Q#15 (Past Exam Paper – November 2015 Paper 13 Q15)
A cross-shaped structure, freely pivoted at O, has arms of lengths 5.0 m, 4.0 m, 3.0 m and 2.0 m. It is acted on by forces of 2.0 N, 3.0 N, 4.0 N and an unknown force F. The structure is in rotational equilibrium.
What is the magnitude of force F ?
A 0.40 N B 2.0 N C 2.6 N D 4.4 N
Solution:
Answer: B.
For rotational equilibrium, the sum of clockwise moment is equal to the sum of anticlockwise moment.
Moment = Force × perpendicular distance from line of action of force to pivot
The line of action of force 4.0 N acts along the pivot. It is NOT perpendicular, but parallel. So, the force 4.0 N does not cause any moment.
The clockwise moment is caused by the force F only.
Clockwise moment = F × 5.0
The 3.0 N force and the 2.0 N cause an anticlockwise moment. However, we need to use the perpendicular component of the 2.0 N force.
Anticlockwise moment = (3.0 × 2.0) + (2.0 sin30 × 4.0)
Clockwise moment = Anticlockwise moment
F × 5.0 = (3.0 × 2.0) + (2.0 sin30 × 4.0)
F = [(3.0 × 2.0) + (2.0 sin30 × 4.0)] / 5.0
F = 2.0 N
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