AQA AS Level Physics Paper 2, June 2025: Question 25

1 mark · Medium difficulty · Multiple Choice

Identify the correct relationship between forces for a suspended mass pulled horizontally by a force at an angle $\theta$ in equilibrium.

Practise this question

Question

A diagram shows a mass m suspended from point P on a vertical wall by a light string under tension T. The string is displaced from the vertical by an angle theta due to a horizontal force F acting to the right on mass m. Four multiple-choice options are given: A: F = T cos(theta), B: (mg)^2 = T^2 + F^2, C: mg = T sin(theta), D: mg sin(theta) = F cos(theta).

Mark scheme

Show the mark scheme Mark scheme table indicating for question number 25 that the correct answer is option D, corresponding to the equation mg sin(theta) = F cos(theta), with assessment objective AO2.

How to answer it

Equilibrium of a Suspended Mass in Coplanar Forces

📋 What This Question Tests

This question assesses your ability to apply conditions of static equilibrium to a three-force system using trigonometric resolution and vector algebra:

  • Identifying all forces acting on a point mass (weight mg , tension T , and applied force F ).
  • Resolving forces into perpendicular horizontal and vertical components.
  • Formulating simultaneous equilibrium equations and manipulating ratios to eliminate unknown variables.
  • Applying alternative resolution directions (e.g. resolving perpendicular to tension).

Question 25: Analysis & Solution

Multiple Choice • 1 Mark • AO2 (Application of Knowledge)

✅ Correct Answer

Option D: mg sin θ = F cos θ

Mark Scheme Breakdown:
1 mark for selecting D.

By establishing standard equilibrium equations:
• Horizontal: T sin θ = F
• Vertical: T cos θ = mg
Dividing horizontal by vertical eliminates T :
tan θ = F / mg ⇒ sin θ / cos θ = F / mg ⇒ mg sin θ = F cos θ .

💡 Key Knowledge

  • Condition for Equilibrium: The resultant force in any direction must equal zero ( ΣF = 0 ).
  • Identifying Angles: The angle θ is defined between the string and the vertical. By alternate interior angles, the angle between tension T and the vertical at the mass is also θ .
  • Components of Tension:
    • Vertical: T cos θ (adjacent to θ )
    • Horizontal: T sin θ (opposite to θ )
  • Closed Vector Triangle: In equilibrium, the three force vectors form a closed right-angled triangle where the hypotenuse is T , the adjacent side is mg , and the opposite side is F .

📐 Step-by-Step Derivation

Method 1: Resolving Horizontally and Vertically (Standard Route)

  1. Identify the vertical balance:
    Upward force = Downward force
    T cos θ = mg   — (Equation 1)
  2. Identify the horizontal balance:
    Leftward force = Rightward force
    T sin θ = F   — (Equation 2)
  3. Eliminate tension T:
    Divide Equation (2) by Equation (1):
    (T sin θ) / (T cos θ) = F / (mg)
    sin θ / cos θ = F / (mg)
  4. Cross-multiply to match the given options:
    mg sin θ = F cos θ (which is Option D).

Method 2: Resolving Perpendicular to the String (Direct Route)

Because tension T acts purely along the string, resolving forces perpendicular to the string eliminates T immediately:

  • Component of mg perpendicular to string = mg sin θ (acting down and left).
  • Component of F perpendicular to string = F cos θ (acting up and right).
  • Equating opposing components: mg sin θ = F cos θ .

🧠 Exam Technique: Process of Elimination

  • Test Option A: F = T cos θ . Since θ is to the vertical, horizontal force must involve sin θ ( F = T sin θ ). Eliminate A.
  • Test Option B: Using Pythagoras on the right-angled force triangle, T is the hypotenuse: T² = (mg)² + F² , which rearranges to (mg)² = T² - F² . Option B incorrectly states (mg)² = T² + F² . Eliminate B.
  • Test Option C: mg = T sin θ . The vertical component adjacent to the angle is T cos θ , so mg = T cos θ . Eliminate C.
  • Result: Only Option D can be correct.

❌ Common Errors & Pitfalls

  • Swapping Sine and Cosine: Many students default to associating horizontal with cos and vertical with sin . This is only true when θ is measured from the horizontal. Always locate the angle relative to the axis!
  • Misidentifying the Hypotenuse: Mistaking the weight mg for the hypotenuse in Pythagoras led students to choose Option B. Tension T must support both F and mg , meaning T > mg and T > F .
  • Giving Up on Algebra: Students who successfully found tan θ = F / mg often failed to spot that rewrite tan θ as sin θ / cos θ leads directly to Option D.

Topics

Physics · 3.4 Mechanics and materials

Question and mark scheme from the AQA AS Level Physics examination, Paper 2, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.