A non-uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in figure. 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

Suppose bar AB is suspended at rest by two strings OA and O.B of negligible weight. The angles made by the strings with the vertical are `36.9^(@)and53.1^(@)` respectively.
`angleOA A.=53.1^(@)andangleO.BB=36.9^(@)`
`AB=2m` and suppose distance of centre of mass from end A is d.

`T_(1)andT_(2)` are the tensions produced in the left and right strings respectively and their mutual components are shown as in figure.
At translational equilibrium,
`-T_(1)cos53.1^(@)=-T_(2)cos36.9^(@)`
`:.T_(1)cos53.1^(@)=T_(2)cos36.9^(@)and....(1)`
`T_(1)sin53.1^(@)+T_(2)cos36.9^(@)=W....(2)`
Sum of torque at point A
`:.-T_(2)sin36.9^(@)xxAB+W.d=0`
`:.-T_(2)sin36.9^(@)xx2+W.d=0`
`:.T_(2)=(Wd)/(2sin36.9^(@))....(3)`
From eqn. (2) and (3)
`T_(1)sin53.1^(@)=W-T_(2)sin36.9^(@)`
`=W-(Wdsin36.9^(@))/(2sin36.9^(@))`
`=W-(Wd)/(2)`
`:.T_(1)=(W(1-(d)/(2)))/(sin53.1^(@))....(4)`
From eqn. (1)
`T_(1)cos53.1^(@)=T_(2)cos36.9^(@)`
Putting the value of eqn. (2) and (3)
`(W(1-(d)/(2))cos53.1^(@))/(sin53.1^(@))=(Wdcos36.9^(@))/(2sin36.9^(@))`
`:.(1-(d)/(2))/(tan53.1^(@))=(d)/(2tan36.9^(@))`
`:.(1-(d)/(2))/(1.3319)=(d)/(2xx0.7508)`
`:.1-(d)/(2)=(d xx1.3319)/(1.5016)`
`:.1=0.5d+0.887d`
`:.1=1.387d`
`:.d=(1)/(1.387)=0.72098`
`:.d~~0.721m`
`:.d~~72.1cm`