Question #8b2b6

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Answer 1 · 2016-12-17T15:17:49

drawn

$AB-> "Meter scale"$

$D-> "Position of support where AD"=20 cm$

$C-> "Mid point of the scale i.e. AC"=50cm=>CD=30cm$

$P->"Position of center of mass of the box "$

As the balanced position is achieved, the moments of two forces

$500gwt" at P" and 200gwt " at C"$ about D will be equal

Hence $PDxx500=CDxx200=30xx200$

$PD=6000/500=12cm$

So the position $P$ of center of mass of the box is such that

$AP = AD-PD=20-12=8 cm$