AQA A-Level Mathematics Paper 2, June 2025: Question 18
6 marks · Medium difficulty · Multi-step Problem
Find the normal reaction force at a support by taking moments for a horizontal platform in equilibrium, and explain qualitatively how the forces change as a person moves across the platform.
Practise this questionQuestion
Question text
18 In this question use g = 9.81 m s–2
A uniform platform AD has length 2.5 metres and weight 400 newtons.
The platform is attached to a chain at A. The other end of the chain is fixed to the floor.
The platform rests on a support at a point B which is 0.6 metres from A
The support exerts an upwards reaction force on the platform.
A child of mass 30 kilograms stands at a point C
The point C is 0.2 metres from D, as shown in the diagram.
A B C D
0.2 m
The system is in equilibrium with the platform resting horizontally.
18 (a) By taking moments about A, find the reaction force at B
[4 marks]
18 (b) (i) The child moves from C to B
(28)State what happens to the reaction force at B during this movement.
[1 mark]
18 (b) (ii) Having reached B, the child steps off the platform.
State what happens to the tension in the chain during this movement.
[1 mark]
Mark scheme
Show the mark scheme
Q Marking instructions AO Marks Typical solution
Obtains one of the following
18(a) moments about A. clockwise:2 3. 30 9 81.+1 25. 400
2 3. 30g OE anticlockwise:0 6.R
3.3
1 25. 400 OE B1 0 6.R = 2 3. 30 9 81.+1 25. 400
23.30 9 81.+1 25. 400
PI AWRT 1960 or AWRT 1983 R =
0 6.
Forms a three term moments
R = 1960 N
equation about A with at least
one correct term.
3.3 M1
Must be dimensionally correct.
PI 0.6R = AWFW [1176, 1177]
or 0.6R = 1190
Forms a fully correct moments
equation about A.
1.1b A1
PI AWRT 1960 or AWRT 1983
Obtains 1960 N
3.2a A1
Must include units
Subtotal 4
18(b)(i) States that the reaction force 3.5a E1 As the child walks towards B the
decreases with no other reaction force at B decreases.
incorrect reasoning.
Subtotal 1
18(b)(ii) States that the tension in the 3.5a E1 The tension in the chain does not
chain does not change. change.
Subtotal 1
Question 18 Total 6
How to answer it
Equilibrium of a Rigid Rod: Moments, Reactions & Chains
This question evaluates your ability to apply the conditions of static equilibrium to a rigid body resting on a pivot and restrained by a chain:
- Taking Moments: Selecting an appropriate pivot point (here point A) to eliminate unknown forces (the tension in the chain).
- Force Identification & Distances: Distinguishing between mass ( 30 kg ) and weight ( W = mg ), while using rod geometry to establish correct perpendicular distances.
- Conceptual Equilibrium Analysis: Analysing how internal support reactions and tensions vary dynamically when a mass moves along the beam.
Visualising the Setup (Force Diagram)
- Beam AD: Horizontal line of total length 2.5 m .
- Point A (x = 0 m): Tension force T acts vertically downwards (chain prevents the left end from tipping upwards).
- Point B (x = 0.6 m): Normal reaction force R acts vertically upwards from the support.
- Centre of Mass (x = 1.25 m): Platform weight 400 N acts vertically downwards (uniform rod: midpoint = 2.5 / 2).
- Point C (x = 2.3 m): Child weight 30 × 9.81 = 294.3 N acts vertically downwards ( 2.5 - 0.2 = 2.3 m from A).
Finding the Reaction Force at B by Taking Moments About A
Calculate normal contact force R at B using rotational equilibrium
💡 Key Knowledge
- Principle of Moments: For rotational equilibrium:
Σ Moments (anticlockwise) = Σ Moments (clockwise) - Elimination of Unknowns: Moments about point A eliminate the chain tension T completely because its line of action passes through A (perpendicular distance = 0).
- Mass vs Weight: Platform weight is given as 400 N (already a force). The child has a mass of 30 kg , so force = 30 × 9.81 N .
🧠 Exam Technique
- Always state which point you are taking moments about: "Taking moments about A ⟳".
- Double-check lengths from A:
• Support B: 0.6 m
• Rod midpoint: 1.25 m
• Child at C: 2.5 - 0.2 = 2.3 m - AQA requires appropriate significant figures (typically 3 s.f. when using g = 9.81 ) and explicitly requires units in the final answer.
📐 Step-by-Step Calculation
• Anticlockwise moment = R × 0.6
• Clockwise moment from rod weight = 400 × 1.25 = 500 N m
• Clockwise moment from child = (30 × 9.81) × 2.3 = 294.3 × 2.3 = 676.89 N m
0.6 × R = (400 × 1.25) + (30 × 9.81 × 2.3)
0.6 × R = 500 + 676.89
0.6 × R = 1176.89
R = 1176.89 / 0.6 = 1961.483... N
Rounding to 3 significant figures: R = 1960 N
✅ Correct Answer
R = 1960 N (or 1961 N / awrt 1960 N)
• B1: One correct moment term about A ( ±1.25 × 400 or ±2.3 × 30g ).
• M1: Three-term moments equation formed about A with at least one correct term.
• A1: Fully correct moments equation about A.
• A1: Correct final value 1960 N (must include unit N ).
❌ Common Errors to Avoid
- Multiplying weight by g again: The platform is given as 400 N , not 400 kg ! Writing 400 × 9.81 loses accuracy marks.
- Distance confusion: Using 0.2 m instead of 2.3 m as the distance from A to the child.
- Missing units: The mark scheme explicitly states "Must include units"—writing just 1960 forfeits the final mark.
Effect on Reaction Force at B as Child Moves from C to B
Qualitative analysis of varying distances
💡 Key Knowledge
Recall the moment equation about A for any position x (distance from A) of the child:
As the child moves from C ( x = 2.3 m ) to B ( x = 0.6 m ), the distance x decreases continuously. Therefore, the clockwise moment decreases, so R must decrease.
✅ Correct Answer
The reaction force at B decreases.
• E1: States that the reaction force decreases with no contradictory or incorrect reasoning.
Effect on Chain Tension When Child Steps Off from B
Moments about the pivot point B
💡 Key Knowledge
To see what happens to the chain tension T , take moments about B:
- Anticlockwise moment about B = T × 0.6 (from chain at A)
- Clockwise moment about B = 400 × (1.25 - 0.6) = 400 × 0.65 (from rod weight)
- Child at B: When the child is standing exactly at B, their line of action passes directly through the pivot B (perpendicular distance = 0 ).
- Thus, the child exerts zero moment about B whether they are standing on B or step off B!
- Therefore, 0.6 T = 400 × 0.65 remains completely unchanged.
🧠 Exam Insight & Misconceptions
✅ Correct Answer
The tension in the chain does not change (remains constant).
• E1: States that the tension in the chain does not change / remains the same.
Summary: Full Mark Strategy Checklist
- ✔ Check units on values given: Mass in kg needs multiplying by 9.81 ; weight in N does not.
- ✔ Always state pivot choice clearly to keep moment terms transparent.
- ✔ For conceptual "what happens" questions, set up the governing equation algebraically before guessing.
- ✔ Always include units ( N ) on final calculated force values.
Topics
Mechanics · S: Moments
Question and mark scheme from the AQA A-Level Mathematics examination, Paper 2, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.