Vectors in 2D — Science Sage

Scalars vs Vectors

A scalar has magnitude only (e.g. speed, mass, temperature). A vector has both magnitude and direction (e.g. force, velocity, displacement, acceleration).

Vectors are written in bold (F) or with an arrow overhead (F⃗). The magnitude is written as |F| or simply F (italic).

Direction conventions used in CAPS: Bearing from positive x-axis (measured anti-clockwise), compass bearing (N/S/E/W), or angle stated relative to a named direction (e.g. 30° N of E).

Resolving a Vector into Components

Any vector F at angle θ from the positive x-axis can be broken into perpendicular components:

Fx = F cosθ
Fy = F sinθ

The components are scalars (can be positive or negative). Always draw a sketch first to identify which quadrant θ falls in.

Tip: cos goes with the x-axis; sin goes with the y-axis. This only holds when θ is measured from the positive x-axis.

Adding Vectors Using Components

To add multiple vectors algebraically:

Resolve each vector into x and y components.
Sum all x-components: ΣFx
Sum all y-components: ΣFy
Find the magnitude of the resultant: R = √((ΣFx)² + (ΣFy)²)
Find the direction: θ = arctan(ΣFy / ΣFx) — then adjust for the correct quadrant.

R = √((ΣFx)² + (ΣFy)²)
θ = arctan(|ΣFy| / |ΣFx|) [adjust quadrant using signs of ΣFx and ΣFy]

Symbol	Meaning	Unit
F	Magnitude of force vector	N
Fx	x-component of force	N
Fy	y-component of force	N
R	Magnitude of resultant	N
θ	Direction angle from positive x-axis	°
ΣFx	Sum of all x-components	N
ΣFy	Sum of all y-components	N

Triangle and Parallelogram Laws

Triangle law: Draw vectors head-to-tail (one after the other). The resultant is the vector drawn from the tail of the first to the head of the last, closing the triangle.

Parallelogram law: Draw both vectors from the same point. Complete the parallelogram. The resultant is the diagonal from the common point.

Scale diagrams: Choose an appropriate scale (e.g. 1 cm = 10 N), draw accurately with a ruler and protractor, and measure the resultant. State the scale used.

Equilibrium and the Equilibrant

An object is in equilibrium when the net force on it is zero. For 2D problems this means:

ΣFx = 0 AND ΣFy = 0

A closed vector diagram (the tail of the last vector meets the head of the first) indicates equilibrium.

The equilibrant is the single force that would bring a system into equilibrium. It is equal in magnitude but opposite in direction to the resultant.

Worked Example — Three Forces in 2D

Three forces act on an object: F₁ = 40 N at 0°, F₂ = 30 N at 90°, F₃ = 50 N at 210°. Find the resultant.

F₁x = 40cos0° = 40 N, F₁y = 40sin0° = 0 N
F₂x = 30cos90° = 0 N, F₂y = 30sin90° = 30 N
F₃x = 50cos210° = −43.3 N, F₃y = 50sin210° = −25 N

ΣFx = 40 + 0 − 43.3 = −3.3 N
ΣFy = 0 + 30 − 25 = 5 N

R = √(3.3² + 5²) = √(10.9 + 25) = √35.9 ≈ 5.99 N
θ = arctan(5/3.3) ≈ 56.6° above negative x-axis → ≈ 123.4° from positive x-axis

IEB Extension — Lami's Theorem

When three concurrent coplanar forces are in equilibrium, Lami's Theorem states:

F₁/sinα = F₂/sinβ = F₃/sinγ

where α, β, γ are the angles opposite each force (i.e. the angle between the other two forces). This is a quick method when three forces meet at a point and are in equilibrium — no need to resolve into components.

IEB Extension — Introduction to 3D Components

In three dimensions, a vector has three mutually perpendicular components: Fx, Fy, Fz. The magnitude is R = √(Fx² + Fy² + Fz²). Direction is specified by angles to each axis. IEB may present introductory problems where a force in a plane tilted in 3D must be resolved into horizontal and vertical components first, then further resolved.

Vector Addition Sandbox

Mode

Vector 1 ●

Magnitude60 N

Angle30°

Vector 2 ●

Magnitude40 N

Angle120°

ΣFx

—N

ΣFy

—N

|R|

—N

θ of R

—°

Question 1 of 8

A force of 50 N acts at 37° above the positive x-axis. What is its x-component? (Use cos 37° ≈ 0.80)

Question 2 of 8

Two forces act on an object: 30 N due East and 40 N due North. What is the magnitude of the resultant?

Question 3 of 8

The equilibrant of a resultant force of 80 N at 45° NE is:

Question 4 of 8

A vector diagram forms a closed polygon. This means the object is in:

Question 5 of 8

ΣFx = −20 N and ΣFy = 20 N. In which quadrant does the resultant lie?

Question 6 of 8

A force of 100 N is directed at a compass bearing of 30° East of North (i.e. 060° bearing). Its northward component is closest to:

Question 7 of 8

Two forces act on an object: F₁ = 60 N at 0° and F₂ = 45 N at 90°. What is the magnitude of the third force, F₃, that must be added for the object to be in equilibrium?

Question 8 of 8

A boat can travel at 4 m·s⁻¹ in still water and is steered directly across a river (i.e. its heading is perpendicular to the banks). The river current flows at 3 m·s⁻¹, parallel to the banks. Calculate the magnitude of the boat's resultant velocity relative to the ground (an observer standing on the bank).

IEB Extended Questions

IEB Question 1

Three concurrent forces are in equilibrium. Force A = 60 N, Force B = 80 N, and the angle between A and B is 90°. Using Lami's Theorem, the magnitude of Force C is closest to:

IEB Question 2

For three concurrent forces in equilibrium, Lami's Theorem states that each force divided by the sine of the angle opposite to it is constant. Which angle is "opposite" to a force?

Answer all questions using the component method unless a scale diagram is specifically requested. Show all working including formulae, substitution, and units.

Question 1 — Scale Diagram (6 marks)

Two forces act on a bolt: 45 N at 0° (East) and 35 N at 60° (NE). Using a scale of 1 cm = 5 N, construct a scale diagram using the triangle law. Measure and state (a) the magnitude of the resultant and (b) the direction of the resultant as an angle measured from East.

Question 2 — Component Method (8 marks)

Three forces act on an object: F₁ = 80 N at 0°, F₂ = 60 N at 135°, F₃ = 50 N at 270°. (a) Resolve each force into x and y components. (b) Calculate ΣFx and ΣFy. (c) Find the magnitude of the resultant. (d) Find the direction of the resultant (angle from the positive x-axis, correctly adjusted for quadrant).

Question 3 — Equilibrant (5 marks)

A hanging sign is held in place by two cables. Cable 1 exerts 200 N at 120° from the positive x-axis; Cable 2 exerts 200 N at 60°. (a) Show that these two forces and the weight of the sign form a closed triangle (equilibrium). (b) Calculate the weight of the sign.

Question 4 — Bearing Problem (7 marks)

A ship travels 50 km on a bearing of 030° (N30°E), then 80 km on a bearing of 150° (S30°E). Using the component method, find (a) the total displacement (magnitude) and (b) the bearing of the final position from the starting point.

Question 5 — Equilibrium Design (4 marks)

An object of weight 120 N is supported by two forces F₁ and F₂. F₁ is horizontal (0°). F₂ acts at 30° above the horizontal. For the system to be in equilibrium: (a) write the two equilibrium equations (ΣFx = 0 and ΣFy = 0). (b) Solve for F₁ and F₂.

Question 6 — Four Cables on a Ring (6 marks)

Four cables pull on a metal ring, each at a different angle measured anticlockwise from the positive x-axis:

Cable	Force (N)	Angle (° from +x-axis)
A	100	0°
B	80	90°
C	60	200°
D	40	300°

(a) Copy the table and calculate the x-component and y-component of each cable's force (to 2 decimal places). (4 marks)
(b) Calculate ΣFx and ΣFy, and hence the magnitude of the resultant force on the ring. (2 marks)