AQA AS Level Physics Paper 1, June 2018: Question 4

✅ Correct Answer

• Resultant force is zero (or forces up equal forces down, and forces left equal forces right).
• Resultant torque/moment is zero (or sum of anticlockwise moments equals sum of clockwise moments about any point).

💡 Key Knowledge

For an extended rigid body to be completely at rest (static equilibrium), both translational and rotational motion must be prevented.

✅ Correct Answer

The single point through which the entire weight/mass of the body can be considered to act (or where a force has no turning effect).

❌ Common Errors

Students often incorrectly state it is "where gravity acts" instead of specifying that it is the point where the entire mass/weight can be considered concentrated.

✅ Correct Answer

• Point A acts as the pivot.
• The lines of action of tensions T₁ and T₂ both pass through point A, creating zero moment about A.
• Therefore, the weight (acting through the centre of mass) must also pass vertically below A so that its moment about A is zero, keeping the beam in equilibrium with no resultant turning moment.

🧠 Exam Technique

Always look for a smart pivot choice! Taking moments about point A eliminates tensions T₁ and T₂ from the turning moment equation because their lines of action intersect directly at A (distance = 0).

📐 Calculation Steps (Resolution Method)

1. Recognise a closed triangle of forces: Since the beam is in equilibrium under three forces (Weight = 12000 N downwards, T₁ and T₂), they form a closed vector triangle.
2. Resolve vertically: T₁ cos(53) = 12000 cos(37) or use triangle geometry. Looking at the vertical components: T₁ = 12000 cos(53) = 7220 N (or 7200 N to 2 s.f.).
3. Resolve horizontally / use sine components: T₂ = 12000 sin(53) = 9583 N (or 9600 N to 2 s.f.).

✅ Correct Answers

T₁ = 7200 N (or 7221 N)

T₂ = 9600 N (or 9583 N)

📐 Step-by-Step Calculation

1. Recall formula: Δl = Fl / (A E)
2. Calculate cross-sectional area (A): Diameter d = 1.5 × 10⁻² m , so radius r = 0.75 × 10⁻² m .
   A = π r² = π × (0.75 × 10⁻²)² = 1.767 × 10⁻⁴ m²
3. Substitute values: Force F = 12000 N (weight of beam supported by the single main cable), length l = 12 m , E = 2.0 × 10¹¹ Pa .
   Δl = (12000 × 12) / (1.767 × 10⁻⁴ × 2.0 × 10¹¹ )
4. Evaluate extension: Δl = 4.08 × 10⁻³ m = 4.1 × 10⁻³ m (or 4.0 × 10⁻³ m )

❌ Common Calculation Traps

Using diameter instead of radius in the area formula ( π d² instead of π r² ) limits you to a maximum of 1 mark. Forgetting to convert cm/mm or diameter-to-radius correctly will ruin the power of ten.