45^{\circ} = 0 \Longrightarrow T_2 = T_1 \frac{\sin 30^{\circ}}{\sin 45^{\circ}} = T_1 \times \frac{0.5}{0.7071} = 0.7071\,T_1.$

$y$ -direction:

$T_1\cos 30^{\circ} + T_2\cos 45^{\circ} - 100 = 0.$

Substituting $T_2 = 0.7071\,T_1$ :

$T_1\times 0.8660 + 0.7071\,T_1 \times 0.7071 = 100,$

$0.866\,T_1 + 0.500\,T_1 = 100,$

$1.366\,T_1 = 100 \Longrightarrow T_1 = 73.2\,\text{N}.$

Then $T_2 = 0.7071 \times 73.2 = 51.8\,\text{N}$ .

Check: $T_1\cos 30^{\circ} + T_2\cos 45^{\circ} = 73.2 \times 0.866 + 51.8 \times 0.7071 = 63.4 + 36.6 = 100.0\,\text{N}$ . ✓

The $30^{\circ}$ string carries more tension because it is closer to vertical and therefore takes a larger share of the vertical weight.

The Component Method as a Universal Tool

The component method works for any number of vectors at any angles:

Choose coordinate axes (usually horizontal and vertical, but rotate to parallel/perpendicular for inclined-plane problems).
For each vector, compute its components using $\cos$ and $\sin$ of the angle from the chosen $x$ -axis.
Sum with signs all $x$ -components to get $R_x$ .
Sum with signs all $y$ -components to get $R_y$ .
Magnitude: $|\mathbf{R}| = \sqrt{R_x^{2} + R_y^{2}}$ .
Direction: $\theta = \arctan(R_y/R_x)$ — with a quadrant check from the signs of $R_x$ and $R_y$ .

Quadrant care with arctan

The calculator's $\arctan$ function returns angles in $(-90^{\circ}, +90^{\circ})$ only. This corresponds to the first and fourth quadrants (positive $R_x$ ). If $R_x$ is negative — i.e. the resultant points into the second or third quadrant — you must adjust:

$R_x$ sign	$R_y$ sign	Quadrant	True angle
$+$	$+$	1st (NE)	$\arctan(R_y/R_x)$
$-$	$+$	2nd (NW)	$180^{\circ} + \arctan(R_y/R_x)$ (= $180^{\circ}$ − absolute arctan)
$-$	$-$	3rd (SW)	$180^{\circ} + \arctan(R_y/R_x)$
$+$	$-$	4th (SE)	$\arctan(R_y/R_x)$ (already negative)

In OCR mark schemes the direction is usually stated as " $X$ N of E" or " $Y$ S of W" — the directional descriptor handles the quadrant ambiguity, so always sketch the resultant before quoting.

Worked Example 6 — four-force problem

Four forces act on a particle: $10\,\text{N}$ east, $15\,\text{N}$ north, $8\,\text{N}$ at $30^{\circ}\,\text{S of W}$ , $12\,\text{N}$ at $45^{\circ}\,\text{above W}$ (i.e. $45^{\circ}\,\text{N of W}$ ). Find the resultant.

Solution. Set $+x =$ east, $+y =$ north.

Force	$F_x$	$F_y$
$10\,\text{N E}$	$+10$	$0$
$15\,\text{N N}$	$0$	$+15$
$8\,\text{N}$ at $30^{\circ}\,\text{S of W}$	$-8\cos 30^{\circ} = -6.928$	$-8\sin 30^{\circ} = -4.00$
$12\,\text{N}$ at $45^{\circ}\,\text{N of W}$	$-12\cos 45^{\circ} = -8.485$	$+12\sin 45^{\circ} = +8.485$

Sum:

$R_x = 10 + 0 - 6.928 - 8.485 = -5.413\,\text{N}.$

$R_y = 0 + 15 - 4.00 + 8.485 = +19.485\,\text{N}.$

Magnitude:

$|\mathbf{R}| = \sqrt{5.413^{2} + 19.485^{2}} = \sqrt{29.30 + 379.66} = \sqrt{408.96} \approx 20.2\,\text{N}.$

Direction: $R_x < 0$ , $R_y > 0$ — second quadrant (NW). The angle from the negative $x$ -axis is $\arctan(19.485/5.413) = \arctan(3.600) = 74.5^{\circ}$ . So the resultant lies $74.5^{\circ}\,\text{N of W}$ .

Answer: $\mathbf{R} \approx 20\,\text{N}$ at $74.5^{\circ}\,\text{N of W}$ .

The method handles four vectors at arbitrary angles in a few lines — no scale drawing, no cosine rule. Scaling to twenty vectors would be just as quick.

Worked Example 7 — Resolving Across a Smooth Pulley

A $2.0\,\text{kg}$ block on a horizontal table is connected via a string over a smooth pulley to a hanging $1.0\,\text{kg}$ mass. The pulley is slightly raised, so the string makes $15^{\circ}$ above the horizontal as it leaves the block. Find the horizontal and vertical components of the tension on the block, assuming static equilibrium of the hanging mass.

Solution. Static equilibrium of the hanging mass: $T = mg = 1.0 \times 9.81 = 9.81\,\text{N}$ . The block experiences this $T$ pulling at $15^{\circ}$ above horizontal.

$T_x = T\cos 15^{\circ} = 9.81 \times 0.9659 = 9.48\,\text{N}.$

$T_y = T\sin 15^{\circ} = 9.81 \times 0.2588 = 2.54\,\text{N}\,(\text{upward}).$

The $2.54\,\text{N}$ upward reduces the block's effective weight on the table (and hence the normal reaction, and hence the friction). The $9.48\,\text{N}$ horizontal is what accelerates the block. A common Mid-band slip is to use the full $9.81\,\text{N}$ horizontally without resolving — the $0.33\,\text{N}$ difference can swing a calculation.

A Mermaid Diagram — Resolution and Reconstruction

graph TD
    V[Vector F<br/>magnitude F, angle θ from x-axis] --> Resolve{Resolve}
    Resolve --> X[F_x = F cos θ]
    Resolve --> Y[F_y = F sin θ]
    X --> Sum[Sum F_x with other x-components<br/>to get R_x]
    Y --> Sum2[Sum F_y with other y-components<br/>to get R_y]
    Sum --> Recon{Reconstruct}
    Sum2 --> Recon
    Recon --> Mag["|R| = √(R_x² + R_y²)"]
    Recon --> Dir["θ = arctan(R_y / R_x)<br/>with quadrant check"]

Resolution and reconstruction are inverse operations. Any vector can be split into two perpendicular components; any two perpendicular components can be recombined into a single vector.

Specific Applications Across A-Level

The resolving technique appears everywhere. Some highlights:

Resolving Vectors

Resolving Vectors

The Idea: Any Vector Equals the Sum of Its Components

Where do cos and sin come from?

Sign conventions

Worked Example 1 — Basic Resolution

Worked Example 2 — A Projectile's Initial Velocity

Worked Example 3 — Resolving Weight on an Inclined Plane

Worked Example 4 — Non-Perpendicular Addition via Components

Worked Example 5 — Three Forces in Equilibrium

The Component Method as a Universal Tool

Quadrant care with arctan

Worked Example 6 — four-force problem

Worked Example 7 — Resolving Across a Smooth Pulley

A Mermaid Diagram — Resolution and Reconstruction

Specific Applications Across A-Level

More in Physics