✅ Moments equal
The total clockwise moment equals the total anticlockwise moment. The object is balanced. It does not start to turn.
Forces, Movement, Shape and Momentum » The Principle of Moments
What you'll learn this session
Study time: 30 minutes
Edexcel spec: 1.32P
- What the principle of moments says about a balanced object
- How to find an unknown force or distance on a see-saw or a balanced ruler
- How to deal with three forces, including the weight of a uniform beam
🔒 Unlock Full Course Content
Sign up to access the complete lesson and track your progress!
Unlock This CourseWhen turning effects cancel out
This lesson is examined in Paper 2 only.
In Moments and Centre of Gravity you worked out the moment of one force. Most real objects have more than one force trying to turn them. Think of two children on a see-saw. One child tries to turn it clockwise. The other tries to turn it anticlockwise. If the see-saw stays level and does not turn, their turning effects must cancel out exactly.
This idea is called the principle of moments:
For a balanced object, the sum of the clockwise moments = the sum of the anticlockwise moments about the pivot.
"Sum" means total. If two forces both try to turn the object clockwise, add their moments together.
In this lesson every force is a weight pulling straight down on a level beam. Forces like these, whose lines are parallel to each other, are called parallel forces. They can point the same way or in exactly opposite directions. The exam only asks about simple systems of parallel forces.
Key terms:
- Principle of moments: for a balanced object, the sum of the clockwise moments equals the sum of the anticlockwise moments about the pivot.
- Balanced: not turning, because the turning effects cancel out.
- Parallel forces: forces that act along parallel lines, so they point the same way or in exactly opposite directions, such as weights pulling straight down and a pivot pushing straight up.
Balanced or not?
❌ Moments not equal
The object turns in the direction of the larger total moment. If the clockwise moments add up to 12 Nm and the anticlockwise moments to 9 Nm, it turns clockwise.
Notice that a small force can balance a large force, as long as the small force is further from the pivot. It is the moments that must be equal, not the forces.
A method that always works
- Find the pivot. Describe or sketch the beam as a straight line with a small triangle under it for the pivot.
- For each force, find its perpendicular distance from the pivot. Read the question carefully: is the distance given from the pivot or from the end of the beam?
- Decide whether each force gives a clockwise or an anticlockwise moment. A weight on the right of the pivot turns the beam clockwise. A weight on the left turns it anticlockwise.
- Write: sum of clockwise moments = sum of anticlockwise moments.
- Put in the numbers and solve for the unknown force or distance.
The pivot also pushes up on the beam. This force acts at the pivot itself, so its perpendicular distance from the pivot is zero and its moment is zero. You can leave it out.
Two forces
Worked example 1: finding a force
A uniform metre ruler is balanced on a pivot at its 50 cm mark. A weight of 2.0 N hangs from the 20 cm mark. An unknown weight W hangs from the 90 cm mark. Calculate W.
Distance of the 2.0 N weight from the pivot = 50 − 20 = 30 cm = 0.30 m (anticlockwise)
Distance of W from the pivot = 90 − 50 = 40 cm = 0.40 m (clockwise)
sum of clockwise moments = sum of anticlockwise moments
W × 0.40 = 2.0 × 0.30
W × 0.40 = 0.60
W = 0.60 / 0.40
W = 1.5 N
Check that the answer makes sense. W is further from the pivot than the 2.0 N weight, so it should be the smaller force. It is.
Worked example 2: finding a distance
A girl of weight 300 N sits on a see-saw, 2.0 m to the left of the pivot. Her brother has a weight of 400 N. Calculate how far from the pivot he must sit, on the right, to balance the see-saw.
sum of clockwise moments = sum of anticlockwise moments
400 × d = 300 × 2.0
400 × d = 600
d = 600 / 400
d = 1.5 m
Three forces
With three forces, two of them are usually on the same side of the pivot. Work out each moment by itself, then add the two that turn the same way.
Worked example 3
Two children sit on the left of a see-saw. One has a weight of 250 N and sits 2.0 m from the pivot. The other has a weight of 350 N and sits 1.0 m from the pivot. An adult of weight 680 N sits on the right. Calculate the adult's distance from the pivot when the see-saw is balanced.
Sum of anticlockwise moments = (250 × 2.0) + (350 × 1.0) = 500 + 350 = 850 Nm
sum of clockwise moments = sum of anticlockwise moments
680 × d = 850
d = 850 / 680
d = 1.25 m
Never add the forces together and then multiply by one distance. Each force has its own distance, so each one needs its own moment.
The weight of the beam itself
A beam has weight too. Remember that the weight of a uniform beam acts at its centre. What this does depends on where the pivot is.
- Pivot at the centre of the beam. The weight acts at the pivot. Its distance from the pivot is zero, so its moment is zero. You can ignore it. This is why the ruler's weight was left out of worked example 1.
- Pivot not at the centre. The weight of the beam has a moment about the pivot. Treat it as one more force, acting at the middle of the beam.
Worked example 4
A uniform metre ruler is balanced on a pivot at its 30 cm mark. A weight of 0.80 N hangs from the 5 cm mark. Calculate the weight of the ruler.
Picture it: the pivot is left of the middle, so the long side of the ruler is on the right. The hanging weight on the left stops the long side from dropping.
Distance of the 0.80 N weight from the pivot = 30 − 5 = 25 cm = 0.25 m (anticlockwise)
The ruler's weight W acts at the 50 cm mark. Distance from the pivot = 50 − 30 = 20 cm = 0.20 m (clockwise)
sum of clockwise moments = sum of anticlockwise moments
W × 0.20 = 0.80 × 0.25
W × 0.20 = 0.20
W = 0.20 / 0.20
W = 1.0 N
This is a neat way to weigh a ruler without a balance. All you need is one known weight and a pivot.
Common mistakes
- Measuring distances from the end of the beam. Every distance must be from the pivot. A weight at the 20 cm mark is 30 cm from a pivot at the 50 cm mark, not 20 cm.
- Balancing the forces and not the moments. The weights on each side do not need to be equal. The moments do.
- Mixing units. Use the same distance unit on both sides of the equation. Changing every distance to metres is the safest way.
- Forgetting the weight of the beam. If the beam is uniform and the pivot is not at its centre, the beam's weight has a moment.
- Putting a force on the wrong side of the equation. Decide clockwise or anticlockwise for every force before you write any numbers.
Exam-style question
A uniform plank is balanced on a pivot at its centre. A bag of sand of weight 120 N rests on the plank, 0.80 m to the left of the pivot. A box rests on the plank, 0.60 m to the right of the pivot. The plank is level and balanced.
(a) State the principle of moments. [2 marks]
(b) Calculate the weight of the box. [3 marks]
(c) The box is replaced with a heavier box of weight 200 N at the same position. Determine which way the plank turns. [2 marks]
(d) Explain why the weight of the plank was not needed in part (b). [2 marks]
Model answer
(a) For a balanced object (1), the sum of the clockwise moments equals the sum of the anticlockwise moments about the pivot (1).
(b) Anticlockwise moment = 120 × 0.80 = 96 Nm (1)
W × 0.60 = 96 (1)
W = 96 / 0.60 = 160 N (1)
(c) Clockwise moment = 200 × 0.60 = 120 Nm (1). This is larger than the anticlockwise moment of 96 Nm, so the plank turns clockwise, with the right-hand end moving down (1).
(d) The plank is uniform, so its weight acts at its centre, which is at the pivot (1). The perpendicular distance from the pivot is zero, so the weight has no moment (1).
Exam tip
In part (a), the word "balanced" is easy to miss out and can cost a mark. Always finish the statement with "about the pivot". In part (c), a "determine" question needs numbers: work out both moments and compare them before you give the direction.