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Question

The diagram shows a uniform meter rule weighing 100 gf, pivoted at its centre O. Two weights 150gf and 250gf hang from the point A and B respectively of the metre rule such that OA = 40 cm and OB = 20 cm.
Calculate:
(i) the total anticlockwise moment about O,
(ii) the total clockwise moment about O,
(iii) the difference of anticlockwise and clockwise moment, and
(iv) the distance from O where a 100gf weight should be placed to balance the metre rule.

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Answer

(i) The total anticlockwise moment about the centre o

= 150 × 40

= 6000 gf cm

(ii) The total clockwise moment about the centre o

= 250 × 20

= 5000  gf cm

(iii) The difference of anticlockwise and clockwise moment

= 6000 – 5000

= 1000 gf cm

(iv) As we know, the principle of moments states that

Anticlockwise moment = Clockwise moment

150 × 40 = 250 × 20 + 100 × d

⇒ 100d = 1000

⇒ d = 1000\over 100 = 10 cm

So d = 10 cm on the right side of o

 

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