Mathematics (9660) | Forces and Newton’s Laws

Welcome to Forces and Newton's Laws!

Hello! This chapter is the absolute heart of Mechanics (M1). If you understand how forces work, you understand why things move—or why they stay still! It's the foundation for everything else you'll study in this topic.
Don't worry if this seems tricky at first. We will break down Sir Isaac Newton's famous laws and show you exactly how to apply them step-by-step. Let's get started!

Key Prerequisite Concept: Modelling Assumptions

In M1, we often simplify real-world objects into models to make the maths manageable. When tackling a question, always be aware of the key assumptions:

• Particle: The object has mass but negligible size. We can treat all the forces as acting at a single point.
• Light String/Rod: The string or rod has zero mass.
• Inextensible String: The string cannot stretch. This means that all connected particles must have the same magnitude of acceleration.
• Smooth Surface/Pulley/Peg: There is no friction present.

1. Understanding the Fundamental Forces

Before we use Newton's laws, we need to know the basic forces we encounter in M1 problems.

1.1. Weight (Force of Gravity)

The weight of an object is the force exerted on it by gravity. It always acts vertically downwards.

Formula:
$$W = mg$$

Where:

• \(W\) is the Weight (measured in Newtons, N).
• \(m\) is the mass of the object (measured in kilograms, kg).
• \(g\) is the acceleration due to gravity. You must use the standard value:
  $$g = 9.8 \text{ ms}^{-2}$$

Analogy: Your weight is how hard the Earth is pulling you down.

1.2. Normal Reaction (R)

When an object is resting on a surface, the surface pushes back against the object. This is called the Normal Reaction force, \(R\).

• 'Normal' means perpendicular. This force always acts perpendicular (at 90°) to the surface of contact.
• If an object is resting on a flat, horizontal surface and is not accelerating vertically, the Normal Reaction \(R\) will exactly balance the weight \(W\).

Example: If you press your hand against a wall, the wall pushes back on your hand with a Normal Reaction force.

1.3. Tension (T) and Thrust

These are forces transmitted through ropes, strings, or rods:

• Tension (T): The pulling force transmitted by a string or rope. Tension always acts away from the object.
• Thrust/Compression: The pushing force transmitted by a rigid rod (or when two objects are pressed together). Thrust always acts towards the object.

1.4. Resistive Forces (Friction)

Resistive forces, like air resistance or friction, always act in the direction opposite to the motion or intended motion.

The syllabus focuses on dynamic friction (when the object is sliding).

Dynamic Friction Formula:
$$F = \mu R$$

Where:

• \(F\) is the Friction Force (N).
• \(\mu\) (the Greek letter 'mu') is the coefficient of friction (a unitless number, typically between 0 and 1).
• \(R\) is the Normal Reaction (N).

Key Takeaway: If a surface is described as smooth, then \(\mu = 0\), and thus the friction force \(F\) is zero. If the surface is rough, friction is present, and we use \(F = \mu R\).

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2. Newton's Three Laws of Motion

These are the fundamental rules governing all movement in Mechanics.

2.1. Newton's First Law (The Law of Inertia)

A body remains at rest or moves with constant velocity unless acted upon by a resultant force.

This law defines Equilibrium. If an object is in equilibrium, it means it is either:

• Totally at rest, OR
• Moving at a constant velocity (i.e., zero acceleration).

In mathematical terms, if a body is in equilibrium, the Net Force (\(F_{net}\)) acting on it is zero.
$$F_{net} = 0$$

2.2. Newton's Second Law (The Law of Acceleration)

The resultant force acting on a body is equal to the product of its mass and its acceleration, and acts in the direction of the acceleration.

This is the most crucial formula in this chapter:

Formula:
$$F = ma$$

Where:

• \(F\) must be the Net Force or Resultant Force (N).
• \(m\) is the mass (kg).
• \(a\) is the acceleration (\(ms^{-2}\)).

Memory Aid: If you want to accelerate (change velocity), you need a Net Force! The bigger the mass (\(m\)), the bigger the force (\(F\)) needed for the same acceleration (\(a\)).

2.3. Newton's Third Law (The Law of Action and Reaction)

For every action, there is an equal and opposite reaction.

When Object A exerts a force on Object B, Object B simultaneously exerts a force of equal magnitude and opposite direction back on Object A.

Important Distinction to Avoid Common Mistakes:
The two forces in an action-reaction pair act on different bodies. They never cancel each other out because they aren't acting on the same object.

Did you know? When you jump, you push the Earth down with a certain force, and the Earth pushes you up with an equal force! Because the Earth is so massive, its acceleration is negligible (we just can’t see it move).

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3. Applying Newton’s Second Law: F = ma

The core skill in this chapter is correctly setting up and solving the equation of motion for a single particle moving in a straight line (1D dynamics).

3.1. Step-by-Step Procedure

1. Draw a Free Body Diagram (FBD): This is essential. Draw the particle as a dot and show all external forces acting on that particle, labelling them clearly (W, R, T, F, etc.).

2. Choose the Positive Direction: This should usually be the direction in which the object is accelerating, \(a\).

3. Resolve Perpendicularly (If applicable for R): Write down the equation for forces perpendicular to the motion. In 1D motion (horizontal or vertical), the acceleration perpendicular to motion is zero. This step is usually used to find the Normal Reaction \(R\).

4. Apply \(F = ma\) Parallel to Motion: Write down the equation for forces parallel to the motion (in the direction of \(a\)).

$$(\text{Forces in positive direction}) - (\text{Forces opposing motion}) = ma$$

Analogy: Think of it like a tug-of-war. The winning team's force minus the losing team's force equals the mass being dragged multiplied by the resulting acceleration.

Example: Box being pulled horizontally

A 5 kg box is pulled horizontally by a force of 30 N along a rough surface. The coefficient of friction is 0.4. Find the acceleration.

Step 1 & 3 (Vertical Forces - finding R):
The box is not accelerating vertically, so \(F_{net} = 0\).
Forces up = Forces down
$$R = W$$
$$R = mg = 5 \times 9.8 = 49 \text{ N}$$

Calculate Friction:
$$F_{friction} = \mu R = 0.4 \times 49 = 19.6 \text{ N}$$

Step 2 & 4 (Horizontal Forces - using \(F=ma\)):
Positive direction = direction of 30 N pull.
$$(\text{Pulling Force}) - (\text{Friction Force}) = ma$$
$$30 - 19.6 = 5a$$
$$10.4 = 5a$$
$$a = 2.08 \text{ ms}^{-2}$$

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4. Connected Particle Problems (1D)

Connected particle problems involve two or more objects linked together (usually by a string) and moving together. The key is recognizing that they share a common acceleration and tension.

4.1. The Car and Trailer System (Horizontal Motion)

Imagine a car (mass \(M\)) pulling a trailer (mass \(m\)) using a tow-bar (which exerts a Tension, T, or Thrust).

Approach 1: Treat as a Single System

If the string/tow-bar is light and inextensible, the system moves as one unit (Total Mass \(M+m\)). The internal force (Tension \(T\)) is ignored.

$$(\text{Driving Force}) - (\text{Total Resistance}) = (M+m)a$$

Approach 2: Treat Particles Individually

We write an equation for each mass, using the common acceleration \(a\) and common tension \(T\). This is necessary if you need to find the value of \(T\).

For the Car (M):
$$F_{Drive} - T - R_{Car} = Ma$$

For the Trailer (m):
$$T - R_{Trailer} = ma$$

You now have two simultaneous equations to solve for \(T\) and \(a\).

4.2. Particles Connected over a Smooth Pulley/Peg (Vertical Motion)

This classic setup involves two masses, \(M_1\) and \(M_2\), connected by a light inextensible string passing over a smooth pulley or peg.

Assumptions: The pulley is smooth and light, meaning the tension \(T\) in the string is constant throughout, and the magnitudes of the accelerations are equal.

Let \(M_1\) be heavier than \(M_2\). \(M_1\) accelerates down (\(a\)) and \(M_2\) accelerates up (\(a\)).

Equation for \(M_1\) (Accelerating Downwards):
We choose down as positive.
$$W_1 - T = M_1 a$$
$$M_1 g - T = M_1 a \hspace{1cm} (1)$$

Equation for \(M_2\) (Accelerating Upwards):
We choose up as positive.
$$T - W_2 = M_2 a$$
$$T - M_2 g = M_2 a \hspace{1cm} (2)$$

By adding Equation (1) and Equation (2), the tension \(T\) cancels out, allowing you to solve for the acceleration \(a\).

$$(M_1 g - T) + (T - M_2 g) = M_1 a + M_2 a$$
$$g(M_1 - M_2) = a(M_1 + M_2)$$

Key Takeaway: Always draw separate FBDs for each connected particle and write two distinct equations of motion. Remember that acceleration and tension link them together!

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Summary of Forces and Laws (M1.3)

• Weight: \(W = mg\) (always vertical down, \(g=9.8\)).
• Friction: \(F = \mu R\) (opposes motion, zero on 'smooth' surfaces).
• Equilibrium (1st Law): \(F_{net} = 0\).
• Dynamics (2nd Law): \(F_{net} = ma\).
• For all dynamics problems, resolve forces in the direction of motion and apply \(F_{net} = ma\). Resolution of forces diagonally is not required for this section of the syllabus.