Moments (A2 Unit 4: Applied Mathematics B) — WJEC A-Level Mathematics
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Moments (A2 Unit 4: Applied Mathematics B) explained
For a rigid body (such as a uniform beam, rod, or ladder) to remain in static equilibrium, two conditions must be satisfied simultaneously: 1.
Read the full explanation
Translational equilibrium: the vector sum of all external forces is zero (ΣF_x = 0 and ΣF_y = 0). 2. Rotational equilibrium: the resultant turning effect about any chosen pivot is zero (Principle of Moments: ΣM_clockwise = ΣM_anticlockwise). The moment of a force is Force × perpendicular distance. For a uniform rod of mass m and length L, weight mg acts at the geometric midpoint L/2; for a non-uniform rod, weight acts at the centre of mass. Taking moments about an unknown support eliminates its reaction.
Your focus
- Apply the Principle of Moments (ΣM_clockwise = ΣM_anticlockwise) to rigid bodies in static equilibrium.
- Combine moment equations with vertical force equilibrium to determine multiple support reactions.
- Solve tilting problems by identifying support lift-off conditions where reaction forces become zero.
Moments (A2 Unit 4: Applied Mathematics B) exam tips
Marking Points
- stating or applying the Principle of Moments: total clockwise moments = total anticlockwise moments
- calculating moments using force multiplied by perpendicular distance to the chosen pivot
- placing the weight of a uniform rod at its geometric midpoint L/2
- formulating force equilibrium (e.g. R_A + R_B = total weight) and solving for unknown reactions
Examiner Tips
- 💡Take moments about a pivot with an unknown reaction force to eliminate that force from your equation.
- 💡If a beam is 'on the point of tilting' about support B, the reaction at the other support A drops to zero (R_A = 0).
Common Mistakes
- taking moments about a point without using the perpendicular distance from the pivot to the line of action
- placing the weight at the midpoint for a non-uniform rod where the centre of mass is unknown