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ICSE Class 10 Physics Question 14 of 20

Force — Question 16

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Question 16

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.

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. Force, Concise Physics Class 10 Solutions.
Answer

(i) The total anticlockwise moment about the centre o

=150gf×40cm=6000gfcm= 150gf \times 40cm \\ = 6000gf cm

(ii) The total clockwise moment about the centre o

=250gf×20cm=5000gfcm= 250gf \times 20cm \\[0.5em] = 5000gf cm \\[0.5em]

(iii) The difference of anticlockwise and clockwise moment

=6000−5000=1000gfcm= 6000 - 5000 \\[0.5em] = 1000gf cm \\[0.5em]

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

Anticlockwise moment = Clockwise moment

150gf×40cm=250gf×20cm+100gf×d6000gf=5000gf+100gf×dd=1000100d=10cm150gf \times 40cm = 250gf \times 20cm + 100gf \times d \\[0.5em] 6000gf = 5000gf +100gf \times d \\[0.5em] d = \dfrac {1000}{100} \\[0.5em] d = 10cm

So d = 10 cm on the right side of o