Question Number 66527 by mr W last updated on 16/Aug/19
Commented by mr W last updated on 16/Aug/19
$${the}\:{perimeter}\:{of}\:{a}\:{rope}\:{loop}\:{is}\:{L}. \\ $$$${now}\:{it}\:{is}\:{hanged}\:{on}\:{two}\:{pins}\:{A}\:{and}\:{B}. \\ $$$${the}\:{distance}\:{between}\:{the}\:{pins}\:{is}\:{b}.\: \\ $$$${all}\:{contact}\:{is}\:{frictionless}. \\ $$$$\left({L}>\mathrm{2}{b}\right) \\ $$$${if}\:{the}\:{lowest}\:{point}\:{of}\:{the}\:{loop}\:{is}\:{h} \\ $$$${below}\:{the}\:{pins}. \\ $$$${find}\:{h}\:{in}\:{terms}\:{of}\:{b}\:{and}\:{L}. \\ $$
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
$${when}\:{the}\:{rope}\:{loop}\:{is}\:{hanging}\:{on}\:{the} \\ $$$${pins},\:{it}\:{consists}\:{of}\:\mathrm{2}\:{parts}\:{S}_{\mathrm{1}} \:{and}\:{S}_{\mathrm{2}} . \\ $$$${S}_{\mathrm{1}} +{S}_{\mathrm{2}} ={L} \\ $$$$ \\ $$$${generally}\:{for}\:{a}\:{rope}\:{between}\:{A}\:{and}\:{B}: \\ $$$${mass}\:{of}\:{rope}\:{of}\:{unit}\:{length}=\:\rho \\ $$$${tension}\:{in}\:{rope}\:{at}\:{point}\:{A}\:\left({B}\right)=\:{T} \\ $$$${length}\:{of}\:{rope}\:{from}\:{A}\:{to}\:{B}\:={S} \\ $$$${T}_{\mathrm{0}} ={T}\:\mathrm{cos}\:\theta \\ $$$${a}=\frac{{T}_{\mathrm{0}} }{\rho{g}}=\frac{{T}\:\mathrm{cos}\:\theta}{\rho{g}} \\ $$$${y}={a}\:\mathrm{cosh}\:\frac{{x}}{{a}} \\ $$$${y}'=\mathrm{sinh}\:\frac{{x}}{{a}} \\ $$$${at}\:{point}\:{B}: \\ $$$${y}'=\mathrm{sinh}\:\frac{{b}}{\mathrm{2}{a}}=\mathrm{tan}\:\theta={t} \\ $$$$\Rightarrow\frac{{b}}{\mathrm{2}{a}}=\mathrm{sinh}^{−\mathrm{1}} \:{t}=\mathrm{ln}\:\left({t}+\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\right) \\ $$$$\frac{{S}}{\mathrm{2}}={a}\:\mathrm{sinh}\:\frac{{b}}{\mathrm{2}{a}}={a}\:\mathrm{tan}\:\theta={at} \\ $$$$\Rightarrow{S}=\mathrm{2}{at}=\frac{\mathrm{2}{a}}{{b}}×{bt}=\frac{{bt}}{\mathrm{sinh}^{−\mathrm{1}} \:{t}}=\frac{{bt}}{\mathrm{ln}\:\left({t}+\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\right)} \\ $$$$\Rightarrow\frac{{T}}{\rho{g}}=\frac{{a}}{\mathrm{cos}\:\theta}=\frac{\mathrm{2}{a}}{{b}}×\frac{{b}\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\theta}}{\mathrm{2}}=\frac{{b}\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}{\mathrm{2}\:\mathrm{ln}\:\left({t}+\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\right)} \\ $$$${h}={a}\left(\mathrm{1}+\mathrm{cosh}\:\frac{{b}}{\mathrm{2}{a}}\right)=\frac{{b}}{\mathrm{2}}×\frac{\mathrm{2}{a}}{{b}}\left(\mathrm{1}+\mathrm{cosh}\:\frac{{b}}{\mathrm{2}{a}}\right) \\ $$$$\Rightarrow\frac{\mathrm{2}{h}}{{b}}=\frac{\mathrm{2}{a}}{{b}}\left(\mathrm{1}+\mathrm{cosh}\:\frac{{b}}{\mathrm{2}{a}}\right)=\frac{\mathrm{1}+\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}{\mathrm{ln}\:\left({t}+\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\right)} \\ $$$$ \\ $$$${for}\:{rope}\:{part}\:\mathrm{1}: \\ $$$${t}_{\mathrm{1}} =\mathrm{tan}\:\theta_{\mathrm{1}} \\ $$$${S}_{\mathrm{1}} =\frac{{bt}_{\mathrm{1}} }{\mathrm{ln}\:\left({t}_{\mathrm{1}} +\sqrt{\mathrm{1}+{t}_{\mathrm{1}} ^{\mathrm{2}} }\right)} \\ $$$$\frac{{T}_{\mathrm{1}} }{\rho{g}}=\frac{{b}\sqrt{\mathrm{1}+{t}_{\mathrm{1}} ^{\mathrm{2}} }}{\mathrm{2}\:\mathrm{ln}\:\left({t}_{\mathrm{1}} +\sqrt{\mathrm{1}+{t}_{\mathrm{1}} ^{\mathrm{2}} }\right)} \\ $$$${for}\:{rope}\:{part}\:\mathrm{2}: \\ $$$${t}_{\mathrm{2}} =\mathrm{tan}\:\theta_{\mathrm{2}} \\ $$$${S}_{\mathrm{2}} =\frac{{bt}_{\mathrm{2}} }{\mathrm{ln}\:\left({t}_{\mathrm{2}} +\sqrt{\mathrm{1}+{t}_{\mathrm{2}} ^{\mathrm{2}} }\right)} \\ $$$$\frac{{T}_{\mathrm{2}} }{\rho{g}}=\frac{{b}\sqrt{\mathrm{1}+{t}_{\mathrm{2}} ^{\mathrm{2}} }}{\mathrm{2}\:\mathrm{ln}\:\left({t}_{\mathrm{2}} +\sqrt{\mathrm{1}+{t}_{\mathrm{2}} ^{\mathrm{2}} }\right)} \\ $$$$ \\ $$$${T}_{\mathrm{1}} ={T}_{\mathrm{2}} \\ $$$$\frac{{b}\sqrt{\mathrm{1}+{t}_{\mathrm{1}} ^{\mathrm{2}} }}{\mathrm{2}\:\mathrm{ln}\:\left({t}_{\mathrm{1}} +\sqrt{\mathrm{1}+{t}_{\mathrm{1}} ^{\mathrm{2}} }\right)}=\frac{{b}\sqrt{\mathrm{1}+{t}_{\mathrm{2}} ^{\mathrm{2}} }}{\mathrm{2}\:\mathrm{ln}\:\left({t}_{\mathrm{2}} +\sqrt{\mathrm{1}+{t}_{\mathrm{2}} ^{\mathrm{2}} }\right)} \\ $$$$\Rightarrow\frac{\sqrt{\mathrm{1}+{t}_{\mathrm{1}} ^{\mathrm{2}} }}{\mathrm{ln}\:\left({t}_{\mathrm{1}} +\sqrt{\mathrm{1}+{t}_{\mathrm{1}} ^{\mathrm{2}} }\right)}=\frac{\sqrt{\mathrm{1}+{t}_{\mathrm{2}} ^{\mathrm{2}} }}{\mathrm{ln}\:\left({t}_{\mathrm{2}} +\sqrt{\mathrm{1}+{t}_{\mathrm{2}} ^{\mathrm{2}} }\right)}\:\:\:\:…\left({i}\right) \\ $$$${S}_{\mathrm{1}} +{S}_{\mathrm{2}} ={L} \\ $$$$\frac{{bt}_{\mathrm{1}} }{\mathrm{ln}\:\left({t}_{\mathrm{1}} +\sqrt{\mathrm{1}+{t}_{\mathrm{1}} ^{\mathrm{2}} }\right)}+\frac{{bt}_{\mathrm{2}} }{\mathrm{ln}\:\left({t}_{\mathrm{2}} +\sqrt{\mathrm{1}+{t}_{\mathrm{2}} ^{\mathrm{2}} }\right)}={L} \\ $$$${let}\:\lambda=\frac{{L}}{{b}} \\ $$$$\Rightarrow\:\frac{{t}_{\mathrm{1}} }{\mathrm{ln}\:\left({t}_{\mathrm{1}} +\sqrt{\mathrm{1}+{t}_{\mathrm{1}} ^{\mathrm{2}} }\right)}+\frac{{t}_{\mathrm{2}} }{\mathrm{ln}\:\left({t}_{\mathrm{2}} +\sqrt{\mathrm{1}+{t}_{\mathrm{2}} ^{\mathrm{2}} }\right)}=\lambda\:\:\:…\left({ii}\right) \\ $$$$ \\ $$$${t}_{\mathrm{1}} ={t}_{\mathrm{2}} ={t}\:{is}\:{always}\:{a}\:{solution}\:{of}\:\left({i}\right), \\ $$$$\Rightarrow\:\frac{{t}}{\mathrm{ln}\:\left({t}+\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\right)}=\frac{\lambda}{\mathrm{2}} \\ $$$${in}\:{this}\:{case}\:{both}\:{rope}\:{parts}\:{are}\:{equal}:\: \\ $$$${S}_{\mathrm{1}} ={S}_{\mathrm{2}} =\frac{{L}}{\mathrm{2}} \\ $$$$ \\ $$$${in}\:{the}\:{diagram}\:{the}\:{green}\:{curve}\:{shows} \\ $$$${the}\:{equation}\:\left({i}\right).\:{the}\:{red}\:{curve}\:{shows} \\ $$$${the}\:{equation}\:\left({ii}\right)\:{for}\:{different}\:{values} \\ $$$${of}\:\lambda.\:{we}\:{can}\:{see}\:{that}\:{there}\:{is}\:{only}\:{one} \\ $$$${solution}\:{for}\:{the}\:{shape}\:{of}\:{the}\:{rope}\:{if} \\ $$$$\lambda\leqslant\mathrm{2}.\mathrm{5155},\:{with}\:{t}_{\mathrm{1}} ={t}_{\mathrm{2}} ,\:{i}.{e}.\:{S}_{\mathrm{1}} ={S}_{\mathrm{2}} . \\ $$$${but}\:{for}\:\lambda>\mathrm{2}.\mathrm{5155}\:{there}\:{are}\:{two} \\ $$$${possible}\:{shapes},\:{one}\:{with}\:{t}_{\mathrm{1}} ={t}_{\mathrm{2}} ,\:{i}.{e}. \\ $$$${S}_{\mathrm{1}} ={S}_{\mathrm{2}} \:{and}\:{the}\:{other}\:{one}\:{with}\:{t}_{\mathrm{1}} \neq{t}_{\mathrm{2}} , \\ $$$${i}.{e}.\:{S}_{\mathrm{1}} \neq{S}_{\mathrm{2}} . \\ $$$${for}\:{a}\:{given}\:{value}\:{of}\:\lambda=\frac{{L}}{{b}},\:{numerical} \\ $$$${solutions}\:{of}\:{eqn}.\:\left({i}\right)\:{and}\:\left({ii}\right)\:{can}\:{be} \\ $$$${obtained}. \\ $$
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
$${please}\:{i}\:{want}\:{the}\:{name}\: \\ $$$$\:{of}\:{that}\:{application} \\ $$
Commented by mr W last updated on 17/Aug/19
$${for}\:{the}\:{graph}\:{i}\:{used}\:{an}\:{app}\:{called} \\ $$$${Grapher}. \\ $$