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Question Number 61762 by mr W last updated on 08/Jun/19

Commented by mr W last updated on 08/Jun/19

$(s e e a l s o Q 61612)$ $F i n d t h e r e l a t i o n b e t w e e n h, d a n d L$ $o f a h a n g i n g r o p e .$
	Answered by mr W last updated on 09/Jun/19
	$I. form of rope as catenary y=a cosh (x/a) at x=(d/2): h+a=a cosh (d/(2a)) ⇒(1+(h/a))=cosh (d/(2a)) ...(i) (L/2)=a sinh (d/(2a)) ⇒(L/(2a))=sinh (d/(2a)) ...(ii) ⇒(1+(h/a))^2 −((L/(2a)))^2 =1 ⇒8ha+4h^2 −L^2 =0 ⇒a=((L^2 −4h^2 )/(8h)) from (i): sinh (d/(2a))=((4Lh)/(L^2 −4h^2 )) ⇒d={(((L^2 −4h^2 )/(4Lh))) sinh^(−1) (((4Lh)/(L^2 −4h^2 )))}L with κ=((L^2 −4h^2 )/(4Lh)) ⇒d=(κ sinh^(−1) (1/κ))L example: L=25m, h=7.5m ⇒κ=((25^2 −4×7.5^2 )/(4×25×7.5))=(8/(15)) ⇒d≈0.73936L=18.48 m let η=((2h)/d), λ=(L/d) η×(d/(2a))=cosh (d/(2a))−1 ⇒η=((cosh (d/(2a))−1)/(d/(2a))) λ×(d/(2a))=sinh (d/(2a)) ⇒λ=((sinh (d/(2a)))/(d/(2a))) with parameter t=(d/(2a)) we get { ((η=((cosh t−1)/t))),((λ=((sinh t)/t))) :}$
	$I . f o r m o f r o p e a s c a t e n a r y$ $y = a \cosh \frac{x}{a}$ $a t x = \frac{d}{2} :$ $h + a = a \cosh \frac{d}{2 a}$ $\Rightarrow (1 + \frac{h}{a}) = \cosh \frac{d}{2 a} . . . (i)$ $\frac{L}{2} = a \sinh \frac{d}{2 a}$ $\Rightarrow \frac{L}{2 a} = \sinh \frac{d}{2 a} . . . (i i)$ $\Rightarrow {(1 + \frac{h}{a})}^{2} - {(\frac{L}{2 a})}^{2} = 1$ $\Rightarrow 8 h a + 4 h^{2} - L^{2} = 0$ $\Rightarrow a = \frac{L^{2} - 4 h^{2}}{8 h}$ $f r o m (i) :$ $\sinh \frac{d}{2 a} = \frac{4 L h}{L^{2} - 4 h^{2}}$ $\Rightarrow d = {(\frac{L^{2} - 4 h^{2}}{4 L h}) \sinh^{- 1} (\frac{4 L h}{L^{2} - 4 h^{2}})} L$ $w i t h κ = \frac{L^{2} - 4 h^{2}}{4 L h}$ $\Rightarrow d = (κ \sinh^{- 1} \frac{1}{κ}) L$ $e x a m p l e : L = 25 m, h = 7.5 m$ $\Rightarrow κ = \frac{25^{2} - 4 \times 7 . 5^{2}}{4 \times 25 \times 7.5} = \frac{8}{15}$ $\Rightarrow d \approx 0.73936 L = 18.48 m$ $l e t η = \frac{2 h}{d}, λ = \frac{L}{d}$ $η \times \frac{d}{2 a} = \cosh \frac{d}{2 a} - 1$ $\Rightarrow η = \frac{\cosh \frac{d}{2 a} - 1}{\frac{d}{2 a}}$ $λ \times \frac{d}{2 a} = \sinh \frac{d}{2 a}$ $\Rightarrow λ = \frac{\sinh \frac{d}{2 a}}{\frac{d}{2 a}}$ $w i t h p a r a m e t e r t = \frac{d}{2 a} w e g e t$ ${\begin{cases} η = \frac{\cosh t - 1}{t} \\ λ = \frac{\sinh t}{t} \end{cases}$
	Commented by Tawa1 last updated on 09/Jun/19

	$wow, God bless you sir$
	Answered by mr W last updated on 08/Jun/19

	$I I . f o r m o f r o p e a s p a r a b o l a$ $y = h {(\frac{2 x}{d})}^{2}$ $y^{'} = \frac{8 h x}{d^{2}}$ $\frac{L}{2} = \int_{0}^{\frac{d}{2}} \sqrt{1 + y^{' 2}} d x = \int_{0}^{\frac{d}{2}} \sqrt{1 + {(\frac{8 h x}{d^{2}})}^{2}} d x$ $L = \frac{d^{2}}{4 h} \int_{0}^{\frac{d}{2}} \sqrt{1 + {(\frac{8 h x}{d^{2}})}^{2}} d (\frac{8 h x}{d^{2}})$ $L = \frac{d^{2}}{4 h} \int_{0}^{\frac{4 h}{d}} \sqrt{1 + t^{2}} d t$ $L = 2 h {(\frac{d}{4 h})}^{2} {[t \sqrt{1 + t^{2}} + \ln (t + \sqrt{1 + t^{2}})]}_{0}^{\frac{4 h}{d}}$ $w i t h μ = \frac{4 h}{d}$ $\Rightarrow L = \frac{2 h}{μ^{2}} [μ \sqrt{1 + μ^{2}} + \ln (μ + \sqrt{1 + μ^{2}})]$ $\Rightarrow \frac{L}{2 h} = \frac{1}{μ^{2}} [μ \sqrt{1 + μ^{2}} + \ln (μ + \sqrt{1 + μ^{2}})]$ $e x a m p l e : L = 25 m, h = 7.5 m$ $\frac{1}{μ^{2}} [μ \sqrt{1 + μ^{2}} + \ln (μ + \sqrt{1 + μ^{2}})] = \frac{5}{3}$ $\Rightarrow μ = 1.6008$ $\Rightarrow d = \frac{4 h}{μ} = 18.74 m$ $l e t η = \frac{2 h}{d}, λ = \frac{L}{d}$ $\Rightarrow λ = \frac{1}{4 η} [2 η \sqrt{1 + 4 η^{2}} + \ln (2 η + \sqrt{1 + 4 η^{2}})]$
	Commented by mr W last updated on 08/Jun/19

	Commented by mr W last updated on 08/Jun/19

	$C = c a t e n a r y, P = p a r a b o l a$ $w e c a n s e e t h e d i f f e r e n c e b e t w e e n b o t h$ $i s v e r y s m a l l .$
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