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A non-uniform bar of weight W and length...

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 left end is.
.

A

`L(tantheta_1 + tantheta_2/tantheta_1)`

B

`L(tantheta_1/tantheta_1 + tantheta_2)`

C

`L(tantheta_2/tantheta_1 + tantheta_2)`

D

`L(tantheta_1 + tantheta_2/tantheta_2)`

Text Solution

Verified by Experts

The correct Answer is:
B
Let `T_(1)` and `T_(2)` be the tensions in two strings as shown in the figure.

For translation equilibrium along the horizontal direction, we get
`T_(1)sintheta_(1)=T_(2)sintheta_(2)`
For rotational equilibrium about G
`-T_(1)sintheta_(1)d+T_(2)costheta_(2)(L-d)=0`
`T_(1)costheta_(1)d=T_(2)costheta_(2)(L-d)`
Dividing equation (i) by (ii), we get
`(tantheta_(1))/(d) = (tantheta_(2))/(L-d)` or `(L-d)/(d) = (tantheta_(2))/(tantheta_(1))`
`(L/d-1) = (tantheta_(2))/(tantheta_(1)) rArr L/d= (tantheta_(2))/(tantheta_(1))+1`
`L/d= (tantheta_(2)+tantheta_(1))/(tantheta_(1)) rArr d=L(tantheta_(1))/(tantheta_(1)+tantheta_(2))`
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