A uniform chain of total length L is at rest, | vedclass

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(C) Let $\lambda$ be the linear mass density of the chain. Total length is $L$. Let $x$ be the length on the incline.For the chain to be at rest,the f

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$\frac{7}{5}$

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(C) Let $\lambda$ be the linear mass density of the chain. Total length is $L$. Let $x$ be the length on the incline.For the chain to be at rest,the forces must balance.$1$. For $L_{\max}$ (chain tends to slide down): The friction acts upwards.$(\lambda L_{\max}) g \sin 	heta = (\lambda L - \lambda L_{\max}) g + \mu (\lambda L_{\max}) g \cos 	heta$$L_{\max} \sin 	heta = L - L_{\max} + \mu L_{\max} \cos 	heta$$L_{\max} (\sin 	heta + 1 - \mu \cos 	heta) = L$Given $	heta = 30^{\circ}$,$\sin 30^{\circ} = 0.5$,$\cos 30^{\circ} = \frac{\sqrt{3}}{2}$,$\mu = \frac{1}{2\sqrt{3}}$.$L_{\max} (0.5 + 1 - \frac{1}{2\sqrt{3}} \cdot \frac{\sqrt{3}}{2}) = L \implies L_{\max} (1.5 - 0.25) = L \implies 1.25 L_{\max} = L \implies L_{\max} = \frac{L}{1.25} = 0.8 L$.$2$. For $L_{\min}$ (chain tends to slide up): The friction acts downwards.$(\lambda L - \lambda L_{\min}) g = (\lambda L_{\min}) g \sin 	heta + \mu (\lambda L_{\min}) g \cos 	heta$$L - L_{\min} = L_{\min} (\sin 	heta + \mu \cos 	heta)$$L = L_{\min} (1 + \sin 	heta + \mu \cos 	heta)$$L = L_{\min} (1 + 0.5 + 0.25) = 1.75 L_{\min} \implies L_{\min} = \frac{L}{1.75} = \frac{L}{7/4} = \frac{4}{7} L$.Ratio $\frac{L_{\max}}{L_{\min}} = \frac{0.8 L}{(4/7) L} = \frac{4/5}{4/7} = \frac{7}{5}$.

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