Answer
$F_A$ is 2940N and $F_B$ is 14,700N
Work Step by Step
We will sum the torques about point B and then sum the forces vertically to solve for the unknown forces;
$(+ \circlearrowleft) \sum \tau_B =0$
$F_A*20m-m*g*5m=0$
$F_A=\frac{m*g*5m}{20m}$
$F_A=\frac{1200kg*9.8m/s^2*5m}{20m}$
$F_A\approx2940N$
$(\uparrow +) \sum \overrightarrow{F}_{y} =0 $
$F_B+F_A-m*g=0$
$F_B=F_A+m*g$
$F_B=2940N+1200kg*9.8m/s^2$
$F_B\approx14,700N$
