Moments and Centre of Gravity
PhysicsΒ· 1.5.2, 1.5.3 (2026-2028 Syllabus)Β· 25 min read
1. 1. Moments: The Turning Effect of a Forceβ β ββββ± 8 min
Moment of a force
The turning effect of a force around a pivot, equal to the product of the force and the perpendicular distance from the pivot to the line of action of the force.
Example:
A 10 N force applied 2 m perpendicular to a pivot creates a moment of 20 Nm.
The SI unit for moment is the newton-metre (Nm). Moments have a direction: either clockwise or anticlockwise around the pivot.
= force in newtons (N)
= perpendicular distance from pivot in metres (m)
Calculate the moment produced by a 25 N force applied perpendicularly 0.4 m from the pivot of a spanner. State the unit of your answer.
- 1
- Identify given values: N, m
- 2
- Substitute into the moment formula:
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- Add the correct unit: Final answer = 10 Nm
Exam tip:
Always confirm the distance you use is perpendicular to the line of action of the force before substituting into the moment formula.
2. 2. Principle of Momentsβ β β βββ± 10 min
Principle of Moments
For a system in equilibrium (balanced and not rotating), the sum of the clockwise moments around any pivot equals the sum of the anticlockwise moments around the same pivot.
This rule applies to all rigid non-rotating objects including seesaws, beams, levers and balance scales. Core questions include up to two forces on each side of the pivot; Extended questions may include multiple forces or pivots not at the centre of the beam.
A seesaw is balanced with a 400 N child sitting 1.5 m to the left of the pivot. A second child sits 2 m to the right of the pivot. Calculate the weight of the second child.
- 1
- Anticlockwise moment (left side) = clockwise moment (right side)
- 2
- Substitute known values into the equation:
- 3
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A uniform 1 m long beam of weight 50 N is pivoted 0.3 m from its left end. A 100 N weight is hung from the left end of the beam. Calculate the weight that must be hung from the right end to balance the beam.
- 1
- Anticlockwise moments = 100 N weight; Clockwise moments = beam weight + unknown weight
- 2
- Perpendicular distances: 100 N = 0.3 m left of pivot, beam CoG = 0.2 m right of pivot, = 0.7 m right of pivot
- 3
- Apply principle of moments:
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3. 3. Centre of Gravityβ β ββββ± 5 min
Centre of Gravity (CoG)
The point on an object where the entire weight of the object appears to act. For regular uniform objects, this is at their geometric centre.
For irregularly shaped objects, you can find the centre of gravity experimentally using the plumb line method:
Hang the object freely from a pivot point
Hang a weighted plumb line from the same pivot and draw a line along the string on the object
Repeat for two more pivot points
The intersection of the three lines is the centre of gravity
State the position of the centre of gravity of a uniform 20 cm long ruler.
- 1
- The ruler is a regular, uniform object, so its CoG is at its geometric centre.
- 2
- Midpoint of 20 cm is 10 cm, so CoG is at the 10 cm mark along the ruler.
Exam tip:
You may be asked to describe the plumb line experiment in written exams, so memorize all four steps clearly.
4. 4. Stability of Objectsβ β β βββ± 7 min
Stability is a measure of how likely an object is to topple over when tilted. Three factors affect stability:
Height of centre of gravity: lower CoG = more stable
Width of the base: wider base = more stable
Total weight: heavier objects are more stable for the same CoG height and base width
Explain why a racing car is more stable than a tall, narrow van.
- 1
- The racing car has a much lower centre of gravity, as heavy components (engine, fuel tank) are mounted close to the ground.
- 2
- The racing car also has a wider wheel base than the van, increasing its base area.
- 3
- Both factors mean the racing car is far less likely to topple over when cornering at speed.
5. Common Pitfalls
Wrong move:
Using straight-line distance between pivot and force application point instead of perpendicular distance
Why:
Moments only depend on the perpendicular component of distance, as non-perpendicular components do not contribute to turning force
Correct move:
Explicitly confirm the distance is perpendicular to the line of action of the force before substituting into the formula
Wrong move:
Forgetting to include the moment of the beam's own weight in Extended problems where the pivot is not at the beam centre
Why:
The weight of a uniform beam acts at its midpoint, so if that point is not on the pivot, it will produce a moment that affects equilibrium
Correct move:
Locate the beam's CoG at its midpoint, calculate its moment, and add it to the appropriate clockwise/anticlockwise side of your equation
Wrong move:
Assigning moments to the wrong rotation direction (clockwise vs anticlockwise)
Why:
This leads to incorrect addition/subtraction when applying the principle of moments, resulting in wrong final answers
Correct move:
Imagine pushing the force in the given direction, trace the rotation around the pivot, and label each moment as clockwise/anticlockwise before setting up your equation
Wrong move:
Assuming the centre of gravity is always inside the object
Why:
For hollow or irregular objects (e.g. a ring, a chair), the CoG can be in empty space inside or outside the object's material
Correct move:
For regular uniform objects use the geometric centre; for irregular objects reference the plumb line method result instead of making assumptions
Wrong move:
Stating that stability only depends on the height of the centre of gravity
Why:
Width of the base and total weight also directly affect how likely an object is to topple over
Correct move:
When explaining stability, reference all relevant factors for the object in question, not just CoG height
6. Quick Reference Cheatsheet
Concept | Formula / Rule | Key Note / Unit |
|---|---|---|
Moment of a force | Unit: Nm; = perpendicular distance from pivot | |
Principle of Moments | Sum of clockwise moments = Sum of anticlockwise moments | Applies only to non-rotating balanced systems |
Centre of Gravity | Geometric centre (regular uniform objects) | Found via plumb line method for irregular objects |
Stability Rule | Lower CoG + wider base = higher stability | Object topples if CoG line falls outside base (Extended) |
7. Frequently Asked
Do I have to use perpendicular distance when calculating moments?
Yes, you must use the perpendicular distance between the line of action of the force and the pivot, not the straight-line distance between the force application point and pivot. Core exam questions almost always provide the perpendicular distance directly to simplify calculations.
How do I distinguish between stable, unstable and neutral equilibrium?
- Stable equilibrium: When tilted, the centre of gravity rises, and the object returns to its original position (e.g. a book lying flat)
- Unstable equilibrium: When tilted, the centre of gravity falls, and the object topples (e.g. a pencil balanced on its tip)
- Neutral equilibrium: When tilted, the centre of gravity stays at the same height, and the object stays in its new position (e.g. a ball on a flat surface)
Going deeper
What's Next
Now that you have mastered moments and centre of gravity, you can move on to studying types of levers and mechanical advantage, which build directly on the principle of moments to explain how simple machines work. You will also apply these concepts to work, energy and power calculations in later units of the CIE IGCSE Physics 0625 syllabus. Make sure to practice past paper structured questions to reinforce your understanding of both Core and Extended problem types for this topic, as moments are frequently tested in both Paper 3 (Core) and Paper 4 (Extended) written exams. Pay special attention to Extended questions involving non-central pivots, as these are common high-mark questions that test both your knowledge of moments and centre of gravity.
