Answer
$F = (-671~\hat{j})~N$
Work Step by Step
To find the gravitational force on the beam, we can consider the torque about the rotation axis at the hinge:
$\sum \tau = 0$
$(L~cos~\theta)(400~N)-(\frac{L}{2}~sin~\theta)~(mg) = 0$
$(\frac{L}{2}~sin~\theta)~(mg) = (L~cos~\theta)(400~N)$
$mg = (2)(cot~\theta)(400~N)$
$mg = (2)(cot~50.0^{\circ})(400~N)$
$mg = 671~N$
The gravitational force on the beam is $~~671~N~~$ and this is directed downward.
We can express the gravitational force on the beam in unit-vector notation:
$F = (-671~\hat{j})~N$
