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 $
$F_C+F_R-320kg*9.8m/s^2-110kg*9.8m/s^2=0$
$F_C=-F_R+320kg*9.8m/s^2+110kg*9.8m/s^2=0$
$F_C\approx2890N$