Answer
The horizontal component of the force on the strut from the hinge is $~~5740~N$
Work Step by Step
To find $F_x$, the horizontal component of the force on the strut from the hinge, we can consider the horizontal forces on the strut.
In part (a), we found that $T = 6630~N$
We can find $F_x$:
$F_{net,x} = 0$
$T~cos~\phi-F_x = 0$
$F_x = T~cos~\phi$
$F_x = (6630~N)~cos~30.0^{\circ}$
$F_x = 5740~N$
The horizontal component of the force on the strut from the hinge is $~~5740~N$
