Question

A non-uniform bar of weight W and length L is suspended by two strings of negligible 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 its left end is

• A. $L\left( \frac{\tan\theta_{1} + \tan\theta_{2}}{\tan\theta_{1}} \right)$
• B. $L\left( \frac{\tan\theta_{1}}{\tan\theta_{1} + \tan\theta_{2}} \right)$
• C. $L\left( \frac{\tan\theta_{2}}{\tan\theta_{1} + \tan\theta_{2}} \right)$
• D. $L\left( \frac{\tan\theta_{1} + \tan\theta_{2}}{\tan\theta_{2}} \right)$

Answer 1

Answer: (B)

$L\left( \frac{\tan\theta_{1}}{\tan\theta_{1} + \tan\theta_{2}} \right)$

Answer 2

Let $T_{1}$and $T_{2}$be the tensions in two strings as shown in the figures.

(

For translations equilibrium along the horizontal directions, we get

$T_{1}\sin\theta_{1} = T_{2}\sin\theta_{2}$ ….. (i)

For rotational equilibrium about G

$- T_{1}\cos\theta_{1}d + T_{2}\cos\theta_{2}(L - d) = 0$

$T_{1}\cos\theta_{1}d = T_{2}\cos\theta_{2}(L - d)$

Dividing equations (i) by (ii), we get

$\frac{\tan\theta_{1}}{d} = \frac{\tan\theta_{2}}{(L - d)}or\frac{L - d}{d} = \frac{\tan\theta_{2}}{\tan\theta_{1}}$

$\left( \frac{L}{d} - 1 \right) = \frac{\tan\theta_{2}}{\tan\theta_{1}} \Rightarrow \frac{L}{d} = \left( \frac{\tan\theta_{2}}{\tan\theta_{1}} + 1 \right)$

$\frac{L}{d} = \frac{\tan\theta_{2} + \tan\theta_{1}}{\tan\theta_{1}}$

$d = L\left( \frac{\tan\theta_{1}}{\tan\theta_{1} + \tan\theta_{2}} \right)$