# Question #8b2b6

Dec 17, 2016

$A B \to \text{Meter scale}$

$D \to \text{Position of support where AD} = 20 c m$

$C \to \text{Mid point of the scale i.e. AC} = 50 c m \implies C D = 30 c m$

$P \to \text{Position of center of mass of the box }$

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

$500 g w t \text{ at P" and 200gwt " at C}$ about D will be equal

Hence $P D \times 500 = C D \times 200 = 30 \times 200$

$P D = \frac{6000}{500} = 12 c m$

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

$A P = A D - P D = 20 - 12 = 8 c m$