An iron sphere weighing 10, N rests in a V shaped smooth trough whose sides form an angle of 60^o as

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4-1.Newton's Laws of Motion

normal

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

$R_A = 10 N$ & $R_B = 0$ in case $(i)$

$R_A = 10 \,N$ & $R_B = 10\, N$ in case $(ii)$

$R_A = \frac{{20}}{{\sqrt 3 }} N$ & $R_B = \frac{{10}}{{\sqrt 3 }} N$ in case $(iii)$

All of the above

since the sphere is not moving $\sum F_{H}=0$

$R_{B} \sin 60=0$

$\therefore R_{B}=0$

$\& R_{A}=W=10 N$

$\sum F_{H}=0$

$\therefore R_{A} \sin 60=R_{B} \sin 60 \Rightarrow R_{A}=R_{B}=R$

Now $\sum F_{U}=0 \therefore 2 R \cos 60-W=0$

$R=W=10 N$

$\sum F v=0 \therefore R_{A} \sin 60=W$

$\Rightarrow R_{A}=\frac{20}{\sqrt{3}} N$ also $\sum F_{H}=0$

$R_{\frac{A}{2}}-R_{B}=0$ So $R_{B}=\frac{10}{\sqrt{3}} N$

Standard 11

Physics

medium

easy

hard

easy

(AIIMS-2005)

medium