Solve this please.

Solution,
Weight of the metre scale, W = 100 gf
Centre of mass of the metre scale is at the point O.
(i) Total anticlockwise moment about O:
$$\begin{gathered} =150gf\times OA \\ =150gf\times 0.40m \\ =150\times 0.0098\times 0.40 \\ =0.588 Nm \end{gathered}$$

(ii) Total clockwise moment about O:
$$\begin{gathered} =250gf\times OB \\ =250gf\times 0.20m \\ =250\times 0.0098\times 0.20 \\ =0.490 Nm \end{gathered}$$

(iii) The difference of anticlockwise and clockwise moments:
$$\begin{gathered} =0.588-0.490 \\ =0.098 Nm \end{gathered}$$

In order to balance the metre rule, the 100gf weight must be placed at point O' such that
$$\begin{gathered} 150gf\times O\text{'}A=250\times O\text{'}B \\ and \\ O\text{'}A+O\text{'}B=60cm \\ Now, \\ 150gf\times O\text{'}A=250\times O\text{'}B \\ O\text{'}A=\frac{250\times O\text{'}B}{150}=\frac{5}{3}O\text{'}B \\ Thus, \\ from the relation, \\ O\text{'}A+O\text{'}B=60cm \\ \frac{5}{3}O\text{'}B+O\text{'}B=60cm \\ \frac{8}{3}O\text{'}B=60cm \\ O\text{'}B=\frac{3}{8}\times 60cm=22.5cm \\ O\text{'}A=60cm-22.5cm=37.5cm \\ So, the weight must be displaced 2.5cm towards the load 150gf from the point O. \end{gathered}$$