An iron sphere weighing 10, N rests in a V sh | vedclass

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Question

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

Vedclass · Accepted Answer

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Vedclass · Answer

since the sphere is not moving $\sum F_{H}=0$

$R_{B} \sin 60=0$

$	herefore R_{B}=0$

$\& R_{A}=W=10 N$

$\sum F_{H}=0$

$	herefore R_{A} \sin 60=R_{B} \sin 60 \Rightarrow R_{A}=R_{B}=R$

Now $\sum F_{U}=0 	herefore 2 R \cos 60-W=0$

$R=W=10 N$

$\sum F v=0 	herefore R_{A} \sin 60=W$

$\Rightarrow R_{A}=\frac{20}{\sqrt{3}} N$ also $\sum F_{H}=0$

$R_{\frac{A}{2}}-R_{B}=0$ So $R_{B}=\frac{10}{\sqrt{3}} N$

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

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