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Question Number 66310 by mr W last updated on 12/Aug/19

Commented by mr W last updated on 12/Aug/19

$$allcontactisfrictionless.$$$$findtheminimumlengthofuniform$$$$ropesuchthatitcanstayinequilibrium$$$$asshown.$$

Answered by mr W last updated on 13/Aug/19

Commented by mr W last updated on 14/Aug/19

$$\rho =massofunitlengthofrope$$$$totallengthofropeL=s+2h$$$$tensioninropeatpointAandB:$$$$T=\rho gh$$$$horizontalcomponentoftension:$$$$T_0=T\cos \theta =\rho gh\cos \theta$$$$a=\frac{T_0}{\rho g}=h\cos \theta$$$$y=a\cosh \frac{x}{a}$$$$y^{\prime }=\sinh \frac{x}{a}$$$$atpointB:$$$$y^{\prime }=\sinh \frac{b}{2a}=\tan \theta$$$$\Rightarrow \frac{b}{2a}={\sinh }^{-1}\tan \theta$$$$\frac{s}{2}=a\sinh \frac{b}{2a}=a\tan \theta$$$$\Rightarrow s=2a\tan \theta$$$$h=\frac{a}{\cos \theta }$$$$L=s+2h=2a(\tan \theta +\frac{1}{\cos \theta })$$$$\Rightarrow \frac{L}{b}=\frac{2a}{b}(\tan \theta +\frac{1}{\cos \theta })$$$$\Rightarrow \frac{L}{b}=\frac{1}{{\sinh }^{-1}\tan \theta }(\tan \theta +\frac{1}{\cos \theta })$$$$\Rightarrow \frac{L}{b}=\frac{\tan \theta +\sqrt{1+{\tan }^2\theta }}{{\sinh }^{-1}\tan \theta }$$$$lett=\tan \theta$$$$\Rightarrow \frac{L}{b}=\frac{\mathrm{t}+\sqrt{1+{\mathrm{t}}^2}}{{\sinh }^{-1}\mathrm{t}}=\frac{t+\sqrt{1+t^2}}{\ln (t+\sqrt{1+t^2})}$$$$letp=t+\sqrt{1+t^2}$$$$\Rightarrow \frac{L}{b}=\frac{p}{\ln p}$$$$\frac{d}{dp}\left(\frac{L}{b}\right)=\frac{1}{\ln p}-\frac{1}{{\left(\ln p\right)}^2}=0$$$$\Rightarrow \frac{1}{\ln p}=1\Rightarrow p=e$$$$\Rightarrow \frac{L_{min}}{b}=\frac{e}{\ln e}=e$$$$\Rightarrow L_{min}=eb$$$$ifb=1,L_{min}=e\approx 2.718$$$$$$$$t+\sqrt{1+t^2}=e$$$$1=e^2-2et$$$$\Rightarrow t=\tan \theta =\frac{e^2-1}{2e}$$$$\Rightarrow \theta ={\tan }^{-1}\frac{e^2-1}{2e}=49.605°$$$$$$$$\frac{h}{b}=\frac{2a}{b}\times \frac{1}{2\cos \theta }=\frac{\sqrt{1+{\tan }^2\theta }}{2\ln (\tan \theta +\sqrt{1+{\tan }^2\theta })}$$$$=\frac{e^2+1}{4e}=0.7715$$

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