A uniform rod of $20 kg$ is hanging in a horizontal position with the help of two threads. It also supports a $40 kg$ mass as shown in the figure. Find the tensions developed in each thread.

Free body diagram of the rod is shown in the figure.

Translational equilibrium requires

$\sum _{ }{F}_y=0\Rightarrow T_1+T_2=400+200=600 N$

Rotational equilibrium: Applying the condition about $A$, we get $T_2$.

$$\begin{gathered} \sum _{ }\overrightarrow{{\tau }_A}=\overrightarrow{0}\Rightarrow -400\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.\right)-200\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)+T_{2}1=0 \\ {T}_2=200 N \end{gathered}$$

From above equation, $T_1=400 N$.