Q.
A body of mass $m$ is suspended by two strings making angle $\alpha$ and $\beta$ with the horizontal as shown in figure. Tensions in the two strings are
A) $T_{1} = \frac{mg\cos\beta}{\sin{(\alpha + \beta)}} = T_{2}$
B) $T_{1} = \frac{mg\sin\beta}{\sin{(\alpha + \beta)}}{= T}_{2}$
C) $T_{1} = \frac{mg\cos\beta}{\sin{(\alpha + \beta)}}{;T}_{2} = \frac{mg\cos\alpha}{\sin{(\alpha + \beta)}}$
D) None of the above
Explanation
(c) Applying Lami’s theorem
$\frac{T_{1}}{\sin{(90{^\circ} + \beta)}} = \frac{T_{2}}{\sin{(90{^\circ} + \alpha)}} = \frac{mg}{\sin{\lbrack 180{^\circ} - (\alpha + \beta)\rbrack}}$
or $\frac{T_{1}}{\cos\beta} = \frac{T_{2}}{\cos\alpha} = \frac{mg}{\sin{(\alpha + \beta)}}$
$\therefore\ T_{1} = \frac{mg\cos\beta}{\sin{(\alpha + \beta)}};T_{2} = \frac{mg\cos\alpha}{\sin{(\alpha + \beta)}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.