AQA AS Level Physics Paper 2, June 2024: Question 25

1 mark · Medium difficulty · Multiple Choice

Calculate the tension in cable Y given a mass held stationary by two cables X and Y at angles of 12° and 34° to the horizontal, with the tension in X being 390 N.

Practise this question

Question

A diagram shows a mass held stationary by two cables X and Y going over pulleys. Cable X makes an angle of 12 degrees to the horizontal, and cable Y makes an angle of 34 degrees to the horizontal on the opposite side. Multiple choice options for the tension in cable Y are A: 145 N, B: 380 N, C: 390 N, D: 460 N.
Question text

25 A mass is held stationary by two cables X and Y.

The tension in X is 390 N.

What is the tension in Y?

[1 mark]

A 145 N

B 380 N

C 390 N

D 460 N

Mark scheme

Show the mark scheme The mark scheme table indicates the correct answer for question 25 is option D, which corresponds to 460 N.

25 D 460 N

How to answer it

Equilibrium of a Mass Suspended by Two Cables

What this question tests

This question assesses your understanding of resolving forces into components and applying the principles of static equilibrium (specifically Newton's first law, where the net horizontal force must equal zero) for objects subjected to multiple non-collinear tensions.

Question Part 2.5

Determining the Tension in Cable Y

✅ Correct Answer

Option D: 460 N

💡 Key Knowledge

  • First Condition of Equilibrium: For a stationary object, the resultant force in any direction is zero.
  • Horizontal Equilibrium: The horizontal component of the tension in cable X must balance the horizontal component of the tension in cable Y ( T_X cos(12°) = T_Y cos(34°) ).

🧠 Exam Technique

Multiple-choice questions worth 1 mark require fast, efficient setup. Recognise immediately that vertical forces include an unknown downward weight, but the horizontal axis involves only the two tensions and their respective angles to the horizontal.

❌ Common Errors

  • Using sin instead of cos when resolving relative to the given horizontal angles.
  • Attempting to calculate the vertical weight first, which wastes time since the mass/weight value is not provided.

📐 Step-by-Step Calculation

  1. Identify the condition for horizontal equilibrium:
    F_(horizontal, left) = F_(horizontal, right)
    T_X × cos(12°) = T_Y × cos(34°)
  2. Substitute the known values ( T_X = 390 N ):
    390 × cos(12°) = T_Y × cos(34°)
  3. Rearrange to solve for T_Y :
    T_Y = (390 × cos(12°)) / cos(34°)
  4. Calculate the numerical values:
    cos(12°) ≈ 0.9781
    cos(34°) ≈ 0.8290
    T_Y = (390 × 0.9781) / 0.8290 ≈ 459.9 N
  5. Round to appropriate significant figures:
    Matches 460 N (Option D).
Mark Allocation: [1 mark] awarded for selecting option D.

Topics

Physics · 3.4 Mechanics and materials

Question and mark scheme from the AQA AS Level Physics examination, Paper 2, June 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.