Forces & Newton's Laws

This lesson covers forces and Newton's laws of motion, which form the core of A-Level Mechanics. Newton's laws explain why objects move the way they do and provide the mathematical framework for analysing the effect of forces on motion.

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Types of Forces

Force	Description	Direction
Weight ($W = mg$W=mg)	Gravitational force on an object	Vertically downwards
Normal reaction ($R$R or $N$N)	Contact force perpendicular to a surface	Perpendicular to surface
Tension ($T$T)	Force in a taut string, rope, or cable	Along the string, away from the object
Friction ($F$F)	Force opposing motion or tendency to move	Opposite to direction of motion (or tendency)
Thrust/driving force	Applied force propelling an object	In the direction of motion
Resistance/drag	Forces opposing motion (air resistance, etc.)	Opposite to direction of motion

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Newton's Three Laws

First Law (Equilibrium)

An object remains at rest or continues to move at constant velocity unless acted upon by a resultant (net) force.

If the resultant force is zero: $\sum F = 0$∑F=0 — the object is in equilibrium.

Second Law (F = ma)

The resultant force on an object is equal to its mass times its acceleration:

$F_{\text{net}} = ma$Fnet​=ma

This is the most important equation in mechanics. The acceleration is in the same direction as the resultant force.

Third Law (Action-Reaction)

When object A exerts a force on object B, object B exerts an equal and opposite force on object A. These forces:

• Are equal in magnitude
• Act in opposite directions
• Act on different objects
• Are of the same type

Exam Tip: Newton's Third Law pairs act on different objects. A common mistake is identifying the weight and normal reaction as a Third Law pair — they are not, because they both act on the same object.

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Drawing Force Diagrams

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

1. Draw the object as a dot or simple shape.
2. Draw each force as an arrow from the object, labelled with its name and magnitude.
3. Show the direction clearly.
4. Only include forces acting on the object, not forces the object exerts on other things.

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Applying Newton's Second Law

Steps:

1. Draw a force diagram for each object.
2. Choose a positive direction.
3. Resolve forces in that direction.
4. Apply $F_{\text{net}} = ma$Fnet​=ma in the chosen direction.
5. Solve for the unknown.

Example: A 5 kg object is pushed along a smooth horizontal surface by a force of 20 N. Find the acceleration.

$F_{\text{net}} = 20$Fnet​=20 N, $m = 5$m=5 kg

$a = \frac{F}{m} = \frac{20}{5} = 4$a=mF​=520​=4 m/s$^2$2

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Connected Particles

When objects are connected by a string:

1. Draw separate force diagrams for each object.
2. The tension is the same throughout a light, inextensible string.
3. If connected, the objects have the same acceleration (in magnitude).
4. Write $F = ma$F=ma for each object separately, then solve the simultaneous equations.

Example: Two particles of mass 3 kg and 5 kg are connected by a light string over a smooth pulley (Atwood's machine).

For the 5 kg mass (moving down): $5g - T = 5a$5g−T=5a For the 3 kg mass (moving up): $T - 3g = 3a$T−3g=3a

Adding: $5g - 3g = 8a$5g−3g=8a, so $a = \frac{2g}{8} = \frac{g}{4} = 2.45$a=82g​=4g​=2.45 m/s$^2$2

$T = 3g + 3a = 3(9.8) + 3(2.45) = 36.75$T=3g+3a=3(9.8)+3(2.45)=36.75 N

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Resolving Forces on an Inclined Plane

On a plane inclined at angle $\theta$θ to the horizontal:

• Component of weight parallel to the plane (down the slope): $mg\sin\theta$mgsinθ
• Component of weight perpendicular to the plane: $mg\cos\theta$mgcosθ
• Normal reaction: $R = mg\cos\theta$R=mgcosθ (if no other perpendicular forces)

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Summary

• Newton's First Law: No resultant force means no change in velocity.
• Newton's Second Law: $F_{\text{net}} = ma$Fnet​=ma — the key equation for mechanics problems.
• Newton's Third Law: Equal and opposite forces on different objects.
• Always draw force diagrams before applying equations.
• For connected particles, use the same acceleration and tension throughout.
• On inclined planes, resolve weight into components parallel and perpendicular to the slope.

Exam Tip: Always start by drawing a clear force diagram. Mark all forces, choose a positive direction, and apply $F = ma$F=ma. Many marks are lost by students who skip the diagram and make sign errors.

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

Spec mapping

AQA 7357 specification, Paper 3 — Mechanics, Section U: Forces and Newton's Laws. The content statements demand confidence with: Newton's first, second and third laws; the equation of motion $F = ma$F=ma for a particle of constant mass; force diagrams (free-body diagrams) showing weight, normal reaction, friction, tension, thrust and applied forces; and resolution of forces into perpendicular components in two dimensions. This material sits at the heart of Paper 3 and is examined synoptically with Section R (Kinematics) — every equation-of-motion problem first determines the resultant force, then uses $a$a in $v = u + at$v=u+at, $s = ut + \tfrac{1}{2}at^2$s=ut+21​at2 etc. — and with Section V (Moments) for rigid bodies in equilibrium. AQA's formula booklet gives the SUVAT equations but does not give $F = ma$F=ma, the friction inequality $F \leq \mu R$F≤μR or any resolution formulae; these must be reproduced from memory and applied without prompt.

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Worked example with full mark scheme

Question (8 marks):

A particle of mass $5\,\text{kg}$5kg is held at rest on a rough plane inclined at $30°$30° to the horizontal. The coefficient of friction between the particle and the plane is $\mu = 0.2$μ=0.2. The particle is released and slides down the line of greatest slope. Take $g = 9.8\,\text{m s}^{-2}$g=9.8m s−2.

(a) Draw a labelled force diagram and resolve forces parallel and perpendicular to the plane. (3)

(b) Find the magnitude of the acceleration of the particle down the plane, giving your answer to 3 s.f. (5)

Solution with mark scheme:

(a) Step 1 — identify forces. The particle experiences three forces: weight $W = mg = 5 \times 9.8 = 49\,\text{N}$W=mg=5×9.8=49N acting vertically downwards; normal reaction $R$R acting perpendicular to the plane (away from the surface); and friction $F$F acting up the plane (opposing the impending motion down the slope).

B1 — correct identification of all three forces, each labelled with a clear arrow direction.

Step 2 — resolve weight into components. With the plane inclined at $\theta = 30°$θ=30°, the weight component parallel to the plane (down the slope) is $mg\sin\theta$mgsinθ, and the component perpendicular to the plane (into the surface) is $mg\cos\theta$mgcosθ.

$W_{\parallel} = 49\sin 30° = 24.5\,\text{N}, \qquad W_{\perp} = 49\cos 30° = 42.435...\,\text{N}$W∥​=49sin30°=24.5N,W⊥​=49cos30°=42.435...N

M1 — correct use of $\sin\theta$sinθ for the parallel component and $\cos\theta$cosθ for the perpendicular component. The most common slip is swapping these — the rule is "$\sin\theta$sinθ goes with the component opposite the angle the plane makes with the horizontal", verifiable by considering limits (as $\theta \to 0$θ→0 the parallel component must vanish, ruling out $\cos\theta$cosθ).

A1 — both components correct.

(b) Step 3 — perpendicular equilibrium. The particle does not accelerate perpendicular to the plane, so resolving $\perp$⊥:

$R - mg\cos\theta = 0 \implies R = 49\cos 30° = 42.435\,\text{N}$R−mgcosθ=0⟹R=49cos30°=42.435N

M1 — applying Newton's second law perpendicular to the plane with zero acceleration in that direction.

Step 4 — friction force. Since the particle is sliding, friction is limiting and kinetic, with $F = \mu R$F=μR:

$F = 0.2 \times 42.435 = 8.487\,\text{N}$F=0.2×42.435=8.487N

M1 — using $F = \mu R$F=μR (not the inequality $F \leq \mu R$F≤μR, which applies in static cases).

Step 5 — equation of motion parallel to the plane. Taking down the slope as positive:

$mg\sin\theta - F = ma$mgsinθ−F=ma $24.5 - 8.487 = 5a$24.5−8.487=5a $a = \dfrac{16.013}{5} = 3.2026\,\text{m s}^{-2}$a=516.013​=3.2026m s−2

M1 — correct $F = ma$F=ma along the line of motion with friction subtracted (acting up the slope).

A1 — $a \approx 3.20\,\text{m s}^{-2}$a≈3.20m s−2 (3 s.f.).

Total: 8 marks (B1 M1 A1 M1 M1 M1 A1 + presentation).

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Specimen question modelled on the AQA 7357 Paper 3 format

Question (6 marks): Two particles $A$A (mass $3\,\text{kg}$3kg) and $B$B (mass $2\,\text{kg}$2kg) are connected by a light inextensible string passing over a smooth fixed pulley. The system is released from rest with the string taut.

(a) Find the acceleration of the system. (3)

(b) Find the tension in the string. (2)

(c) State one modelling assumption you have used and explain its effect. (1)

Mark scheme decomposition by AO:

(a)

• M1 (AO3.3) — equation of motion for $A$A (heavier, descends): $3g - T = 3a$3g−T=3a.
• M1 (AO3.3) — equation of motion for $B$B (lighter, ascends): $T - 2g = 2a$T−2g=2a.
• A1 (AO1.1b) — adding to eliminate $T$T: $g = 5a$g=5a, so $a = g/5 = 1.96\,\text{m s}^{-2}$a=g/5=1.96m s−2.

(b)

• M1 (AO1.1b) — substituting $a$a back: $T = 2(g + a) = 2(9.8 + 1.96) = 23.52\,\text{N}$T=2(g+a)=2(9.8+1.96)=23.52N.
• A1 (AO1.1b) — $T \approx 23.5\,\text{N}$T≈23.5N (3 s.f.).

(c)

• B1 (AO3.5) — e.g. "The string is light (massless), so the tension is the same throughout"; or "the pulley is smooth, so no energy is lost to friction at the pulley and the tension is the same on both sides".

Total: 6 marks split AO1 = 3, AO3 = 3. Connected-particle questions are AO3-rich because they require modelling decisions and synthesis of multiple equations.

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Synoptic links

Connects to:

1. Section R — Kinematics: once $a$a is found from $F = ma$F=ma, the SUVAT suite computes velocities, displacements and times. Almost every Paper 3 dynamics question chains a force calculation into a kinematics calculation; failure to compute $a$a correctly cascades into the entire question.

2. Connected particles and pulleys (Section U extension): the same Newton's-second-law analysis, now applied to two bodies linked by a string. The tension is the same throughout a light inextensible string over a smooth pulley — an assumption worth half a mark to state explicitly.

3. Vectors (AS Pure, Section J): resolving forces is vector decomposition. The force $\vec{F} = F_x\hat{i} + F_y\hat{j}$F=Fx​i^+Fy​j^​ on an inclined plane is most efficiently handled by rotating the basis to align with the slope — the same trick used in 3D rigid-body problems at university.

4. Friction (Section U extension): the inequality $F \leq \mu R$F≤μR governs static friction; equality $F = \mu R$F=μR holds at the point of slipping or during motion. Distinguishing these regimes is examined in "is the particle on the point of slipping?" questions.

5. Statics and equilibrium (Section V): when a system is in equilibrium, all the same resolution techniques apply but with $a = 0$a=0 everywhere. Section V questions on three-force equilibrium use the same diagrams with the resultant set to zero — often via the triangle of forces or Lami's theorem.

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Mark-scheme literacy

Forces and Newton's-law questions on Paper 3 typically split AO marks as:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	50–60%	Drawing diagrams correctly, resolving with correct trigonometry, applying $F = ma$F=ma and $F = \mu R$F=μR
AO2 (reasoning)	15–25%	Justifying the direction of friction, choosing axes that simplify the algebra, interpreting whether equilibrium or motion applies
AO3 (problem-solving / modelling)	20–30%	Setting up a multi-body system, stating modelling assumptions and their effect, deciding what is held constant

Examiner-rewarded phrasing: "resolving perpendicular to the plane: $R = mg\cos\theta$R=mgcosθ"; "applying $F = ma$F=ma down the slope: $mg\sin\theta - \mu R = ma$mgsinθ−μR=ma"; "since the string is light and inextensible, the tension is the same throughout and both particles share the same magnitude of acceleration". Phrases that lose marks: "the force on the slope" (which force?); confusing $g$g-values mid-question; writing $F = \mu R$F=μR when the body is stationary and not on the point of slipping.

A specific AQA pattern: questions phrasing "find the acceleration of the system" demand a signed or magnitude answer, with direction stated. "$1.96\,\text{m s}^{-2}$1.96m s−2" alone may earn only the M1 — the A1 requires "in the direction of motion of the heavier mass" or equivalent phrasing. Read the question carefully for the demanded specificity.

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Grade-band model answers

3-mark question

Question: A box of mass $10\,\text{kg}$10kg rests on a horizontal floor. A horizontal force of $30\,\text{N}$30N is applied. The coefficient of friction is $\mu = 0.4$μ=0.4. Determine whether the box moves, and if so, find its acceleration. Take $g = 9.8\,\text{m s}^{-2}$g=9.8m s−2.

Grade C response (~180 words):

The normal reaction $R = mg = 10 \times 9.8 = 98\,\text{N}$R=mg=10×9.8=98N. The maximum friction is $F_{\max} = \mu R = 0.4 \times 98 = 39.2\,\text{N}$Fmax​=μR=0.4×98=39.2N. Since the applied force $30\,\text{N} < 39.2\,\text{N}$30N<39.2N, the box does not move. The acceleration is $a = 0$a=0.