Answer
$F_A$ must be 6300N and $F_B$ must be 6100N.
Work Step by Step
Sum the torques about point B:
$(+ \circlearrowleft) \sum \tau=0$
$-F_A*10m+4300N*8m+280kg*9.8m/s^2*5m+3100N*4m+2200N*1m=0$
$F_A=(4300N*8m+280kg*9.8m/s^2*5m+3100N*4m+2200N*1m)/(10m)$
$F_A\approx6300N$
Now sum the forces vertically:
$(\uparrow +) \sum \overrightarrow{F}_{y} =0 $
$F_A+F_B-4300N-2800N-280kg*9.8m/s^2-3100N-2200N=0$
$F_B=4300N+2800N+280kg*9.8m/s^2+3100N+2200N-F_A$
$F_B=4300N+2800N+280kg*9.8m/s^2+3100N+2200N-6300N$
$F_B\approx6100N$