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Question #5cb57

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Jan 18, 2017

The weight of 100cm uniform bar will at the center of gravity i.e. at the middle point $G$ $G$ 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 $W k g$ $Wkg$ is hanged to balance the system.

Here $X A = 23 c m and X G = (50 - 23) = 27 c m$ $XA=23cm and XG=(50-23)=27cm$

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

$weight of bar \times X G = W \times X A$ $"weight of bar"xxXG=WxxXA$

$\Rightarrow 17.2 \times 27 = W \times \times 23$ $=>17.2xx27=Wxxxx23$

$\Rightarrow W = 20.2 c m$ $=>W=20.2cm$

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