the diagram in fig 1.34 shows a uniform meter rule weighing 100gf,pivoted at its centre o.two weights 150gf and 250gf hang from the points a and brespectively of the metre rule such that oa=40cm and ob=20cm.calculate the distance from o where a 100gf weight should be placed to balance the meter rule.
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Explanation:
Total anticlockwise moment about 0
=150gf×40cm=6000gfcm
(II) Total clockwise moment about 0
=250gf×20cm=5000gfcm
(III) The difference of anti-clockwise and clockwise moment
=6000−5000=1000gfcm
(IV) From the principle of moments,
anti-clockwise moment=clockwise moment
To balance it, 100gf weight should be kept on right hand side so as to produce a clockwise moment about the 0.
Let its distance from the point 0 be dcm
Then, 150gf×40cm=250gf×20cm+100gf×d
6000gfcm=5000gfcm+100gf×d
⇒1000gfcm=100gf×d
∴d=
100gf
1000gfcm
=10cm on the right side of 0
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