OCR section 3.03 is the heart of Mechanics: Newton’s three laws, F = ma, resolving forces in two dimensions, friction with its crucial inequality, inclined planes, and connected particles over pulleys.
Take g = 9.8 m/s² throughout. Work through each screen, answer the questions as you go and collect ⭐ stars. Press Start when you're ready.
A resultant force of 12 N acts on a mass of 3 kg.
a = F/m = 12/3 = 4 m/s², in the direction of the resultant force.
The third-law trap: the book on the table and the table on the book are a third-law pair (equal, opposite, on different bodies). But the book’s weight and the table’s normal reaction are not a third-law pair — they both act on the book. They happen to be equal here only because the book is in equilibrium.
A particle is in equilibrium when the resultant force is zero. In two dimensions this gives you two equations: resolve horizontally and vertically (or, on a slope, parallel and perpendicular to the slope).
A person of mass 60 kg stands in a lift accelerating upwards at 1.5 m/s². Find the normal reaction R from the floor.
Take up as positive. The forces on the person are R (up) and weight mg = 60 × 9.8 = 588 N (down).
F = ma: R − 588 = 60(1.5) = 90.
R = 588 + 90 = 678 N.
Does it make sense? R is greater than the weight — which is exactly why you feel heavier as a lift starts to go up. ✓
On a slope, resolve parallel and perpendicular to the slope — never horizontally and vertically. The weight mg splits into two components:
A particle of mass 2 kg rests on a smooth plane inclined at 30°. Find its acceleration down the slope.
Parallel to the slope: the only force is mg sin 30° = 2 × 9.8 × 0.5 = 9.8 N down the slope.
F = ma: 9.8 = 2a ⇒ a = 4.9 m/s².
Note this equals g sin 30° = 9.8 × 0.5 = 4.9 — the mass cancels. On a smooth slope, every object slides at the same rate. ✓
sin or cos? Check the extremes. At θ = 90° (a vertical drop) the object should be in free fall: g sin 90° = g. ✓ If you had used cos, you would get zero — obviously wrong. That test settles it every time.
Friction opposes motion (or the tendency to move) between rough surfaces. It is not a fixed value — it is only as big as it needs to be, up to a maximum.
Mass 5 kg, coefficient of friction μ = 0.4, g = 9.8.
Normal reaction: R = mg = 5 × 9.8 = 49 N.
Maximum (limiting) friction = μR = 0.4 × 49 = 19.6 N.
Now pull with a horizontal force of 30 N: since 30 > 19.6, the block moves, and friction acts at its maximum 19.6 N.
Resultant = 30 − 19.6 = 10.4 N ⇒ a = 10.4 / 5 = 2.08 m/s².
But if you pulled with only 12 N: 12 < 19.6, so the block does not move and friction is only 12 N — just enough to balance. It is not 19.6 N.
The inequality is the whole point. Automatically writing F = μR when the object is stationary and not on the point of moving is wrong. Ask first: is it slipping, or on the point of slipping?
Two particles joined by a light, inextensible string over a smooth pulley. Those assumptions give you two facts: the tension is the same throughout, and both particles have the same acceleration (in magnitude).
Method: write F = ma separately for each particle, taking the direction of motion as positive for each. Then solve the pair of simultaneous equations.
The 5 kg descends and the 3 kg rises. Let the acceleration be a and the tension T.
For the 5 kg (down positive): 5g − T = 5a ⇒ 49 − T = 5a
For the 3 kg (up positive): T − 3g = 3a ⇒ T − 29.4 = 3a
Add them (this eliminates T): 49 − 29.4 = 8a ⇒ 19.6 = 8a ⇒ a = 2.45 m/s².
Substitute back: T = 3(9.8) + 3(2.45) = 3(12.25) = 36.75 N.
Check with the other equation: 49 − T = 5(2.45) = 12.25 ⇒ T = 49 − 12.25 = 36.75 N. ✓
Sanity check the tension. T = 36.75 N lies between the two weights (29.4 N and 49 N) — it must. If your tension comes out bigger than both weights or smaller than both, you have made a sign error.
Tap a statement, then tap the law it illustrates.
Tap a description on the left, then the correct expression.
First law: no resultant force ⇒ at rest or constant velocity (equilibrium)
Second law: F = ma, where F is the resultant force
Third law: equal and opposite, and acting on different bodies
Resolving: split forces into perpendicular directions; on a slope use parallel and perpendicular
Slopes: down the slope mg sin θ; into the slope mg cos θ = R (when smooth)
Friction: F ≤ μR — equality only at the point of slipping or while sliding
Pulleys: light + inextensible + smooth ⇒ same tension, same acceleration; add the equations to eliminate T
Check: the tension must lie between the two weights
That is OCR 3.03 — and moments extends it from particles to rigid bodies. Press Finish to see your score.
You've worked through Forces and Newton's laws for OCR A-level Mathematics A. 🎉
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