A non-uniform bar of weight 'w' is suspended at rest by two strings of negligible weight. The angles made by the strings with the vertical are `36.9^(@)` and `53.1^(@)C` respectively. The bar is 2m long. Calculate the distance 'd' of the centre of gravity of the bar from its left. (refer fig).-NCERT

Explanation

From the figure,

`T_1cos36.9^(@)+T_2cos53.1^(@)=Mg`
`0.7997T_1+0.6004T_2=Mg`
Also `T_1sin36.9^(@)=T_2sin53.1^(@)`
i.e `0.6004T_1=0.7997T_2`
i.e `T_2=0.7508T_1`
Form the law of moments,
Taking moments about A, `(T_2cos53.1^(@))(L)`=Mgd
`2xx0.6004T_2=Mgd`
`1.2008(0.7508)T_1=mgd,mgd=0.9016T_1`
Using (2) in (1) we get
`0.7997T_1+0.6004(0.7500)_T_1=Mg`
`1.2505T_1=Mg`
Hence using (4) in (3) we write
`(1.2505T1)d_F=0.9016T_1`
`:. d_F=(0.9016)/(1.2505)=0.72`
0.72m from the left end and 1.28 m from the right hand.