Question

A uniform rod of length $l$ and mass $m$ is hung from two strings of equal length from a ceiling as shown in figure. Determine the tensions in the strings?

• A. $T_A=\frac{2mg}{3};T_B=\frac{mg}{3}$
• B. $T_A=\frac{mg}{3};T_B=\frac{2mg}{3}$
• C. $T_A=T_B=\frac{mg}{2}$
• D. $T_A=\frac{mg}{3},T_B=\frac{mg}{4}$

Answer 1

The correct option is B $T_A=\frac{mg}{3};T_B=\frac{2mg}{3}$

Let us assume that tension in left and right string is $T_A$ and $T_B$ respectively.

The rod is in equilibrium, then ${\overrightarrow{F}}_{net}=0and{\overrightarrow{\tau }}_{net}=0$

From, ${\overrightarrow{F}}_{net}=0$
$mg=T_A+T_B$ ...$\left(1\right)$


From ${\tau }_{net}=0$ about $A$,
$mg\frac{l}{2}-\frac{3l}{4}{T}_B=0$
$\Rightarrow T_B=\frac{2mg}{3}$

So, from eq. $\left(1\right)$,
$T_A+\frac{2mg}{3}=mg$
$\Rightarrow T_A=\frac{mg}{3}$