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Question Number 203636 by ajfour last updated on 24/Jan/24

Commented by ajfour last updated on 24/Jan/24

$$Takestringlength\mathit{s}forquestion\left(1\right).$$

Answered by mr W last updated on 24/Jan/24

Commented by mr W last updated on 27/Jan/24

$$F_x=\frac{mgR}{\sqrt{s^2-R^2}}$$$$T=\frac{mgs}{\sqrt{s^2-R^2}}=\frac{mg}{\sqrt{1-\frac{R^2}{s^2}}}$$$${\theta }_k=\frac{2\pi k}{n},k=1...n-1$$$$\phi =\frac{\pi }{2}-\frac{{\theta }_k}{2}=\frac{\pi }{2}-\frac{k\pi }{n}$$$$r_k=2R\sin \frac{{\theta }_k}{2}=2R\sin \frac{k\pi }{n}$$$$F_k=\frac{{cq}^2}{r_k^2}=\frac{{cq}^2}{4R^2{\sin }^2\frac{k\pi }{n}}$$$$F_{k,x}=F_k\cos \phi =\frac{{cq}^2}{4R^2\sin \frac{k\pi }{n}}$$$$F_x=\underset{k=1}{\sum ^{n-1}}{F}_{k,x}=\frac{{cq}^2}{4R^2}\underset{k=1}{\sum ^{n-1}}\frac{1}{\sin \frac{k\pi }{n}}=\frac{mgR}{\sqrt{s^2-R^2}}$$$$\frac{R^3}{\sqrt{s^2-R^2}}=\frac{{cq}^2}{4mg}\underset{k=1}{\sum ^{n-1}}\frac{1}{\sin \frac{k\pi }{n}}={\lambda }^2$$$$with\lambda =\sqrt{\frac{{cq}^2}{4mg}\underset{k=1}{\sum ^{n-1}}\frac{1}{\sin \frac{k\pi }{n}}}$$$$R^6+{\lambda }^4{R}^2-{\lambda }^4{s}^2=0$$$$R^2=\sqrt[3]{{\lambda }^4(\sqrt{\frac{s^4}{4}+\frac{{\lambda }^4}{27}}+\frac{s^2}{2})}-\sqrt[3]{{\lambda }^4(\sqrt{\frac{s^4}{4}+\frac{{\lambda }^4}{27}}-\frac{s^2}{2})}$$$$$$$$specialcase:n=3$$$$\underset{k=1}{\sum ^{n-1}}\frac{1}{\sin \frac{k\pi }{n}}=\frac{1}{\sin \frac{\pi }{3}}+\frac{1}{\sin \frac{2\pi }{3}}=\frac{4}{\sqrt{3}}$$$$\lambda =\sqrt{\frac{{cq}^2}{4mg}\times \frac{4}{\sqrt{3}}}=\frac{\mu }{\sqrt{\sqrt{3}}}with\mu =\sqrt{\frac{{cq}^2}{mg}}$$$$R^2=\sqrt[3]{\frac{{\mu }^4}{3}(\sqrt{\frac{s^4}{4}+\frac{{\mu }^4}{81}}+\frac{s^2}{2})}-\sqrt[3]{\frac{{\mu }^4}{3}(\sqrt{\frac{s^4}{4}+\frac{{\mu }^4}{81}}-\frac{s^2}{2})}$$$$\frac{T}{mg}=\frac{s}{\sqrt{s^2+\sqrt[3]{\frac{{\mu }^4}{3}(\sqrt{\frac{s^4}{4}+\frac{{\mu }^4}{81}}-\frac{s^2}{2})}-\sqrt[3]{\frac{{\mu }^4}{3}(\sqrt{\frac{s^4}{4}+\frac{{\mu }^4}{81}}+\frac{s^2}{2})}}}$$

Commented by ajfour last updated on 24/Jan/24

$$utterlyamazingSir.$$

Commented by mr W last updated on 25/Jan/24

$$thankssir!thequestion2ismuch$$$$harder.haveyoutried?$$

Commented by ajfour last updated on 25/Jan/24

Thats why i havnt tried, but i m starting to think, few hours...

Commented by mr W last updated on 27/Jan/24

$$solutiontoquestion2$$$$seeQ203726$$

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