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Question Number 68941 by ahmadshah last updated on 17/Sep/19

Commented by mr W last updated on 17/Sep/19

$$whatifthecableis100mlong?$$

Commented by MJS last updated on 18/Sep/19

$${\mathrm{there}}^{\prime }\mathrm{s}\mathrm{a}\mathrm{formula}$$$$w=\frac{(4h^2-l^2)}{4h}\ln \frac{l-2h}{l+2h}$$$$\mathrm{in}\mathrm{our}\mathrm{case}h=40;l=80\Rightarrow w=0$$$$\mathrm{for}h=40;l=100\mathrm{we}\mathrm{get}w=45\ln 3\approx 49.438m$$

Commented by mr W last updated on 18/Sep/19

$$veryrightsir!thanks!$$$$igottheformula,seebelow.$$

Answered by ahmadshah last updated on 17/Sep/19

Answered by MJS last updated on 17/Sep/19

$$\mathrm{we}\mathrm{had}\mathrm{this}\mathrm{before}$$$$\mathrm{if}\mathrm{the}\mathrm{cable}\mathrm{is}2l\mathrm{long}\mathrm{and}\mathrm{the}\mathrm{height}\mathrm{of}\mathrm{the}$$$$\mathrm{curve}\mathrm{is}l\Rightarrow \mathrm{the}\mathrm{distance}\mathrm{of}\mathrm{the}\mathrm{poles}=0$$

Answered by mr W last updated on 18/Sep/19

Commented by mr W last updated on 18/Sep/19

$$y=a\cosh \frac{x}{a}$$$$a+h=a\cosh \frac{w}{2a}$$$$\frac{h}{a}+1=\cosh \frac{w}{2a}...\left(i\right)$$$$$$$$\frac{l}{2}=a\sinh \frac{w}{2a}$$$$\frac{l}{2a}=\sinh \frac{w}{2a}...\left(ii\right)$$$${\left(i\right)}^2-{\left(ii\right)}^2:$$$${(\frac{h}{a}+1)}^2-{\left(\frac{l}{2a}\right)}^2=1$$$${(2h+2a)}^2-l^2=4a^2$$$$8ha=l^2-4h^2$$$$\Rightarrow a=\frac{l^2-4h^2}{8h}$$$$\frac{4hl}{l^2-4h^2}=\sinh \frac{w}{2a}$$$$\Rightarrow \frac{w}{2a}={\sinh }^{-1}\frac{4hl}{l^2-4h^2}$$$$\Rightarrow \frac{w}{l}=\frac{l^2-4h^2}{4hl}{\sinh }^{-1}\frac{4hl}{l^2-4h^2}$$$$with\lambda =\frac{4hl}{l^2-4h^2}$$$$\Rightarrow \frac{w}{l}=\frac{{\sinh }^{-1}\lambda }{\lambda }=\frac{\ln (\lambda +\sqrt{1+{\lambda }^2})}{\lambda }$$$$$$$$\lambda +\sqrt{1+{\lambda }^2}=\frac{4hl}{l^2-4h^2}+\sqrt{1+\frac{{\left(4hl\right)}^2}{{(l^2-4h^2)}^2}}$$$$=\frac{4hl+(l^2+4h^2)}{l^2-4h^2}=\frac{{(l+2h)}^2}{(l-2h)(1+2h)}=\frac{l+2h}{l-2h}$$$$\Rightarrow w=\frac{l^2-4h^2}{4h}\ln \frac{l+2h}{l-2h}$$$$$$$$example:$$$$h=40,l=100$$$$\Rightarrow \lambda =\frac{4\times 40\times 100}{{100}^2-4\times {40}^2}=\frac{40}{9}$$$$\frac{w}{l}=\frac{{\sinh }^{-1}\frac{40}{9}}{\frac{40}{9}}=0.4944$$$$\Rightarrow w=0.4944\times 100=49.44m$$

Commented by MJS last updated on 19/Sep/19

$$\mathrm{great}!$$

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