Forces and motion · GCSE Physics
Moments
Teacher-written GCSE Physics revision on moments: moment = force × perpendicular distance, clockwise and anticlockwise balance, centre of mass, and levers as force multipliers.
A moment is a turning effect: M = Fd, with d the perpendicular distance from the pivot. Equilibrium: total clockwise moment = total anticlockwise moment.
The important bits
What you need to know
- 1
Moment of a force M = F × d, where d is the perpendicular distance from the pivot to the line of action of the force. M is in N m (newton-metres).
- 2
The distance must be perpendicular. If the force is at an angle, use the perpendicular component or the perpendicular distance — do not use the slanted length along the beam unless it is already perpendicular.
- 3
A larger force or a larger perpendicular distance gives a larger moment. That is why a long spanner loosens a tight nut.
- 4
Principle of moments: for an object in equilibrium, total clockwise moment about a pivot equals total anticlockwise moment about the same pivot.
- 5
The object must also have zero resultant force. Moments balance turning; forces still have to balance along each line.
- 6
Weight acts at the centre of mass. For a uniform beam, that is the midpoint. Include the beam’s weight if the question gives it.
- 7
Levers are force multipliers when the load is closer to the pivot than the effort: a small effort at a large distance balances a large load at a small distance.
- 8
If the line of action of a force goes through the pivot, d = 0 and the moment is zero. Pushing towards the hinge does not close the door.
Quotations worth analysing
Short evidence. Real method.
“Moment of a force = force × perpendicular distance from the pivot”
Perpendicular is the word that carries the mark. A 2.0 m beam with a force along its length about one end produces no moment.
“For equilibrium, clockwise moments equal anticlockwise moments.”
Pick one pivot, take every force’s moment about that point, and set the two senses equal. Then solve for the unknown.
Go deeper
Choose a pivot that deletes an unknown
If a beam is supported at two ends and you need one support force, take moments about the other end. That support’s distance is zero, so its moment vanishes, and you do not need it yet. Include the weights of people or boxes at their distances, and the beam’s weight at the centre if it is not negligible. Set clockwise = anticlockwise, find the unknown, then use vertical equilibrium (upward forces = weights) to find the other support. Students take moments about a random point and drown in algebra. They also use the length of the beam instead of the distance to the force. Sketch, mark the pivot with a triangle, draw perpendiculars, then write M = Fd for each force on a separate line.
Go deeper
Levers and centre of mass are the same physics
A wheelbarrow puts the load between the wheel (pivot) and the handles. Effort at the handles has a large d, so a modest force balances a heavy load. A crowbar and a see-saw are the same sentence with different pictures. Centre of mass is where the weight arrow goes; if it lies outside the base of a lorry, the weight has a moment that tips the lorry over. Stability: wide base, low centre of mass, so a large tilt is needed before the weight’s line of action leaves the base. Combined and Triple both like a see-saw with a parent and a child: m₁g × d₁ = m₂g × d₂, and g cancels if you are finding a mass, but write it first so the units are honest. Distance still in metres.
See the idea in action
A uniform 4.0 m plank of weight 20 N is pivoted at its centre. A 12 N weight sits 1.5 m to the left of the pivot. For balance, a force F on the right must satisfy clockwise moment = anticlockwise moment. The plank’s own weight acts at the centre, so its moment about the pivot is zero. Anticlockwise: 12 × 1.5 = 18 N m. Clockwise: F × 1.5 if F is also 1.5 m to the right, so F = 18 / 1.5 = 12 N. If instead F is only 0.60 m to the right, F × 0.60 = 18, so F = 30 N. Same moment, smaller distance, larger force. Units: N × m = N m.
Exam technique
Turn knowledge into marks
Write moment = force × perpendicular distance, with metres. State clockwise = anticlockwise about a named pivot. Include the weight of a uniform beam at its centre. Check that upward forces still equal downward forces.
Common mistakes
Do not give these marks away
- 01
Using a non-perpendicular distance, or forgetting to convert centimetres to metres.
- 02
Leaving out the beam’s weight, or placing it at one end instead of the centre of mass.
- 03
Balancing moments but not checking that the resultant force is also zero.
A 5.0 N force acts 0.40 m perpendicular from a pivot. What is the moment?
A2.0 N m
B12.5 N m
C0.080 N m
D5.4 N m
Show the answer
2.0 N m. M = Fd = 5.0 × 0.40 = 2.0 N m. 12.5 N m divided instead of multiplying; 0.080 N m used 4.0 cm as 0.040 m twice or similar.
Quick questions
If this is the bit you searched
What is a moment?
The turning effect of a force about a pivot: force multiplied by the perpendicular distance from the pivot to the force’s line of action, in N m.
What is the principle of moments?
For an object in equilibrium, the total clockwise moment about any point equals the total anticlockwise moment about that point.
Where does weight act on a uniform beam?
At the centre of mass, which for a uniform beam is the midpoint. Draw it vertically down from there.
Why does a long spanner help?
It increases the perpendicular distance, so the same force produces a larger moment on the nut.