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Question
The diagram in Fig. shows a uniform metre rule weighing 100 gf, pivoted at its centre O. Two weights 150 gf and 250 gf hang from the point A and B respectively of the metre rule such that OA = 40 cm and OB = 20 cm. Calculate :
- the total anticlockwise moment about O,
- the total clockwise moment about O,
- the difference of anticlockwise and clockwise moments, and
- the distance from O where a 100 gf weight should be placed to balance the metre rule.
Solution
i. Anticlockwise moments = 150 × 40 = 6000 gf
ii. Clockwise moments = 250 × 20 = 5000 gf
iii. Difference of anticlockwise moments and clockwise moments = 6000 − 5000 = 1000 gf
iv. 150 gf × 40 cm = 250 gf × 20 cm + 100 gf × d
6000 gf cm = 5000 gf cm + 100 gf × d
1000 = 100 × d
d = 10 cm on the right side of O
∴ 100 gf wt. should be placed 10 cm to right of O.
Distance from O = 60 − 50 = 10 cm right side of O.
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