Forces and Newton’s First Law — Edexcel A-level Mathematics (Applied)

SPEC 8.1 • FORCES AND NEWTON'S LAWS

Forces and Newton's First Law

Understanding forces, force diagrams, and the principle of equilibrium

What is a Force?

Definition

A force is a push or pull that can change an object's velocity (speed or direction) or cause deformation.

Forces are vector quantities — they have both magnitude and direction.

Magnitude

Measured in Newtons (N)

1 N = 1 kg m s⁻²

Direction

Must always be specified

Often given as angles or compass bearings

Key Insight

Forces arise from interactions between objects. An isolated object cannot exert a force on itself.

Types of Forces

ForceSymbolDescriptionDirection Diagram

Weight

W

Normal Reaction

R

Tension

T

Thrust

P

Friction

F

Air Resistance

D

Modelling Assumptions

In mechanics, we simplify real-world problems using standard assumptions:

AssumptionMeaningImplication

ParticleObject has mass but no sizeIgnore rotation; forces act at one point

LightString/rod has no massTension is same throughout

InextensibleString doesn't stretchConnected objects have same speed

SmoothNo frictionNo frictional force present

RoughFriction present

F \leq μ R

applies

RigidObject doesn't deformShape remains constant

Exam Tip

Always state which assumptions you are using when setting up a model. Examiners award marks for explicitly identifying assumptions.

Newton's First Law

An object remains at rest or continues to move with constant velocity unless acted upon by a resultant force.

Key Implication

If the resultant force on an object is zero:

A stationary object stays stationary

A moving object continues at constant velocity (same speed, same direction)

Equilibrium Condition

\sum F = 0

The vector sum of all forces acting on the object is zero.

Static Equilibrium

Object at rest with zero resultant force

Dynamic Equilibrium

Object moving at constant velocity with zero resultant force

Force Diagrams

A free body diagram shows all forces acting on a single isolated object.

How to Draw Force Diagrams

Isolate the object of interest

Draw the object as a simple shape (dot or box)

Identify all forces acting ON the object

Draw arrows from the object showing each force

Label each force with its symbol and/or value

Book on a table: Weight

W

acts downward, Normal reaction

R

acts upward

Common Mistake

Only draw forces acting ON the object — not forces the object exerts on other things.

Worked Example01

A book of mass 2 kg rests on a horizontal table. Draw a force diagram and find the normal reaction force. Take $g = 9.8 m s^{- 2}$ .

1

The book is in equilibrium (at rest), so resultant force = 0

2

Resolving vertically (↑ positive):

R - W = 0

3

Therefore:

R = W = 19.6 N

AnswerNormal reaction

R = 19.6 N

(acting upward)

Worked Example02

A particle of mass 5 kg hangs in equilibrium from a light inextensible string attached to a ceiling. Find the tension in the string. Take $g = 9.8 m s^{- 2}$ .

1

Calculate weight:

W = m g = 5 \times 9.8 = 49 N

2

Particle is in equilibrium, so

\sum F = 0

3

Resolving vertically (↑ positive):

T - W = 0

4

Therefore:

T = W = 49 N

AnswerTension

T = 49 N

Worked Example03

A box of mass 8 kg rests on a rough horizontal floor. A horizontal force of 30 N is applied to the box, but it remains stationary. Find the friction force acting on the box.

1

The box remains stationary, so it is in static equilibrium

2

Resultant force = 0 in all directions

3

Resolving horizontally (→ positive):

30 - F = 0

4

Therefore:

F = 30 N

(acting opposite to the applied force)

AnswerFriction force = 30 N (acting to the left)

Worked Example04

A particle is acted upon by forces $(3 i + 4 j)$ N, $(- 5 i + 2 j)$ N, and $F$ N. The particle is in equilibrium. Find $F$ .

1

For equilibrium, the sum of all forces must be zero:

\sum F = 0

2

Sum of given forces:

(3 i + 4 j) + (- 5 i + 2 j) = - 2 i + 6 j

3

For equilibrium:

(- 2 i + 6 j) + F = 0

4

Therefore:

F = - (- 2 i + 6 j) = 2 i - 6 j

Answer

F = (2 i - 6 j) N

Worked Example05

A car of mass 1200 kg travels along a straight horizontal road at a constant velocity of 25 m s⁻¹. The resistance to motion is 800 N. Find the driving force.

1

Constant velocity means the car is in dynamic equilibrium

2

Therefore, resultant horizontal force = 0

3

Horizontally:

D - 800 = 0

4

D = 800 N

Constant velocity ⟹ zero acceleration ⟹ zero resultant force. The mass (1200 kg) is not needed here!

AnswerDriving force = 800 N

Worked Example06

Two particles A (mass 3 kg) and B (mass 2 kg) are connected by a light inextensible string passing over a smooth fixed pulley. The system is held at rest and then released. Draw force diagrams for both particles.

1

Particle A (3 kg): Weight

3 g

downward, Tension

T

upward

2

Particle B (2 kg): Weight

2 g

downward, Tension

T

upward

3

The tension is the same on both sides because the string is light and the pulley is smooth.

4

Since

3 g > 2 g

, particle A will accelerate downward and B will accelerate upward when released.

AnswerEach particle has tension

T

upward and weight downward (as shown).

Practice

Question 1: A crate of mass 50 kg is being pushed across a rough horizontal floor at constant velocity. The pushing force is 200 N horizontally. Draw a force diagram and find the friction force.

Solution:

Constant velocity ⟹ dynamic equilibrium ⟹ resultant force = 0

Horizontally:

200 - F = 0

Therefore:

F = 200 N

AnswerFriction force = 200 N (opposing motion)

Practice

Question 2: A particle is acted upon by forces

(5 i - 3 j)

N,

(- 2 i + 7 j)

N, and

(a i + b j)

N. Given that the particle is in equilibrium, find the values of

a

and

b

.

Solution:

For equilibrium: Sum of all forces =

0

(5 i - 3 j) + (- 2 i + 7 j) + (a i + b j) = 0

(5 - 2 + a) i + (- 3 + 7 + b) j = 0

(3 + a) i + (4 + b) j = 0

Comparing coefficients:

3 + a = 0 \Rightarrow a = - 3

4 + b = 0 \Rightarrow b = - 4

Answer

a = - 3

,

b = - 4

Practice

Question 3: A lamp of mass 4 kg hangs from two light strings. One string makes an angle of 60° with the vertical and has tension

T_{1}

. The other string is horizontal and has tension

T_{2}

. Find

T_{1}

and

T_{2}

. Take

g = 10 m s^{- 2}

.

Solution:

Weight:

W = 4 \times 10 = 40 N

Resolving vertically (↑ positive):

T_{1} cos 60° - 40 = 0

T_{1} \times 0.5 = 40

, so

T_{1} = 80 N

Resolving horizontally (← positive):

T_{2} - T_{1} sin 60° = 0

T_{2} = 80 \times \frac{3}{2} = 403 = 69.3 N

Answer

T_{1} = 80 N

,

T_{2} = 69.3 N

(or

403

N)

Summary

Key Concepts

ConceptDescription

ForceVector quantity (magnitude in N, direction specified)

Newton's First LawZero resultant force ⟹ constant velocity (or rest)

Equilibrium

\sum F = 0

(vector sum of all forces is zero)

Force DiagramShows all forces acting ON a single object

Force Types to Remember

Weight

W = m g

— always acts vertically downward

Normal reaction

R

— perpendicular to contact surface

Tension

T

— along string, away from object

Friction

F

— opposes motion (or tendency to move)

Exam Strategy

Always start by drawing a clear force diagram. State your modelling assumptions. Then resolve forces in two perpendicular directions (usually horizontal and vertical).