AQA A-level Mathematics 7357, section S (Moments), version 1.3 (first assessment 2018). Original teaching notes; objective-level mapping lives in content/teaching/aqa_a_level_mathematics/objective-review.yaml.
Prerequisites: forces and Newton's laws from section R. Each lesson develops its own method before independent transfer.
These are original teaching examples, not official AQA material.
A long spanner loosens a tight nut that a short spanner cannot move, with exactly the same push. Where does the extra turning effect come from?
Start with a prediction. State what the turning effect of a force depends on, and why doubling the handle length doubles it.
Build the model
The moment of a force about a point is the force multiplied by the PERPENDICULAR distance from the point to the line of the force, in N m. A force through the point has zero distance and zero moment — it cannot turn the body about that point.
A body is in equilibrium 平衡 only when BOTH conditions hold: the forces sum to zero (no translation) AND the moments about every point sum to zero (no rotation). Taking moments about an unknown force's pivot 支点 removes it from the equation, because its distance is zero — one equation, one unknown.
moment: Force × perpendicular distance from the point to the force's line of action, in N m; equilibrium: A state with zero resultant force and zero resultant moment, so no translation and no rotation; pivot: The point about which moments are taken or a body turns.
Choose evidence that can test it
Choose the point to take moments about so that the most awkward unknown passes through it. For a beam on two supports, take moments about one support to find the other reaction, then use ΣF = 0 for the first.
The distance is measured to the LINE of the force, not to where the force is drawn: a force at 40° to a 3 m bar has perpendicular distance 3 sin 40°, not 3 m.
Work from known quantities
State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
Known: a light 4 m beam rests on supports A and B at its ends, with a 260 N load 1 m from A. Moments about A: R_B × 4 = 260 × 1, so R_B = 65 N. Vertical forces: R_A + R_B = 260, so R_A = 195 N. The check R_A + R_B = 260 ✓ and moments about B (195 × 4 = 260 × 3 = 780 ✓) both close.
Example:
A uniform 6 m plank of weight 200 N rests on supports at its ends, and a 400 N crate sits 2 m from one end. Find both reactions, using the same sequence: moments about one support → ΣF = 0 → check about the other. (Answer: moments about the near support give 6R_far = 400 × 2 + 200 × 3 = 1,400, so R_far ≈ 233 N and R_near = 600 − 233 ≈ 367 N.)
Check the conclusion and its limits
Take moments about a SECOND point as a check: a sign slip in the first equation shows up immediately. Both reactions of a supported beam must come out positive — a negative reaction means the beam would lift off that support and the model has changed.
Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.
Warn:
Zero resultant force is enough for equilibrium. This claim is false: two equal opposite forces at different points (a couple) give zero resultant force but still spin the body. Equilibrium needs zero resultant moment as well — that is why the spanner's length matters even when the push is unchanged.
Key:
Moments: moment = force × perpendicular distance in N m; equilibrium needs ΣF = 0 AND ΣM = 0; take moments about an unknown force to remove it, and check by re-taking moments about a second point.
