Question and Answers Forum

All Questions      Topic List

Mechanics Questions

Previous in All Question      Next in All Question      

Previous in Mechanics      Next in Mechanics      

Question Number 177446 by mr W last updated on 05/Oct/22

Commented by mr W last updated on 05/Oct/22

$$aropewithlengthLhasuniform$$$$mass.oneendoftheropeisfixedon$$$$thetopofapolewithheigthhabovea$$$$inclinedplane(h<L).theotherend$$$$oftheropecanbefreelylaidonthe$$$$inclinedplane.thefrictioncoefficient$$$$betweenropeandinclinedplaneis\mu .$$$$findthemaximumareaofthe$$$$inclinedplanewhichcanbecovered$$$$bytherope.$$

Answered by mr W last updated on 06/Oct/22

Commented by mr W last updated on 06/Oct/22

$$duetosymmetryweonlyneedtolook$$$$at-\frac{\pi }{2}⩽\phi ⩽\frac{\pi }{2}.$$$$thecases\phi =\frac{\pi }{2}and\phi =-\frac{\pi }{2}$$$$seeQ170601.$$

Commented by mr W last updated on 06/Oct/22

Commented by mr W last updated on 07/Oct/22

$$saythelengthofropepartwhichlies$$$$ontheinclinedplaneisb.$$$$thetensionintheropebetweenthe$$$$hangingpartandthepartlyingon$$$$theinclinedplaneis{T}_1.$$$$\mu b\rho g\cos \theta \sin \alpha =b\rho g\sin \theta \cos \phi$$$$\Rightarrow \sin \alpha =\frac{\tan \theta \cos \phi }{\mu }$$$$\mu b\rho g\cos \theta \cos \alpha =T_1+b\rho g\sin \theta \sin \phi$$$$\Rightarrow \frac{T_1}{b\rho g}=\mu \cos \theta \cos \alpha -\sin \theta \sin \phi$$$$\Rightarrow \frac{T_1}{b\rho g}=\sqrt{{\mu }^2{\cos }^2\theta -{\sin }^2\theta {\cos }^2\phi }-\sin \theta \sin \phi$$$$\Rightarrow \frac{T_1}{b\rho g}=\sqrt{({\mu }^2+1){\cos }^2\theta -1+{\sin }^2\theta {\sin }^2\phi }-\sin \theta \sin \phi$$

Commented by mr W last updated on 07/Oct/22

$$let\eta =\frac{h}{L},\xi =\frac{b}{L}$$$$\sin \varphi =\sin \phi \sin \theta =\lambda ,say$$$$a=\frac{T_0}{\rho g}=\frac{T_1\cos \varphi }{\rho g}$$$$=b\cos \varphi (\sqrt{(1+{\mu }^2){\cos }^2\theta -1+{\lambda }^2}-\lambda )$$$$=\xi L\sqrt{1-{\lambda }^2}(\sqrt{({\mu }^2+1){\cos }^2\theta -1+{\lambda }^2}-\lambda )$$$$let\delta =\sqrt{({\mu }^2+1){\cos }^2\theta -1+{\lambda }^2}-\lambda$$$$\Rightarrow a=\xi L\delta \sqrt{1-{\lambda }^2}$$$$$$$$eqn.ofcatenaryinxy-system:$$$$y=a\cosh \frac{x}{a}$$$$y^{\prime }=\sinh \frac{x}{a}$$$$s=a\sinh \frac{x}{a}$$$$$$$$atpointD:$$$$y^{\prime }=\sinh \frac{d_1}{a}=\tan \varphi$$$$\Rightarrow \frac{d_1}{a}={\sinh }^{-1}\left(\tan \varphi \right)={\sinh }^{-1}\frac{\lambda }{\sqrt{1-{\lambda }^2}}$$$$h_2=a\cosh \frac{d_1}{a}=a\cosh \left({\sinh }^{-1}\frac{\lambda }{\sqrt{1-{\lambda }^2}}\right)$$$$=a\sqrt{1+{\left(\frac{\lambda }{\sqrt{1-{\lambda }^2}}\right)}^2}=\frac{a}{\sqrt{1-{\lambda }^2}}$$$$s_1=a\tan \varphi =\frac{a\lambda }{\sqrt{1-{\lambda }^2}}$$$$$$$$atpointB:$$$$h_3=a\cosh \frac{d_2}{a}$$$$s_2=a\sinh \frac{d_2}{a}$$$$h_1=h-h_3$$$$\frac{h_1}{a}=\frac{h}{a}-\frac{h_3}{a}$$$$h_1+h_2=(d_1+d_2)\tan \varphi$$$$(\frac{h}{a}-\frac{h_3}{a}+\frac{h_2}{a})\frac{\sqrt{1-{\lambda }^2}}{\lambda }=\frac{d_1}{a}+\frac{d_2}{a}$$$$(\frac{h}{a}-\cosh \frac{d_2}{a}+\frac{1}{\sqrt{1-{\lambda }^2}})\frac{\sqrt{1-{\lambda }^2}}{\lambda }={\sinh }^{-1}\frac{\lambda }{\sqrt{1-{\lambda }^2}}+\frac{d_2}{a}$$$$(\frac{h}{a}-\cosh \frac{d_2}{a})\frac{\sqrt{1-{\lambda }^2}}{\lambda }+\frac{1}{\lambda }={\sinh }^{-1}\frac{\lambda }{\sqrt{1-{\lambda }^2}}+\frac{d_2}{a}$$$$\frac{d_2}{a}+\frac{\sqrt{1-{\lambda }^2}}{\lambda }\cosh \frac{d_2}{a}=\frac{1}{\lambda }(\frac{h\sqrt{1-{\lambda }^2}}{a}+1)-{\sinh }^{-1}\frac{\lambda }{\sqrt{1-{\lambda }^2}}$$$$\Rightarrow \lambda \left(\frac{d_2}{a}\right)+\sqrt{1-{\lambda }^2}\cosh \frac{d_2}{a}=\frac{\eta }{\xi \delta }+1-\lambda {\sinh }^{-1}\frac{\lambda }{\sqrt{1-{\lambda }^2}}...\left(i\right)$$$$$$$$s_1+s_2=L-b$$$$\frac{s_1}{a}+\frac{s_2}{a}=\frac{L-b}{a}$$$$\frac{\lambda }{\sqrt{1-{\lambda }^2}}+\sinh \frac{d_2}{a}=\frac{1-\xi }{\xi \delta \sqrt{1-{\lambda }^2}}$$$$\sinh \frac{d_2}{a}=\frac{1}{\sqrt{1-{\lambda }^2}}(\frac{1-\xi }{\xi \delta }-\lambda )$$$$\Rightarrow \frac{d_2}{a}={\sinh }^{-1}\left[\frac{1}{\sqrt{1-{\lambda }^2}}(\frac{1-\xi }{\xi \delta }-\lambda )\right]...\left(ii\right)$$$$insert\left(ii\right)into\left(i\right)wegetanequation$$$$forsolving\xi .$$$$$$$$r=AC=b+\frac{d_1+d_2}{\cos \varphi }$$$$=b+\frac{a}{\cos \varphi }(\frac{d_1}{a}+\frac{d_2}{a})$$$$=\xi L+\xi L\delta \{{\sinh }^{-1}\frac{\lambda }{\sqrt{1-{\lambda }^2}}+{\sinh }^{-1}\left[\frac{1}{\sqrt{1-{\lambda }^2}}(\frac{1-\xi }{\xi \delta }-\lambda )\right]\}$$$$\Rightarrow \frac{r}{L}=\xi +\xi \delta \{{\sinh }^{-1}\frac{\lambda }{\sqrt{1-{\lambda }^2}}+{\sinh }^{-1}\left[\frac{1}{\sqrt{1-{\lambda }^2}}(\frac{1-\xi }{\xi \delta }-\lambda )\right]\}$$$$$$$$\mathit{e}\mathit{x}\mathit{a}\mathit{m}\mathit{p}\mathit{l}\mathit{e}\mathit{s}:$$$$\mu =1.5,\theta =30°,\eta =\frac{h}{L}=\frac{3}{5}=0.6,$$$$\phi =0°\Rightarrow \xi =\frac{b}{L}=0.1975,\frac{r}{L}=0.6556$$$$\phi =30°\Rightarrow \xi =\frac{b}{L}=0.2556,\frac{r}{L}=0.7583$$$$\phi =60°\Rightarrow \xi =\frac{b}{L}=0.3092,\frac{r}{L}=0.8551$$$$\phi =90°\Rightarrow \xi =\frac{b}{L}=0.{3320}^{\ast )},\frac{r}{L}=0.8975$$$$$$$$\ast )thesameresultasinQ170601$$

Commented by mr W last updated on 06/Oct/22

Commented by mr W last updated on 07/Oct/22

Commented by mr W last updated on 07/Oct/22

Commented by mr W last updated on 07/Oct/22

$$\mu b\rho g\cos \theta \sin \alpha =b\rho g\sin \theta \cos \phi$$$$\mu \cos \theta \sin \alpha =\sin \theta \cos \phi$$$$\Rightarrow \sin \alpha =\frac{\tan \theta \cos \phi }{\mu }$$$$T_1=\mu b\rho g\cos \theta \cos \alpha +b\rho g\sin \theta \sin \phi$$$$\frac{T_1}{b\rho g}=\mu \cos \theta \sqrt{1-{\left(\frac{\tan \theta \cos \phi }{\mu }\right)}^2}+\sin \theta \sin \phi$$$$\Rightarrow \frac{T_1}{b\rho g}=\sqrt{({\mu }^2+1){\cos }^2\theta -1+{\sin }^2\theta {\sin }^2\phi }+\sin \theta \sin \phi$$

Commented by mr W last updated on 07/Oct/22

Commented by mr W last updated on 07/Oct/22

$$let\eta =\frac{h}{L},\xi =\frac{b}{L}$$$$\sin \varphi =\sin \phi \sin \theta =\lambda ,say$$$$a=\frac{T_0}{\rho g}=\frac{T_1\cos \varphi }{\rho g}$$$$=b\cos \varphi (\sqrt{(1+{\mu }^2){\cos }^2\theta -1+{\lambda }^2}+\lambda )$$$$=\xi L\sqrt{1-{\lambda }^2}(\sqrt{({\mu }^2+1){\cos }^2\theta -1+{\lambda }^2}+\lambda )$$$$let\delta =\sqrt{({\mu }^2+1){\cos }^2\theta -1+{\lambda }^2}+\lambda$$$$\Rightarrow a=\xi L\delta \sqrt{1-{\lambda }^2}$$$$$$$$eqn.ofcatenaryinxy-system:$$$$y=a\cosh \frac{x}{a}$$$$y^{\prime }=\sinh \frac{x}{a}$$$$s=a\sinh \frac{x}{a}$$$$$$$$atpointD:$$$$y^{\prime }=\sinh \frac{d_1}{a}=\tan \varphi$$$$\Rightarrow \frac{d_1}{a}={\sinh }^{-1}\left(\tan \varphi \right)={\sinh }^{-1}\frac{\lambda }{\sqrt{1-{\lambda }^2}}$$$$h_1=a\cosh \frac{d_1}{a}=a\cosh \left({\sinh }^{-1}\frac{\lambda }{\sqrt{1-{\lambda }^2}}\right)$$$$=a\sqrt{1+{\left(\frac{\lambda }{\sqrt{1-{\lambda }^2}}\right)}^2}=\frac{a}{\sqrt{1-{\lambda }^2}}$$$$s_1=a\tan \varphi =\frac{a\lambda }{\sqrt{1-{\lambda }^2}}$$$$$$$$atpointB:$$$$h_1+d_2\tan \varphi +h=a\cosh \frac{d_1+d_2}{a}$$$$\frac{h_1}{a}+\frac{d_2}{a}\tan \varphi +\frac{h}{a}=\cosh \frac{d_1+d_2}{a}$$$$\frac{1}{\sqrt{1-{\lambda }^2}}+\frac{\eta }{\xi \delta \sqrt{1-{\lambda }^2}}=\cosh (\frac{d_2}{a}+{\sinh }^{-1}\frac{\lambda }{\sqrt{1-{\lambda }^2}})-\frac{d_2}{a}\times \frac{\lambda }{\sqrt{1-{\lambda }^2}}$$$$\Rightarrow \sqrt{1-{\lambda }^2}\cosh (\frac{d_2}{a}+{\sinh }^{-1}\frac{\lambda }{\sqrt{1-{\lambda }^2}})-\lambda \left(\frac{d_2}{a}\right)=1+\frac{\eta }{\xi \delta }...\left(i\right)$$$$$$$$s_1+s=a\sinh \frac{d_1+d_2}{a}$$$$\frac{s_1}{a}+\frac{L-b}{a}=\sinh (\frac{d_2}{a}+\frac{d_1}{a})$$$$\frac{\lambda }{\sqrt{1-{\lambda }^2}}+\frac{1-\xi }{\xi \delta \sqrt{1-{\lambda }^2}}=\sinh (\frac{d_2}{a}+{\sinh }^{-1}\frac{\lambda }{\sqrt{1-{\lambda }^2}})$$$$\Rightarrow \frac{d_2}{a}={\sinh }^{-}\left[\frac{1}{\sqrt{1-{\lambda }^2}}(\frac{1-\xi }{\xi \delta }+\lambda )\right]-{\sinh }^{-1}\frac{\lambda }{\sqrt{1-{\lambda }^2}}...\left(ii\right)$$$$insert\left(ii\right)into\left(i\right)wegetanequation$$$$forsolving\xi .$$$$$$$$r=AC=b+\frac{d_2}{\cos \varphi }$$$$=b+\frac{a}{\cos \varphi }\times \frac{d_2}{a}$$$$=\xi L+\xi L\delta \{{\sinh }^{-1}\left[\frac{1}{\sqrt{1-{\lambda }^2}}(\frac{1-\xi }{\xi \delta }+\lambda )\right]-{\sinh }^{-1}\frac{\lambda }{\sqrt{1-{\lambda }^2}}\}$$$$\Rightarrow \frac{r}{L}=\xi +\xi \delta \{{\sinh }^{-1}\left[\frac{1}{\sqrt{1-{\lambda }^2}}(\frac{1-\xi }{\xi \delta }+\lambda )\right]-{\sinh }^{-1}\frac{\lambda }{\sqrt{1-{\lambda }^2}}\}$$$$$$$$\mathit{e}\mathit{x}\mathit{a}\mathit{m}\mathit{p}\mathit{l}\mathit{e}\mathit{s}:$$$$\mu =1.5,\theta =30°,\eta =\frac{h}{L}=\frac{3}{5}=0.6,$$$$\phi =0°\Rightarrow \xi =\frac{b}{L}=0.1975,\frac{r}{L}=0.6556$$$$\phi =30°\Rightarrow \xi =\frac{b}{L}=0.1521,\frac{r}{L}=0.5733$$$$\phi =60°\Rightarrow \xi =\frac{b}{L}=0.1257,\frac{r}{L}=0.5225$$$$\phi =90°\Rightarrow \xi =\frac{b}{L}=0.1172,\frac{r}{L}=0.5053$$

Commented by mr W last updated on 08/Oct/22

Terms of Service

Privacy Policy

Contact: info@tinkutara.com