An iron sphere weighing $10 \mathrm{N}$ rests in a $\mathrm{V}$ shaped smooth trough whose sides form an angle of $60°$ as shown in the figure. Then the reaction forces are :

Since the sphere is not moving $\sum F_H=0$
$$\begin{gathered} R_B\sin 60=0 \\ \therefore R_B=0 \\ \& R_A=W=10N \end{gathered}$$

Fig. 2

$\sum F_H=0$

$$\begin{gathered} \therefore R_A\sin 60=R_B\sin 60\Rightarrow R_A=R_B=R \\ \text{ Now }\sum F_U=0 \therefore 2R\cos 60-W=0 \\ R=W=10N \end{gathered}$$

figure -3

$$\begin{gathered} \sum F_V=0 \\ \Rightarrow R_A=\frac{20}{\sqrt{3}}N\text{ also }\sum F_H=0; \\ {R}_{A/2}-R_B=0\text{ So }{R}_B=\frac{10}{\sqrt{3}}N \end{gathered}$$