Principle of moments and centre of gravity
Key idea: In equilibrium, clockwise and anticlockwise moments about the same point balance. Include the object's weight acting through its centre of gravity.
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Guided lessons · coming soon
Guided lesson
Guided steps
- Read why both forces and moments must balance.
- Follow how the two people’s moments balance about the pivot.
- Choose the principle and find the balancing distance.
- Read how to choose one pivot and add the moments.
- Find the unknown load, then balance the vertical forces.
- Find both forces and explain the support’s zero moment.
- Add the two moments, then find the balancing and support forces.
- Read where the weight acts on uniform and non-uniform bodies.
- Find the moment and turning direction after moving the pivot.
- Find where the rule’s weight acts and measure from the pivot.
- Find where the weight acts, then its moment and direction.
- Include the plank’s weight when adding the moments.
- Measure from the pivot, include the rule’s weight and find F.
- Find the stone’s weight by balancing the two moments.
- Use the measured centre of gravity to find moment and direction.
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ContinueA heavy adult and a small child can balance on a seesaw if the adult sits closer to the middle. The seesaw stays level because the turning effects on it cancel: the child’s weight turns it one way and the adult’s turns it the other, equally. That balance, the principle of moments, lets you find an unknown force, distance or weight on any level beam, once you know where the beam’s own weight acts.
The principle of moments
A body is in equilibrium when the forces and the moments on it balance. A body at rest in equilibrium stays at rest: it does not start to turn and it does not start to move. A level seesaw with two children sitting still on it is in equilibrium.
For a body in equilibrium, the turning effects cancel. This is the principle of moments:
For a body in equilibrium, the sum of the clockwise moments about a pivot is equal to the sum of the anticlockwise moments about the same pivot.
For a body initially at rest and free to rotate about its pivot, a greater total clockwise moment makes it start to turn clockwise; a greater total anticlockwise moment makes it start to turn anticlockwise.
A balanced body still has forces acting on it. Balanced moments do not mean that individual forces disappear. The pivot also pushes up on the body, and for equilibrium the upward forces must equal the downward forces. The pivot’s force has no moment about the pivot, because its line of action passes through the pivot.
Worked example 1
Where must the adult sit?
Problem
In the seesaw above, the plank is uniform and pivoted at its middle. A 400 N child sits 1.5 m to the left of the pivot. How far to the right of the pivot must a 600 N adult sit for the seesaw to balance?
Show full solution
Sort the moments
Method
The child’s weight is left of the pivot and the adult’s is right of it.Reason
The plank’s own weight acts at its middle, through the pivot, so it has no moment about the pivot.Working
Answer
Anticlockwise: 400 × 1.5 = 600 N m. Clockwise: 600 × d.Apply the principle of moments
Working
Answer
600 × d = 600, so d = 1.0 m to the right of the pivot.
Quick check
The heavier person sits closer to the pivot, as on a real seesaw. The pivot pushes up with 400 + 600 = 1000 N, plus the plank’s own weight.
Check your understanding 1 Apply the principle about one pivot
A uniform seesaw is pivoted at its middle. A 300 N child sits 2.0 m to the left of the pivot. Find that child’s moment, then the distance to the right where a 500 N child must sit for balance.
Show the steps Optional working
Show answer
The principle compares the sums of moments about the same pivot, for a body in equilibrium. It does not require equal forces on the two sides, or the absence of forces. For this uniform seesaw, the plank’s weight has zero moment about its midpoint pivot. The child’s anticlockwise moment is 300 × 2.0 = 600 N m. Set 500 × d = 600, giving d = 1.2 m to the right of the pivot.
Use the principle to solve problems
Follow the same steps for any balanced body:
- Choose one pivot. If a force you do not need acts at a point, taking moments about that point removes it, because its moment there is zero.
- For each force, find its perpendicular distance from that pivot, in metres. Measure from the pivot, not from the end of the beam.
- Label each moment clockwise or anticlockwise.
- Add up each side and write sum of clockwise moments = sum of anticlockwise moments.
- Solve for the unknown, then check the unit and whether the answer makes sense.
Worked example 2
An unknown force and the pivot’s force
Problem
A light horizontal beam is supported at a pivot O. A 36 N downward force acts 0.25 m to the left of O. An unknown downward force F acts 0.60 m to the right of O. The beam is in equilibrium. Find F and the upward force R from the pivot.
Show full solution
Take moments about O
Method
Use O as the pivot.Reason
R acts through O, so it has no moment about O and does not appear in the equation. “Light” means the beam’s weight can be ignored.Working
Answer
Clockwise: F × 0.60. Anticlockwise: 36 × 0.25 = 9.0 N m.Solve for F
Method
Apply the principle of moments.Reason
The beam is in equilibrium.Working
Answer
0.60F = 9.0, so F = 15 N.Balance the forces
Working
Answer
Upward force = total downward force: R = 36 + 15 = 51 N upward.
Quick check
Taking moments about O removed R from the moment equation, not from the beam. Without R, the beam would have a 51 N resultant force downward and would fall, even though the two moments balance.
Check your understanding 2 A beam and its support
A 30 N load hangs 0.20 m to the left of a pivot on a light horizontal beam. Find the downward force needed 0.50 m to the right of the pivot for equilibrium, then the upward force from the pivot. Choose why the support has no moment about the pivot.
Show the steps Optional working
Show the steps Optional working
Show answer
30 × 0.20 = F × 0.50, so F = 12 N. The pivot pushes up with 30 + 12 = 42 N to balance the downward forces. Its line of action passes through the pivot, so its moment about that pivot is zero; this removes it from the moment equation, not from the forces on the beam.
Guided practice 3
Two loads on one side
Problem
A light beam is pivoted at its middle. A 20 N load hangs 0.50 m to the left of the pivot and a 10 N load hangs 0.20 m to the left of the pivot. A downward force F acts 0.40 m to the right of the pivot and the beam is level.
Try this before viewing the solution
Show the steps Optional working
Show the steps Optional working
Show the steps Optional working
Hints
Hint 1
one side
Both loads turn the beam the same way. Work out each moment, then add them.
Hint 2
the pivot force
The beam does not move up or down, so the upward force equals the total of the downward forces.
Show solution step by step
Add the anticlockwise moments
Method
Both loads are left of the pivot.Reason
Each has its own distance, so find each moment before adding.Working
Answer
20 × 0.50 + 10 × 0.20 = 12 N mEquate and solve
Method
F is the only clockwise moment.Reason
The beam is in equilibrium, so the totals are equal.Working
Answer
0.40F = 12, so F = 30 N.Find the pivot force
Working
Answer
R = 20 + 10 + 30 = 60 N upward.
Quick check
Adding the two loads first (30 N) and using one distance gives the wrong answer, because the loads act at different distances from the pivot.
Centre of gravity
Every part of an object has weight. You do not need to add up the moments of all those parts. The centre of gravity of an object is the single point at which its whole weight may be taken to act. When you take moments, draw one downward arrow for the object’s weight, through its centre of gravity.
- A uniform beam or rule has the same weight in every equal length, so its centre of gravity is at its middle. For a metre rule, that is the 50 cm mark.
- A non-uniform rod, such as a tapered rod, need not have its centre of gravity at its middle. Use the position given or measured, rather than assuming the midpoint.
- If the pivot is at the centre of gravity, the weight acts through the pivot and has no moment. That is why a uniform plank balances on its own when pivoted at its middle, and why the balance point of a metre rule on a knife edge is its centre of gravity.
Check your understanding 3 A plank on its own
A uniform horizontal plank weighs 40 N and initially balances on a pivot at its middle. The pivot is moved 10 cm to the left of the middle, with the left end on your left. Find the moment of the plank’s weight about the new pivot and predict the direction it turns. Would doubling its weight restore balance?
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Show answer
The centre of gravity stays at the middle, now 0.10 m to the right of the pivot. The weight gives a moment of 40 × 0.10 = 4.0 N m clockwise, so the right side turns down. Doubling the weight would double this moment to 8.0 N m clockwise, not restore balance. The pivot’s own force still acts through the pivot and cannot balance that moment.
Worked example 4
Balancing a metre rule with its own weight
Problem
A uniform metre rule weighs 1.2 N and is pivoted at the 40 cm mark. What downward force F at the 10 cm mark keeps it level?
Show full solution
Place the rule's weight
Method
The rule is uniform, so its centre of gravity is at the 50 cm mark.
Reason
The pivot is not at that point, so the rule’s weight has a moment and must be included.
Working
Answer
The weight acts 50 - 40 = 10 cm = 0.10 m right of the pivot: clockwise.
Measure F's distance from the pivot
Method
F acts at the 10 cm mark.Reason
Distances are measured from the pivot, not from the end of the rule.
Working
Answer
40 - 10 = 30 cm = 0.30 m left of the pivot: anticlockwise.
Apply the principle of moments
Working
Answer
F × 0.30 = 1.2 × 0.10, so F = 0.12/0.30 = 0.40 N downward.
Quick check
F is smaller than the rule’s weight because it acts three times as far from the pivot. Leaving out the rule’s weight would give F = 0, which cannot be right: the rule would tip clockwise.
Check your understanding 4 Where the weight acts
A uniform horizontal 1.5 m rod weighs 24 N and is pivoted 0.25 m from its left end. Where does its weight act, and what is the moment of the weight about the pivot, viewed with the left end on your left?
Show the steps Optional working
Show answer
The rod is uniform, so its weight acts at its 0.75 m midpoint, 0.75 − 0.25 = 0.50 m right of the pivot. The moment is 24 × 0.50 = 12 N m clockwise.
Worked example 5
A plank that is not pivoted at its middle
Problem
A uniform 2.0 m plank weighs 120 N and is pivoted 0.75 m from its left end. A 200 N load hangs 0.20 m from the left end. Find the downward force F at the right end that keeps the plank level.
Show full solution
Find each distance from the pivot
Method
Change each position along the plank into a distance from the pivot.
Reason
The plank is uniform, so its weight acts at its 1.0 m midpoint.
Working
Answer
Load: 0.75 - 0.20 = 0.55 m left. Weight: 1.0 - 0.75 = 0.25 m right. F: 2.0 - 0.75 = 1.25 m right.
Write the moment equation
Method
The load turns the plank anticlockwise; the weight and F turn it clockwise.
Reason
Both clockwise moments go on the same side.Working
Answer
200 × 0.55 = 120 × 0.25 + F × 1.25
Solve
Working
Answer
110 = 30 + 1.25F, so F = 80/1.25 = 64 N downward.
Quick check
Check: clockwise 30 + 64 × 1.25 = 30 + 80 = 110 N m, equal to the anticlockwise 110 N m. F must push down because the load alone has the larger moment.
Check your understanding 5 Two errors
A uniform metre rule weighing 1.2 N is pivoted at the 30 cm mark. A 3.0 N load hangs at the 10 cm mark, and a downward force F at the 90 cm mark keeps the rule level. A learner writes 3.0 × 0.10 = F × 0.90. Find the load’s distance from the pivot, the rule’s weight moment and the correct force F.
Show the steps Optional working
Show the steps Optional working
Show answer
The learner used scale readings instead of distances from the pivot, and left out the rule’s weight. The load is 0.20 m left: 3.0 × 0.20 = 0.60 N m anticlockwise. The rule’s weight acts at 50 cm, 0.20 m right: 1.2 × 0.20 = 0.24 N m clockwise. F is 0.60 m right, so 0.60 = 0.24 + 0.60F and F = 0.36 ÷ 0.60 = 0.60 N.
Guided practice 6
Weighing a stone with a metre rule
Problem
A uniform metre rule weighs 1.5 N. It is pivoted at the 40 cm mark, and a stone hangs from the 15 cm mark. The rule is level. Find the weight W of the stone.
Try this before viewing the solution
Show the steps Optional working
Show the steps Optional working
Hints
Hint 1
the rule's weight
The rule is the only other thing with a moment. Its weight acts at its centre of gravity.
Hint 2
distances
Measure both distances from the 40 cm mark, and change them to metres.
Show solution step by step
Moment of the rule's weight
Method
The rule’s weight acts at the 50 cm mark.Reason
A uniform rule’s centre of gravity is at its middle.Working
Answer
1.5 × 0.10 = 0.15 N m clockwise.Balance it with the stone
Working
Answer
The stone is 40 - 15 = 25 cm = 0.25 m left of the pivot, so W × 0.25 = 0.15 and W = 0.60 N.
Quick check
The stone is lighter than the rule because it hangs farther from the pivot. The same method gives the weight of the rule itself if you hang a known weight instead.
Check your understanding 6 A non-uniform rod
A horizontal non-uniform rod weighs 20 N. Positions in the table are measured from its left end.
| Feature | Position / m |
|---|---|
| Left end | 0.00 |
| Pivot | 0.50 |
| Measured centre of gravity | 0.80 |
| Right end | 1.20 |
A learner puts the rod’s weight at its 0.60 m midpoint. Correct the placement, then find the moment of the weight about the pivot.
Show the steps Optional working
Show answer
The rod is not uniform, so its weight acts at the measured centre of gravity, 0.80 m, not the midpoint. That is 0.80 − 0.50 = 0.30 m right of the pivot, so the moment is 20 × 0.30 = 6.0 N m clockwise. The midpoint would give 20 × 0.10 = 2.0 N m, which is wrong for this rod.
Common mistakes
- Saying a balanced beam has no forces on it. Several forces act; their moments cancel, and the pivot pushes up to balance the downward forces.
- Leaving out the beam’s own weight. Unless the beam is light, its weight has a moment whenever the pivot is not at its centre of gravity.
- Putting the weight of a non-uniform object at its middle. Use the centre of gravity that is given or measured.
- Using scale readings as distances. A load at the 10 cm mark is 0.30 m from a pivot at the 40 cm mark.
- Adding the loads on one side before multiplying. Loads at different distances have different moments, so find each moment and then add.
- Taking some moments about one point and some about another. Measure every distance from the same pivot.
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Syllabus and review details
- SEC G3 Combined Science Physics component 2027 · 2027
Physics section, PDF pages 8–24
Last reviewed: