Resolving Vectors

The previous lesson taught you how to add vectors using the triangle method. In this final lesson we learn the inverse — and arguably more powerful — operation of resolving a single vector into two perpendicular components. Once a vector is broken 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 straightforward substitution.

This technique underpins everything in the mechanics, electricity, waves and fields modules. Master it, and you will save yourself enormous effort across the entire A-Level course.

---

The Idea: Any Vector Equals the Sum of Its Components

Given any vector F making an angle θ with a chosen horizontal axis, we can draw a right-angled triangle with F as its hypotenuse, one side along the horizontal axis, and the other vertical.

graph TD
    O((O)) -->|"F"| P((P))
    O -->|"F_x"| A((A))
    A -->|"F_y"| P

Using basic trigonometry on this right-angled triangle:

F_x = F cos θ (horizontal component) F_y = F sin θ (vertical component)

And by Pythagoras:

F² = F_x² + F_y²

And by trigonometry for the angle:

tan θ = F_y / F_x

These four relationships are the heart of this lesson. You must know them absolutely.

Where Do cos and sin Come From?

Think of the right-angled triangle. The angle θ is measured between the horizontal and the vector F. The adjacent side to θ is the horizontal (length F cos θ). The opposite side is the vertical (length F sin θ). Remember SOHCAHTOA:

• cos θ = adjacent / hypotenuse ⇒ adjacent = hypotenuse × cos θ ⇒ F_x = F cos θ
• sin θ = opposite / hypotenuse ⇒ opposite = hypotenuse × sin θ ⇒ F_y = F sin θ

If the angle were measured from the vertical instead of the horizontal, the cos and sin would swap — always check which axis the angle is measured from.

Sign Conventions

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

• Positive x = to the right (east)
• Positive y = upwards (north)

A force acting at 135° from the positive x-axis (that is, 45° above the negative x-axis) has:

• F_x = F cos 135° = −F/√2 (negative — points left)
• F_y = F sin 135° = +F/√2 (positive — points up)

You do not need to memorise cos and sin of obtuse angles — simply check the sign from a quick sketch of where the vector points.

---

Worked Example 1 — Basic Resolution

A 50 N force acts at 30° above the horizontal. Find its horizontal and vertical components.

Solution:

F_x = F cos θ = 50 × cos 30° = 50 × 0.8660 = 43.3 N (horizontal) F_y = F sin θ = 50 × sin 30° = 50 × 0.5000 = 25.0 N (vertical)

Check: √(43.3² + 25.0²) = √(1874.9 + 625) = √2499.9 ≈ 50 N ✓

---

Worked Example 2 — A Projectile's Initial Velocity

A ball is launched at 20 m s⁻¹ at 40° above the horizontal. Calculate its initial horizontal and vertical velocities.

Solution:

v_x = v cos θ = 20 × cos 40° = 20 × 0.7660 = 15.3 m s⁻¹ v_y = v sin θ = 20 × sin 40° = 20 × 0.6428 = 12.9 m s⁻¹

These two components evolve independently under gravity: v_x stays constant (no air resistance) while v_y decreases as −g × t. This is the foundation of all projectile motion analysis — see the mechanics module for a full treatment.

---

Worked Example 3 — Resolving Weight on an Inclined Plane

A 5.0 kg block rests on a frictionless slope inclined at 25° to the horizontal. Calculate the components of its weight parallel to and perpendicular to the slope.

Solution:

Weight magnitude: W = mg = 5.0 × 9.81 = 49.05 N, acting vertically downward.

The key move on an incline is to rotate your coordinate system so that the axes are parallel and perpendicular to the slope, rather than horizontal and vertical. The angle of the slope θ is then the angle the weight makes with the normal to the slope.

• Component parallel to slope (down the slope): W sin θ
• Component perpendicular to slope (into the slope): W cos θ

Calculate:

W_∥ = 49.05 × sin 25° = 49.05 × 0.4226 = 20.7 N (down the slope) W_⊥ = 49.05 × cos 25° = 49.05 × 0.9063 = 44.5 N (into the slope)

The normal reaction force from the slope equals W_⊥ = 44.5 N (Newton's third law). The block accelerates down the slope under the unbalanced 20.7 N (Newton's second law), giving a = 20.7 / 5.0 = 4.14 m s⁻².

Exam Tip: The single hardest thing about inclined plane questions is remembering that sin θ goes with the parallel component and cos θ goes with the perpendicular component — the opposite of what you might expect from a vector that is at angle θ from horizontal. The reason is that the slope itself is tilted by θ, so the weight (still vertical) now makes angle θ with the slope's normal, not with the slope itself.

---

Worked Example 4 — Non-Perpendicular Vector Addition Using Components

Return to Lesson 9, Worked Example 3: forces of 10 N and 15 N with 60° between them. Solve using components this time.

Solution:

Place the 15 N force along the x-axis. Then:

• F₁ (15 N along x): F₁_x = 15, F₁_y = 0
• F₂ (10 N at 60° above x-axis): F₂_x = 10 cos 60° = 5.0, F₂_y = 10 sin 60° = 8.66

Sum:

R_x = 15 + 5.0 = 20.0 R_y = 0 + 8.66 = 8.66

Magnitude:

|R| = √(20.0² + 8.66²) = √(400 + 75) = √475 = 21.8 N

Direction:

θ = arctan(8.66 / 20.0) = arctan(0.433) = 23.4° above the 15 N force

This matches the cosine rule answer from Lesson 9 exactly — but with much simpler mathematics. This is why the component method is the universal tool for vector problems at A-Level and beyond.

---

Worked Example 5 — Three Forces in Equilibrium

A mass hangs from two strings, one at 30° to the vertical and the other at 45° to the vertical, on opposite sides. The weight of the mass is 100 N. Find the tensions in the two strings.

Solution:

Let T₁ be the tension in the 30° string and T₂ be the tension in the 45° string.

Resolve both strings and the weight horizontally and vertically. The weight (100 N downward) has components (0, −100).

• T₁ components: (−T₁ sin 30°, +T₁ cos 30°)
• T₂ components: (+T₂ sin 45°, +T₂ cos 45°)

(Here "+x" means to the right; T₁ pulls up-left, T₂ pulls up-right.)

For equilibrium:

x-direction: T₂ sin 45° − T₁ sin 30° = 0

T₂ × 0.7071 = T₁ × 0.5 T₂ = 0.7071 T₁ / 1.0 → hmm, wait, let me redo: T₂ = T₁ × (0.5 / 0.7071) = 0.7071 T₁

y-direction: T₁ cos 30° + T₂ cos 45° − 100 = 0

T₁ × 0.8660 + T₂ × 0.7071 = 100

Substitute T₂ = 0.7071 T₁:

0.8660 T₁ + 0.7071 × 0.7071 T₁ = 100 0.8660 T₁ + 0.5 T₁ = 100 1.366 T₁ = 100 T₁ = 73.2 N

Then:

T₂ = 0.7071 × 73.2 = 51.8 N

Check: vertical components sum to 73.2 × 0.866 + 51.8 × 0.7071 = 63.4 + 36.6 = 100.0 N ✓

The 30° string is under more tension than the 45° string because it carries a larger share of the vertical load.

---

The Component Method as a Universal Tool