A non-uniform bar of weight `W` and weight `L` is suspended by two strings of neigligible weight as shown in figure. The angles made by the strings with the vertical are `theta_1` and `theta_2` respectively.
The distance `d` of the centre of gravity of the bar from left end is.
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Options

A. `L((tan theta_1 + tan theta_2)/(tan theta_1))`
B. `L((tan theta_1)/(tan theta_1 + tan theta_2))`
C. `L((tan theta_2)/(tan theta_1 + tan theta_2))`
D. `L((tan theta_1+ tan theta_2)/( tan theta_2))`

Answer

B

Explanation

(b) Let `T_1` and `T_2` be the tensions in two strings as shown in the figure.
For translational equilibrium along the horizontal direction, we get
`T_1 sin theta_1 = T_2 sin theta_2` …(i)
For ratational equilibrium about `G`
`-T_1 cos theta_1 d + T_2 cos theta_2 (L - d) = 0`
`T_1 cos theta_1 d = T_2 cos theta_2 (L -d)` ...(ii)
Dividing equation (i) by (ii), we get
`(tan theta_1)/(d) =(tan theta_2)/((L - d))` or `(L - d)/(d) = (tan theta_2)/(tan theta_1)`
`((L)/(d) -1) = (tan theta_2)/(tan theta_1) rArr (L)/(d) = ((tan theta_2)/(tan theta_1) +1)`
`(L)/(d) =(tan theta_2 + tan theta_1)/(tan theta_1)`
`d = L((tan theta_1)/(tan theta_1 + tan theta_2))`.
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