#### Answer

The reaction force near the piano is 2890N. The reaction away from the piano is 1320N

#### Work Step by Step

We will sum the torques about the support near the piano and then sum the force vertically to find the reactions in the supports. Let $F_C$ be the reaction force near the piano and $F_A$ be the other reaction.
$(+ \circlearrowleft) \sum \tau_c =0$
$-320kg*9.8m/s^2*1m-110kg*9.8m/s^2*2m+F_R*4m=0$
$F_R=\frac{320kg*9.8m/s^2*1m+110kg*9.8m/s^2*2m}{4m}$
$F_R\approx1320N$
$(\uparrow +) \sum \overrightarrow{F}_{y} =0 $
$F_C+F_R-320kg*9.8m/s^2-110kg*9.8m/s^2=0$
$F_C=-F_R+320kg*9.8m/s^2+110kg*9.8m/s^2=0$
$F_C\approx2890N$