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Question Number 187355 by mr W last updated on 16/Feb/23

Commented by mr W last updated on 17/Feb/23

$$findthemaximumandminimum$$$$possibledeflectioninthemiddle$$$$oftheuniformthinropewithlength$$$$L.$$$$thefrictioncoefficientbetweenthe$$$$ropeandthefixedpulleyswith$$$$radiusris\mu .$$

Commented by Humble last updated on 16/Feb/23

$$Ilookuptomysuperiorsforthesolution$$

Answered by mr W last updated on 19/Feb/23

Commented by mr W last updated on 19/Feb/23

$$T_1=T_2{e}^{\mu (\frac{\pi }{2}+\theta )}=\rho {ghe}^{\mu (\frac{\pi }{2}+\theta )}$$$$T_0=T_1\cos \theta$$$$a=\frac{T_0}{\rho g}={he}^{\mu (\frac{\pi }{2}+\theta )}\cos \theta$$$$x_B=\frac{d}{2}-r\sin \theta$$$$y_B=a+f-(1-\cos \theta )r$$$$s_{AB}=\frac{L}{2}-h-(\frac{\pi }{2}+\theta )r$$$$eqn.ofcatenary:$$$$y=a\cosh \frac{x}{a}$$$$atpointB:$$$$y_B=a\cosh \frac{\frac{d}{2}-r\sin \theta }{a}=a+f-(1-\cos \theta )r$$$$s=a\sinh \frac{\frac{d}{2}-r\sin \theta }{a}=\frac{L}{2}-h-(\frac{\pi }{2}+\theta )r$$$$\tan \theta =\sinh \frac{\frac{d}{2}-r\sin \theta }{a}$$$$\tan \theta =\sinh \frac{\frac{d}{2}-r\sin \theta }{{he}^{\mu (\frac{\pi }{2}+\theta )}\cos \theta }$$$$\Rightarrow h=\frac{\frac{d}{2}-r\sin \theta }{\left[{\sinh }^{-1}\left(\tan \theta \right)\right]e^{\mu (\frac{\pi }{2}+\theta )}\cos \theta }$$$${he}^{\mu (\frac{\pi }{2}+\theta )}\cos \theta \tan \theta =\frac{L}{2}-h-(\frac{\pi }{2}+\theta )r$$$$h[1+e^{\mu (\frac{\pi }{2}+\theta )}\sin \theta ]=\frac{L}{2}-(\frac{\pi }{2}+\theta )r$$$$\Rightarrow h=\frac{\frac{L-\pi r}{2}-\theta r}{1+e^{\mu (\frac{\pi }{2}+\theta )}\sin \theta }$$$$\overline{)\begin{array}{c}\frac{\frac{L-\pi r}{2}-\theta r}{1+e^{\mu (\frac{\pi }{2}+\theta )}\sin \theta }=\frac{\frac{d}{2}-r\sin \theta }{\left[{\sinh }^{-1}\left(\tan \theta \right)\right]e^{\mu (\frac{\pi }{2}+\theta )}\cos \theta }\end{array}}$$$$wegettwosolutionsfor0<\theta <\frac{\pi }{2}.one$$$$for{f}_{max}andtheotherfor{f}_{min}.$$$$f=a(\cosh \frac{\frac{d}{2}-r\sin \theta }{a}-1)+(1-\cos \theta )r$$$$\overline{)\begin{array}{c}f=[\frac{\frac{L-\pi r}{2}-\theta r}{e^{-\mu (\frac{\pi }{2}+\theta )}+\sin \theta }+r](1-\cos \theta )\end{array}}$$

Commented by mr W last updated on 17/Feb/23

Commented by mr W last updated on 17/Feb/23

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