Question Number 61762 by mr W last updated on 08/Jun/19 | ||
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Commented by mr W last updated on 08/Jun/19 | ||
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$$\left({see}\:{also}\:{Q}\mathrm{61612}\right) \\ $$$${Find}\:{the}\:{relation}\:{between}\:{h},\:{d}\:{and}\:{L} \\ $$$${of}\:{a}\:{hanging}\:{rope}. \\ $$ | ||
Answered by mr W last updated on 09/Jun/19 | ||
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$${I}.\:{form}\:{of}\:{rope}\:{as}\:{catenary} \\ $$$${y}={a}\:\mathrm{cosh}\:\frac{{x}}{{a}} \\ $$$${at}\:{x}=\frac{{d}}{\mathrm{2}}: \\ $$$${h}+{a}={a}\:\mathrm{cosh}\:\frac{{d}}{\mathrm{2}{a}} \\ $$$$\Rightarrow\left(\mathrm{1}+\frac{{h}}{{a}}\right)=\mathrm{cosh}\:\frac{{d}}{\mathrm{2}{a}}\:\:\:\:\:...\left({i}\right) \\ $$$$\frac{{L}}{\mathrm{2}}={a}\:\mathrm{sinh}\:\frac{{d}}{\mathrm{2}{a}} \\ $$$$\Rightarrow\frac{{L}}{\mathrm{2}{a}}=\mathrm{sinh}\:\frac{{d}}{\mathrm{2}{a}}\:\:\:\:...\left({ii}\right) \\ $$$$\Rightarrow\left(\mathrm{1}+\frac{{h}}{{a}}\right)^{\mathrm{2}} −\left(\frac{{L}}{\mathrm{2}{a}}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$$\Rightarrow\mathrm{8}{ha}+\mathrm{4}{h}^{\mathrm{2}} −{L}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{a}=\frac{{L}^{\mathrm{2}} −\mathrm{4}{h}^{\mathrm{2}} }{\mathrm{8}{h}} \\ $$$${from}\:\left({i}\right): \\ $$$$\mathrm{sinh}\:\frac{{d}}{\mathrm{2}{a}}=\frac{\mathrm{4}{Lh}}{{L}^{\mathrm{2}} −\mathrm{4}{h}^{\mathrm{2}} } \\ $$$$\Rightarrow{d}=\left\{\left(\frac{{L}^{\mathrm{2}} −\mathrm{4}{h}^{\mathrm{2}} }{\mathrm{4}{Lh}}\right)\:\mathrm{sinh}^{−\mathrm{1}} \:\left(\frac{\mathrm{4}{Lh}}{{L}^{\mathrm{2}} −\mathrm{4}{h}^{\mathrm{2}} }\right)\right\}{L} \\ $$$${with}\:\kappa=\frac{{L}^{\mathrm{2}} −\mathrm{4}{h}^{\mathrm{2}} }{\mathrm{4}{Lh}} \\ $$$$\Rightarrow{d}=\left(\kappa\:\mathrm{sinh}^{−\mathrm{1}} \:\frac{\mathrm{1}}{\kappa}\right){L} \\ $$$$ \\ $$$${example}:\:{L}=\mathrm{25}{m},\:{h}=\mathrm{7}.\mathrm{5}{m} \\ $$$$\Rightarrow\kappa=\frac{\mathrm{25}^{\mathrm{2}} −\mathrm{4}×\mathrm{7}.\mathrm{5}^{\mathrm{2}} }{\mathrm{4}×\mathrm{25}×\mathrm{7}.\mathrm{5}}=\frac{\mathrm{8}}{\mathrm{15}} \\ $$$$\Rightarrow{d}\approx\mathrm{0}.\mathrm{73936}{L}=\mathrm{18}.\mathrm{48}\:{m} \\ $$$$ \\ $$$${let}\:\eta=\frac{\mathrm{2}{h}}{{d}},\:\lambda=\frac{{L}}{{d}} \\ $$$$\eta×\frac{{d}}{\mathrm{2}{a}}=\mathrm{cosh}\:\frac{{d}}{\mathrm{2}{a}}−\mathrm{1} \\ $$$$\Rightarrow\eta=\frac{\mathrm{cosh}\:\frac{{d}}{\mathrm{2}{a}}−\mathrm{1}}{\frac{{d}}{\mathrm{2}{a}}} \\ $$$$\lambda×\frac{{d}}{\mathrm{2}{a}}=\mathrm{sinh}\:\frac{{d}}{\mathrm{2}{a}} \\ $$$$\Rightarrow\lambda=\frac{\mathrm{sinh}\:\frac{{d}}{\mathrm{2}{a}}}{\frac{{d}}{\mathrm{2}{a}}} \\ $$$${with}\:{parameter}\:{t}=\frac{{d}}{\mathrm{2}{a}}\:{we}\:{get}\: \\ $$$$\begin{cases}{\eta=\frac{\mathrm{cosh}\:{t}−\mathrm{1}}{{t}}}\\{\lambda=\frac{\mathrm{sinh}\:{t}}{{t}}}\end{cases} \\ $$ | ||
Commented by Tawa1 last updated on 09/Jun/19 | ||
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$$\mathrm{wow},\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$ | ||
Answered by mr W last updated on 08/Jun/19 | ||
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$${II}.\:{form}\:{of}\:{rope}\:{as}\:{parabola} \\ $$$${y}={h}\left(\frac{\mathrm{2}{x}}{{d}}\right)^{\mathrm{2}} \\ $$$${y}'=\frac{\mathrm{8}{hx}}{{d}^{\mathrm{2}} } \\ $$$$\frac{{L}}{\mathrm{2}}=\int_{\mathrm{0}} ^{\frac{{d}}{\mathrm{2}}} \sqrt{\mathrm{1}+{y}'^{\mathrm{2}} }\:{dx}=\int_{\mathrm{0}} ^{\frac{{d}}{\mathrm{2}}} \sqrt{\mathrm{1}+\left(\frac{\mathrm{8}{hx}}{{d}^{\mathrm{2}} }\right)^{\mathrm{2}} }\:{dx} \\ $$$${L}=\frac{{d}^{\mathrm{2}} }{\mathrm{4}{h}}\int_{\mathrm{0}} ^{\frac{{d}}{\mathrm{2}}} \sqrt{\mathrm{1}+\left(\frac{\mathrm{8}{hx}}{{d}^{\mathrm{2}} }\right)^{\mathrm{2}} }\:{d}\left(\frac{\mathrm{8}{hx}}{{d}^{\mathrm{2}} }\right) \\ $$$${L}=\frac{{d}^{\mathrm{2}} }{\mathrm{4}{h}}\int_{\mathrm{0}} ^{\frac{\mathrm{4}{h}}{{d}}} \sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\:{dt} \\ $$$${L}=\mathrm{2}{h}\left(\frac{{d}}{\mathrm{4}{h}}\right)^{\mathrm{2}} \left[{t}\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }+\mathrm{ln}\:\left({t}+\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\right)\right]_{\mathrm{0}} ^{\frac{\mathrm{4}{h}}{{d}}} \\ $$$${with}\:\mu=\frac{\mathrm{4}{h}}{{d}} \\ $$$$\Rightarrow{L}=\frac{\mathrm{2}{h}}{\mu^{\mathrm{2}} }\left[\mu\sqrt{\mathrm{1}+\mu^{\mathrm{2}} }+\mathrm{ln}\:\left(\mu+\sqrt{\mathrm{1}+\mu^{\mathrm{2}} }\right)\right] \\ $$$$\Rightarrow\frac{{L}}{\mathrm{2}{h}}=\frac{\mathrm{1}}{\mu^{\mathrm{2}} }\left[\mu\sqrt{\mathrm{1}+\mu^{\mathrm{2}} }+\mathrm{ln}\:\left(\mu+\sqrt{\mathrm{1}+\mu^{\mathrm{2}} }\right)\right] \\ $$$$ \\ $$$${example}:\:{L}=\mathrm{25}{m},\:{h}=\mathrm{7}.\mathrm{5}{m} \\ $$$$\frac{\mathrm{1}}{\mu^{\mathrm{2}} }\left[\mu\sqrt{\mathrm{1}+\mu^{\mathrm{2}} }+\mathrm{ln}\:\left(\mu+\sqrt{\mathrm{1}+\mu^{\mathrm{2}} }\right)\right]=\frac{\mathrm{5}}{\mathrm{3}} \\ $$$$\Rightarrow\mu=\mathrm{1}.\mathrm{6008} \\ $$$$\Rightarrow{d}=\frac{\mathrm{4}{h}}{\mu}=\mathrm{18}.\mathrm{74}{m} \\ $$$$ \\ $$$${let}\:\eta=\frac{\mathrm{2}{h}}{{d}},\:\lambda=\frac{{L}}{{d}} \\ $$$$\Rightarrow\lambda=\frac{\mathrm{1}}{\mathrm{4}\eta}\left[\mathrm{2}\eta\sqrt{\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} }+\mathrm{ln}\:\left(\mathrm{2}\eta+\sqrt{\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} }\right)\right] \\ $$ | ||
Commented by mr W last updated on 08/Jun/19 | ||
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Commented by mr W last updated on 08/Jun/19 | ||
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$${C}={catenary},\:{P}={parabola} \\ $$$${we}\:{can}\:{see}\:{the}\:{difference}\:{between}\:{both} \\ $$$${is}\:{very}\:{small}. \\ $$ | ||