Edexcel A-level Mathematics (9MA0) · Forces and Newton's laws
Mini-Lesson
Forces and Newton's laws
This mini-lesson covers Topic 8 of Edexcel A-level Mathematics (9MA0): the concept of a force and Newton's three laws; F = ma; weight and motion under gravity with g = 9.8 m s⁻²; equilibrium and resultant forces; connected particles, lifts and smooth pulleys; resolving forces in two dimensions; and the F ≤ μR model for friction, including motion on an inclined plane.
Resolve the weight parallel (mg sin θ) and perpendicular (mg cos θ) to the slope. The normal reaction on a slope is R = mg cos θ, not mg.
Work through each screen, answer the questions as you go (some are wordy, some are calculations) and collect ⭐ stars. Throughout, take g = 9.8 m s−2 unless a question says otherwise. Press Start when you are ready.
Forces · Newton's laws
Newton's three laws
First law: a body stays at rest, or moves with constant velocity, unless a resultant force acts on it. So "in equilibrium" means resultant force = 0 — it does not mean "at rest": constant velocity counts too.
Second law:F = ma, where F is the resultant force. Acceleration is in the same direction as the resultant force.
Third law: for every action there is an equal and opposite reaction. Crucially, the two forces of the pair act on different bodies — which is why they never cancel out on a single body.
Common forces you will meet: weight (W = mg, always vertically down), the normal reaction R (perpendicular to the surface), tension (a string pulls), thrust / compression (a rod pushes), friction and air resistance.
Method that never fails: draw the force diagram, choose a positive direction, resolve, then write F = ma in that direction. Where there is no acceleration (e.g. perpendicular to a surface), the equation becomes resultant = 0.
Sort it
Which of Newton's laws?
Tap a statement, then tap the law it belongs to.
1️⃣ First law
2️⃣ Second law
3️⃣ Third law
Calculate
F = ma
1A resultant force of 12 N acts on a particle of mass 3 kg. Find its acceleration.
m s⁻²
Hint: F = ma → a = F ÷ m = 12 ÷ 3.
Quick check
Constant velocity
?A car travels along a straight road at a constant 30 m s⁻¹. What is the resultant force on the car?
Forces · connected particles
Lifts and connected particles
A person in a lift feels "heavier" when the lift accelerates upwards. That feeling is the normal reaction R from the floor, and it is not equal to the weight.
Worked example — a lift accelerating upwards
A person of mass 60 kg stands in a lift accelerating upwards at 2 m s⁻². Find R.
Taking up as positive, the forces on the person are R (up) and mg (down):
R − mg = ma → R = m(g + a) = 60 × (9.8 + 2) = 60 × 11.8 = 708 N
(At rest or at constant velocity, R would be just 60 × 9.8 = 588 N. Accelerating downwards at 2 m s⁻² would give R = 60 × 7.8 = 468 N.)
Two-body problems: you can treat the two connected bodies as one system to find the acceleration (internal forces cancel), then look at one body alone to find the tension or contact force.
Calculate
The lift problem
2A person of mass 60 kg stands in a lift accelerating upwards at 2 m s⁻². Taking g = 9.8 m s⁻², find the normal reaction R from the floor.
N
Hint: Up is positive: R − mg = ma, so R = m(g + a) = 60 × (9.8 + 2).
Forces · pulleys
Smooth pulleys & light inextensible strings
The standard modelling assumptions unlock the whole problem:
Smooth pulley ⇒ the tension is the same throughout the string.
Light string ⇒ the string has no mass, so the tension does not change along it.
Inextensible string ⇒ both particles have the same magnitude of acceleration.
A smooth pulley means the tension is the same on both sides; an inextensible string means both particles have the same acceleration.
Worked example — 5 kg and 3 kg over a smooth pulley
Take the direction of motion as positive for each particle (5 kg down, 3 kg up).
5 kg: 5g − T = 5a 3 kg: T − 3g = 3a
Add: 5g − 3g = 8a → a = 2 × 9.8 ÷ 8 = 2.45 m s⁻²
Tension: T = 3(g + a) = 3 × (9.8 + 2.45) = 3 × 12.25 = 36.75 N
Check with the other equation: 5(g − a) = 5 × 7.35 = 36.75 ✔
Sanity check: the tension always lies between the two weights (3g = 29.4 N and 5g = 49 N). If your T is bigger than both or smaller than both, you have a sign error.
Calculate
Pulley — acceleration
3Particles of mass 5 kg and 3 kg hang either side of a smooth pulley on a light inextensible string. Taking g = 9.8 m s⁻², find the acceleration of the system.
m s⁻²
Hint: Adding the two equations: (5 − 3)g = (5 + 3)a, so a = 2 × 9.8 ÷ 8.
Calculate
Pulley — tension
4For the same pulley system (5 kg and 3 kg, a = 2.45 m s⁻²), find the tension in the string.
N
Hint: Use the 3 kg particle: T − 3g = 3a, so T = 3(9.8 + 2.45) = 3 × 12.25.
Quick check
Which assumption?
?Why do the two particles connected over a smooth pulley have the same magnitude of acceleration?
Forces · friction
Friction: the F ≤ μR model
Friction opposes the direction the object is moving or would tend to move. Its size depends on how hard the surfaces are pressed together — the normal reaction R:
F ≤ μRμ = coefficient of friction · F = μR only when the object is moving, or on the point of moving (limiting)
In equilibrium, friction takes whatever value it needs (up to μR) to prevent motion — so here F ≤ μR is an inequality.
On the point of sliding (limiting equilibrium) or already moving: F = μR exactly.
On a horizontal surface with no other vertical force, R = mg. On a slope, R = mg cos θ.
Worked example — horizontal
A block of mass 5 kg rests on a rough horizontal plane with μ = 0.4.
R = mg = 5 × 9.8 = 49 N. Maximum (limiting) friction = μR = 0.4 × 49 = 19.6 N.
So any horizontal pull under 19.6 N will not move it; friction simply matches the pull.
Calculate
Limiting friction
5A block of mass 5 kg rests on a rough horizontal plane with coefficient of friction μ = 0.4. Taking g = 9.8 m s⁻², find the maximum (limiting) friction.
N
Hint: R = mg = 5 × 9.8 = 49 N. Then F = μR = 0.4 × 49.
Quick check
How much friction?
?A crate sits in equilibrium on a rough horizontal floor. It is pulled horizontally with a force of 10 N, and the limiting friction is 25 N. What is the actual friction force acting?
Forces · inclined planes
Resolving on an inclined plane
On a slope, tilt your axes: take one direction along the slope and one perpendicular to it. Then resolve the weight:
Perpendicular to the slope (no acceleration): R = mg cos θ
Along the slope: mg sin θ − F = ma (with friction F = μR opposing the motion)
Worked example — smooth slope
A particle of mass 2 kg on a smooth slope at 30°: there is no friction, so ma = mg sin θ.
a = g sin 30° = 9.8 × 0.5 = 4.9 m s⁻² down the slope. (Note: the mass cancels.)
Worked example — rough slope, μ = 0.2
Same 2 kg particle, same 30° slope, but now μ = 0.2.
R = mg cos 30° = 2 × 9.8 × 0.8660 = 16.97 N, so F = μR = 0.2 × 16.97 = 3.395 N.
Along the slope: mg sin 30° − F = ma → 9.8 − 3.395 = 2a → a = 6.405 ÷ 2 = 3.20 m s⁻² (3 s.f.)
The classic error: writing R = mg on a slope. It is R = mg cos θ — the surface only has to support the component of the weight pressing into it.
Calculate
Rough inclined plane
6A particle of mass 2 kg slides down a rough slope inclined at 30°, with μ = 0.2. Taking g = 9.8 m s⁻², find its acceleration down the slope, to 3 significant figures.
m s⁻²
Hint: R = 2 × 9.8 × cos30° = 16.97 N → F = 0.2 × 16.97 = 3.395 N. Then 2 × 9.8 × sin30° − 3.395 = 2a.
Match it
Expression → what it describes
Tap an item on the left, then its partner on the right.
Expression
Description
Quick check
Normal reaction on a slope
?A particle rests on a rough plane inclined at angle θ. What is the normal reaction R?
Recap
The big ideas to know
N1: no resultant force ⇔ at rest or constant velocity (equilibrium)
N2: F = ma, with F the resultant · N3: equal and opposite, acting on different bodies
Weight W = mg with g = 9.8 m s⁻² · lift: R = m(g + a) when accelerating upwards
Pulleys: smooth ⇒ same tension; inextensible ⇒ same acceleration
Friction: F ≤ μR — equality only when moving or in limiting equilibrium
Slope: R = mg cos θ · driving component = mg sin θ · mg sin θ − μR = ma
That is the whole of 9MA0 Topic 8 — Forces and Newton's laws. Press Finish to see your score.
🏆
Mini-lesson complete!
⭐⭐⭐
You have worked through Forces and Newton's laws for Edexcel A-level Mathematics (9MA0). 🎉
Your stars: 0 / 0
Next: test yourself in the Evaluate stage Confidence Quiz, then lock it in with Verify.
📣 Smashed it? Share your score
Challenge a mate to beat your stars, or show a parent how you got on.