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Question Number 175949 by mr W last updated on 10/Sep/22

Commented by mr W last updated on 10/Sep/22

$$amassmisfixedatthemidpoint$$$$ofanuniformthinropewithunit$$$$mass\rho .findtheminimumlengthof$$$$theropesuchthatitcanbekeptin$$$$equilibriumasshown.$$$$\left(nofriction\right)$$$$asexample:m=\rho d$$

Answered by mr W last updated on 14/Sep/22

Commented by mr W last updated on 11/Sep/22

Commented by mr W last updated on 13/Sep/22

$$letL=totallengthoftherope$$$$saylengthofropepartBCish,$$$$thenthelengthofpartAB=\frac{L}{2}-h$$$$ABisasegmentofacatenary.$$$$T_2=h\rho g$$$$T_1\sin {\theta }_1=\frac{mg}{2}$$$$T_0=T_2\cos {\theta }_2=T_1\cos {\theta }_1$$$$$$$$witha=\frac{T_0}{\rho g}theeqn.ofthecatenaryis:$$$$y=a\cosh \frac{x}{a}$$$$\tan \theta =y^{\prime }=\sinh \frac{x}{a}$$$$s=a\sinh \frac{x}{a}$$$$$$$$atpointA:$$$$\tan {\theta }_1=\sinh \frac{d_1}{a}$$$$s_1=a\sinh \frac{d_1}{a}$$$$$$$$atpointB:$$$$\tan {\theta }_2=\sinh \frac{d_1+\frac{d}{2}}{a}$$$$s_2=a\sinh \frac{d_1+\frac{d}{2}}{a}$$$$$$$$T_0=T_1\cos {\theta }_1$$$$a\rho g=\frac{mg}{2\tan {\theta }_1}$$$$\tan {\theta }_1=\frac{m}{2\rho a}$$$$\sinh \frac{d_1}{a}=\frac{m}{2\rho a}$$$$\Rightarrow \frac{d_1}{a}={\sinh }^{-1}\frac{m}{2\rho a}$$$$$$$$T_0=T_2\cos {\theta }_2$$$$a\rho g=h\rho g\cos {\theta }_2$$$$\cos {\theta }_2=\frac{a}{h}\Rightarrow \tan {\theta }_2=\frac{\sqrt{1-{\left(\frac{a}{h}\right)}^2}}{\frac{a}{h}}$$$$\tan {\theta }_2=\sinh \frac{d_1+\frac{d}{2}}{a}=\frac{\sqrt{1-{\left(\frac{a}{h}\right)}^2}}{\frac{a}{h}}$$$$\Rightarrow \frac{d_1+\frac{d}{2}}{a}={\sinh }^{-1}\frac{\sqrt{1-{\left(\frac{a}{h}\right)}^2}}{\frac{a}{h}}$$$$\Rightarrow \frac{d}{2a}+{\sinh }^{-1}\frac{m}{2\rho a}={\sinh }^{-1}\frac{\sqrt{1-{\left(\frac{a}{h}\right)}^2}}{\frac{a}{h}}$$$$letk=\frac{m}{\rho d},\eta =\frac{a}{h},\xi =\frac{d}{2a}$$$$\Rightarrow \xi +{\sinh }^{-1}\left(k\xi \right)={\sinh }^{-1}\frac{\sqrt{1-{\eta }^2}}{\eta }$$$$\Rightarrow \frac{\sqrt{1-{\eta }^2}}{\eta }=\sinh [\xi +{\sinh }^{-1}\left(k\xi \right)]$$$$\Rightarrow \eta =\frac{1}{\sqrt{1+{\sinh }^2[\xi +{\sinh }^{-1}\left(k\xi \right)]}}$$$$$$$$s_2-s_1=\frac{L}{2}-h$$$$a(\sinh \frac{d_1+\frac{d}{2}}{a}-\sinh \frac{d_1}{a})=\frac{L}{2}-h$$$$a(\frac{\sqrt{1-{\left(\frac{a}{h}\right)}^2}}{\frac{a}{h}}-\frac{m}{2\rho a})=\frac{L}{2}-h$$$$L+\frac{m}{\rho }=2h[1+\sqrt{1-{\left(\frac{a}{h}\right)}^2}]$$$$\mathrm{\Phi }=\frac{L}{d}+k=\frac{1+\sqrt{1-{\eta }^2}}{\eta \xi }$$$$\mathrm{\Phi }=\frac{\sinh [\xi +{\sinh }^{-1}\left(k\xi \right)]+\sqrt{1+{\sinh }^2[\xi +{\sinh }^{-1}\left(k\xi \right)]}}{\xi }$$$$minimumL\iff \frac{d\mathrm{\Phi }}{d\xi }=0$$$$$$$$\lambda =\frac{h}{d}=\frac{1}{2\eta \xi }=\frac{\sqrt{1+{\sinh }^2[\xi +{\sinh }^{-1}\left(k\xi \right)]}}{2\xi }$$$$$$$$\underset{―}{examples:}$$$$m=0\Rightarrow k=0:$$$${\mathrm{\Phi }}_{min}=e\Rightarrow L_{min}=ed$$$$$$$$m=\rho d\Rightarrow k=1:$$$${\mathrm{\Phi }}_{min}=5.3268\Rightarrow L_{min}=4.3268d$$$$$$$$m=2\rho d\Rightarrow k=2:$$$${\mathrm{\Phi }}_{min}=7.7568\Rightarrow L_{min}=5.7568d$$

Commented by mr W last updated on 11/Sep/22

Commented by mr W last updated on 11/Sep/22

Commented by Tawa11 last updated on 15/Sep/22

$$\mathrm{Great}\mathrm{sir}.$$

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