# Question #5cb57

Jan 18, 2017

The weight of 100cm uniform bar will at the center of gravity i.e. at the middle point $G$ of the bar.

A support is at the point X on the bar and this point is at a distance of 23 cm from one end A from which an unknown weight $W k g$ is hanged to balance the system.

Here $X A = 23 c m \mathmr{and} X G = \left(50 - 23\right) = 27 c m$

Considering the moments of the weight $W k g$ and weight of the bar about X we can write

$\text{weight of bar} \times X G = W \times X A$

$\implies 17.2 \times 27 = W \times \times 23$

$\implies W = 20.2 c m$