In the diagram,BAC is a rigid fixed rough wir | vedclass

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(A) The angle $\angle BAC = 60^o$. Let $	heta$ be the angle the wire makes with the vertical. Since the system is symmetric,$	heta = 30^o$.The exten

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) The angle $\angle BAC = 60^o$. Let $	heta$ be the angle the wire makes with the vertical. Since the system is symmetric,$	heta = 30^o$.The extension in the string is $x = PQ - 2a$. In $	riangle APQ$,$PQ = 2(3a) \sin(30^o) = 3a$. So,$x = 3a - 2a = a$.The spring force is $F_s = kx = (\frac{mg}{a}) \cdot a = mg$.For equilibrium of ring $P$ along the wire: $mg \cos(30^o) - f - F_s \sin(30^o) = 0$.$f = mg \frac{\sqrt{3}}{2} - mg \cdot \frac{1}{2} = mg \frac{\sqrt{3}-1}{2}$.For equilibrium perpendicular to the wire: $N - mg \sin(30^o) - F_s \cos(30^o) = 0$.$N = mg \cdot \frac{1}{2} + mg \cdot \frac{\sqrt{3}}{2} = mg \frac{1+\sqrt{3}}{2}$.For limiting friction,$f = \mu N$,so $\mu = \frac{f}{N} = \frac{\sqrt{3}-1}{\sqrt{3}+1} = \frac{(\sqrt{3}-1)^2}{3-1} = \frac{4-2\sqrt{3}}{2} = 2 - \sqrt{3}$.Thus,$\mu + \sqrt{3} = (2 - \sqrt{3}) + \sqrt{3} = 2$.

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