Newton's Laws of Motion

Understanding the fundamental principles that govern all motion and forces

SPEC 3.4.1.5 • MECHANICS AND MATERIALS

Why Newton's Laws Matter

Isaac Newton published his three laws of motion in 1687. These laws form the foundation of classical mechanics and explain how forces affect the motion of objects.

Key Concept

Newton's laws apply to all objects as long as speeds are much less than the speed of light and objects are much larger than atoms.

First Law

Objects remain at rest or in uniform motion unless acted upon by a resultant force

Second Law

F = ma — Force equals mass times acceleration

Third Law

If a force acts on one body, an equal and opposite force acts on another body

Newton's First Law

First Law

A body remains at rest or in a state of uniform motion unless acted on by a resultant force.

This law introduces the concept of inertia — the tendency of an object to resist changes to its motion.

At Rest

If an object is stationary and there is no resultant force, it stays stationary. The forces are balanced.

Uniform Motion

If moving with no resultant force, the object continues at constant velocity — same speed, same direction.

Key Implication

The velocity of an object will only change if a resultant force is acting on the object. No resultant force means no acceleration.

Newton's Second Law

Second Law

Newton's second law states that:

Acceleration is directly proportional to the resultant force

Acceleration is inversely proportional to the mass

F = ma

QuantitySymbolSI Unit

Resultant ForceFNewton (N)

MassmKilogram (kg)

Accelerationam s⁻²

Definition of the Newton

One newton is the force that will give a mass of 1 kilogram an acceleration of 1 m s⁻².

Weight as a Special Case

Weight is the gravitational force acting on an object. Using Newton's second law:

Weight Formula

W = m g

where g = 9.81 m s⁻² near Earth's surface

Mass

Scalar quantity

Measured in kg

Same everywhere

Measure of inertia

Weight

Vector (acts downward)

Measured in N

Varies with location

Gravitational force

Exam Tip

Note that it is mass that is used in F = ma, measured in kg. Do not convert mass to weight!

Newton's Third Law

Third Law

If a force acts on one body, an equal and opposite force acts on another body.

or: Action and reaction are equal and opposite.

Properties of a Newton Pair

Act on two different bodies — this is crucial!

Same type of force — both gravitational, both contact, etc.

Equal in magnitude

Opposite in direction

Car pulls caravan forward; caravan pulls car backward — equal and opposite forces

Exam Tip

If one force of the pair is removed, the other vanishes too. You cannot have one without the other.

Identifying Newton Pairs

A book resting on a table involves TWO separate Newton pairs:

Newton Pair 1: Gravitational Forces

Earth pulls book downward (weight)

Book pulls Earth upward

Both gravitational, equal magnitude, opposite directions, different objects

Newton Pair 2: Contact Forces

Table pushes book upward (normal)

Book pushes table downward

Both contact forces, equal magnitude, opposite directions, different objects

Misconception Alert

The weight of the book and the normal force from the table are NOT a Newton pair!

They act on the same object (the book) and are different types of force. They only happen to be equal because the book is in equilibrium.

Worked Example01

A 250 kg box is pulled along a smooth road (no friction) using a force of 2000 N. What is the acceleration of the block?

1

Identify: F = 2000 N, m = 250 kg

2

Apply F = ma:

a = \frac{F}{m} = \frac{2000}{250} = 8 m s^{- 2}

Answer: a = 8 m s⁻²

Worked Example02

At the start of a 100 m race, a sprinter exerts a force of 1150 N with the rear foot and 800 N with the front foot on the starting blocks. If the sprinter has mass 83 kg, what is the initial acceleration?

1

Total horizontal force:

F = 1150 + 800 = 1950 N

2

Apply F = ma:

a = \frac{F}{m} = \frac{1950}{83} = 23.5 m s^{- 2}

Answer: a = 23 m s⁻² (2 s.f.)

Worked Example03

A rocket has a mass of 50,000 kg and its motors exert a thrust of 550,000 N. What is the initial vertical acceleration of the rocket? (g = 9.8 m s⁻²)

1

Weight:

W = m g = 50000 \times 9.8 = 490000 N

2

Resultant upward force:

F_{n e t} = 550000 - 490000 = 60000 N

3

Acceleration:

a = \frac{60000}{50000} = 1.2 m s^{- 2}

Answer: a = 1.2 m s⁻² upward

Worked Example04

A 1200 kg car experiences a driving force of 4500 N and a resistive force of 900 N. Calculate the acceleration.

1

Resultant force:

F_{n e t} = 4500 - 900 = 3600 N

2

Apply F = ma:

a = \frac{3600}{1200} = 3.0 m s^{- 2}

Answer: a = 3.0 m s⁻²

Worked Example05

A 2.0 kg block slides down a frictionless incline at 30° to the horizontal. Calculate the acceleration. (g = 9.81 m s⁻²)

1

Force parallel to slope:

F = m g sin θ = m g sin 30°

2

Apply F = ma:

ma = m g sin 30°

→

a = g sin 30°

3

Calculate:

a = 9.81 \times 0.5 = 4.91 m s^{- 2}

Answer: a = 4.9 m s⁻² (Note: independent of mass!)

Worked Example06

Two masses of 3.0 kg and 5.0 kg are connected by a light string over a frictionless pulley. Calculate the acceleration and tension. (g = 9.81 m s⁻²)

1

For 3.0 kg (rises):

T - 3.0 g = 3.0 a

2

For 5.0 kg (falls):

5.0 g - T = 5.0 a

3

Add:

5.0 g - 3.0 g = 8.0 a

→

a = \frac{2.0 \times 9.81}{8.0} = 2.45 m s^{- 2}

4

Find T:

T = 3.0 (9.81 + 2.45) = 36.8 N

Answer: a = 2.5 m s⁻², T = 37 N

Worked Example07

A 60 kg person stands on scales in a lift. Calculate the scale reading when the lift accelerates: (a) upward at 2.0 m s⁻², (b) downward at 2.0 m s⁻². (g = 9.81 m s⁻²)

1

Scales measure normal force N. Apply N - W = ma (up positive)

2

(a) Up:

N = m (g + a) = 60 (9.81 + 2.0) = 709 N

3

(b) Down:

N = m (g - a) = 60 (9.81 - 2.0) = 469 N

Answer: (a) 710 N (feels heavier), (b) 470 N (feels lighter)

Free-Body Diagrams

A free-body diagram shows all forces acting on a single object. Essential for applying F = ma.

Common Forces

W — Weight (always downward)

N — Normal force (⊥ to surface)

T — Tension (along string)

F — Applied/driving force

f — Friction (opposes motion)

D — Drag/air resistance

Object with applied force

Drawing Steps

Identify the object of interest

Draw as a simple shape

Add all forces as arrows from the object

Label each force; arrow length shows relative magnitude

Forces on Inclined Planes

When on a slope, resolve the weight into components parallel and perpendicular to the surface.

Key Equations

Parallel to slope (down): m g sin θ

Perpendicular to slope: m g cos θ

Normal force:

N = m g cos θ

(for equilibrium perpendicular to slope)

Practice

Question 1: A car of mass 1000 kg pulls a caravan of mass 800 kg. The driving wheels exert a force of 8000 N and total resistance is 3000 N. (a) What is the net accelerating force? (b) What is the acceleration? (c) What is the force of the car on the caravan?

a

Net force:

F_{n e t} = 8000 - 3000 = 5000 N

b

Total mass = 1800 kg.

a = \frac{5000}{1800} = 2.78 m s^{- 2}

c

For caravan:

T = 800 \times 2.78 = 2220 N

Answer: (a) 5000 N, (b) 2.8 m s⁻², (c) 2200 N

Practice

Question 2: A 4.0 kg block rests on a plane inclined at 25°. The coefficient of friction is 0.35. Calculate the acceleration when released. (g = 9.81 m s⁻²)

1

Weight parallel:

m g sin 25° = 4.0 \times 9.81 \times 0.423 = 16.6 N

2

Normal:

N = m g cos 25° = 4.0 \times 9.81 \times 0.906 = 35.6 N

3

Friction:

f = 0.35 \times 35.6 = 12.5 N

4

Resultant:

16.6 - 12.5 = 4.1 N

.

a = 4.1/4.0 = 1.0 m s^{- 2}

Answer: a = 1.0 m s⁻²

Practice

Question 3: A 2.0 kg mass on a frictionless table is connected by string over a pulley to a 3.0 kg hanging mass. Find: (a) the acceleration, (b) the tension. (g = 9.81 m s⁻²)

1

For 2.0 kg (horizontal):

T = 2.0 a

2

For 3.0 kg (vertical):

3.0 g - T = 3.0 a

→

29.43 - T = 3.0 a

3

Substitute:

29.43 = 5.0 a

→

a = 5.89 m s^{- 2}

4

Tension:

T = 2.0 \times 5.89 = 11.8 N

Answer: (a) 5.9 m s⁻², (b) 12 N

Summary

Newton's Three Laws

LawStatementApplication

FirstA body remains at rest or in uniform motion unless acted on by a resultant forceEquilibrium, terminal velocity

SecondF = maForce/acceleration calculations

ThirdForces come in pairs: equal, opposite, same type, different objectsIdentifying Newton pairs

Key Exam Skills

Draw free-body diagrams showing all forces on ONE object

Resolve forces on inclines using sin θ and cos θ

Newton pairs act on DIFFERENT objects and are the SAME type

Use simultaneous equations for connected systems