A non-uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in Fig.7.39. The angles made by the strings with the vertical are `36.9^(@)` and `53.1°` respectively. The bar is 2 m long. Calculate the distance d of the centre of gravity of the bar from its left end.

Explanation

By Lami.s theorem
`(T_(1))/(sin127^(@))=(T_(2))/(sin143^(@))=(W)/(sin90^(@))`
Thus `T_(1)=(4W)/(5)` and `T_(2)=(3W)/(5)`
Taking torques about CG
`T_(1)cos37xxd=T_(2)cos53xx(2-d)`
On solving `d=(18)/(25)m`