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9702 · 4.1

Turning effects of forces — practice questions

Practice and worked examples for 9702 Turning effects of forces. Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

A uniform beam, 3.0 m long and weighing 50 N, is pivoted at its centre. A 20 N weight is placed 1.0 m from the left end. Where should a 30 N weight be placed to balance the beam?

Show solution outline
  1. Identify pivot and forces: The pivot is at the centre of the beam (1.5 m from either end). The beam's own weight acts at the pivot, so it creates no moment.
  2. Calculate anticlockwise moment: The 20 N weight is 1.0 m from the left end, which is 1.5 m - 1.0 m = 0.5 m from the pivot. This creates an anticlockwise moment. M_ACW = Force × Distance = 20 N × 0.5 m = 10 Nm.
  3. Apply Principle of Moments: For the beam to balance, the clockwise moment must equal the anticlockwise moment. Let the 30 N weight be placed at a distance 'd' from the pivot on the other side. M_CW = 30 N × d.
  4. Set moments equal: M_CW = M_ACW, so 30 N × d = 10 Nm.
  5. Solve for d: d = 10 Nm / 30 N = 0.333... m ≈ 0.33 m.
  6. Final Answer: The 30 N weight should be placed 0.33 m to the right of the pivot.

Worked example 2

A uniform 4.0 m long rod with a weight of 120 N is hinged to a vertical wall. It is held horizontally by a cable attached to the end of the rod and to a point on the wall. The cable makes an angle of 30° with the rod. Calculate the tension (T) in the cable.

Show solution outline
  1. Identify the pivot and forces: The pivot is the hinge. The forces creating moments are the weight of the rod (acting downwards at its centre) and the tension in the cable (acting upwards at an angle).
  2. Calculate the clockwise moment (from the rod's weight): The rod is uniform, so its weight acts at its centre, which is 4.0 m / 2 = 2.0 m from the hinge. This creates a clockwise moment. Moment_cw = Force × Perpendicular Distance = 120 N × 2.0 m = 240 Nm.
  3. Calculate the anticlockwise moment (from the cable's tension): The tension T acts at an angle. We need the component of the tension that is perpendicular to the rod. This component creates the anticlockwise moment. T_perpendicular = T × sin(30°). This force acts at the end of the rod, 4.0 m from the pivot. Moment_acw = T_perpendicular × Distance = (T × sin(30°)) × 4.0 m.
  4. Apply the Principle of Moments: For the rod to be in equilibrium, the sum of clockwise moments must equal the sum of anticlockwise moments. Sum of clockwise moments = Sum of anticlockwise moments 240 Nm = (T × sin(30°)) × 4.0 m.
  5. Solve for Tension (T): We know sin(30°) = 0.5. 240 = (T × 0.5) × 4.0 240 = T × 2.0 T = 240 / 2.0 T = 120 N.
  6. Final Answer: The tension in the cable is 120 N.
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