OCR section 3.04 is the last piece of Mechanics. A force does not just push a body — it can turn it. The moment measures that turning effect, and it lets you solve rigid body problems: seesaws, beams on supports, and rods on the point of tilting.
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.
The moment of a force about a point measures its turning effect.
A force of 12 N acts at a perpendicular distance of 0.5 m from a pivot.
Moment = 12 × 0.5 = 6 N m.
Moments have a sense: clockwise or anticlockwise. Choose one as positive and be consistent.
Perpendicular distance is the whole subtlety. If the force acts at an angle θ to the rod, the moment is Fd sin θ, not Fd. And a force whose line of action passes through the pivot has zero moment about it — no matter how large it is. That is why the reaction at a pivot never appears when you take moments about that pivot.
For a particle, equilibrium needed only one condition: zero resultant force. A rigid body can also rotate, so it needs two:
The strategy: take moments about a point where an unknown force acts — that unknown then has zero moment and vanishes from the equation.
A light rod is pivoted at its centre. A downward force of 40 N acts 1.5 m to the left. What downward force F, placed 2 m to the right, balances it?
Anticlockwise moment (the 40 N on the left) = 40 × 1.5 = 60 N m.
Clockwise moment (F on the right) = F × 2.
Balance: 2F = 60 ⇒ F = 30 N.
Why ‘light’ matters: a light rod has no weight, so there is no weight moment to include. If the rod has weight, you must add it — see the next screen.
A uniform rod has its mass evenly distributed, so its weight acts at the midpoint (its centre of mass). Draw that weight as a single downward arrow at the centre.
A uniform rod AB of length 4 m and weight 60 N rests horizontally on supports at A and B.
The weight acts at the midpoint, 2 m from each end. By symmetry the reactions are equal:
RA + RB = 60, and RA = RB ⇒ RA = RB = 30 N.
Confirm with moments about A: RB × 4 = 60 × 2 = 120 ⇒ RB = 30 N. ✓
A uniform beam AB of length 6 m and weight 200 N rests on supports at A and at C, where AC = 4 m. A load of 60 N hangs at B. Find the reaction at C.
Take moments about A (this eliminates RA):
Clockwise: weight 200 N at 3 m ⇒ 200 × 3 = 600. Load 60 N at 6 m ⇒ 60 × 6 = 360. Total = 960.
Anticlockwise: RC × 4.
4RC = 960 ⇒ RC = 240 N.
Check vertically: RA + RC = 200 + 60 = 260, so RA = 260 − 240 = 20 N. Both reactions are positive, so the beam really does rest on both supports. ✓
A non-uniform rod has its centre of mass somewhere other than the midpoint. Moments let you find exactly where.
A non-uniform rod AB of length 4 m rests on supports at A and B. The reactions are RA = 20 N and RB = 40 N. Find the distance d of the centre of mass from A.
Vertically: the weight W = RA + RB = 20 + 40 = 60 N.
Moments about A: W × d = RB × 4 ⇒ 60d = 40 × 4 = 160.
d = 160/60 = 2.67 m (3 s.f.).
Sensible? The centre of mass is past the midpoint (2 m), which is right — B is carrying the larger reaction, so the mass must lie nearer B. ✓
On the point of tilting. As a load moves outwards, one reaction falls. The instant the body is about to tilt about one support, the reaction at the other support becomes zero. Setting that reaction to zero is how you solve tilting problems.
The tilting condition in one line: “about to tilt about C” ⇒ RA = 0. Substitute that in and everything falls out.
A rod is pivoted at a point O. Tap each force, then the sense of its moment about O.
Tap a description on the left, then the correct statement.
Moment: force × perpendicular distance; units N m
Zero moment: a force through the pivot has no turning effect — use this to eliminate unknowns
Equilibrium: resultant force = 0 and total moment about any point = 0
Strategy: take moments about a point where an unknown force acts, to make it disappear
Uniform rod: weight acts at the midpoint
Non-uniform rod: use moments to locate the centre of mass
Tilting: on the point of tilting about one support ⇒ the other reaction is zero
Check: reactions must be positive, and should sum to the total downward force
That is OCR 3.04 — and with it, the whole of Mechanics for H240. Press Finish to see your score.
You've worked through Moments for OCR A-level Mathematics A. 🎉
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