Moment: M = Fd (perpendicular distance)
Moment of a force = the turning effect about a pivot; , where is the force in N and the perpendicular distance from the pivot to the force's line of action (m); unit newton metre, N m. Both elements needed: N and a perpendicular distance in m. For an angled force, use the shortest distance from the pivot to the force line. Write the unit "N m" (a space), never N/m (a spring constant) or N m². E.g. 80 N at 0.35 m: N m.
Principle of moments and equilibrium
Principle of moments: in equilibrium the sum of clockwise moments = the sum of anticlockwise moments about any pivot. Procedure: choose a pivot; find each ; label it CW or ACW; set ; solve for the unknown. Full equilibrium needs BOTH (a) resultant force = 0 AND (b) resultant moment = 0 — stating only "forces balance" misses condition (b) and loses a mark.
Centre of gravity and stability
Centre of gravity = the point through which the whole weight of an object acts. It equals the geometric centre ONLY for a uniform, symmetrical body; for an irregular object it shifts towards the heavier side. Stability: an object is stable if a small tilt lets it return rather than topple. It increases with (1) a lower centre of gravity and (2) a wider base. An object topples when the vertical line through its centre of gravity passes outside its base.
Drawn from real examiner reports.
Moment unit written N/m or N m²
A moment = force × distance, so its unit is N × m = N m. "N/m" (newtons per metre) is the unit of a spring constant and is dimensionally wrong here; "N m²" is also wrong. If you work in centimetres the unit is N cm — a slash between the units is never correct. Correct numeric working still loses the unit mark if the unit is written with a slash.
June 2024 Paper 2P Q2(a) — sizable minority of candidates wrote N/cm or N/m as the unit instead of N cm or N m, losing the unit mark despite correct numeric working.
Slant distance used, not perpendicular
In , is the perpendicular distance from the pivot to the force's line of action, not the straight-line length along the object. For a 10 N force at 60° on a 0.5 m bar, the perpendicular distance is m, not 0.5 m — using 0.5 m overestimates the moment. Ask: "what is the shortest distance from the pivot to the force line?"
June 2024 Paper 2P Q2 general: candidates who misidentified the perpendicular distance in a force diagram calculated a larger, incorrect moment.
Centre of gravity called "the middle"
"The centre of gravity is the middle / geometric centre" is only true for a uniform object. The mark-scheme form is "the point through which the weight acts". For a non-uniform object (e.g. clay on one end of a rule) it shifts towards the heavier side. Lead with "the point through which the weight acts", then add "= geometric centre" only if the object is uniform.
General examiner guidance across multiple sittings: vague descriptions of centre of gravity as "the middle" or "the centre" without reference to where weight acts score zero for definition marks.
Adding forces instead of moments
On a balanced lever, balance the MOMENTS (), not the forces. Adding the two weights (e.g. 120 N + ) ignores that they act at different distances from the pivot. A small force far out can balance a large force close in. Always compute each and apply .
Two forces on one side not summed
When two forces act on the same side, add their individual moments — do NOT add the forces then multiply by one distance. For 40 N at 0.30 m and 60 N at 0.80 m: N m, not N m. Each force-distance pair is a separate moment.
November 2024 Paper 2P Q6(a): candidates who evaluated a single moment scored partial credit; those who correctly handled both forces (doubled the moment for equal forces at equal distances) scored full marks.
Beam's own weight ignored or mis-placed
A uniform beam's weight acts at its centre of gravity (its geometric centre). If the pivot is at the centre, that weight has zero moment; if the pivot is off-centre, the weight has a moment at its distance from the pivot. Forgetting the beam's weight, or placing it at the pivot when it is not there, gives a wrong balance equation.
Label CW/ACW, then write the equation
For a lever-balance question: label each force CW or ACW about the chosen pivot; write explicitly; substitute each moment as ; then solve. Writing the equation before the numbers earns the method mark.
Choose the pivot to remove an unknown
Take moments about the point where an unknown force acts: that force then has zero distance, so zero moment, and drops out — leaving one unknown to solve. Remember a uniform beam pivoted at its centre also contributes no moment from its own weight.
Write the moment unit as N m (no slash)
Give a moment's unit as N m (a space between N and m), or N cm if you kept centimetres — never with a slash, which means "per" and turns it into a spring-constant unit. Convert cm to m first if the answer must be in N m.
| Quantity | Symbol | Formula | Unit |
|---|---|---|---|
| Moment of a force | newton metre, N m | ||
| Principle of moments | — | N m = N m |
Where = force (N), = perpendicular distance from pivot to line of action of force (m), = moment (N m).
Critical: is ALWAYS the perpendicular distance — the shortest distance from the pivot to the line along which the force acts. If a force is applied at an angle, the perpendicular distance is shorter than the physical arm length.
Two conditions for equilibrium (both must be satisfied):
Define the moment of a force.
A mechanic applies a force of 80 N to a wrench at a perpendicular distance of 0.35 m from the centre of a bolt.
Calculate the moment of the force about the centre of the bolt. Give the unit of your answer.
