Forces and Newton's Laws

This lesson covers forces and Newton's three laws of motion as required by the Edexcel 9MA0 A-Level Mathematics specification. You need to be able to identify forces, draw force diagrams, and apply Newton's laws to solve problems.

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Types of Force

Force	Symbol	Description
Weight	W = mg	Gravitational force acting downwards; g ≈ 9.8 m s⁻²
Normal reaction	R or N	Perpendicular contact force from a surface
Tension	T	Pulling force in a taut string or rope
Friction	F	Opposes motion (or attempted motion) along a surface
Thrust/compression	T	Pushing force in a rigid rod
Driving force	D	Force propelling a vehicle
Resistance	R	Forces opposing motion (e.g. air resistance)

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Newton's First Law

A body remains at rest or moves with constant velocity unless acted on by a resultant (net) force.

In mathematical terms: If the resultant force is zero, the acceleration is zero (the body is in equilibrium).

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Newton's Second Law

The resultant force on a body is equal to the mass times the acceleration:

F = ma

Where F is the net (resultant) force in newtons (N), m is mass in kilograms (kg), and a is acceleration in m s⁻².

This is the most important equation in mechanics. Always resolve forces parallel to the direction of motion (or the direction of acceleration) before applying F = ma.

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Newton's Third Law

When two bodies interact, the forces they exert on each other are equal in magnitude and opposite in direction. These forces act on different bodies.

Exam Tip: Newton's Third Law pairs act on different objects. The weight of a book on a table and the normal reaction from the table are NOT a Newton's Third Law pair (they both act on the book). The correct pair is: the book pushes down on the table, and the table pushes up on the book.

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Drawing Force Diagrams

A force diagram (or free body diagram) shows all forces acting on a single object. Steps:

1. Isolate the object.
2. Draw and label the weight (always vertically downwards from the centre of mass).
3. Draw the normal reaction (perpendicular to any surface in contact).
4. Draw any tension, friction, driving force or resistance.
5. Mark the direction of acceleration.

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Applying Newton's Second Law

Horizontal Motion

Example: A 5 kg block is pushed along a rough horizontal surface by a force of 30 N. The frictional force is 10 N. Find the acceleration.

Resolving horizontally (direction of motion): F = ma 30 - 10 = 5a 20 = 5a a = 4 m s⁻²

Vertical Motion

Example: A lift of mass 800 kg accelerates upwards at 1.5 m s⁻². Find the tension in the cable.

Taking upwards as positive: T - mg = ma T - 800(9.8) = 800(1.5) T = 7840 + 1200 = 9040 N

Motion on an Inclined Plane

On a plane inclined at angle θ to the horizontal:

• The component of weight along the slope (down the slope) = mg sin θ
• The component of weight perpendicular to the slope = mg cos θ
• The normal reaction R = mg cos θ (if there is no acceleration perpendicular to the slope)

Example: A 2 kg block slides down a smooth plane inclined at 30°. Find the acceleration.

Resolving along the slope (down is positive): mg sin θ = ma a = g sin 30° = 9.8 x 0.5 = 4.9 m s⁻²

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Connected Bodies

When two or more objects are connected (e.g. by a string), you can either:

1. Treat the whole system as one object to find the acceleration (the internal forces cancel).
2. Then consider each object separately to find the internal forces (e.g. tension).

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Assumptions in Mechanics Models

Common modelling assumptions:

• Particle: The object has mass but no size (ignore rotational effects).
• Light string: The string has no mass; tension is the same throughout.
• Inextensible string: The string does not stretch; connected objects have the same acceleration.
• Smooth surface: No friction.
• Rigid body: The object does not deform.

Exam Tip: If asked to state assumptions, list the relevant ones and explain briefly how each simplifies the model (e.g. "The string is light, so the tension is the same at both ends").

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Summary

• Newton's Second Law: F = ma (resultant force = mass x acceleration).
• Always draw a clear force diagram before setting up equations.
• On inclines, resolve weight into components parallel and perpendicular to the slope.
• For connected bodies, use the whole system to find acceleration, then individual bodies to find tensions.
• State modelling assumptions clearly.

A-Level Deep Dive: Forces and Newton's Laws

Spec mapping

Edexcel 9MA0-03 specification section 10 — Forces and Newton's laws. Candidates must "understand the concept of a force; understand and use Newton's first law; understand and use Newton's second law for motion in a straight line (restricted to forces in two perpendicular directions or simple cases of forces given as 2-D vectors); understand and use weight and motion in a straight line under gravity; gravitational acceleration, $g$g, and its value in S.I. units to varying degrees of accuracy; understand and use Newton's third law; equilibrium of a particle under coplanar forces; application to problems involving smooth or rough inclined planes". The topic is examined throughout Paper 3 Section B and feeds into Section 11 (further kinematics, projectiles), Section 12 (moments) and the AS-level dynamics applications. Note: $g = 9.8\,\text{m s}^{-2}$g=9.8m s−2 is the convention unless stated otherwise; the formula booklet does not list Newton's laws.

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Worked example with full mark scheme

Question (8 marks): A particle $P$P of mass $5\,\text{kg}$5kg rests on a rough plane inclined at $30°$30° to the horizontal. The coefficient of friction between $P$P and the plane is $\mu = 0.2$μ=0.2. A force of $40\,\text{N}$40N acts on $P$P up the line of greatest slope. Find the acceleration of $P$P up the plane, giving your answer to 2 significant figures. Take $g = 9.8\,\text{m s}^{-2}$g=9.8m s−2.

Solution with mark scheme:

Step 1 — draw the force diagram and resolve.

Forces acting on $P$P: weight $5g\,\text{N}$5gN vertically down; normal reaction $R$R perpendicular to the plane; applied force $40\,\text{N}$40N up the slope; friction $F$F down the slope (opposing motion up the slope).

M1 — diagram with all four forces correctly placed and labelled. Examiners reward an explicit free-body diagram even when not asked: it secures the modelling marks and prevents sign errors downstream.

Step 2 — resolve perpendicular to the plane.

Perpendicular equilibrium (no acceleration into the plane):

$R = 5g\cos(30°) = 5 \times 9.8 \times \dfrac{\sqrt{3}}{2} = 24.5\sqrt{3} \approx 42.44\,\text{N}$R=5gcos(30°)=5×9.8×23​​=24.53​≈42.44N

M1 — resolving weight perpendicular using $\cos(30°)$cos(30°) (not $\sin(30°)$sin(30°) — the angle between the weight vector and the perpendicular-to-plane axis is $30°$30°).

A1 — correct value of $R$R.

Step 3 — compute friction at limit of motion.

$F = \mu R = 0.2 \times 42.44 \approx 8.488\,\text{N}$F=μR=0.2×42.44≈8.488N

M1 — correct application of $F = \mu R$F=μR, noting that here motion is occurring (the applied force is large enough), so kinetic friction acts at its maximum value down the plane.

Step 4 — apply Newton's second law along the plane.

Taking up the plane as positive:

$40 - 5g\sin(30°) - F = 5a$40−5gsin(30°)−F=5a $40 - 5 \times 9.8 \times 0.5 - 8.488 = 5a$40−5×9.8×0.5−8.488=5a $40 - 24.5 - 8.488 = 5a$40−24.5−8.488=5a $7.012 = 5a$7.012=5a

M1 — Newton's second law equation along the plane with all three force components (applied, weight component, friction) and correct signs.

M1 — substituting numerical values consistently.

A1 — $a = 1.4024\,\text{m s}^{-2}$a=1.4024m s−2.

A1 — final answer $a \approx 1.4\,\text{m s}^{-2}$a≈1.4m s−2 up the plane, to 2 significant figures, with units stated.

Total: 8 marks (M5 A3).

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Specimen question modelled on the Edexcel 9MA0 Paper 3 format

Question (6 marks): Two forces $\mathbf{F}_1 = (3\mathbf{i} + 4\mathbf{j})\,\text{N}$F1​=(3i+4j)N and $\mathbf{F}_2 = (-5\mathbf{i} + 2\mathbf{j})\,\text{N}$F2​=(−5i+2j)N act on a particle of mass $2\,\text{kg}$2kg initially at rest.

(a) Find the magnitude of the resultant force. (2)

(b) Find the acceleration vector of the particle. (2)

(c) Find the speed of the particle after $3\,\text{s}$3s. (2)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1a) — summing components: $\mathbf{F} = (3 - 5)\mathbf{i} + (4 + 2)\mathbf{j} = -2\mathbf{i} + 6\mathbf{j}$F=(3−5)i+(4+2)j=−2i+6j.
• A1 (AO1.1b) — magnitude $|\mathbf{F}| = \sqrt{(-2)^2 + 6^2} = \sqrt{40} = 2\sqrt{10} \approx 6.32\,\text{N}$∣F∣=(−2)2+62​=40​=210​≈6.32N.

(b)

• M1 (AO1.1b) — applying $\mathbf{F} = m\mathbf{a}$F=ma in vector form: $\mathbf{a} = \mathbf{F}/m$a=F/m.
• A1 (AO1.1b) — $\mathbf{a} = (-\mathbf{i} + 3\mathbf{j})\,\text{m s}^{-2}$a=(−i+3j)m s−2.

(c)

• M1 (AO2.5) — applying $\mathbf{v} = \mathbf{u} + \mathbf{a}t$v=u+at with $\mathbf{u} = \mathbf{0}$u=0: $\mathbf{v} = 3(-\mathbf{i} + 3\mathbf{j}) = -3\mathbf{i} + 9\mathbf{j}$v=3(−i+3j)=−3i+9j.
• A1 (AO1.1b) — speed $|\mathbf{v}| = \sqrt{9 + 81} = \sqrt{90} = 3\sqrt{10} \approx 9.49\,\text{m s}^{-1}$∣v∣=9+81​=90​=310​≈9.49m s−1.

Total: 6 marks split AO1 = 5, AO2 = 1. This is the synoptic dynamics—kinematics—vectors triple that Edexcel favours on Paper 3: F=ma combined with constant-acceleration kinematics in vector form.

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Synoptic links

Connects to:

1. Section 9 — Kinematics: Once Newton's second law gives $\mathbf{a}$a, the SUVAT or vector kinematic equations take over. Every dynamics question on Paper 3 ends with a kinematics step (find velocity, displacement, time). Confidence with both is non-negotiable.

2. Connected particles and pulleys (Section 10 sub-strand): A string over a smooth pulley transmits tension equally; Newton's second law applied to each particle separately, plus the constraint that they share acceleration magnitude, yields a pair of simultaneous equations. The "treat the system as one body" trick gives $a$a quickly; the individual-body equations give the tension.

3. Vectors in mechanics (Section 11 in Pure, and $\mathbf{i}$i, $\mathbf{j}$j form here): Force diagrams resolve into perpendicular components precisely because vectors decompose along any orthogonal basis. The $\sin\theta$sinθ/$\cos\theta$cosθ split on an inclined plane is the same operation as projecting a vector onto two axes.

4. Friction (Section 10 sub-strand): $F \leq \mu R$F≤μR in static cases, with equality at the point of slipping; $F = \mu R$F=μR once kinetic. The direction always opposes motion (or impending motion). Mis-signing friction is the single most common Paper 3 error.

5. Statics and equilibrium (Section 10 plus Section 12 moments): A particle in equilibrium has zero resultant force — both $\sum F_x = 0$∑Fx​=0 and $\sum F_y = 0$∑Fy​=0. For rigid bodies, equilibrium also requires $\sum M = 0$∑M=0 about any point (Section 12).

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Mark-scheme literacy

Forces and Newton's laws questions on 9MA0 Paper 3 split AO marks across the spectrum:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	50–60%	Drawing correct force diagrams, resolving forces along chosen axes, applying $F = ma$F=ma with correct signs
AO2 (reasoning / interpretation)	20–30%	Justifying choice of axes (along/perpendicular to plane), interpreting "smooth" vs "rough", explaining why a particle is/is not in equilibrium
AO3 (problem-solving / modelling)	15–25%	Selecting modelling assumptions (light string, smooth pulley, particle), commenting on validity, multi-step problems with implicit steps

Examiner-rewarded phrasing: "resolving perpendicular to the plane, $R - mg\cos\theta = 0$R−mgcosθ=0"; "applying Newton's second law along the plane, taking up the plane as positive"; "since the string is light and inextensible, both particles share the same acceleration magnitude"; "modelling the box as a particle assumes air resistance is negligible". Phrases that lose marks: omitting units on the final answer; using $g = 10$g=10 when the question specifies $9.8$9.8; failing to state the direction of acceleration (just a number, no "up the plane").

A specific Edexcel pattern: questions phrased "find the acceleration" demand both magnitude and direction (or sign relative to a stated positive axis). Numerical answer alone — even if correct in magnitude — risks losing the final A1.

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Grade-band model answers

3-mark question

Question: A particle of mass $4\,\text{kg}$4kg is acted on by a single horizontal force of $20\,\text{N}$20N. Find the acceleration of the particle.