Newton's Laws

Newton’s Laws provide the fundamentals for understanding forces and the relationship between forces and motion.

Use this page to revise the following concepts within Newton's Laws:

Force and motion basics
Vector representation
Newton’s First Law
Newton’s Second Law
Newton’s Third Law

Force and motion basics

Forces are everyday interactions. They are pushes and pulls that act on objects and are responsible for a change in either an object's motion, shape or both. Forces are measured in newtons $(\mathrm{N})$ and are a vector quantity, which means they have both a magnitude (size) and a direction.

Types of forces

There are many types of forces which fall into two broad categories: contact forces – forces that occur between objects that physically touch; and non-contact forces – forces that occur between objects from a noticeable distance.

Contact forces

There are four main contact forces to consider (ignoring air resistance):

Applied forces are forces exerted by one object on another through direct physical contact.
For example, when you push your hand against a door, you apply a force that acts on the door, causing it to move.
Normal forces are the perpendicular forces exerted between surfaces in contact.
For example, when a book rests on a desk, the desk exerts an upward normal force that counteracts the downward pull of gravity, keeping the book in equilibrium.
For an object resting on a flat surface, the magnitude of the normal force is equal to the gravitational force.
When objects are on an inclined surface, the normal force can be calculated using the perpendicular vector components.
Friction, or frictional force, is the resistance between two surfaces that oppose the motion of an object.
For example, pushing a heavy box across a floor can be challenging because friction acts in the opposite direction of the applied force, resisting movement.
Tension, or tensile force, is an internal force within a material that resists being stretched and is used to pull an object.
For example, when a rope is used to lift a weight, the tension within the rope transmits the pulling force.

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Non-contact forces

There are three main types of non-contact forces to consider:

Gravitational force is the force on an object due to the acceleration of gravity. At the surface of the Earth, it is calculated as a product of the mass of the object and the acceleration due to gravity:
\[F_{g} = mg\]
Where:
- $F_{g}$ = force due to gravity
- $m$ = mass
- $g$ = acceleration due to gravity
Gravity is the attractive force between objects with mass. As the distance from a mass increases, the gravitational force decreases rapidly. At substantial distances, the force can no longer be modelled by the equation above, requiring a more comprehensive approach that will be explored in the context of gravitational fields.
Electromagnetic forces govern the interactions between charged particles, electric fields, and magnetic fields. These forces arise through various mechanisms, including electric forces between charges, magnetic forces on moving charges, electromagnetic induction, and the propagation of electromagnetic waves. A deeper exploration of these interactions is covered in the sections on electric and magnetic fields .
There are two types of nuclear forces:
- Strong Nuclear Force
  The strong nuclear force, or strong interaction, is the force that holds protons and neutrons together in the nucleus of an atom. It is the strongest of the fundamental forces but acts only over very small distances. This force overcomes the repulsion between positively charged protons, ensuring the stability of atomic nuclei.
- Weak Nuclear Force
  The weak nuclear force, or weak interaction, is responsible for certain types of radioactive decay, such as beta decay. It acts over even smaller distances than the strong nuclear force and can change the type of quarks inside subatomic particles, leading to the transformation of one particle into another.
This brief introduction to nuclear forces provides completeness to the section but will not be explored further.

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Vector representation

Vector basics

Vectors are mathematical objects that represent quantities with both magnitude (size) and direction. They are typically depicted as arrows, where the arrow's length corresponds to the magnitude, and its orientation indicates the direction.

Some examples of vector quantities include:

force
velocity
displacement

These differ to scalar quantities such as mass or temperature which have a magnitude, but no direction.

Vector Notation

Vectors are typically depicted as arrows but may also be represented in written form. Common notation for vectors in written from, using $v$ as an example, include:

Bold letters: $\pmb{v}$
An arrow above the symbol: $\vec{v}$
The magnitude of a vector may also be represented by vertical bars: $\left|v\right|$

Vectors may not always be represented using special notation, but they can still be identified by their units, as they represent quantities with both magnitude and direction.

Basic vector operations

Vector addition and subtraction

Vectors can be added together using the head-to-tail method, creating a resultant vector. The resultant vector is drawn from the tail of the first vector to the head of the last vector, representing the sum of the two vectors.

For example, vector $a$ can be added to vector $b$, head-to-tail, to produce a resultant vector $c$:

A diagram illustrating vector addition. The left side shows two separate vectors, a and b. The right side shows the vector sum using the tip-to-tail method: vector a is followed by vector b, forming a triangle. The resultant vector c (in red) represents their sum, with the equation c = a + b written below

The same head-to-tail method can also be used for the subtraction of vectors.

Scalar multiplication of vectors

Vectors can also be multiplied by a scalar , which increases or decreases their magnitude without changing their direction (unless the scalar is negative, in which case the direction is reversed).

For example, if vector $\pmb{b}$ is multiplied by $2$, the result is $\pmb{2b}$, a vector that is twice the length of $\pmb{b}$ but in the same direction:

A diagram illustrating scalar multiplication of a vector. The left side shows vector b, and x 2, while the right side shows the result of multiplying b by 2, labelled as 2b (in red). The new vector is in the same direction as b but twice as long. The equation b × 2 = 2b is written below.>

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Vector components

Just as vectors can be represented by the sum of other vectors, a single vector can also be decomposed into its horizontal and vertical components.

For example, vector $\pmb{a}$ below can be represented as the sum of $\pmb{a}_{\pmb{H}}$ the horizontal component of $\pmb{a}$ and $\pmb{a}_{\pmb{V}}$ the vertical component of $\pmb{a}$.

A diagram illustrating the decomposition of a vector a into its horizontal and vertical components. The diagonal vector a (blue) is represented as the hypotenuse of a right triangle. The horizontal component a_H and the vertical component a_v are shown in red, forming a right angle at their intersection.

Since the horizontal and vertical components of a vector are perpendicular, they can be expressed in terms of the original vector using basic trigonometry, provided the angle to the horizontal or vertical is known.

A diagram showing the decomposition of a vector a into horizontal and vertical components using trigonometry. The vector a (blue) forms the hypotenuse of a right triangle with angle θ at the base. The horizontal component is labelled a_H = a cos(θ), and the vertical component is labelled a_v = a sin(θ). Both components are shown in red, forming a right-angled triangle.

Worked Example

A cart is being pulled to the right with an $80 \mathrm{~N}$ force applied through a rope angled $30^\circ$ above the horizontal.

person pulling a cart through a rope with a force of 80 N which is at an angle of 30° to the horizontal

Determine the horizontal component of the force acting on the cart.

The force applied can be broken into it’s respective horizontal and vertical components:

A right-angled triangle illustrating the decomposition of an 80 N force vector. The force vector (blue) forms the hypotenuse, with a 30° angle at the base. The horizontal and vertical components (red) form a right angle, representing the force’s decomposition into its perpendicular components

The horizontal component can be calculated as:

\[80 \cos(30) = 69.28 \mathrm{~N}\]

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Force diagrams

Drawing forces

Often diagrams are used to illustrate forces acting on a system. There are some general drawing conventions that should be adhered to depending on the type of force.

Type of force	Arrows	Example
Contact forces	Arrows drawn at the point where it touches the object
Non-contact forces	Arrows drawn from the centre of mass

In all cases, the magnitude of arrows should be represented appropriately. This means they do not need to be drawn to scale, but their relative sizes must be accurate in comparison to other forces in the diagram. For example, if the normal force and gravitational force are equal in magnitude, their arrows should be the same length, ensuring a correct visual representation of force balance.

Forces should always be correctly labeled in diagrams. Standard labels exist for common forces, such as:

Normal force: $F_N$
Gravitational force: $F_g$
Frictional force: $F_f$
Tension force: $F_T$

Any other forces should include an appropriate subscript to clearly describe their source or effect. For example:

$F_{\text{push}}$ for a pushing force
$F_{\text{engine}}$ for an engine force from a car

Free body diagrams

Drawing forces on a diagram can sometimes make it difficult to clearly understand how they interact to create motion in an object. This can be simplified using a free body diagram (FBD).
A free body diagram represents the object as a point, with all forces acting on it drawn as arrows originating from that point.

For example, the diagram below shows a person pulling a box that is resting on a surface.

A person pulling a cart through a rope with a force of 80 N which is at an angle of 30° to the horizontal. Diagram also illustrates normal force, gravitational force, and friction force

A free body diagram to represent the forces acting on object in a diagram (the box) can be illustrated below.

A free-body diagram showing an 80 N force acting at a 30° angle above the horizontal, which is represented by a dashed line. Three other forces are present: F_N (normal force) acting upward, F_g (gravitational force) acting downward, and F_f (friction force) acting leftward.

This shows all the forces acting on the object with a focus on their magnitude and direction.

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Motion concepts

Motion describes the movement of an object in space. To understand how forces interact to cause motion and how to describe complex motion, the basic language and tools must be reviewed.

Key terms

Motion is inherently relative as it is always measured in relation to a point of reference. An object's position in space is the reference point for measuring motion. Generally, the starting point of an object is used for this reference, where all other measurements are derived.
Position can also be described within a plane or in three-dimensional space, but this requires a more advanced understanding of mathematics.
Distance and displacement both describe the change in an object's position, but they differ in how they are measured. The SI unit for both distance and displacement is the metre $(\mathrm{m})$.
Distance
- A scalar quantity that represents the total path travelled
- Typically denoted by $d$ to distinguish from displacement.
Displacement
- A vector quantity that describes the difference between the initial and final positions
- Typically denoted by $s$.
\[
\begin{align*}
s &= \Delta x \\
&= x_2 - x_1
\end{align*}
\]
Where:
- $s$ = displacement $(\mathrm{m})$
- $\Delta{x}$ = change in position
- $x_2$ = final position
- $x_1$ = initial position
Speed and velocity describe the rate of change in total distance or displacement over time respectively.
Speed
- A scalar quantity
- Typically denoted by $v$.
\[\begin{align}v &= \frac{d}{\Delta{t}} \\ &=\frac{d}{t_2 - t_1}\end{align}\]
Where
- $v$ = speed $\left(\mathrm{ms^{-1}}\right)$
- $d$ = total distance travelled $(\mathrm{m})$
- $\Delta{t}$ = time interval $(\mathrm{s})$
- $t_1$ = initial time $(\mathrm{s})$
- $t_2$ = final time $(\mathrm{s})$
Velocity
- Describes the change in displacement over time.
- A vector quantity
- Typically denoted by $v$ as well.
\[
\begin{align*}
v &= \frac{\Delta s}{\Delta t} \\
&= \frac{s_2 - s_1}{t_2 - t_1}
\end{align*}
\]
Where:
- $v=$ velocity $(m s^{-1})$
- $\Delta{s}=$ change in displacement $(\mathrm{m})$
- $s_1=$ initial displacement $(\mathrm{m})$
- $s_2=$ final displacement $(\mathrm{m})$
Acceleration describes the rate of change of velocity over time. It is denoted by $a$, and is derived from velocity, making it a vector quantity. It can be calculated using the formula:
\[\begin{align}a &= \frac{\Delta{v}}{\Delta{t}} \\ &= \frac{v_2 - v_1}{t_2 - t_1}\end{align}\]
Where:
- $a$ = acceleration $\left(\mathrm{m s}^{-2}\right)$
- $\Delta{v}$ = change in velocity $\left(\mathrm{m s}^{-1}\right)$
- $v_1$ = initial velocity $\left(\mathrm{m s}^{-1}\right)$
- $v_2$ = final velocity $\left(\mathrm{m s}^{-1}\right)$

Straight line motion

Straight line motion problems can be solved using a set of kinematic equations that describe motion along a straight path. These equations are based on two key assumptions:

Acceleration is constant
Motion is linear.

The equations involve five fundamental motion variables:

Displacement $(s)$
Initial velocity $(u)$
Final velocity $(v)$
Acceleration $(a)$
Time $(t)$.

There are five kinematic equations that describe the relationships between these variables. Each equation includes four of the five variables, allowing for the determination of an unknown value when any three are known. These equations are derived from the definitions of acceleration and average velocity:

Acceleration is the rate of change of velocity over time. Using the variables above, it can be described as:

\[a = \frac{v - u}{t}\]

Average velocity can be described in two ways:

\[
\begin{align*}
v_{\text{average}} &= \frac{u + v}{2} \\
v_{\text{average}} &= \frac{s}{t}
\end{align*}
\]

Substitutions and rearrangements of these equations give us the following kinematic equations:

Equation	Missing variable
$v = u + at$	$s$
$s = vt - \frac{1}{2}at^2$	$u$
$s = ut + \frac{1}{2}at^2$	$v$
$s = \frac{u +v}{2}t$	$a$
$v^2 = u^2 + 2as$	$t$

Worked Example

A car is travelling at $20 \mathrm{~ms}^{-1}$ when the driver suddenly applies the brakes, bringing the car to a complete stop in $2 \mathrm{~s}$. Assuming the car undergoes constant acceleration, determine the acceleration of the car and the distance the car travels before stopping.

Identify the variables:

\[ \begin{aligned} s &= ? \\ u &= 20 \mathrm{~ms}^{-1} \\ v &= 0 \mathrm{~ms}^{-1} \text{, since the car comes to a complete stop} \\ a &= ? \\ t &= 2 \mathrm{~s} \end{aligned} \]

To determine the unknown variables, use the equation that involves the unknown and the three known variables.

To find the acceleration $a$, with variables $u$, $v$, and $t$, use the equation without $s$:

\[ v = u + at \]

Substitute known variables:

\[ 0 = 20 + a(2) \]

Rearrange for $a$:

\[ \begin{aligned} a &= -\frac{20}{2} \\ &= -10 \mathrm{~ms}^{-2} \end{aligned} \]

The negative acceleration indicates that the object is decelerating, meaning it is slowing down.

Since there are now four known variables, any equation involving $s$ can be used. Using the equation that excludes $a$:

\[ s = \frac{u + v}{2}t \]

Substitute known variables:

\[ \begin{aligned} s &= \frac{20 + 0}{2}(2) \\ &= 20 \mathrm{~m} \end{aligned} \]

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Newton’s First Law

Have you ever felt your body move forwards when a car suddenly brakes? This happens because of inertia, an object's natural resistance to changes in motion. The greater an object's mass, the greater its inertia, meaning more force is required to alter its motion.

This principle is captured in Newton’s First Law of Motion, also known as the Law of Inertia, which states that an object will remain at rest or continue moving at a constant velocity unless acted upon by an unbalanced external force. This forms the foundation for understanding how forces influence motion.

While it is intuitive that an object at rest stays at rest unless acted upon, the idea that a moving object will continue moving forever is less obvious. In everyday life, it may seem like moving objects naturally slow down and stop on their own. However, this is actually caused by external forces like air resistance and friction. Have you ever kicked a ball across a field? At first, it moves quickly, but over time, it slows down and stops. This happens because of friction from the ground and air resistance pushing against it. If these external forces weren’t present, the ball would keep rolling forever at a constant speed.

A clearer way to visualise true inertia is to imagine an object floating through space far from gravitational influences. In the near vacuum of space, with little to no particles exerting resistance, the object would continue moving at a constant velocity indefinitely.

For simplicity in calculations, air resistance will be considered negligible unless explicitly stated otherwise.

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Newton’s Second Law

Newton’s Second Law of Motion describes the relationship between an object's acceleration , net force , and mass . It is mathematically expressed as:

\[ F_{\text{net}} = ma \]

Where:

$F_{\text{net}}$ = net force $(\mathrm{~N})$
$m$ = mass $(\mathrm{~kg})$
$a$ = acceleration $(\mathrm{~ms}^{-2})$

Since acceleration is a vector , net force is also a vector, as detailed in the figure below.

The net force $F_{\text{net}}$ is the sum of all forces acting on an object. Forces in opposite directions partially or completely oppose each other. For example, a $5 \mathrm{~N}$ force to the right and a $3 \mathrm{~N}$ force to the left result in a $2 \mathrm{~N}$ net force to the right.

This law tells us:

An object accelerates when a net force is applied.
The larger the force, the greater the acceleration, if mass is constant.
The larger the mass, the smaller the acceleration, if net force is constant.

In summary, Newton's second law tells us:

An object accelerates when a net force is applied.
The larger the force, the greater the acceleration (directly proportional).
The larger the mass, the smaller the acceleration (inversely proportional).

Worked example

A $2 \mathrm{~kg}$ block is pushed across a frictionless surface by a force of $10 \mathrm{~N}$. Calculate the acceleration of the block.

Identify the known variables:

$m = 2 \mathrm{~kg}$

$F_{\text{applied}} = 10 \mathrm{~N}$

$F_f = 0 \mathrm{~N}$

As there is no friction, $F_{\text{net}} = F_{\text{applied}} = 10 \mathrm{~N}$.

By Newton’s Second Law, $F_{\text{net}} = ma$.

Rearrange the equation for $a$:

\[ a = \frac{F_{\text{net}}}{m} \]

Substitute values:

\[ \begin{aligned} a &= \frac{10}{2} \\ &= 5 \mathrm{~ms}^{-2} \end{aligned} \]

The acceleration is in the same direction as the applied force.

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Forces on an inclined plane

On a flat surface, Newton's Second Law shows that objects accelerate only when there is an unbalanced force. When pushing a box with $100 \mathrm{~N}$ against a frictional force of $80 \mathrm{~N}$, a net force of $20 \mathrm{~N}$ causes acceleration.

Place that same box on a slope, and it may slide without any push. This occurs because gravity's effect on motion requires breaking down the force into two components - one perpendicular to the slope, one parallel to it. This component analysis reveals how objects behave on any inclined plane .

The following forces are acting on an object which remains stationary on an inclined plane:

Gravitational force $\left(F_g\right)$
Normal force $\left(F_N\right)$
Frictional force$ \left(F_f\right)$

$A box rests on an inclined plane that makes an angle θ with the horizontal. Three force vectors are drawn on the box: gravitational force (\mathbit{F}_\mathbit{g}) pointing vertically downward, normal force (\mathbit{F}_\mathbit{N}) pointing perpendicular and outward from the inclined surface, and friction force (\mathbit{F}_\mathbit{f}) pointing upward along the inclined surface (parallel to the slope).$

Unlike a flat surface where gravitational force directly opposes normal force, on an inclined plane these forces are not equal and opposite. The gravitational force splits into two components - one parallel to the slope and one perpendicular to it. The perpendicular component balances the normal force. The components of the gravitational force are illustrated below:

For this box to remain stationary on this slope, the frictional force must equal the parallel $(\parallel)$ component of the gravitational force, and the normal force must equal the perpendicular $(\perp)$ component of the gravitational force.

Mathematically, we can calculate these two components as:

\[
\begin{aligned}
F_{g,\perp} &= mg\cos(\theta) \\
F_{g,\parallel} &= mg\sin(\theta)
\end{aligned}
\]

Worked Example

A box with a mass of $20 \mathrm{~kg}$ is at rest on a ramp. The ramp makes an angle of $15^\circ$ with the horizontal. Calculate the friction force acting on the box, using $g = 9.8 \mathrm{~ms}^{-2}$.

Identify the known variables:

$m = 20 \mathrm{~kg}$

$\theta = 15^\circ$

$g = 9.8 \mathrm{~ms}^{-2}$

If the box is stationary, the friction force is equal in magnitude to the parallel component of the gravitational force on the box:

\[ \begin{aligned} F_f &= F_{g,\parallel} \\ &= mg\sin(\theta) \end{aligned} \]

Substitute values into the equation:

\[ \begin{aligned} F_f &= (20)(9.8)\sin(15^\circ) \\ &\approx 50.73 \mathrm{~N} \end{aligned} \]

As the box would naturally tend to slide down the ramp, the friction force acts in the opposite direction, up the ramp, to prevent this motion.
The friction force is $50.73 \mathrm{~N}$, parallel to and up the ramp.

When an object is moving up or down a slope, friction always opposes the direction of motion:

For an object sliding down, friction points up the slope
For an object sliding up, friction points down the slope

The net force is calculated by adding these forces in your chosen direction (usually down the slope as positive).

Worked example

A $50 \mathrm{~kg}$ box is pulled up a ramp that makes a $30^\circ$ angle with the horizontal. A constant force of $300 \mathrm{~N}$ pulls the box parallel to and up the ramp's surface. Assuming the ramp is frictionless, calculate the box's acceleration, using $g = 9.8 \mathrm{~ms}^{-2}$.

Identify the known variables:
$m = 50 \mathrm{~kg}$
$\theta = 30^\circ$
$g = 9.8 \mathrm{~ms}^{-2}$
$F_{\text{applied}} = 300 \mathrm{~N}$ up the ramp

Draw a diagram showing all forces on the box:

Specify a positive direction: up the ramp.

To calculate the acceleration, the net force is required:

\[ \begin{aligned} F_{\text{net}} &= F_{\text{applied}} - F_{g,\parallel} \\ &= 300 - (50)(9.8)\sin(30^\circ) \\ &= 300 - 245 \\ &= 55 \mathrm{~N} \end{aligned} \]

Substitute values into Newton’s second law:

\[ \begin{aligned} F_{\text{net}} &= ma \\ a &= \frac{F_{\text{net}}}{m} \\ &= \frac{55}{50} \\ &= 1.1 \mathrm{~ms}^{-2} \end{aligned} \]

The acceleration is positive; therefore, the box is accelerating at $1.1 \mathrm{~ms}^{-2}$ parallel to and up the ramp.

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The net force is equal to the force down the slope minus the friction force:
\[ F_{\text{net}} = F_{g,\parallel} - F_f \]
This can be solved using:
\[ F_{\text{net}} = 5(9.8)\sin(30^\circ) - 8 = 16.525 \mathrm{~N} \]
The acceleration can then be calculated using Newton’s second law:
\[ a = \frac{F_{\text{net}}}{m} = \frac{16.525}{5} \approx 3.3 \mathrm{~ms}^{-2} \]
Since the box is moving in a straight line, a kinematics equation can be used to determine the velocity at $s = 3 \mathrm{~m}$:
\[ \begin{aligned} v^2 &= u^2 + 2as \\ &= 0^2 + 2(3.3)(3) \\ &= 19.8 \\ v &= \sqrt{19.8} \approx 4.45 \mathrm{~ms}^{-1} \end{aligned} \]

Newton’s Third Law

Have you ever stubbed your toe on a table leg? That sharp pain you feel isn't just from hitting the table - it's evidence of an important physics principle. While your toe exerts a force on the table, the table pushes back on your toe with equal strength. This explains why it hurts!

Newton’s Third Law of Motion states that for every action (force) there is an equal and opposite reaction (force). This means forces always exist in pairs.

The size of the objects doesn't affect the magnitude of the forces. Take gravity as an example: Earth pulls down on you with a force, and you pull up on Earth with an equal force in the opposite direction. Even though Earth is much larger, both forces have the same strength!

By identifying and labelling forces using the convention of "force on object A by object B", it becomes easier to find the opposite force pair. For example, if we identify a force as "force on book by table" (upward), its opposite pair must be "force on table by book" (downward). These forces are equal in strength but act in opposite directions.

This can be simplified as:

\[F_{\text{on A by B}} = -F_{\text{on B by A}}\]

This principle is also what makes rockets work. A rocket engine expels gas downward (action), and in response the rocket is pushed upward with equal force (reaction). Even in the vacuum of space, where there’s nothing to "push against", the rocket can utilise these action-reaction pairs to move.

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Back Next

Equation	Missing variable
\(v = u + at\)	\(s\)
\(s = vt - \frac{1}{2}at^2\)	\(u\)
\(s = ut + \frac{1}{2}at^2\)	\(v\)
\(s = \frac{u +v}{2}t\)	\(a\)
\(v^2 = u^2 + 2as\)	\(t\)

Student Academic Success

Student Academic Success

Newton's Laws

Force and motion basics

Types of forces

Contact forces

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Non-contact forces

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Vector representation

Vector basics

Vector Notation

Basic vector operations

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Vector components

Worked Example

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Force diagrams

Drawing forces

Free body diagrams

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Motion concepts

Key terms

Straight line motion

Worked Example

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Newton’s First Law

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Newton’s Second Law

Worked example

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Forces on an inclined plane

Worked Example

Worked example

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Newton’s Third Law

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