Centre of gravity and stability
Why does a tall lamp tip more easily than a wide, low one? Find where weight acts, investigate a suspended card, and predict whether a tilted object will return or topple.
This is Cambridge IGCSE Physics 0625, separate from Egyptian Baccalaureate / Thanaweya Amma and Year 3. All three objectives in this lesson are Core and also needed by Extended candidates.
Imagine the weight acting at one point
Gravity pulls on every part of an object. To describe its overall turning effect, we can treat the whole weight as acting through one point: the centre of gravity (c.g.). This is a useful model, not a claim that gravity acts only there.
Worked example
A uniform 80 cm rule balances at its 40 cm mark. Its weight acts through that midpoint, so it has no turning moment about a support there.
Think first: Does supporting an object at its centre of gravity make its weight disappear?
No. The weight remains; its line passes through the support, so its moment about that support is zero.
Practical method: turn the card, record another vertical
- Suspend thin irregular card freely from a small hole using a pin in a secure stand.
- Hang a plumb line from the same pin and wait until card and thread settle.
- Mark two points along the thread on the card, then join them with a thin ruled line.
- Rehang from another well-separated hole and repeat. The intersection locates the c.g.
- Use a third suspension to check the crossing. Handle pins and scissors carefully.
The card changes orientation each time: every thread is vertical during its own hanging, but the lines recorded on the card point in different directions. Two distinct lines locate the point; a third checks it. Small disagreements indicate uncertainty, not three different centres of gravity.
Explain why, rather than just list apparatus
A freely suspended card settles with its c.g. below the pin, giving zero moment of its weight about the pin. The plumb line shows that vertical. A second hanging supplies a second line through the same c.g.; their intersection finds its position. A tight clamp would spoil this reasoning by applying a turning effect.
Where does the weight line go?
Predict, tilt, explain
A 150 cm tall model block is held at the chosen tilt about its right bottom edge. The red point is its centre of gravity; the dashed vertical line shows where its weight acts. Assume a rigid body, flat ground, sufficient friction to prevent sliding, and release from rest. This is a geometry model, not measured experimental data.
What does the weight tend to do when the block is released?
Try this investigation: keep height and tilt fixed, and change only width. Then reset width and lower only the centre of gravity. Describe what changes. You do not need a trigonometry formula for this syllabus objective.
Practise, then explain
Six multiple-choice questions, three numerical responses and seven written tasks. Numerical answers are checked automatically; written answers use a self-assessment guide. The suggested marks are for this original practice, not an official Cambridge mark scheme.
Practice stays only while this page is open. Refreshing or leaving clears these responses. Copy useful written work into your notes before leaving.
What does the centre of gravity represent?
Optional hint
Think of replacing all the small downward gravitational forces with one resultant force.
Source alignment: supplied Physics-0625 Volume A, PDF pp. 84–89, and official Cambridge 2026–2028 syllabus, section 1.5.3. Questions, explanations and model are original course practice. Clarifications to the book: support area includes the space between feet; a plumb line gives a measured vertical with uncertainty; greater mass alone does not guarantee a larger toppling angle. No quantitative toppling-angle formula is required here.