Two identical rollers each of weight W = 500 N are supported by an inclined plane and a vertical wall as shown in diagram. Assuming all surfaces are smooth , A,B,C and D are contact points
  

 $F_A=250\sqrt{3}N,F_B=\frac{1250}{\sqrt{3}}Nand{F}_C=\frac{1000}{\sqrt{3}}N$
 
Let $F_D$  be the force of interaction (reaction) between the rollers. Consider the  equilibrium of upper roller: we have
 ${\sum }^{\text{}} F_x=0. or F_D-500\text{sin}{30}^{\circ }=0$
or  $F_D=500\text{sin}{30}^{\circ }=250 \text{N}$
and  ${\sum }^{\text{}} F_y=0. or F_A-500\text{cos}{30}^{\circ }=0$
 
or   $F_A=500\text{cos}{30}^{\circ }=\frac{500\sqrt{3}}{2}=250\sqrt{3} \text{N}$
Now consider the equilibrium of lower roller ;
 ${\sum }^{\text{}} F_x=0, or F_C\text{cos}{30}^{\circ }-F_D-500\text{sin}{30}^{\circ }=0$
or  $F_C\times \frac{\sqrt{3}}{2}-250-500\times \frac{1}{2}=0$
or  $F_C=\frac{1000}{\sqrt{3}} \text{N}$
and  ${\sum }^{\text{}} F_y=0$
or  $F_B-F_C\text{sin}{30}^{\circ }-500\text{cos}{30}^{\circ }=0$
or  $F_B-\frac{1000}{\sqrt{3}}\times \frac{1}{2}-500\times \frac{\sqrt{3}}{2}=0$
or  $F_B=\frac{1250}{\sqrt{2}} \text{N}$