Forces and Newton's laws

Mathematics 2210 · CCEA AS Level · 8 min read · Updated: 31 Aug 2026

Forces and Newton's Laws

Welcome to one of the most exciting and fundamental topics in AS 2: Applied Mathematics (Section A: Mechanics)! Whether you are aiming for top marks or just trying to get your head around mechanics, this chapter gives you the tools to describe why and how things move.

Mechanics is simply the mathematics of motion in the real world. By the end of this guide, you will be able to draw clear diagrams, break down forces into simple components, and use Newton's Laws to solve problems involving moving cars, sliding boxes, and pulley systems.

1. Understanding Forces and Modelling Assumptions

What is a Force?

A force is a vector quantity (meaning it has both magnitude and direction) that pushes or pulls an object, changing or tending to change its state of rest or motion. Forces are measured in Newtons ($\text{N}$), where $1\text{ N} = 1\text{ kg}\cdot\text{m}\cdot\text{s}^{-2}$.

In two dimensions, forces are written either as column vectors $\begin{pmatrix} x \\ y \end{pmatrix}$ or in Cartesian component form $\mathbf{F} = x\mathbf{i} + y\mathbf{j}$, or described using a magnitude and an angle/bearing.

Key Types of Forces

In your exam, you will encounter a standard set of contact and non-contact forces:

Weight ($W$): The gravitational pull of the Earth acting vertically downwards on an object. It is calculated using $W = mg$, where $m$ is mass in kilograms ($\text{kg}$) and $g = 9.8\text{ m}\cdot\text{s}^{-2}$ (the CCEA standard acceleration due to gravity).
Normal Reaction ($R$ or $N$): The contact push exerted perpendicular ($90^\circ$) to the surface an object rests upon. If you press down on a table, the table pushes right back up on you!
Tension ($T$): The pulling force transmitted through a taut string, rope, or cable. Tension always pulls away from the object along the line of the string.
Thrust / Compression ($T$ or $C$): The pushing force transmitted through a rigid rod.
Friction ($F$ or $F_r$): A resistive contact force that opposes the relative motion (or the tendency of motion) between two rough surfaces in contact.
Driving Force / Engine Tractive Force ($P$ or $F_D$): The forward force produced by an engine to propel a vehicle forward.

Modelling Assumptions

In mechanics questions, real-life objects are simplified using standard mathematical models. Knowing what these words mean will help you set up your equations correctly:

Particle: The mass is treated as being concentrated at a single point. This means we can ignore the size, shape, air resistance due to body dimensions, and rotational effects (spinning).
Light string / Light rod: The mass of the string or rod is negligible ($m = 0$). This means the tension is uniform (the same) throughout its entire length.
Inextensible string: The string does not stretch. This means connected objects must move with the exact same magnitude of acceleration and speed.
Smooth surface / Smooth pulley: There is zero friction between surfaces, or across the pulley. For a smooth pulley, tension is identical on both sides.
Rough surface: Friction acts along the surface to oppose motion.

Key Takeaway: Always start every problem by identifying all the forces acting on the particle and sketch a neat Free-Body Diagram showing the direction of each force.

2. Newton's Three Laws of Motion

Sir Isaac Newton formulated three fundamental laws that govern the motion of objects. Let's look at each one in detail.

Newton's First Law (N1L)

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

In mathematical terms: $\Sigma \mathbf{F} = \mathbf{0} \iff \mathbf{a} = \mathbf{0}$.

If the forces balance out to zero, the object does not speed up, slow down, or change direction. It is in a state of equilibrium.

Newton's Second Law (N2L)

The resultant force acting on a body is equal to the rate of change of its momentum. For a constant mass $m$, this gives the famous equation:

$$\Sigma \mathbf{F} = m\mathbf{a}$$

$\Sigma \mathbf{F}$ is the resultant force (vector sum of all forces in the direction of motion) in Newtons ($\text{N}$).
$m$ is the mass of the object in kilograms ($\text{kg}$).
$\mathbf{a}$ is the acceleration in $\text{m}\cdot\text{s}^{-2}$.

Analogy: Imagine pushing a shopping trolley. A bigger push ($\Sigma F$) gives a bigger acceleration ($a$). But if you fill the trolley with heavy groceries (increasing mass $m$), you will need a much bigger push to achieve the same acceleration!

Newton's Third Law (N3L)

When two bodies interact, the force exerted by the first body on the second is equal in magnitude and opposite in direction to the force exerted by the second body on the first.

In mathematical notation: $F_{A\text{ on }B} = -F_{B\text{ on }A}$.

Quick Review Box:

N1L: Balanced forces ($\Sigma F = 0$) $\implies$ constant velocity or rest ($a = 0$).
N2L: Unbalanced forces ($\Sigma F \neq 0$) $\implies$ acceleration occurs ($\Sigma F = ma$).
N3L: Action and reaction are equal and opposite.

3. Resolving Forces and Equilibrium

How to Resolve a Force

When a force acts at an angle, it pulls partly horizontally and partly vertically. To work with it, we break (or resolve) the force into two perpendicular components:

Adjacent to the angle $\theta$: Component $= F\cos\theta$
Opposite to the angle $\theta$: Component $= F\sin\theta$

Memory Trick: "Close to the angle? Cosine! Far away? Sine!"

Equilibrium in 2D

When a particle is in equilibrium, the forces are completely balanced in every direction. To solve equilibrium problems:

Resolve horizontally: $\Sigma F_x = 0$ (Forces to the right = Forces to the left)
Resolve vertically: $\Sigma F_y = 0$ (Forces upwards = Forces downwards)

Motion on an Inclined Plane

When an object sits or slides on a slope inclined at an angle $\alpha$ to the horizontal, it is easiest to resolve forces parallel ($\parallel$) and perpendicular ($\perp$) to the slope:

The weight ($mg$) acts vertically downwards, which is at an angle $\alpha$ to the perpendicular of the slope.
Component of weight down the slope: $mg\sin\alpha$
Component of weight into the slope: $mg\cos\alpha$

Equations for an inclined plane:

Perpendicular to slope ($\perp$): Because the object doesn't fly off or sink into the slope, forces balance:
$R = mg\cos\alpha$ (assuming no other perpendicular forces act).
Parallel to slope ($\parallel$): Apply $\Sigma F = ma$ down or up the slope:
$\Sigma F_{\parallel} = ma$

Key Takeaway: On an inclined plane, remember that weight splits into $mg\sin\alpha$ down the plane and $mg\cos\alpha$ into the plane.

4. Connected Particles

Don't worry if connected particle problems look intimidating at first! They follow a very predictable step-by-step process.

Case A: Connected Vehicles (Cars, Trailers, Trains)

When two objects are connected by a light, inextensible towbar or cable:

Method 1 (Whole System): Look at both masses combined: total mass $= (m_1 + m_2)$. The internal tension/thrust cancels out, allowing you to quickly find the overall acceleration $a$ using $\Sigma F_{\text{external}} = (m_1 + m_2)a$.
Method 2 (Separate Free-Body Diagrams): Isolate just one object (e.g., the trailer) and apply $\Sigma F = ma$ to find the internal tension $T$.

Case B: Pulleys

Consider two masses $m_1$ and $m_2$ (with $m_2 > m_1$) connected over a smooth, fixed pulley by a light, inextensible string:

Since the string is inextensible, both particles experience the same acceleration $a$.
Since the pulley is smooth and the string is light, the tension $T$ is the same on both sides.

Write an equation of motion for each particle in its direction of movement:

For the lighter particle $m_1$ (moving upwards):
$T - m_1 g = m_1 a$
For the heavier particle $m_2$ (moving downwards):
$m_2 g - T = m_2 a$

Adding these two equations eliminates $T$ instantly, letting you solve for $a$! Once you have $a$, substitute it back into either equation to calculate $T$.

Force on the Pulley

The total force exerted on the pulley is the resultant of the two tension vectors acting on it. For a standard vertical pulley with parallel strings, the force exerted by the string on the pulley is $2T$ downwards.

5. Common Pitfalls & Examiner Tips

Make sure you avoid these common traps in your CCEA AS 2 exam:

Mixing up Mass and Weight: Mass is in $\text{kg}$, weight is a force in $\text{N}$. Never write $F = m$ or use $m$ as a force. Always multiply by $g = 9.8\text{ m}\cdot\text{s}^{-2}$ to get weight ($W = mg$).
Assuming $R = mg$ always: The normal reaction $R$ is only equal to $mg$ on a flat, horizontal surface with no angled forces. On a slope, $R = mg\cos\alpha$. If an external force pulls upwards at an angle, $R = mg - F\sin\theta$. Always resolve perpendicularly to find $R$!
Confusing Sine and Cosine on Slopes: Weight down the slope is $mg\sin\alpha$, while weight perpendicular to the slope is $mg\cos\alpha$.
Forgetting to Define a Positive Direction: Before writing $\Sigma F = ma$, decide which direction is positive (e.g., in the direction of acceleration). Any force pointing against motion must have a minus sign.
Premature Rounding: Keep unrounded values in your calculator during intermediate steps. In CCEA examinations, give your final answers to an appropriate degree of accuracy (typically 3 significant figures unless otherwise stated, using $g = 9.8\text{ m}\cdot\text{s}^{-2}$).

Summary Checklist

Before sitting your exam, make sure you can confidently:

[ ] State and apply Newton's First, Second, and Third Laws of motion.
[ ] Draw accurate Free-Body Diagrams with all forces clearly labelled.
[ ] Resolve forces in horizontal/vertical directions and parallel/perpendicular to inclined planes.
[ ] Form and solve simultaneous equations for connected particles and pulley systems using $\Sigma F = ma$.
[ ] Calculate the resultant force exerted on a pulley.

Quick check

Can you answer these now?

Open each question to check the key ideas from this chapter.

What is a force, and what are the standard units and notations used to represent it in 2D?

A force is a vector quantity (with magnitude and direction) measured in Newtons ($\text{N}$). It can be written as a column vector $\begin{pmatrix} x \\ y \end{pmatrix}$ or in Cartesian component form $\mathbf{F} = x\mathbf{i} + y\mathbf{j}$.

What is the difference between mass and weight, and how is weight calculated?

Mass ($m$) is the amount of matter in $\text{kg}$, whereas weight ($W$) is the downward gravitational force in $\text{N}$, calculated using $W = mg$ (where $g = 9.8\text{ m}\cdot\text{s}^{-2}$).

What physical implications do the modelling assumptions "light", "inextensible", and "smooth" have?

"Light" means zero mass (uniform tension throughout), "inextensible" means no stretching (connected bodies share identical acceleration), and "smooth" means zero friction.

What does Newton's First Law of Motion state about an object in equilibrium?

An object remains at rest or continues at a constant velocity unless acted upon by a resultant non-zero force ($\Sigma \mathbf{F} = \mathbf{0} \iff \mathbf{a} = \mathbf{0}$).

What equation expresses Newton's Second Law of Motion for a body of constant mass?

$\Sigma \mathbf{F} = m\mathbf{a}$, where $\Sigma \mathbf{F}$ is the resultant force ($\text{N}$), $m$ is mass ($\text{kg}$), and $\mathbf{a}$ is acceleration ($\text{m}\cdot\text{s}^{-2}$).

What is Newton's Third Law of Motion?

When two bodies interact, the force exerted by the first body on the second is equal in magnitude and opposite in direction to the force exerted by the second on the first ($F_{A\text{ on }B} = -F_{B\text{ on }A}$).

How is a force $F$ resolved into components adjacent to and opposite to an angle $\theta$?

The adjacent component is $F\cos\theta$ and the opposite component is $F\sin\theta$.

How are weight components broken down on a plane inclined at an angle $\alpha$ to the horizontal?

Weight resolves into $mg\sin\alpha$ acting parallel (down) the slope and $mg\cos\alpha$ acting perpendicular (into) the slope.

How are connected particle systems with vehicles or trailers analysed?

Find overall acceleration by considering the whole system (mass $= m_1 + m_2$), then find internal tension by applying $\Sigma F = ma$ to an isolated single body.

How are equations of motion set up for two masses $m_1$ and $m_2$ hanging over a smooth vertical pulley?

For the lighter mass moving up: $T - m_1 g = m_1 a$; for the heavier mass moving down: $m_2 g - T = m_2 a$.

What is the total force exerted by a vertical string with tension $T$ on both sides of a fixed pulley?

The total force on the pulley is $2T$ acting vertically downwards.

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