Answer
The force at support B is 11,000N. The reaction at the wheel A is 13,000N.
Work Step by Step
1) Sum the torques about point A
$(+ \circlearrowleft) \sum \tau_A=0$
$-2.5m*Mg+B*5.5m=0$
$B=\frac{2.5m*Mg}{5.5m}$
$B=\frac{2.5m*2500kg*9.8m/s^2}{5.5m}$
$B\approx11000N$
The force at support B is 11,000N
2) Sum the forces vertically
$(\uparrow +) \sum \overrightarrow{F}_{y} =0 $
$A+B-Mg=0$
$A=-B+Mg$
$A\approx-11000N+2500kg*9.8m/s^2$
$A\approx13000N$
The reaction at the wheel A is 13,000N.
