Cambridge IGCSE Physics · topic 1.5.3
Centre of gravity
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Extra challenging questions
Work through the linked stages and explain what each result means.
The load that tips a mobile platform
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. A 20 kg platform has supports at x = 0 and x = 1.2 m. Its centre of mass is at x = 0.60 m. A 30 kg load is placed at variable x. Use g = 10 N kg⁻¹. The graph supplies the right-support reaction while both supports remain in contact. Supplied data (Load position x / m; Right-support reaction / N): (0, 100), (0.4, 200), (0.8, 300), (1.2, 400), (1.6, 500). (a) Find both support forces for x = 1.0 m and determine the load position at which tipping begins. [3 marks] (b) Find the combined centre of mass at the tipping threshold and assess lowering the platform's centre of mass on this ideal level-surface threshold. [3 marks]
Right-support reaction
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Right-support reaction / N
Horizontal axis (x): Load position x / m
Read the exact data as a table
| Load position x / m | Right-support reaction / N |
|---|---|
| 0 | 100 |
| 0.4 | 200 |
| 0.8 | 300 |
| 1.2 | 400 |
| 1.6 | 500 |
These are the supplied values, not a worked answer. Lines join the supplied values; they are not a fitted model.
Show a hint
Begin with part (a): Find both support forces for x = 1.0 m and determine the load position at which tipping begins.
Reveal the worked solution
- Taking moments about the left support: RR(1.2) = 200(0.60)+300x. At x = 1.0, RR = 350 N.
- Vertical force balance gives RL = 500−RR = 150 N.
- Tipping begins when the left reaction reaches zero, so RR = 500 N and 500(1.2) = 120+300x, giving x = 1.6 m. Beyond that point the two-contact static model fails.
- At xload = 1.6 m, xCM = [20(0.60)+30(1.6)]/(20+30) = 1.2 m, directly above the right support.
- At the threshold, the combined weight's line of action reaches the support boundary. A support cannot provide a downward reaction to maintain this contact model.
- Lowering the centre of mass can improve resistance to tipping during tilt or lateral acceleration, but it does not change the horizontal moment arms of vertical weights in this strictly level static setup.
Answer (a) At x = 1.0 m: right support = 350 N, left = 150 N; tipping threshold x = 1.6 m. (b) Combined centre of mass lies at x = 1.2 m; lowering it does not change this ideal horizontal overhang threshold.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.