Answer
The tension in the wire is $~~192~N$
Work Step by Step
To find the tension $T$ in the wire, we can consider the net torque about a rotation axis at the hinge.
There is a clockwise torque from the weight of the beam, and an opposing counterclockwise torque from the tension $T$ in the wire.
Let $L$ be the length of the beam.
We can find $T$:
$\tau_{net} = 0$
$(L)(T)~sin~150^{\circ}-(\frac{L}{2})(mg)~sin~60^{\circ} = 0$
$(L)(T)~sin~150^{\circ} = (\frac{L}{2})(mg)~sin~60^{\circ}$
$T~sin~150^{\circ} = (\frac{1}{2})(mg)~sin~60^{\circ}$
$T = \frac{(\frac{1}{2})(mg)~sin~60^{\circ}}{sin~150^{\circ}}$
$T = \frac{(\frac{1}{2})(222~N)~sin~60^{\circ}}{sin~150^{\circ}}$
$T = 192~N$
The tension in the wire is $~~192~N$
