A uniform metre rule is pivoted at its midpoint. a weight of 15 gf is suspended at one end of it should a where should
a weight of 100 gf b e suspended to keep the rule
horizontal
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let the metre rule be AB and it's Centre be O
let 15 gf is suspended at A OA=50cm
torque1 =15*50=750gfcm
let 100gf is suspended at x cm from O
torque 2=x*100
torque1=torque2
750=100*x
x=7.5cm
I.e 100gf should be placed at 57.5mark
let 15 gf is suspended at A OA=50cm
torque1 =15*50=750gfcm
let 100gf is suspended at x cm from O
torque 2=x*100
torque1=torque2
750=100*x
x=7.5cm
I.e 100gf should be placed at 57.5mark
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