Answer
$F = (30~\hat{i}+97~\hat{j})~N$
Work Step by Step
To find the tension $F_T$ in the cable, we can consider the net torque about a rotation axis at the hinge:
$\sum \tau = 0$
$h~F_T~cos~\theta-L~F = 0$
$h~F_T~cos~\theta = L~F$
$F_T = \frac{L~F}{h~cos~\theta}$
$F_T = \frac{(3.2~m)(50~N)}{(2.0~m)~cos~25^{\circ}}$
$F_T = 88.27~N$
We can find the horizontal component of the force on the beam from the hinge:
$F_x+F-F_T~cos~25^{\circ} = 0$
$F_x = F_T~cos~25^{\circ}-F$
$F_x = (88.27~N)~cos~25^{\circ}-(50~N)$
$F_x = 30~N$
We can find the vertical component of the force on the beam from the hinge:
$F_y-mg-F_T~sin~25^{\circ} = 0$
$F_y = mg + F_T~sin~25^{\circ}$
$F_y = (60~N) + (88.27~N)~sin~25^{\circ}$
$F_y = 97~N$
We can express the force on the beam from the hinge in unit-vector notation:
$F = (30~\hat{i}+97~\hat{j})~N$
