# Question-68941

Question Number 68941 by ahmadshah last updated on 17/Sep/19
Commented by mr W last updated on 17/Sep/19
$${what}\:{if}\:{the}\:{cable}\:{is}\:\mathrm{100}\:{m}\:{long}? \\$$
Commented by MJS last updated on 18/Sep/19
$$\mathrm{there}'\mathrm{s}\:\mathrm{a}\:\mathrm{formula} \\$$$${w}=\frac{\left(\mathrm{4}{h}^{\mathrm{2}} −{l}^{\mathrm{2}} \right)}{\mathrm{4}{h}}\mathrm{ln}\:\frac{{l}−\mathrm{2}{h}}{{l}+\mathrm{2}{h}} \\$$$$\mathrm{in}\:\mathrm{our}\:\mathrm{case}\:{h}=\mathrm{40};\:{l}=\mathrm{80}\:\Rightarrow\:{w}=\mathrm{0} \\$$$$\mathrm{for}\:{h}=\mathrm{40};\:{l}=\mathrm{100}\:\mathrm{we}\:\mathrm{get}\:{w}=\mathrm{45ln}\:\mathrm{3}\:\approx\mathrm{49}.\mathrm{438}{m} \\$$
Commented by mr W last updated on 18/Sep/19
$${very}\:{right}\:{sir}!\:{thanks}! \\$$$${i}\:{got}\:{the}\:{formula},\:{see}\:{below}. \\$$
$$\mathrm{we}\:\mathrm{had}\:\mathrm{this}\:\mathrm{before} \\$$$$\mathrm{if}\:\mathrm{the}\:\mathrm{cable}\:\mathrm{is}\:\mathrm{2}{l}\:\mathrm{long}\:\mathrm{and}\:\mathrm{the}\:\mathrm{height}\:\mathrm{of}\:\mathrm{the} \\$$$$\mathrm{curve}\:\mathrm{is}\:{l}\:\Rightarrow\:\mathrm{the}\:\mathrm{distance}\:\mathrm{of}\:\mathrm{the}\:\mathrm{poles}\:=\mathrm{0} \\$$
$${y}={a}\:\mathrm{cosh}\:\frac{{x}}{{a}} \\$$$${a}+{h}={a}\:\mathrm{cosh}\:\frac{{w}}{\mathrm{2}{a}} \\$$$$\frac{{h}}{{a}}+\mathrm{1}=\mathrm{cosh}\:\frac{{w}}{\mathrm{2}{a}}\:\:\:…\left({i}\right) \\$$$$\\$$$$\frac{{l}}{\mathrm{2}}={a}\:\mathrm{sinh}\:\frac{{w}}{\mathrm{2}{a}}\: \\$$$$\frac{{l}}{\mathrm{2}{a}}=\mathrm{sinh}\:\frac{{w}}{\mathrm{2}{a}}\:\:\:…\left({ii}\right) \\$$$$\left({i}\right)^{\mathrm{2}} −\left({ii}\right)^{\mathrm{2}} : \\$$$$\left(\frac{{h}}{{a}}+\mathrm{1}\right)^{\mathrm{2}} −\left(\frac{{l}}{\mathrm{2}{a}}\right)^{\mathrm{2}} =\mathrm{1} \\$$$$\left(\mathrm{2}{h}+\mathrm{2}{a}\right)^{\mathrm{2}} −{l}^{\mathrm{2}} =\mathrm{4}{a}^{\mathrm{2}} \\$$$$\mathrm{8}{ha}={l}^{\mathrm{2}} −\mathrm{4}{h}^{\mathrm{2}} \\$$$$\Rightarrow{a}=\frac{{l}^{\mathrm{2}} −\mathrm{4}{h}^{\mathrm{2}} }{\mathrm{8}{h}} \\$$$$\frac{\mathrm{4}{hl}}{{l}^{\mathrm{2}} −\mathrm{4}{h}^{\mathrm{2}} }=\mathrm{sinh}\:\frac{{w}}{\mathrm{2}{a}} \\$$$$\Rightarrow\frac{{w}}{\mathrm{2}{a}}=\mathrm{sinh}^{−\mathrm{1}} \:\frac{\mathrm{4}{hl}}{{l}^{\mathrm{2}} −\mathrm{4}{h}^{\mathrm{2}} } \\$$$$\Rightarrow\frac{{w}}{{l}}=\frac{{l}^{\mathrm{2}} −\mathrm{4}{h}^{\mathrm{2}} }{\mathrm{4}{hl}}\mathrm{sinh}^{−\mathrm{1}} \:\frac{\mathrm{4}{hl}}{{l}^{\mathrm{2}} −\mathrm{4}{h}^{\mathrm{2}} } \\$$$${with}\:\lambda=\frac{\mathrm{4}{hl}}{{l}^{\mathrm{2}} −\mathrm{4}{h}^{\mathrm{2}} } \\$$$$\Rightarrow\frac{{w}}{{l}}=\frac{\mathrm{sinh}^{−\mathrm{1}} \:\lambda}{\lambda}=\frac{\mathrm{ln}\:\left(\lambda+\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }\right)}{\lambda} \\$$$$\\$$$$\lambda+\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }=\frac{\mathrm{4}{hl}}{{l}^{\mathrm{2}} −\mathrm{4}{h}^{\mathrm{2}} }+\sqrt{\mathrm{1}+\frac{\left(\mathrm{4}{hl}\right)^{\mathrm{2}} }{\left({l}^{\mathrm{2}} −\mathrm{4}{h}^{\mathrm{2}} \right)^{\mathrm{2}} }} \\$$$$=\frac{\mathrm{4}{hl}+\left({l}^{\mathrm{2}} +\mathrm{4}{h}^{\mathrm{2}} \right)}{{l}^{\mathrm{2}} −\mathrm{4}{h}^{\mathrm{2}} }=\frac{\left({l}+\mathrm{2}{h}\right)^{\mathrm{2}} }{\left({l}−\mathrm{2}{h}\right)\left(\mathrm{1}+\mathrm{2}{h}\right)}=\frac{{l}+\mathrm{2}{h}}{{l}−\mathrm{2}{h}} \\$$$$\Rightarrow{w}=\frac{{l}^{\mathrm{2}} −\mathrm{4}{h}^{\mathrm{2}} }{\mathrm{4}{h}}\mathrm{ln}\:\frac{{l}+\mathrm{2}{h}}{{l}−\mathrm{2}{h}} \\$$$$\\$$$${example}: \\$$$${h}=\mathrm{40},\:{l}=\mathrm{100} \\$$$$\Rightarrow\lambda=\frac{\mathrm{4}×\mathrm{40}×\mathrm{100}}{\mathrm{100}^{\mathrm{2}} −\mathrm{4}×\mathrm{40}^{\mathrm{2}} }=\frac{\mathrm{40}}{\mathrm{9}} \\$$$$\frac{{w}}{{l}}=\frac{\mathrm{sinh}^{−\mathrm{1}} \:\frac{\mathrm{40}}{\mathrm{9}}}{\frac{\mathrm{40}}{\mathrm{9}}}=\mathrm{0}.\mathrm{4944} \\$$$$\Rightarrow{w}=\mathrm{0}.\mathrm{4944}×\mathrm{100}=\mathrm{49}.\mathrm{44}{m} \\$$
$$\mathrm{great}! \\$$