Question

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

Answer 1

Free body diagram of the rod is shown in the figure.
Translational equilibrium requires
$\sum F_y=0\Rightarrow T_1+T_2=400+200=600N....\left(i\right)$
Rotational equilibrium: Applying the condition about $A$, we get $T_2$.
$\sum \overrightarrow{{\tau }_A}=\overrightarrow{0}\Rightarrow -400\left(l/4\right)-200\left(l/2\right)+T_{2}l=0$
$T_2=200N$
From equation (i)
$T_1=400N$.