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.

Question 15

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

Reference: Past Exam Paper – November 2015 Paper 13 Q15

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