AQA A-Level Physics Paper 1, June 2025: Question 24

📌 What this question tests

Principle of Moments: Setting the sum of clockwise moments equal to the sum of anticlockwise moments about a pivot for rotational equilibrium.
Centre of Gravity: Identifying that the weight of a uniform object acts directly at its midpoint.
Perpendicular Distances: Using basic trigonometry to resolve distances perpendicular to the line of action of each force.
Superfluous Information Handling: Recognising when a given variable (such as rod length L) cancels out in the final algebra.

Question 24 • Multiple Choice [1 Mark]

D – 4.7 N

Mark Scheme Reference: Option D (4.7 N) [AO2 – 1 mark]

Moment of a Force: Moment = Force × Perpendicular distance to line of action
Pivot Choice: Taking moments about the frictionless hinge eliminates the unknown reaction forces acting at the hinge.
Uniform Body: Weight W = mg acts downwards at distance L / 2 from the hinge.

Calculate the weight of the rod:
W = m × g = 0.55 kg × 9.81 m s⁻² = 5.3955 N
Identify the perpendicular distance for the weight:
The weight acts vertically downwards from the centre of mass (distance L / 2 along the rod).
The horizontal distance from the hinge to this line of action is:
d_W = (L / 2) × sin(60.0°)
This exerts a clockwise moment about the hinge.
Identify the perpendicular distance for force F:
Force F acts horizontally to the right at the tip (distance L along the rod).
The vertical distance from the hinge down to the line of action of F is:
d_F = L × cos(60.0°)
This exerts an anticlockwise moment about the hinge.
Equate moments (Principle of Moments):
Clockwise Moments = Anticlockwise Moments
W × (L / 2) × sin(60.0°) = F × L × cos(60.0°)
Simplify and solve for F:
Notice that the length L cancels out from both sides:
F = ½ × W × [sin(60.0°) / cos(60.0°)] = ½ × W × tan(60.0°)
F = 0.5 × 5.3955 N × tan(60.0°)
F = 2.69775 × 1.73205 = 4.673 N ≈ 4.7 N

Option B (1.6 N): Inverting the trigonometric ratio by using cos(60.0°) / sin(60.0°) = 1 / tan(60.0°) :
F = 2.70 / 1.732 = 1.56 N ≈ 1.6 N .
Option C (2.7 N): Forgetting trigonometry altogether or mistakenly calculating half the weight:
F = ½ × W = 2.7 N .
Option A (0.95 N): Mixing up sine/cosine and dividing by the length 1.3 m or multiplying inappropriately.
Missing the midpoint: Using full length L for the weight gives F = 9.35 N (avoided here by recognising the word uniform).

Always take moments about an unknown hinge/support: The hinge exerts an unknown contact force on the rod. Choosing the hinge as your pivot means the distance to this force is 0, completely removing it from your equation!
Check "Force × Distance" orientation:
• If a force is vertical (weight), multiply by the horizontal distance ( sin θ from the vertical).
• If a force is horizontal (F), multiply by the vertical distance ( cos θ from the vertical).
Do not panic if data seems redundant: The question gives 1.3 m , but it naturally cancels. You can plug in 1.3 m or keep it algebraic—either way works!

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