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# Question-66310

Question Number 66310 by mr W last updated on 12/Aug/19
Commented by mr W last updated on 12/Aug/19
$${all}\:{contact}\:{is}\:{frictionless}. \\$$$${find}\:{the}\:{minimum}\:{length}\:{of}\:{uniform} \\$$$${rope}\:{such}\:{that}\:{it}\:{can}\:{stay}\:{in}\:{equilibrium} \\$$$${as}\:{shown}. \\$$
Answered by mr W last updated on 13/Aug/19
Commented by mr W last updated on 14/Aug/19
$$\rho={mass}\:{of}\:{unit}\:{length}\:{of}\:{rope} \\$$$${total}\:{length}\:{of}\:{rope}\:{L}={s}+\mathrm{2}{h} \\$$$${tension}\:{in}\:{rope}\:{at}\:{point}\:{A}\:{and}\:{B}: \\$$$${T}=\rho{gh} \\$$$${horizontal}\:{component}\:{of}\:{tension}: \\$$$${T}_{\mathrm{0}} ={T}\:\:\mathrm{cos}\:\theta=\rho{gh}\:\mathrm{cos}\:\theta \\$$$${a}=\frac{{T}_{\mathrm{0}} }{\rho{g}}={h}\:\mathrm{cos}\:\theta \\$$$${y}={a}\:\mathrm{cosh}\:\frac{{x}}{{a}} \\$$$${y}'=\mathrm{sinh}\:\frac{{x}}{{a}} \\$$$${at}\:{point}\:{B}: \\$$$${y}'=\mathrm{sinh}\:\frac{{b}}{\mathrm{2}{a}}=\mathrm{tan}\:\theta \\$$$$\Rightarrow\frac{{b}}{\mathrm{2}{a}}=\mathrm{sinh}^{−\mathrm{1}} \:\mathrm{tan}\:\theta \\$$$$\frac{{s}}{\mathrm{2}}={a}\:\mathrm{sinh}\:\frac{{b}}{\mathrm{2}{a}}={a}\:\mathrm{tan}\:\theta \\$$$$\Rightarrow{s}=\mathrm{2}{a}\:\mathrm{tan}\:\theta \\$$$${h}=\frac{{a}}{\mathrm{cos}\:\theta} \\$$$${L}={s}+\mathrm{2}{h}=\mathrm{2}{a}\left(\mathrm{tan}\:\theta+\frac{\mathrm{1}}{\mathrm{cos}\:\theta}\right) \\$$$$\Rightarrow\frac{{L}}{{b}}=\frac{\mathrm{2}{a}}{{b}}\left(\mathrm{tan}\:\theta+\frac{\mathrm{1}}{\mathrm{cos}\:\theta}\right) \\$$$$\Rightarrow\frac{{L}}{{b}}=\frac{\mathrm{1}}{\mathrm{sinh}^{−\mathrm{1}} \:\mathrm{tan}\:\theta}\left(\mathrm{tan}\:\theta+\frac{\mathrm{1}}{\mathrm{cos}\:\theta}\right) \\$$$$\Rightarrow\frac{{L}}{{b}}=\frac{\mathrm{tan}\:\theta+\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\theta}}{\mathrm{sinh}^{−\mathrm{1}} \:\mathrm{tan}\:\theta} \\$$$${let}\:{t}=\mathrm{tan}\:\theta \\$$$$\Rightarrow\frac{{L}}{{b}}=\frac{\mathrm{t}+\sqrt{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }}{\mathrm{sinh}^{−\mathrm{1}} \:\mathrm{t}}=\frac{{t}+\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}{\mathrm{ln}\:\left({t}+\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\right)} \\$$$${let}\:{p}={t}+\sqrt{\mathrm{1}+{t}^{\mathrm{2}} } \\$$$$\Rightarrow\frac{{L}}{{b}}=\frac{{p}}{\mathrm{ln}\:{p}} \\$$$$\frac{{d}}{{dp}}\left(\frac{{L}}{{b}}\right)=\frac{\mathrm{1}}{\mathrm{ln}\:{p}}−\frac{\mathrm{1}}{\left(\mathrm{ln}\:{p}\right)^{\mathrm{2}} }=\mathrm{0} \\$$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{ln}\:{p}}=\mathrm{1}\:\Rightarrow{p}={e} \\$$$$\Rightarrow\frac{{L}_{{min}} }{{b}}=\frac{{e}}{\mathrm{ln}\:{e}}={e} \\$$$$\Rightarrow{L}_{{min}} ={eb} \\$$$${if}\:{b}=\mathrm{1},\:{L}_{{min}} ={e}\:\approx\mathrm{2}.\mathrm{718} \\$$$$\\$$$${t}+\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }={e} \\$$$$\mathrm{1}={e}^{\mathrm{2}} −\mathrm{2}{et} \\$$$$\Rightarrow{t}=\mathrm{tan}\:\theta=\frac{{e}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}{e}} \\$$$$\Rightarrow\theta=\mathrm{tan}^{−\mathrm{1}} \frac{{e}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}{e}}=\mathrm{49}.\mathrm{605}° \\$$$$\\$$$$\frac{{h}}{{b}}=\frac{\mathrm{2}{a}}{{b}}×\frac{\mathrm{1}}{\mathrm{2cos}\:\theta}=\frac{\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\theta}}{\mathrm{2ln}\:\left(\mathrm{tan}\:\theta+\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\theta}\right)} \\$$$$=\frac{{e}^{\mathrm{2}} +\mathrm{1}}{\mathrm{4}{e}}=\mathrm{0}.\mathrm{7715} \\$$