Answer
$T = 6630~N$
Work Step by Step
We can consider the net torque on the strut about a rotation axis at the hinge.
The tension holding the block, the weight of the strut, and the vertical component of $T$ contribute to the clockwise torque. There is an opposing counterclockwise torque from the horizontal component of $T$.
Let $d$ be the length of the strut.
We can find $T$:
$\tau_{net} = 0$
$(d~sin~\theta)(T~cos~\phi)-(d~cos~\theta)(T~sin~\phi)-(\frac{d}{2}~cos~\theta)(45.0~kg)(9.8~m/s^2)-(d~cos~\theta)(225~kg)(9.8~m/s^2) = 0$
$(sin~45^{\circ})(T~cos~\phi)-(cos~45^{\circ})(T~sin~\phi) = (\frac{1}{2}~cos~45^{\circ})(45.0~kg)(9.8~m/s^2)+(cos~45^{\circ})(225~kg)(9.8~m/s^2)$
$T~(cos~\phi-sin~\phi) = (\frac{1}{2})(45.0~kg)(9.8~m/s^2)+(225~kg)(9.8~m/s^2)$
$T = \frac{(\frac{1}{2})(45.0~kg)(9.8~m/s^2)+(225~kg)(9.8~m/s^2)}{cos~\phi-sin~\phi}$
$T = \frac{(\frac{1}{2})(45.0~kg)(9.8~m/s^2)+(225~kg)(9.8~m/s^2)}{cos~30.0^{\circ}-sin~30.0^{\circ}}$
$T = 6630~N$
