Scalars and Vectors

Understanding quantities with and without direction, and how to combine them mathematically

SPEC 3.4.1.1 • FORCE, ENERGY AND MOMENTUM

Scalars vs Vectors

Scalar

A quantity that has magnitude only.

Examples:

Mass

Speed

Distance

Energy

Temperature

Vector

A quantity that has magnitude and direction.

Examples:

Force / Weight

Velocity

Displacement

Acceleration

Momentum

Key Distinction

Speed is how fast something moves (scalar). Velocity is how fast and in what direction (vector).

Distance is total path length (scalar). Displacement is straight-line distance in a specific direction (vector).

Representing Vectors

Vectors are represented by arrows where:

Length represents the magnitude

Direction of the arrow shows the direction of the quantity

Notation

Vectors are written in bold (F) or with an arrow above (

F

). The magnitude is written in italics without bold:

F

or

∣ F ∣

.

Vectors in the Same Line

When vectors act along the same line, they can be added like scalars (but with direction):

Rules for Collinear Vectors

Same direction: Add the magnitudes

Opposite directions: Subtract the magnitudes; resultant points in direction of larger force

Adding Vectors: Scale Drawing

Example: Find the resultant of a 400 N force acting horizontally and a 300 N force acting at 37° to the horizontal.

Method

Choose a suitable scale (e.g., 1 cm = 50 N)

Draw first vector A to scale using a ruler

Draw second vector B from the tip of A (use a protractor for angles)

The resultant R goes from start of A to end of B — measure its length

Adding Vectors at Right Angles

When two vectors are perpendicular, use Pythagoras and trigonometry:

Formulas

R = A^{2} + B^{2}

tan θ = \frac{B}{A}

Important

These formulas only work for vectors at right angles (90°). For other angles, use scale drawings or the cosine rule.

Worked Example01

A boat travels 40 m east and then 30 m north. Calculate the magnitude and direction of the resultant displacement.

1

Identify the perpendicular vectors: East = 40 m, North = 30 m

2

Calculate magnitude using Pythagoras:

R = 4 0^{2} + 3 0^{2} = 1600 + 900 = 2500 = 50 m

3

Calculate direction:

tan θ = \frac{30}{40} = 0.75

4

θ = tan^{- 1} (0.75) = 36.9°

north of east

Answer: 50 m at 36.9° north of east

Resolving Vectors

Any vector can be split into two perpendicular components:

Resolution Formulas

F_{x} = F cos θ

(horizontal component)

F_{y} = F sin θ

(vertical component)

Method

Always draw a right-angled triangle with the original vector as the hypotenuse, then use trigonometry to find each component.

Worked Example02

A force of 50 N acts at 30° above the horizontal. Calculate the horizontal and vertical components of this force.

1

Identify: F = 50 N, θ = 30°

2

Horizontal component:

F_{x} = F cos θ = 50 \times cos 30° = 50 \times 0.866 = 43.3 N

3

Vertical component:

F_{y} = F sin θ = 50 \times sin 30° = 50 \times 0.5 = 25.0 N

Answer: Horizontal = 43.3 N, Vertical = 25.0 N

Vectors on Inclined Planes

For objects on slopes, resolve weight into components parallel and perpendicular to the slope:

Weight Components on Slope

Parallel to slope (down):

m g sin θ

Perpendicular to slope:

m g cos θ

where θ is the angle of the slope to the horizontal

Common Error

Students often mix up sin and cos. The component down the slope uses sin θ (opposite to the angle at the base).

Worked Example03

A 5.0 kg box rests on a slope inclined at 25° to the horizontal. Calculate the component of the box's weight acting parallel to the slope. (g = 9.8 m s⁻²)

1

Calculate weight:

W = m g = 5.0 \times 9.8 = 49 N

2

Component parallel to slope:

W_{∥} = W sin θ

3

W_{∥} = 49 \times sin 25° = 49 \times 0.423 = 20.7 N

Answer: 20.7 N (or 21 N to 2 s.f.)

Equilibrium of Forces

Equilibrium

An object is in equilibrium when the resultant force on it is zero.

The object is either at rest OR moving with constant velocity.

Two Forces

For equilibrium, two forces must be:

Equal in magnitude

Opposite in direction

Acting along the same line

Three Forces

For equilibrium, three forces must:

Form a closed triangle (tip-to-tail)

OR: Sum of components in each direction = 0

Problem Solving

For three-force problems, you can use either:

Closed triangle method (scale drawing or calculation)

Resolution method (resolve all forces, set ΣFx = 0 and ΣFy = 0)

Worked Example04

A 20 N weight is suspended by two strings. One string makes an angle of 40° with the vertical and has tension T₁. The other string is horizontal with tension T₂. Find T₁ and T₂.

1

Resolve T₁ into components: Vertical = T₁ cos 40°, Horizontal = T₁ sin 40°

2

Vertical equilibrium: T₁ cos 40° = 20 N

3

T_{1} = \frac{20}{c o s 40°} = \frac{20}{0.766} = 26.1 N

4

Horizontal equilibrium: T₂ = T₁ sin 40° = 26.1 × 0.643 = 16.8 N

Answer: T₁ = 26.1 N, T₂ = 16.8 N

Closed Triangle Method

For three forces in equilibrium, they form a closed triangle when drawn tip-to-tail:

Using the Triangle

Once you draw the triangle to scale, you can:

Measure unknown forces with a ruler

Measure unknown angles with a protractor

Or use trigonometry (sine rule, cosine rule) for calculation

Equilibrium: Worked Example

Three forces of 100 N, 150 N and 200 N act on a body as shown. Show that the resultant of these three forces is zero and that therefore the body is in equilibrium.

Conclusion

The three forces form a closed triangle when drawn tip-to-tail, proving the resultant is zero and the body is in equilibrium.

Worked Example05

Three coplanar forces act on a point: 8 N due east, 6 N due north, and an unknown force F. The system is in equilibrium. Find the magnitude and direction of F.

1

For equilibrium, F must balance the resultant of the other two forces

2

Resultant of 8 N east and 6 N north:

R = 8^{2} + 6^{2} = 100 = 10 N

3

Direction of resultant:

θ = tan^{- 1} (6/8) = 36.9°

north of east

4

F must be equal and opposite: F = 10 N at 36.9° south of west

Answer: F = 10 N at 36.9° south of west (or 216.9° from east)

Worked Example06

A particle is in equilibrium under three forces: 12 N acting horizontally to the right, 5 N acting vertically upward, and force P acting at angle θ below the horizontal to the left. Find P and θ.

1

Resolve P: Horizontal component = P cos θ (left), Vertical component = P sin θ (down)

2

Horizontal equilibrium: P cos θ = 12 N

3

Vertical equilibrium: P sin θ = 5 N

4

Divide equations:

tan θ = \frac{5}{12}

, so

θ = 22.6°

5

Find P:

P = \frac{12}{c o s 22.6°} = \frac{12}{0.923} = 13.0 N

Answer: P = 13.0 N at θ = 22.6° below horizontal

Worked Example07

Two forces of 15 N and 20 N act at 90° to each other. Using a scale drawing (1 cm = 5 N), find the magnitude and direction of the resultant.

1

Draw 15 N as 3 cm horizontally

2

From the tip, draw 20 N as 4 cm vertically upward

3

Draw resultant from start to end: measure length = 5 cm

4

Convert: 5 cm × 5 N/cm = 25 N

5

Measure angle with protractor: 53.1° from horizontal

Answer: 25 N at 53.1° above horizontal

Check: $1 5^{2} + 2 0^{2} = 625 = 25$ N ✓

Practice

Question 1: A velocity vector has magnitude 25 m s⁻¹ at 60° above the horizontal. Calculate the horizontal and vertical components.

1

Horizontal:

v_{x} = 25 cos 60° = 25 \times 0.5 = 12.5 m s^{- 1}

2

Vertical:

v_{y} = 25 sin 60° = 25 \times 0.866 = 21.7 m s^{- 1}

Answer: Horizontal = 12.5 m s⁻¹, Vertical = 21.7 m s⁻¹

Practice

Question 2: An object of mass 8.0 kg is placed on a frictionless slope inclined at 35° to the horizontal. Calculate the acceleration of the object down the slope. (g = 9.8 m s⁻²)

1

Force down slope:

F = m g sin θ = 8.0 \times 9.8 \times sin 35°

2

F = 78.4 \times 0.574 = 45.0 N

3

Acceleration:

a = \frac{F}{m} = \frac{45.0}{8.0} = 5.62 m s^{- 2}

4

Or directly:

a = g sin θ = 9.8 \times sin 35° = 5.62 m s^{- 2}

Answer: 5.6 m s⁻² (2 s.f.)

Practice

Question 3: A lamp of weight 15 N hangs from two cables. Each cable makes an angle of 50° with the vertical. Find the tension in each cable.

1

By symmetry, both tensions T are equal

2

Vertical component of each tension:

T cos 50°

3

Vertical equilibrium:

2 T cos 50° = 15

4

T = \frac{15}{2 c o s 50°} = \frac{15}{2 \times 0.643} = \frac{15}{1.286} = 11.7 N

Answer: T = 11.7 N in each cable

Summary

Key Concepts

ConceptKey Points

ScalarsMagnitude only (mass, speed, distance, energy)

VectorsMagnitude + direction (force, velocity, displacement, acceleration)

Adding vectorsTip-to-tail method; at 90°:

R = A^{2} + B^{2}

Resolving

F_{x} = F cos θ

,

F_{y} = F sin θ

Inclined planesParallel:

m g sin θ

, Perpendicular:

m g cos θ

EquilibriumResultant force = 0; closed triangle or ΣF = 0

Exam Skills

Always draw a clear diagram showing all forces

Choose sensible axes for resolution (often parallel/perpendicular to slope)

State direction as well as magnitude for vector answers

Check equilibrium conditions: ΣFx = 0 AND ΣFy = 0