Question and Answers Forum

All Questions      Topic List

Geometry Questions

Previous in All Question      Next in All Question      

Previous in Geometry      Next in Geometry      

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 100 m long?

$${what}\:{if}\:{the}\:{cable}\:{is}\:\mathrm{100}\:{m}\:{long}? \\ $$

Commented by MJS last updated on 18/Sep/19

there′s a formula  w=(((4h^2 −l^2 ))/(4h))ln ((l−2h)/(l+2h))  in our case h=40; l=80 ⇒ w=0  for h=40; l=100 we get w=45ln 3 ≈49.438m

$$\mathrm{there}'\mathrm{s}\:\mathrm{a}\:\mathrm{formula} \\ $$$${w}=\frac{\left(\mathrm{4}{h}^{\mathrm{2}} −{l}^{\mathrm{2}} \right)}{\mathrm{4}{h}}\mathrm{ln}\:\frac{{l}−\mathrm{2}{h}}{{l}+\mathrm{2}{h}} \\ $$$$\mathrm{in}\:\mathrm{our}\:\mathrm{case}\:{h}=\mathrm{40};\:{l}=\mathrm{80}\:\Rightarrow\:{w}=\mathrm{0} \\ $$$$\mathrm{for}\:{h}=\mathrm{40};\:{l}=\mathrm{100}\:\mathrm{we}\:\mathrm{get}\:{w}=\mathrm{45ln}\:\mathrm{3}\:\approx\mathrm{49}.\mathrm{438}{m} \\ $$

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

very right sir! thanks!  i got the formula, see below.

$${very}\:{right}\:{sir}!\:{thanks}! \\ $$$${i}\:{got}\:{the}\:{formula},\:{see}\:{below}. \\ $$

Answered by ahmadshah last updated on 17/Sep/19

Answered by MJS last updated on 17/Sep/19

we had this before  if the cable is 2l long and the height of the  curve is l ⇒ the distance of the poles =0

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

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

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

y=a cosh (x/a)  a+h=a cosh (w/(2a))  (h/a)+1=cosh (w/(2a))   ...(i)    (l/2)=a sinh (w/(2a))   (l/(2a))=sinh (w/(2a))   ...(ii)  (i)^2 −(ii)^2 :  ((h/a)+1)^2 −((l/(2a)))^2 =1  (2h+2a)^2 −l^2 =4a^2   8ha=l^2 −4h^2   ⇒a=((l^2 −4h^2 )/(8h))  ((4hl)/(l^2 −4h^2 ))=sinh (w/(2a))  ⇒(w/(2a))=sinh^(−1)  ((4hl)/(l^2 −4h^2 ))  ⇒(w/l)=((l^2 −4h^2 )/(4hl))sinh^(−1)  ((4hl)/(l^2 −4h^2 ))  with λ=((4hl)/(l^2 −4h^2 ))  ⇒(w/l)=((sinh^(−1)  λ)/λ)=((ln (λ+(√(1+λ^2 ))))/λ)    λ+(√(1+λ^2 ))=((4hl)/(l^2 −4h^2 ))+(√(1+(((4hl)^2 )/((l^2 −4h^2 )^2 ))))  =((4hl+(l^2 +4h^2 ))/(l^2 −4h^2 ))=(((l+2h)^2 )/((l−2h)(1+2h)))=((l+2h)/(l−2h))  ⇒w=((l^2 −4h^2 )/(4h))ln ((l+2h)/(l−2h))    example:  h=40, l=100  ⇒λ=((4×40×100)/(100^2 −4×40^2 ))=((40)/9)  (w/l)=((sinh^(−1)  ((40)/9))/((40)/9))=0.4944  ⇒w=0.4944×100=49.44m

$${y}={a}\:\mathrm{cosh}\:\frac{{x}}{{a}} \\ $$$${a}+{h}={a}\:\mathrm{cosh}\:\frac{{w}}{\mathrm{2}{a}} \\ $$$$\frac{{h}}{{a}}+\mathrm{1}=\mathrm{cosh}\:\frac{{w}}{\mathrm{2}{a}}\:\:\:...\left({i}\right) \\ $$$$ \\ $$$$\frac{{l}}{\mathrm{2}}={a}\:\mathrm{sinh}\:\frac{{w}}{\mathrm{2}{a}}\: \\ $$$$\frac{{l}}{\mathrm{2}{a}}=\mathrm{sinh}\:\frac{{w}}{\mathrm{2}{a}}\:\:\:...\left({ii}\right) \\ $$$$\left({i}\right)^{\mathrm{2}} −\left({ii}\right)^{\mathrm{2}} : \\ $$$$\left(\frac{{h}}{{a}}+\mathrm{1}\right)^{\mathrm{2}} −\left(\frac{{l}}{\mathrm{2}{a}}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$$\left(\mathrm{2}{h}+\mathrm{2}{a}\right)^{\mathrm{2}} −{l}^{\mathrm{2}} =\mathrm{4}{a}^{\mathrm{2}} \\ $$$$\mathrm{8}{ha}={l}^{\mathrm{2}} −\mathrm{4}{h}^{\mathrm{2}} \\ $$$$\Rightarrow{a}=\frac{{l}^{\mathrm{2}} −\mathrm{4}{h}^{\mathrm{2}} }{\mathrm{8}{h}} \\ $$$$\frac{\mathrm{4}{hl}}{{l}^{\mathrm{2}} −\mathrm{4}{h}^{\mathrm{2}} }=\mathrm{sinh}\:\frac{{w}}{\mathrm{2}{a}} \\ $$$$\Rightarrow\frac{{w}}{\mathrm{2}{a}}=\mathrm{sinh}^{−\mathrm{1}} \:\frac{\mathrm{4}{hl}}{{l}^{\mathrm{2}} −\mathrm{4}{h}^{\mathrm{2}} } \\ $$$$\Rightarrow\frac{{w}}{{l}}=\frac{{l}^{\mathrm{2}} −\mathrm{4}{h}^{\mathrm{2}} }{\mathrm{4}{hl}}\mathrm{sinh}^{−\mathrm{1}} \:\frac{\mathrm{4}{hl}}{{l}^{\mathrm{2}} −\mathrm{4}{h}^{\mathrm{2}} } \\ $$$${with}\:\lambda=\frac{\mathrm{4}{hl}}{{l}^{\mathrm{2}} −\mathrm{4}{h}^{\mathrm{2}} } \\ $$$$\Rightarrow\frac{{w}}{{l}}=\frac{\mathrm{sinh}^{−\mathrm{1}} \:\lambda}{\lambda}=\frac{\mathrm{ln}\:\left(\lambda+\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }\right)}{\lambda} \\ $$$$ \\ $$$$\lambda+\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }=\frac{\mathrm{4}{hl}}{{l}^{\mathrm{2}} −\mathrm{4}{h}^{\mathrm{2}} }+\sqrt{\mathrm{1}+\frac{\left(\mathrm{4}{hl}\right)^{\mathrm{2}} }{\left({l}^{\mathrm{2}} −\mathrm{4}{h}^{\mathrm{2}} \right)^{\mathrm{2}} }} \\ $$$$=\frac{\mathrm{4}{hl}+\left({l}^{\mathrm{2}} +\mathrm{4}{h}^{\mathrm{2}} \right)}{{l}^{\mathrm{2}} −\mathrm{4}{h}^{\mathrm{2}} }=\frac{\left({l}+\mathrm{2}{h}\right)^{\mathrm{2}} }{\left({l}−\mathrm{2}{h}\right)\left(\mathrm{1}+\mathrm{2}{h}\right)}=\frac{{l}+\mathrm{2}{h}}{{l}−\mathrm{2}{h}} \\ $$$$\Rightarrow{w}=\frac{{l}^{\mathrm{2}} −\mathrm{4}{h}^{\mathrm{2}} }{\mathrm{4}{h}}\mathrm{ln}\:\frac{{l}+\mathrm{2}{h}}{{l}−\mathrm{2}{h}} \\ $$$$ \\ $$$${example}: \\ $$$${h}=\mathrm{40},\:{l}=\mathrm{100} \\ $$$$\Rightarrow\lambda=\frac{\mathrm{4}×\mathrm{40}×\mathrm{100}}{\mathrm{100}^{\mathrm{2}} −\mathrm{4}×\mathrm{40}^{\mathrm{2}} }=\frac{\mathrm{40}}{\mathrm{9}} \\ $$$$\frac{{w}}{{l}}=\frac{\mathrm{sinh}^{−\mathrm{1}} \:\frac{\mathrm{40}}{\mathrm{9}}}{\frac{\mathrm{40}}{\mathrm{9}}}=\mathrm{0}.\mathrm{4944} \\ $$$$\Rightarrow{w}=\mathrm{0}.\mathrm{4944}×\mathrm{100}=\mathrm{49}.\mathrm{44}{m} \\ $$

Commented by MJS last updated on 19/Sep/19

great!

$$\mathrm{great}! \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com