Answer
The force on the beam from roller A is $~~196~N$
Work Step by Step
Let $L$ be the length of the beam.
To find the force on the beam from roller A, we can consider the torque about a rotation axis at roller B:
$\sum~\tau = 0$
$(\frac{L}{4})(mg) - (\frac{L}{2})~F_A = 0$
$(\frac{L}{2})~F_A = (\frac{L}{4})(mg)$
$F_A = \frac{mg}{2}$
$F_A = \frac{(40.0~kg)(9.8~m/s^2)}{2}$
$F_A = 196~N$
The force on the beam from roller A is $~~196~N$
