IGCSE · 1 · extended response
The load that tips a mobile platform
Motion, forces and energy · Original GioPhysics question with a detailed, mark-by-mark answer guide.
- Demand
- discriminating
- Marks
- 6
- Topics
- 2
- Answer
- Complete
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.
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
Ready to self-mark?Reveal the detailed answer guide
(a)
Find both support forces for x = 1.0 m and determine the load position at which tipping begins.
Build a chain of claim, evidence and physics reasoning. Address more than one relevant factor, identify a limitation or assumption, and finish with a conclusion that is conditional on the evidence rather than absolute.
- 1
Taking moments about the left support: RR(1.2) = 200(0.60)+300x. At x = 1.0, RR = 350 N.
- 2
Vertical force balance gives RL = 500−RR = 150 N.
- 3
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.
(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.
Build a chain of claim, evidence and physics reasoning. Address more than one relevant factor, identify a limitation or assumption, and finish with a conclusion that is conditional on the evidence rather than absolute.
- 1
At xload = 1.6 m, xCM = [20(0.60)+30(1.6)]/(20+30) = 1.2 m, directly above the right support.
- 2
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.
- 3
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.
Complete marking guidance
Use these worked steps alongside the part-by-part answers. They explain the method and mark allocation; they do not add extra marks.
- 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.
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