A uniform beam XY is 100 cm long and weighs 4.0N. The beam rests on a pivot 60 cm from end X. A load of 8.0 N hangs from the beam 10 cm from end X. The beam is kept balanced by a force F acting on the beam 80 cm from end X. What is the magnitude of force F ? A 8.0N B 18N C 22N D 44N. Show full working.

Explanation:

The magnitude of the force in the set up can be solved for using the principle of moment. According to the principle, the sum of clockwise moment  is equal to the sum of anticlockwise moments.

Moment = Force * perpendicular distance

Clockwise moments;

The force that acts clockwise is the unknown Force F and 4N force. If the  beam rests on a pivot 60 cm from end X and a Force F acts on the beam 80 cm from end X, the perpendicular distance of the force F from the pivot is 80-60 = 20cm and the perpendicular distance of the 4N force from the pivot is 60-50 = 10cm

Moment of force F about the pivot = F * 20

Moment of 4N force about the pivot = 4*10 = 40Nm

Sum of clockwise moment = 40+20F...(1)

Anticlockwise moment;

The  8N will act anticlockwisely about the pivot.

The distance between the 8N force and the pivot is 60-10 = 50cm

Moment of the 8N force = 8*50

=400Nm...(1)

Equating 1 and 2 we have;

40+20F = 400

20F = 400-40

20F = 360

F = 18N

The magnitude of force F is 18N

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