Answer
$F_1 = (1160~N)~\hat{j}$
Work Step by Step
To find the force on the beam due to support 1, we can consider the torque about a rotation axis at the location of support 2:
$\sum \tau = 0$
$(1.96~m)~(250~kg)(9.8~m/s^2) - (3.92~m)~F_1 - (0.540~m)(46.0~kg)~(9.8~m/s^2) = 0$
$(3.92~m)~F_1 = (1.96~m)~(250~kg)(9.8~m/s^2) - (0.540~m)(46.0~kg)~(9.8~m/s^2)$
$F_1 = \frac{(1.96~m)~(250~kg)(9.8~m/s^2) - (0.540~m)(46.0~kg)~(9.8~m/s^2)}{3.92~m}$
$F_1 = 1160~N$
The force on the beam due to support 1 is $~~1160~N$ and this force is directed upward.
We can express the force on the beam due to support 1 in unit-vector notation:
$F_1 = (1160~N)~\hat{j}$
