Newton's Laws of Motion

This lesson covers AQA Spec 3.4.1.4 (Newton's laws of motion) in full depth. You will master free body diagrams, connected-body problems, pulley systems, and motion on inclined planes.

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1 Newton's Three Laws

First Law (The Law of Inertia)

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

Implications: An object moving at constant velocity has zero resultant force. A stationary object has zero resultant force. A change in speed OR direction requires a resultant force.

Second Law

The rate of change of momentum of an object is directly proportional to the resultant force acting on it, and takes place in the direction of that force.

For constant mass: F = ma

More generally: F = Δp/Δt where p = mv

The unit of force (newton) is defined such that 1 N = 1 kg m s⁻².

Third Law

When two objects interact, they exert equal and opposite forces on each other. These Newton's Third Law force pairs:

• Act on different objects
• Are of the same type (e.g., both gravitational, both contact)
• Are equal in magnitude
• Are opposite in direction
• Act along the same line

Common Misconception: Weight and normal contact force are NOT a Newton's Third Law pair — they act on the SAME object and are of different types. The Third Law pair of the weight of a book on a table is the gravitational pull of the book on the Earth.

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2 Free Body Diagrams

A free body diagram shows all forces acting on a single object. Forces are drawn as arrows from the object's centre, with length proportional to magnitude.

Described diagram — Block on a rough inclined plane at angle θ:

The block sits on a slope. Three forces act:

1. Weight W = mg acting vertically downward from the centre
2. Normal contact force N acting perpendicular to the slope surface (outward from the surface)
3. Friction F acting along the slope surface (up the slope, opposing motion/tendency to slide)

Resolving Forces on an Inclined Plane

Choose axes parallel and perpendicular to the slope:

• Parallel to slope (down): Component of weight = mg sin θ
• Perpendicular to slope (into surface): Component of weight = mg cos θ

For equilibrium on the slope:

• N = mg cos θ
• F = mg sin θ

Worked Example 1 — Block on a Slope

A 5.0 kg block is placed on a smooth slope inclined at 30° to the horizontal. Find the acceleration of the block down the slope.

Solution: The only force component along the slope is mg sin θ (the slope is smooth, so no friction).

F = ma along the slope:

mg sin θ = ma

a = g sin θ = 9.81 × sin 30° = 9.81 × 0.50 = 4.9 m s⁻²

Worked Example 2 — Block on a Rough Slope

A 5.0 kg block is on a rough slope at 30°. The coefficient of friction is μ = 0.30. Find the acceleration.

Solution:

Normal force: N = mg cos 30° = 5.0 × 9.81 × 0.866 = 42.5 N

Friction force: F = μN = 0.30 × 42.5 = 12.7 N (up the slope)

Net force down the slope = mg sin 30° − F = 5.0 × 9.81 × 0.50 − 12.7 = 24.5 − 12.7 = 11.8 N

Acceleration: a = 11.8 / 5.0 = 2.4 m s⁻²

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3 Connected Bodies

When two or more objects are connected (e.g., by a string), they move with the same acceleration (assuming an inextensible string). Treat each body separately with its own free body diagram.

Worked Example 3 — Two Blocks Connected on a Surface

Two blocks, A (3.0 kg) and B (5.0 kg), are connected by a light inextensible string on a smooth horizontal surface. A horizontal force of 24 N is applied to block B. Find (a) the acceleration and (b) the tension in the string.

Solution:

(a) Consider the whole system: F = (mₐ + m_b)a

24 = (3.0 + 5.0)a → a = 24/8.0 = 3.0 m s⁻²