# Moments and Centre of Gravity

> Physics · CIE IGCSE 0625
> Source: https://www.owlsprep.com/study/cie-0625-u1-moments-and-centre-of-gravity/

This guide covers the turning effect of forces (moments), the principle of moments for balanced systems, centre of gravity measurement, and stability concepts for Core and Extended CIE IGCSE Physics 0625 exams.

**Prerequisites:** [Forces and their vector properties](https://www.owlsprep.com/study/cie-0625-u1-forces-vectors/); [Calculations involving weight ($W=mg$)](https://www.owlsprep.com/study/cie-0625-u1-mass-weight/)

## Learning objectives

- Define the moment of a force and calculate its value using the correct formula
- Apply the principle of moments for balanced systems to solve Core and Extended problems
- Describe how to determine the centre of gravity of regular and irregular objects
- Explain how centre of gravity position and base width affect object stability

## 1. Moments: The Turning Effect of a Force

**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.

*Notation:* $M$

*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.

$$M = F \times d$$

- $F$ = force in newtons (N)
- $d$ = perpendicular distance from pivot in metres (m)

**Worked example:** 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. 1. Identify given values: $F = 25$ N, $d = 0.4$ m
2. 2. Substitute into the moment formula:
3. $$M = 25 \times 0.4 = 10$$
4. 3. 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. Principle of Moments

**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.

**Worked example:** 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. 1. Anticlockwise moment (left side) = clockwise moment (right side)
2. 2. Substitute known values into the equation:
3. $$400 \times 1.5 = W \times 2$$
4. $$600 = 2W$$
5. $$W = 300 N$$

> **Extended only**
>
> For problems where the pivot is not at the centre of a uniform beam, you must include the moment produced by the weight of the beam acting at its centre of gravity.

**Worked example:** 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. 1. Anticlockwise moments = 100 N weight; Clockwise moments = beam weight + unknown weight $W$
2. 2. Perpendicular distances: 100 N = 0.3 m left of pivot, beam CoG = 0.2 m right of pivot, $W$ = 0.7 m right of pivot
3. 3. Apply principle of moments:
4. $$100 \times 0.3 = (50 \times 0.2) + (W \times 0.7)$$
5. $$30 = 10 + 0.7W$$
6. $$20 = 0.7W$$
7. $$W = 28.6 \text{ N (3 s.f.)}$$

## 3. Centre of Gravity

**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:

1. Hang the object freely from a pivot point
2. Hang a weighted plumb line from the same pivot and draw a line along the string on the object
3. Repeat for two more pivot points
4. The intersection of the three lines is the centre of gravity

**Worked example:** State the position of the centre of gravity of a uniform 20 cm long ruler.

1. 1. The ruler is a regular, uniform object, so its CoG is at its geometric centre.
2. 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. Stability of Objects

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

**Worked example:** Explain why a racing car is more stable than a tall, narrow van.

1. 1. The racing car has a much lower centre of gravity, as heavy components (engine, fuel tank) are mounted close to the ground.
2. 2. The racing car also has a wider wheel base than the van, increasing its base area.
3. 3. Both factors mean the racing car is far less likely to topple over when cornering at speed.

> **Extended only**
>
> An object will topple over if the vertical line drawn down from its centre of gravity falls outside its base area. You may be asked to draw this line on diagrams in exam questions.

## Common pitfalls

- **Wrong:** Using straight-line distance between pivot and force application point instead of perpendicular distance
  - Why it fails: Moments only depend on the perpendicular component of distance, as non-perpendicular components do not contribute to turning force
  - Correct: Explicitly confirm the distance is perpendicular to the line of action of the force before substituting into the $M=Fd$ formula
- **Wrong:** Forgetting to include the moment of the beam's own weight in Extended problems where the pivot is not at the beam centre
  - Why it fails: 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: Locate the beam's CoG at its midpoint, calculate its moment, and add it to the appropriate clockwise/anticlockwise side of your equation
- **Wrong:** Assigning moments to the wrong rotation direction (clockwise vs anticlockwise)
  - Why it fails: This leads to incorrect addition/subtraction when applying the principle of moments, resulting in wrong final answers
  - Correct: 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:** Assuming the centre of gravity is always inside the object
  - Why it fails: 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: For regular uniform objects use the geometric centre; for irregular objects reference the plumb line method result instead of making assumptions
- **Wrong:** Stating that stability only depends on the height of the centre of gravity
  - Why it fails: Width of the base and total weight also directly affect how likely an object is to topple over
  - Correct: When explaining stability, reference all relevant factors for the object in question, not just CoG height

## Cheatsheet

| Concept | Formula / Rule | Key Note / Unit |
| --- | --- | --- |
| Moment of a force | $M = F \times d$ | Unit: Nm; $d$ = 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) |

## 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.

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