Mathematics (9709) | Newton’s laws of motion

Newton’s Laws of Motion (Mechanics 9709, Paper 4)

Welcome to one of the most fundamental chapters in Mechanics! Newton's three laws of motion explain why objects move, why they stop, and how forces interact. Mastering these laws is absolutely essential, as they form the basis for solving almost every problem involving motion in this section of the syllabus.

Don't worry if applying these laws seems tricky at first—it requires careful drawing and logical thinking. We will break down each law and look at common scenarios step-by-step.

1. Understanding the Three Laws of Motion

Newton’s First Law (N1L): The Law of Inertia

Concept: An object will remain at rest or continue to move with a constant velocity (zero acceleration) unless acted upon by an external resultant force.

In simple terms, N1L explains inertia. Things are "lazy" and resist change in their motion.

• If an object is stationary, it stays stationary.
• If an object is moving at a constant speed in a straight line, it keeps moving that way.
• This state (rest or constant velocity) is called Equilibrium. In equilibrium, the Resultant Force \(F\) is zero (\(F=0\)).

Analogy: When you slam on the brakes in a car, your body continues moving forward until the seatbelt applies a resultant force to stop you.

Newton’s Second Law (N2L): The Powerhouse Equation

Concept: The resultant force acting on an object is equal to the rate of change of momentum, or, more simply for a constant mass particle, the resultant force is directly proportional to the acceleration it produces.

This is the equation you will use constantly in Mechanics problems:

$$F = ma$$

• \(F\) is the resultant force (or net force), measured in Newtons (N).
• \(m\) is the constant mass of the particle, measured in kilograms (kg).
• \(a\) is the acceleration, measured in \(\text{m\,s}^{-2}\).

Key Takeaway for N2L: The acceleration is always in the same direction as the resultant force. If \(a\) is zero, N2L simplifies to N1L (\(F=0\)).

Newton’s Third Law (N3L): Action and Reaction

Concept: When object A exerts a force on object B (the 'action' force), object B simultaneously exerts an equal and opposite force on object A (the 'reaction' force).

Crucial Distinction: The forces in an N3L pair always act on different objects.

Syllabus Example: The force exerted by a particle on the ground is equal and opposite to the force exerted by the ground on the particle.

• Force 1 (Action): Weight of particle pulling down on Earth.
• Force 2 (Reaction): Gravitational force of Earth pulling up on particle.

Did you know? The normal reaction force (\(R\)) exerted by the ground on a block sitting on it is not the N3L pair to the block's weight (\(W\)). \(R\) and \(W\) both act on the block, but they cancel out if the block is in equilibrium (N1L).

2. Mass, Weight, and the Acceleration due to Gravity (\(g\))

Before using \(F=ma\), we must correctly identify mass and weight.

Mass (\(m\))

Mass is the amount of matter in an object. It is a scalar quantity (it has magnitude only) and remains constant regardless of location (unless travelling near the speed of light, which we don't worry about here!).

Weight (\(W\))

Weight is the force exerted on an object due to gravity. It is a vector quantity (magnitude and direction) and always acts vertically downwards.

$$W = mg$$

Where:

• \(W\) is Weight (Force, N).
• \(m\) is Mass (kg).
• \(g\) is the acceleration due to gravity. The syllabus expects the use of the approximate numerical value: \(g = 10 \, \text{m\,s}^{-2}\), unless otherwise specified in the question.

3. Modelling Common Forces in Mechanics

In most problems, you will encounter forces like Weight, Normal Reaction, Tension, Friction, and Thrust.

A. Normal Reaction (\(R\))

This is the force exerted by a surface perpendicular to the surface, pushing back on the object resting on it.

B. Friction (\(F_f\)) and Contact Forces

A contact force between two surfaces can be split into two components: the Normal component (\(R\)) and the Frictional component (\(F_f\)), which is parallel to the surface and opposes motion (or the tendency to move).

Smooth vs. Rough Contacts

• Smooth Contact: This is an idealised model where friction is zero (\(F_f = 0\)). This simplifies calculations significantly.
• Rough Contact: Friction is present.

Limiting Equilibrium (Limiting Friction)

When an object is stationary but is 'about to slip', or is moving, the frictional force reaches its maximum possible value, known as Limiting Friction.

The relationship defining this limit is:

$$F = \mu R$$

• \(\mu\) (mu) is the coefficient of friction (a dimensionless constant).
• \(R\) is the Normal Reaction force.

General Rule for Friction:

• If the object is moving or in limiting equilibrium (about to slip): \(F = \mu R\).
• If the object is stationary and not about to slip: The actual friction force \(F\) is less than the maximum possible friction: \(F < \mu R\).

C. Tension (\(T\)) in an Inextensible String

Tension is the force transmitted through a string, rope, or cable.

• Light String: Assumed to have zero mass, so its weight is ignored.
• Inextensible String: Assumed not to stretch. This means that particles connected by the string must have the same magnitude of acceleration.

D. Thrust or Compression (\(C\)) in a Connecting Rod

When two objects (like a car and a trailer) are pushed together by a rigid rod, the force transmitted is called thrust (or sometimes compression). This force acts away from the object being pushed.

4. Applying Newton’s Second Law (\(F=ma\)) Step-by-Step

Applying N2L involves two main steps: drawing a diagram and resolving forces.

Step 1: Draw a Force Diagram

Always isolate the particle you are analyzing and draw every force acting on that particle (a Free Body Diagram). Indicate the direction of acceleration (\(a\)).

Step 2: Resolve Forces

Choose a set of perpendicular directions to resolve the forces. Usually, these are:

• Perpendicular to the line of motion (often where \(a=0\), used to find \(R\)).
• Parallel to the line of motion (used to calculate the resultant force \(F\) and find \(a\)).

Scenario 1: Vertical Motion

When a particle moves strictly vertically (like a lift or a falling object, ignoring air resistance unless stated), the forces are typically just the weight \(W\) and the supporting/pulling force \(T\) or \(R\).

Example: A lift moving upwards with acceleration \(a\).

Forces (upwards is positive): \(T\) (tension in the cable, up) and \(W\) (weight \(mg\), down).

Applying \(F=ma\): $$T - mg = ma$$

If the lift was accelerating downwards: $$mg - T = ma$$

Scenario 2: Motion on an Inclined Plane

This scenario is common and often causes confusion. We must resolve forces parallel and perpendicular to the slope, not horizontally and vertically.

The key trigonometry trick: If the plane is inclined at angle \(\alpha\) to the horizontal, the weight vector \(mg\) (which is vertical) can be split:

• Perpendicular to the plane: \(mg \cos \alpha\)
• Parallel to the plane (downwards): \(mg \sin \alpha\)

Step-by-step resolution:

1. Perpendicular to the plane (\(a = 0\)): Forces balance. $$R - mg \cos \alpha = 0 \quad \implies \quad R = mg \cos \alpha$$
2. Parallel to the plane (\(F=ma\)): Let the object slide down. Forces acting down are \(mg \sin \alpha\) and maybe friction \(F_f\).
   If the plane is smooth and the block slides down: $$mg \sin \alpha = ma \quad \implies \quad a = g \sin \alpha$$

Encouragement: Remember the components of weight: \(\sin\) slides, \(\cos\) crushes. The sine component makes it slide down, the cosine component determines the reaction force (crushing).

Scenario 3: Motion of Connected Particles

Particles can be connected via strings over pulleys (simple Atwood machines) or towed by rods/ropes (car and trailer).

A. Car Towing a Trailer (Thrust or Tension)

When a car of mass \(M\) tows a trailer of mass \(m\) using a light, rigid tow-bar (thrust) or rope (tension \(T\)), they accelerate together (\(A\)).

Method 1: Treat as a single system.

The total mass is \((M+m)\). The internal forces (Tension/Thrust) cancel out. If the car provides a driving force \(P\) and there is a total resistance \(R_{tot}\): $$P - R_{tot} = (M+m)A$$

Method 2: Analyze each particle separately.

We use this method to find the internal force (Tension \(T\) or Thrust \(C\)):

• For the Trailer (mass \(m\)): If \(T\) is the tension and \(R_m\) is its resistance: $$T - R_m = mA$$
• For the Car (mass \(M\)): If \(P\) is the engine force and \(R_M\) is its resistance: $$P - T - R_M = MA$$

B. Particles Connected over a Smooth Pulley

If two masses, \(m_1\) and \(m_2\) (\(m_1 > m_2\)), are connected by a light inextensible string passing over a smooth pulley, the system moves with a single acceleration magnitude \(a\).

We analyze each mass separately:

• Mass 1 (moving down): Forces are \(m_1g\) (down) and \(T\) (up). $$m_1g - T = m_1a \quad (i)$$
• Mass 2 (moving up): Forces are \(T\) (up) and \(m_2g\) (down). $$T - m_2g = m_2a \quad (ii)$$

(Note: By adding \((i)\) and \((ii)\), the tension \(T\) cancels out, allowing quick calculation of \(a\).)

Summary: Key Takeaways for Newton's Laws

The mechanics section focuses heavily on applying N2L (\(F=ma\)) under constant forces, resulting in constant acceleration. Remember that:

• If velocity is constant or zero, \(a=0\) and \(F_{net}=0\) (N1L).
• Weight is \(W=mg\), use \(g=10 \, \text{m\,s}^{-2}\).
• Friction is \(F=\mu R\) only when the object is moving or about to move.
• Always ensure forces are resolved in the correct direction—especially on inclined planes, where the weight component \(mg \sin \alpha\) drives the motion.