A-Level Physics October/November 2024 Q3(c)(i): By taking moments about A, determine the tension in the wire.

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  By the principle of moments, for the rod to be in equilibrium, the sum of the clockwise moments about any point must equal the sum of the anticlockwise moments about the same point. We will take moments about pivot A.

  Sum of clockwise moments = Sum of anticlockwise moments

  The clockwise moments are caused by the weight of the rod (acting at its centre) and the additional 1.5 N weight. The anticlockwise moment is caused by the perpendicular component of the tension in the wire.

  Clockwise moment from rod's weight = Force × Perpendicular distance $= 33 \times \frac{0.65}{2} = 10.725$ N m

  Clockwise moment from 1.5 N weight = Force × Perpendicular distance $= 1.5 \times (0.65 - 0.12) = 1.5 \times 0.53 = 0.795$ N m

  Anticlockwise moment from tension = Perpendicular component of force × Distance $= (T \sin 50^\circ) \times \frac{0.65}{2} = T \times 0.766 \times 0.325 = 0.2490 \times T$

  Equating the moments: $(33 \times \frac{0.65}{2}) + (1.5 \times (0.65 - 0.12)) = T \sin(50^\circ) \times \frac{0.65}{2}$ $10.725 + 0.795 = 0.2490 \times T$ $11.52 = 0.2490 \times T$ $T = \frac{11.52}{0.2490}$ $T = 46.265...$ N

  $T = 46$ N (to 2 s.f.)

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  How the marks are awarded

  • C1 — Correctly calculating any one of the moments about pivot A. This is shown by writing 33 × 0.65/2, 1.5 × (0.65 – 0.12), or T sin 50° × (0.65/2).
  • C1 — Correctly applying the principle of moments by setting the sum of clockwise moments equal to the sum of anticlockwise moments, as shown in the full equation: (33 × 0.65/2) + (1.5 × (0.65 – 0.12)) = T sin(50°) × (0.65/2).
  • A1 — Obtaining the correct final answer of 46 N, correctly calculated and rounded to an appropriate number of significant figures.

  Common mistakes

  • Using the incorrect trigonometric component for the tension, for example using T cos(50°) instead of T sin(50°), when finding the component perpendicular to the rod.
  • Using an incorrect distance from the pivot for a force, such as using 0.12 m instead of (0.65 – 0.12) m for the 1.5 N weight, or forgetting to use the centre of mass (0.65/2) for the rod's weight.
  • Mixing up clockwise and anticlockwise moments, for example by adding the moment from the 1.5 N weight to the tension's side of the equation.
  • Calculation errors, such as having the calculator in radians mode instead of degrees, or rounding intermediate values too early, leading to an inaccurate final answer.

  Examiner tip: For any moments problem, always resolve forces into components that are perpendicular to the distance from the pivot, or find the perpendicular distance to the line of action of the force.

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