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

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$R_{A} = \frac{20}{\sqrt{3}} N$ and $R_{B} = \frac{10}{\sqrt{3}} N$ in case (iii)

$R_{A} = 10 N$ and $R_{3} = 10 N$ in case (ii)

$R_{A} = 10 N$ and $R_{B} = 0$ in case (i)

$R_{A} = 10 N$ and $R_{B} = 10 N$ in all the $3$ cases

answer is A, B, C.

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Detailed Solution

Since the sphere is not moving $\sum F_{H} = 0$
$R_{B} \sin 60 = 0 ∴ R_{B} = 0 & R_{A} = W = 10 N$

Fig. 2

$\sum F_{H} = 0$

$∴ R_{A} \sin 60 = R_{B} \sin 60 \Rightarrow R_{A} = R_{B} = R Now \sum F_{U} = 0 ∴ 2 R \cos 60 - W = 0 R = W = 10 N$

figure -3

$\sum F_{V} = 0 \Rightarrow R_{A} = \frac{20}{\sqrt{3}} N also \sum F_{H} = 0; R_{A / 2} - R_{B} = 0 So R_{B} = \frac{10}{\sqrt{3}} N$

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