Answer
The reaction at the right support is 5760N. The reaction at the left support is 8060N.
Work Step by Step
1) Sum the torques about the left end of the table
$(+ \circlearrowleft) \sum \tau=0$
$-.5Mg*.25l-Mg*.5l+R_R*l=0$
$-.5Mg*.25-Mg*.5+R_R*=0$
$R_R=.5Mg*.25+Mg*.5$
$R_R=.5*940kg*9.8m/s^2*.25+940kg*9.8m/s^2*.5$
$R_R\approx5760N$
The reaction at the right support is 5760N.
2) Sum the forces vertically
$(\uparrow +) \sum \overrightarrow{F}_{y} =0 $
$R_L+R_R-.5Mg-Mg=0$
$R_L=-R_R+.5Mg+Mg$
$R_L=-5760N+.5*940kg*9.8m/s^2+940kg*9.8m/s^2$
$R_L\approx8060N$
The reaction at the left support is 8060N.
