AQA AS Level Physics Paper 1, June 2022: Question 6
9 marks · Hard difficulty · Extended Answer
Calculate the position of a worker on a uniform platform given cable tensions, determine distance from strain, calculate tensile stress, and sketch a new extension-distance graph based on modified cable properties.
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Question text
06 Figure 5 shows a worker of weight 750 N on a uniform platform. The weight of the
worker is acting at a horizontal distance d from end A.
Throughout this question, assume that the platform is horizontal and that all cables
obey Hooke’s law.
Figure 5
The platform weighs 1800 N and is suspended by vertical cables P and Q.
Each cable has an unstretched length of 3.0 m.
The horizontal distance between P and Q is 3.6 m.
06.1 The worker moves to a position where the tension in the left-hand cable P is 1150 N.
Calculate d for this position.
[3 marks]
d = m
Figure 6 shows how the extension of P varies with d as the worker walks slowly along
the platform from A to B.
*12* Figure 6
The worker moves to a position X where the strain in P is 6.0 × 10−5.
06.2 Determine d for position X.
[2 marks]
d = m
06.3 The cable material has a Young modulus of 1.9 × 1011 N m−2.
Calculate the tensile stress in P when the worker is at X.
[1 mark]
tensile stress14 = N m−2
06.4 The original cables P and Q are replaced.
Table 2 shows how the properties of the original cables compare with the
replacement cables.
Table 2
Young modulus of
Unstretched length Radius
cable material
Original cables L r E
r
Replacement cables L 15 2E
After the cables have been replaced, the worker walks slowly from A to B.
Draw on Figure 7 a line to show the variation of the extension of the replacement
left-hand cable with d.
The original line from Figure 6 is shown on Figure 7 as a dashed line to help you.
[3 marks]
Figure 7
Mark scheme
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Question Answers Additional Comments/Guidance Mark AO
06.1 attempted use of principle of moments: examples of acceptable responses for MP1 3 2 x AO1
1150 × 3.6 or 1400 × 3.6 or 1800 × 1.8 + 750(3.6 –
seen by one correct side of an attempted principle of moments 1 x AO2
d) or 1800 × 1.8 + 750 x or 750 × d + 1800 × 1.8
equation.
Condone one error in distance or signs or force in
full use of principle of moments
an attempted use of principle of moments (must
have 3 forces multiplied by 3 distances)
For moments about B (or Q):
(d =) 2.4 (m) 1150 × 3.6 = 1800 × 1.8 + 750(3.6 – d) /
1150 × 3.6 = 1800 × 1.8 + 750 x
Alternative
x seen (with appropriate working) as 1.2 m or 2.4 m
Finds component of tension in P due to worker’s weight = 250 N /
(even when not answer line) gains MP1 and MP2
Finds tension in P (due to weight of worker) by dividing weight of
platform by 2 and subtracts from 1150 N Moments about A (or P):
OR 750 × d + 1800 × 1.8 = 1400 × 3.6
Finds component of tension in Q due to worker’s weight = 500 N / Alternative for MP1 and MP2:
Finds tension in Q (due to weight of worker) by dividing weight of
Moments about worker’s centre of gravity:
platform by 2 and subtracts from 1400 N
1150 × d + 1800(1.8 – d) = 1400 (3.6 – d)
Recognises the ratio of weight distribution to worker position relative
to cables P and Q MP1 for one correct side of equation seen.
250 N : 500 N = 3.6 − d : d (principle of moments) MP2 all correct terms seen (condone one error)
(d =) 2.4 (m) d = 1.2 m with supporting working gains MP1
and MP2 (need principle of moments)
= 0.18 mm ∆L d ∆L
06.2 Extension or use of ε = or reads off correctly Use of ε = is by rearrangement to make ∆𝐿𝐿 2 1 x AO1
L L
for their extension (+/- half a square) (where working for the subject and 6 x 10-5 × 3 seen (condone 1 x AO2
extension seen) use of L=3.6 m here).
Condone POT error on extension
Allow range of 1.75 m to 1.85 m
(d =) 1.8 m Some supporting use of graph for read-off
seen
06.3 σ =) 1.1(4) × 107 −2 1 AO1
( (N m ) c.a.o
Penalise double and thick lines (limit on
06.4 Straight line with negative gradient that intercepts extension 3 3 x AO3
axis and has a d range of 3.5 m to 3.7 m thickness of line: must be less than half square
thick)
Straight line passes through (0, 0.46)
Within 1/2 square
Straight line passes through (3.6, 0.26)
Within 1/2 square
Condone accuracy within a square max 1
for MP2 and MP3
Total 9
How to answer it
Moments, Materials & Hooke's Law Study Guide
What this question tests
This multi-part question tests your ability to apply the principle of moments for bodies in rotational equilibrium, connect mechanical strain and extension using ε = ΔL / L , calculate tensile stress ( σ = F / A ), and interpret graphical data to predict how changes to physical cable dimensions and material properties affect structural extension.
Question 0.6.1
Calculating position d using the principle of moments
✅ Correct Answer
d = 2.4 m
💡 Key Knowledge
- Principle of Moments: For a system in equilibrium, the sum of clockwise moments about any point equals the sum of anticlockwise moments.
- Uniform Platform: The weight of the uniform platform ( 1800 N ) acts precisely at its geometric centre ( 1.8 m from either end).
🧠 Exam Technique
Take moments about one of the cable supports (e.g., end A or cable Q) to eliminate one unknown force immediately. Always state your pivot point clearly!
❌ Common Errors
- Failing to include all three downward/upward forces (worker weight, platform weight, cable tension).
- Forgetting that the platform weight acts at the midpoint ( 1.8 m ), not at the end.
📐 Step-by-Step Calculation
- Choose a pivot: Take moments about cable Q (right-hand end):
Clockwise moments = Anticlockwise moments - Set up equation: (1150 × 3.6) = (1800 × 1.8) + (750 × (3.6 - d))
- Evaluate numbers: 4140 = 3240 + 2700 - 750d
- Rearrange for d: 750d = 3240 + 2700 - 4140 = 1800 → d = 1800 / 750 = 2.4 m
Question 0.6.2
Determining position d from strain
✅ Correct Answer
d = 1.8 m (Acceptable range: 1.75 m to 1.85 m )
💡 Key Knowledge
- Strain definition: ε = ΔL / L , where ΔL is extension and L is unstretched length ( 3.0 m ).
🧠 Exam Technique
Calculate the target extension ΔL first using the strain formula, then use Figure 6 graph axes to read off the corresponding value of d accurately.
📐 Step-by-Step Calculation
- Rearrange strain formula: ΔL = ε × L
- Substitute values: ΔL = (6.0 × 10⁻⁵) × 3.0 = 1.8 × 10⁻⁴ m = 0.18 mm
- Read graph: Locate 0.18 mm on the extension axis of Figure 6, trace across to the line, and read down to find d = 1.8 m .
Question 0.6.3
Calculating tensile stress in cable P
✅ Correct Answer
1.1 × 10⁷ N m⁻² (or Pa )
💡 Key Knowledge
- Hooke's Law & Young Modulus: E = σ / ε , therefore tensile stress σ = E × ε .
📐 Step-by-Step Calculation
- Apply formula: σ = (1.9 × 10¹¹) × (6.0 × 10⁻⁵)
- Calculate value: 11,400,000 = 1.14 × 10⁷ N m⁻²
Question 0.6.4
Predicting replacement cable extension characteristics
✅ Correct Answer
A straight line starting at (0, 0.46) mm on the extension axis and terminating at (3.6, 0.26) mm at end B with a negative gradient.
💡 Key Knowledge
- Scaling Extension: Since ΔL = F L / (A E) and cross-sectional area A = πr² , halving the radius r quarters the area ( (r/2)² = r²/4 ), multiplying extensions by 4. Doubling Young modulus E halves the extension. Combined factor: ×4 / 2 = ×2 overall extension increase.
🧠 Exam Technique
Calculate the new boundary points at d = 0 m (end A) and d = 3.6 m (end B) by doubling the original dashed line values, then draw a clean, single straight line with a sharp pencil.
❌ Common Errors
- Drawing double-lined or thick fuzzy lines (penalised by examiners).
- Incorrect scaling of radius change (forgetting that area depends on r² ).
Topics
Physics · 3.4 Mechanics and materials
Question and mark scheme from the AQA AS Level Physics examination, Paper 1, June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.