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Question Number 66527 by mr W last updated on 16/Aug/19

Commented by mr W last updated on 16/Aug/19

$$theperimeterofaropeloopisL.$$$$nowitishangedontwopinsAandB.$$$$thedistancebetweenthepinsisb.$$$$allcontactisfrictionless.$$$$(L>2b)$$$$ifthelowestpointoftheloopish$$$$belowthepins.$$$$findhintermsofbandL.$$

Answered by mr W last updated on 17/Aug/19

Commented by mr W last updated on 17/Aug/19

Commented by mr W last updated on 17/Aug/19

$$whentheropeloopishangingonthe$$$$pins,itconsistsof2parts{S}_{1}and{S}_2.$$$$S_1+S_2=L$$$$$$$$generallyforaropebetweenAandB:$$$$massofropeofunitlength=\rho$$$$tensioninropeatpointA\left(B\right)=T$$$$lengthofropefromAtoB=S$$$$T_0=T\cos \theta$$$$a=\frac{T_0}{\rho g}=\frac{T\cos \theta }{\rho g}$$$$y=a\cosh \frac{x}{a}$$$$y^{\prime }=\sinh \frac{x}{a}$$$$atpointB:$$$$y^{\prime }=\sinh \frac{b}{2a}=\tan \theta =t$$$$\Rightarrow \frac{b}{2a}={\sinh }^{-1}t=\ln (t+\sqrt{1+t^2})$$$$\frac{S}{2}=a\sinh \frac{b}{2a}=a\tan \theta =at$$$$\Rightarrow S=2at=\frac{2a}{b}\times bt=\frac{bt}{{\sinh }^{-1}t}=\frac{bt}{\ln (t+\sqrt{1+t^2})}$$$$\Rightarrow \frac{T}{\rho g}=\frac{a}{\cos \theta }=\frac{2a}{b}\times \frac{b\sqrt{1+{\tan }^2\theta }}{2}=\frac{b\sqrt{1+t^2}}{2\ln (t+\sqrt{1+t^2})}$$$$h=a(1+\cosh \frac{b}{2a})=\frac{b}{2}\times \frac{2a}{b}(1+\cosh \frac{b}{2a})$$$$\Rightarrow \frac{2h}{b}=\frac{2a}{b}(1+\cosh \frac{b}{2a})=\frac{1+\sqrt{1+t^2}}{\ln (t+\sqrt{1+t^2})}$$$$$$$$forropepart1:$$$$t_1=\tan {\theta }_1$$$$S_1=\frac{{bt}_1}{\ln (t_1+\sqrt{1+t_1^2})}$$$$\frac{T_1}{\rho g}=\frac{b\sqrt{1+t_1^2}}{2\ln (t_1+\sqrt{1+t_1^2})}$$$$forropepart2:$$$$t_2=\tan {\theta }_2$$$$S_2=\frac{{bt}_2}{\ln (t_2+\sqrt{1+t_2^2})}$$$$\frac{T_2}{\rho g}=\frac{b\sqrt{1+t_2^2}}{2\ln (t_2+\sqrt{1+t_2^2})}$$$$$$$$T_1=T_2$$$$\frac{b\sqrt{1+t_1^2}}{2\ln (t_1+\sqrt{1+t_1^2})}=\frac{b\sqrt{1+t_2^2}}{2\ln (t_2+\sqrt{1+t_2^2})}$$$$\Rightarrow \frac{\sqrt{1+t_1^2}}{\ln (t_1+\sqrt{1+t_1^2})}=\frac{\sqrt{1+t_2^2}}{\ln (t_2+\sqrt{1+t_2^2})}...\left(i\right)$$$$S_1+S_2=L$$$$\frac{{bt}_1}{\ln (t_1+\sqrt{1+t_1^2})}+\frac{{bt}_2}{\ln (t_2+\sqrt{1+t_2^2})}=L$$$$let\lambda =\frac{L}{b}$$$$\Rightarrow \frac{t_1}{\ln (t_1+\sqrt{1+t_1^2})}+\frac{t_2}{\ln (t_2+\sqrt{1+t_2^2})}=\lambda ...\left(ii\right)$$$$$$$$t_1=t_2=tisalwaysasolutionof\left(i\right),$$$$\Rightarrow \frac{t}{\ln (t+\sqrt{1+t^2})}=\frac{\lambda }{2}$$$$inthiscasebothropepartsareequal:$$$$S_1=S_2=\frac{L}{2}$$$$$$$$inthediagramthegreencurveshows$$$$theequation\left(i\right).theredcurveshows$$$$theequation\left(ii\right)fordifferentvalues$$$$of\lambda .wecanseethatthereisonlyone$$$$solutionfortheshapeoftheropeif$$$$\lambda ⩽2.5155,with{t}_1=t_2,i.e.S_1=S_2.$$$$butfor\lambda >2.5155therearetwo$$$$possibleshapes,onewith{t}_1=t_2,i.e.$$$$S_1=S_{2}andtheotheronewith{t}_1\ne t_2,$$$$i.e.S_1\ne S_2.$$$$foragivenvalueof\lambda =\frac{L}{b},numerical$$$$solutionsofeqn.\left(i\right)and\left(ii\right)canbe$$$$obtained.$$

Commented by mr W last updated on 17/Aug/19

Commented by mr W last updated on 17/Aug/19

Commented by Cmr 237 last updated on 17/Aug/19

$$pleaseiwantthename$$$$ofthatapplication$$

Commented by mr W last updated on 17/Aug/19

$$forthegraphiusedanappcalled$$$$Grapher.$$

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