AQA A-level Mathematics (7357) · Forces and Newton's laws
Mini-Lesson
Forces and Newton's laws
This mini-lesson covers AQA section R: Newton's three laws, weight and motion under gravity, resultant forces and resolving in two dimensions, equilibrium, connected particles and smooth pulleys, and the F ≤ μR friction model on horizontal and inclined planes.
Connected particles: write F = ma for each mass separately, taking the direction of motion as positive.
Work through each screen, answer the questions as you go (some are wordy, some are calculations) and collect ⭐ stars. Throughout, g = 9.8 m s⁻² (AQA convention), strings are light and inextensible and pulleys are smooth. Press Start when you're ready.
Section R1–R3 · the laws
Newton's three laws
First law: a body stays at rest, or moves with constant velocity, unless acted on by a resultant force. So "constant velocity" and "at rest" both mean resultant force = 0 (equilibrium).
Second law:F = ma, where F is the resultant force, in the direction of the acceleration.
Third law: if A exerts a force on B, then B exerts an equal and opposite force on A. Crucially the two forces act on different bodies, so they never cancel each other out on the same body.
F = ma · W = mg (g = 9.8 m s⁻²)resolve in two perpendicular directions and write F = ma in each
Worked example
A car of mass 1200 kg has a resultant forward force of 3000 N.
a = F ÷ m = 3000 ÷ 1200 = 2.5 m s⁻²
The classic third-law trap: a book rests on a table. Its weight (Earth pulls book down) pairs with the book pulling the Earth up — not with the normal reaction. The weight and the normal reaction both act on the book, so they are a first-law (equilibrium) pair, not a third-law pair.
Calculate
Newton's second law
1A car of mass 1200 kg experiences a resultant force of 3000 N. Calculate its acceleration.
m s⁻²
Hint: F = ma → a = F ÷ m = 3000 ÷ 1200.
Quick check
Third law pairs
?A book of weight W rests on a table. Which force is the Newton's third law pair of the book's weight?
Section R6 · friction
The F ≤ μR friction model
Friction opposes relative motion (or the tendency to move). AQA's model is:
F ≤ μRμ = coefficient of friction · R = normal reaction · equality F = μR only at the point of slipping, or while sliding
If the body is static and not on the point of moving, friction takes whatever value is needed to balance the applied force — which is less than μR.
Limiting friction (F = μR) occurs when the body is on the point of moving, or is already sliding.
On a smooth surface μ = 0, so there is no friction.
Worked example — horizontal plane
A 5 kg block rests on a rough horizontal floor with μ = 0.4. (g = 9.8)
Resolve vertically: R = mg = 5 × 9.8 = 49 N
Maximum friction = μR = 0.4 × 49 = 19.6 N
A horizontal force of 30 N is applied. Since 30 > 19.6, the block moves, and friction is at its maximum.
Resultant = 30 − 19.6 = 10.4 N, so a = 10.4 ÷ 5 = 2.08 m s⁻²
Calculate
Limiting friction
2A block of mass 5 kg rests on a rough horizontal plane with coefficient of friction μ = 0.4 (g = 9.8). Calculate the maximum friction force available.
N
Hint: Resolve vertically: R = mg = 5 × 9.8 = 49 N. Then F(max) = μR = 0.4 × 49.
Calculate
Friction and F = ma
3That same 5 kg block is now pushed by a horizontal force of 30 N. Calculate its acceleration, to 2 d.p.
m s⁻²
Hint: Resultant = 30 − 19.6 = 10.4 N. Then a = F ÷ m = 10.4 ÷ 5.
Quick check
How big is the friction?
?A 3 kg block sits on a rough horizontal table with μ = 0.5 (so μR = 14.7 N). A horizontal force of 10 N is applied and the block does not move. What is the friction force?
Sort it
Which law is it?
Tap a statement, then tap the law of motion it illustrates.
1️⃣ First law
2️⃣ Second law
3️⃣ Third law
Section R4 · connected particles
Pulleys and connected particles
For a light inextensible string over a smooth pulley: the tension is the same throughout, and both particles have the same magnitude of acceleration.
Worked example — 3 kg and 5 kg over a smooth pulley (g = 9.8)
The 5 kg mass falls, the 3 kg mass rises. Take the direction of motion as positive for each.
Substitute back: T = 3(g + a) = 3(9.8 + 2.45) = 3 × 12.25 = 36.75 N
Check with the other equation: T = 5(g − a) = 5(9.8 − 2.45) = 5 × 7.35 = 36.75 ✓
Sanity check: the tension must lie between the two weights (3g = 29.4 N and 5g = 49 N). 36.75 N does — if your T is bigger than both or smaller than both, you have a sign error.
Calculate
Pulley: acceleration
4Masses of 3 kg and 5 kg hang either side of a smooth pulley on a light inextensible string (g = 9.8). Calculate the magnitude of the acceleration.
m s⁻²
Hint: Adding 5g − T = 5a and T − 3g = 3a gives 2g = 8a, so a = 19.6 ÷ 8.
Calculate
Pulley: tension
5For the same pulley system (a = 2.45 m s⁻²), calculate the tension in the string.
N
Hint: Use the 3 kg mass: T − 3g = 3a → T = 3(9.8 + 2.45) = 3 × 12.25.
Section R2 · resolving
Inclined planes and resolving
On a slope, resolve parallel and perpendicular to the plane — never horizontally and vertically.
Component of weight down the slope = mg sin θ
Component of weight into the slope = mg cos θ, so on a slope R = mg cos θ (not mg!)
Maximum friction on the slope = μR = μ mg cos θ
Worked example — smooth slope
A particle slides down a smooth plane inclined at 30° (g = 9.8).
Down the slope: mg sin 30° = ma → a = g sin 30° = 9.8 × 0.5 = 4.9 m s⁻²
Notice the mass cancels — on a smooth slope, every mass accelerates at the same rate.
Rough slope: a block slides down only if mg sin θ > μ mg cos θ, i.e. if tan θ > μ. The angle at which it just begins to slip satisfies tan θ = μ.
Quick check
Down a smooth slope
?A particle slides down a smooth plane inclined at 30° to the horizontal (g = 9.8 m s⁻²). What is its acceleration?
Match it
Force diagram symbols
Tap a symbol on the left, then what it stands for on the right.
Symbol
Meaning
Quick check
Constant velocity
?A parachutist falls at a constant velocity. What can you say about the forces on her?
Section R2 · lifts
Bodies in a lift (apparent weight)
A person standing in a lift feels the normal reaction R from the floor, not their weight. Write F = ma vertically, taking the direction of the acceleration as positive.
Worked example (g = 9.8)
A person of mass 70 kg stands in a lift accelerating upwards at 2 m s⁻².
Upwards positive: R − mg = ma
R = m(g + a) = 70 × (9.8 + 2) = 70 × 11.8 = 826 N
That is more than their weight (686 N) — they feel heavier.
Sense check: accelerating up ⇒ R > mg (feel heavy). Accelerating down ⇒ R < mg (feel light). In free fall (a = g) ⇒ R = 0 — weightlessness.
Calculate
The lift problem
6A person of mass 70 kg stands in a lift accelerating upwards at 2 m s⁻² (g = 9.8). Calculate the normal reaction from the floor.
N
Hint: R − mg = ma → R = m(g + a) = 70 × (9.8 + 2).
Quick check
Will it slide?
?A block rests on a rough plane inclined at 20°, with coefficient of friction μ = 0.5. Does it slide? (tan 20° = 0.364)
Recap
The big ideas to know
1st law: resultant force = 0 ⇔ at rest or constant velocity (equilibrium)
2nd law: F = ma, using the resultant force in the direction of the acceleration
3rd law: equal and opposite forces on different bodies (weight pairs with the pull on the Earth)
Friction: F ≤ μR; F = μR only when sliding or on the point of sliding; smooth ⇒ μ = 0
Slopes: resolve along and perpendicular to the plane: mg sin θ down, R = mg cos θ
Pulleys: same T throughout, same |a|; write F = ma for each particle and add to eliminate T
That is the whole of section R — Forces and Newton's laws for AQA A-level Mathematics (7357). Press Finish to see your score.
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