AQA A-Level Physics Paper 1, June 2024: Question 19

1 mark · Medium difficulty · Multiple Choice

Calculate the angle theta of the rope supporting a uniform bar of weight W with an object of weight W attached to its end, given a rope tension of 4W.

Practise this question

Question

A multiple-choice physics question featuring a diagram of a horizontal uniform bar pivoted at one end. A rope attached to the center of the bar goes up to a vertical wall at an angle theta to the vertical, with a tension of 4W. An object of weight W hangs from the right-hand end of the bar, and the bar's own weight W acts at its center. Four multiple choice options for angle theta are provided: A 41 degrees, B 45 degrees, C 60 degrees, and D 71 degrees.
Question text

19 The weight of a uniform bar is W.

An object also of weight W is attached to one end.

The bar is pivoted at the other end and held horizontal by a rope attached to its centre.

The tension in the rope is 4W.

What is angle θ?

[1 mark]

A 41°

B 45°

C 60°

D 71°

Mark scheme

Show the mark scheme The mark scheme indicating that the correct multiple-choice answer for question 19 is A, corresponding to an angle of 41 degrees.

19 A 41° AO2

How to answer it

Equilibrium and Moments of Forces

What this question tests

This question assesses your ability to apply the Principle of Moments to an object in rotational equilibrium. You must identify all relevant forces, determine their perpendicular distances (or resolve force components), set up a balanced moments equation about a pivot, and use basic trigonometry to find an unknown angle.

Question 19

Part (a) — Multiple Choice Calculation of Angle θ

✅ Correct Answer

A (41 degrees)

Mark allocation: 1 mark (AO2)

💡 Key Knowledge

  • Principle of Moments: For a system in rotational equilibrium, the sum of clockwise moments about any point equals the sum of anticlockwise moments about that same point.
  • Moment of a Force: Force multiplied by the perpendicular distance from the pivot to the line of action of the force ( Moment = F × d ).
  • Uniform Bar: The weight of the uniform bar ( W ) acts precisely at its geometric centre.

🧠 Exam Technique

  • Choose the pivot at the end of the bar (where the vertical wall / hinge is) to eliminate the unknown reaction force at the pivot from your calculations, since its distance from the pivot is zero.
  • Always define lengths in terms of a variable (e.g., let total length be 2L ) so that fractions cancel out cleanly.

❌ Common Errors

  • Using the wrong trigonometric ratio (e.g., using sin(θ) instead of cos(θ) ) by confusing which angle is given relative to the vertical or horizontal.
  • Forgetting to account for the position of the bar's weight acting at the centre rather than the end.

📐 Step-by-Step Calculation

  1. Define the lengths: Let the total length of the uniform bar be 2L .
    • The weight of the bar ( W ) acts downwards at distance L from the pivot.
    • The rope is attached at the centre, so its distance from the pivot is also L .
    • The object of weight W is attached at the far end, at distance 2L from the pivot.
  2. Set up the Principle of Moments about the pivot:
    Anticlockwise moment = Clockwise moments
    The tension force acts upwards at an angle θ to the vertical. The perpendicular distance from the pivot to the line of action of the tension is L sin(90 - θ) or more simply, considering horizontal and vertical components, the vertical component of tension provides the turning effect at distance L .
    Alternatively, resolve the tension perpendicular to the bar: Tension component perpendicular to bar = 4W cos(θ) (since θ is measured from the vertical, the angle between the rope and the vertical is θ, meaning the angle with the vertical wall is θ, so the component perpendicular to the horizontal bar is 4W cos(θ) ).
  3. Formulate the equation:
    Clockwise moments = (W × L) + (W × 2L) = 3WL
    Anticlockwise moment = (4W cos(θ)) × L
  4. Equate and solve:
    4W × L × cos(θ) = 3W × L
    Cancel W × L from both sides:
    4 cos(θ) = 3
    cos(θ) = 3 / 4 = 0.75
    θ = cos⁻¹(0.75) ≈ 41.4 degrees
  5. Match with options: Rounding to the nearest whole degree gives 41 degrees (Option A).

Topics

Physics · 3.4 Mechanics and materials

Question and mark scheme from the AQA A-Level Physics examination, Paper 1, June 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.