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Question Number 46344 by MrW3 last updated on 24/Oct/18

Commented by MrW3 last updated on 25/Oct/18

$A r o d o f m a s s M a n d l e n g t h l i s s u p p o r t e d$ $a t o n e e n d t h r o u g h a h i n g e . t h e o t h e r f r e e$ $e n d i s h o l d h o r i z o n t a l l y b y a r o p e$ $a s s h o w n . t h e r o p e h a s a m a s s ρ p e r$ $u n i t l e n g t h .$ $F i n d t h e m a x . t e n s i o n i n t h e r o p e .$
Commented by ajfour last updated on 24/Oct/18

Commented by ajfour last updated on 24/Oct/18

$T d θ = ρ g (\frac{d y}{\sin θ}) \cos θ . . . . (i)$ $T \cos θ = (T + d T) \cos (θ + d θ) . . . (i i)$ $F r o m e q . (i i)$ $T \cos θ = T_{0} \cos β = T_{e} \cos α = c$ $________________________$ $\Rightarrow \cos α = \frac{c}{T_{e}}, \cos β = \frac{c}{T_{0}} . . (I)$ $________________________$ $u s i n g (i i) i n (i)$ $\int_{0}^{h} ρ g d y = c \int_{α}^{β} \sec θ \tan θ d θ$ $\Rightarrow c (\sec β - \sec α) = ρ g h$ $________________________$ $\Rightarrow T_{0} - T_{e} = ρ g h . . . . . . (I I)$ $________________________$ $T o r q u e a b o u t l e f t e n d o n l y o n$ $b r i d g e (e x c l u d i n g r o p e) :$ $T_{e} L \sin α = \frac{M g L}{2}$ $\Rightarrow \sqrt{T_{e}^{2} - c^{2}} = \frac{M g}{2}$ $\Rightarrow T_{e} = \sqrt{c^{2} + {(\frac{M g}{2})}^{2}} . . . . (A)$ $A n d u s i n g (A) i n (I I)$ $T_{0} = ρ g h + \sqrt{c^{2} + {(\frac{M g}{2})}^{2}} . . . (B)$ $T o r q u e a b o u t l e f t e n d o f b r i d g e,$ $o n b r i d g e + r o p e :$ $_____________________$ $T_{0} h \cos β = \int_{0}^{l} (ρ g d l) x + \frac{M g L}{2} . . (i i i)$ $_____________________$ $b u t f r o m (i)$ $ρ g d x = T d θ = c \sec θ d θ$ $\Rightarrow \int_{0}^{x} ρ g d x = c \int_{β}^{θ} \sec θ d θ$ $\Rightarrow ρ g x = c \ln (\frac{\sec θ + \tan θ}{\sec β + \tan β}) . . (i i i)$ $f r o m (i i i)$ $\Rightarrow c h = \int_{β}^{α} (T \sec θ d θ) x + \frac{M g L}{2}$ $u s i n g (i i i)$ $c h = c^{2} \int_{β}^{α} \frac{(\sec^{2} θ d θ) \ln (. .)}{ρ g} + \frac{M g L}{2}$ $\Rightarrow c h - \frac{M g L}{2} = \frac{c^{2}}{ρ g} [(\tan θ) \ln (\frac{\sec θ + \tan θ}{\sec β + \tan β}) ∣_{β}^{α}$ $- \int_{β}^{α} \tan θ \sec θ d θ]$ $\Rightarrow \frac{ρ g}{c^{2}} (c h - \frac{M g L}{2}) = (\tan α) \ln (\frac{\sec α + \tan α}{\sec β + \tan β})$ $+ \sec β - \sec α$ $b u t$ $c (\sec β - \sec α) = ρ g h$ $\tan α = \sqrt{\sec^{2} α - 1} = \sqrt{\frac{T_{e}^{2}}{c^{2}} - 1}$ $= \frac{M g}{2 c}; h e n c e$ $\Rightarrow - \frac{ρ g}{2 c^{2}} (M g L) = (\frac{M g}{2 c}) \ln [\frac{\cos β}{\cos α} \times \frac{1 + \sin α}{1 + \sin β}]$ $\Rightarrow \ln [\frac{T_{e} + T_{e} \sin α}{T_{0} + T_{0} \sin β}] = - \frac{ρ g L}{c}$ $\Rightarrow \ln [\frac{T_{0} + \sqrt{T_{0}^{2} - c^{2}}}{T_{e} + \sqrt{T_{e}^{2} - c^{2}}}] = \frac{ρ g L}{c}$ $________________________$ $A n s w e r :$ $\ln (\frac{T_{0} + \sqrt{T_{0}^{2} - c^{2}}}{T_{e} + \frac{M g}{2}}) = \frac{ρ g L}{c}$ $- - - - - - - - - - - - - - -$ $w h e r e [s e e (A) & (B)]$ $T_{e} = \sqrt{c^{2} + {(\frac{M g}{2})}^{2}}$ $T_{0} = ρ g h + \sqrt{c^{2} + {(\frac{M g}{2})}^{2}}$ $________________________.$
Commented by ajfour last updated on 24/Oct/18

$T_{0} = ρ g h + \sqrt{c^{2} + {(\frac{M g}{2})}^{2}}$ $A n d f o r c$ $ρ g h + \sqrt{c^{2} + {(\frac{M g}{2})}^{2}} +$ $\sqrt{{(ρ g h + \sqrt{c^{2} + {(\frac{M g}{2})}^{2}})}^{2} - c^{2}}$ $= [\sqrt{c^{2} + {(\frac{M g}{2})}^{2}} + \frac{M g}{2}] e^{\frac{ρ g L}{c}}$ $I f f u r t h e r i l e t$ $λ = \frac{c}{(M g / 2)}; μ = \frac{ρ g h}{(M g / 2)}; m = \frac{L}{h}$ $\Rightarrow λ \ln (\frac{μ + \sqrt{λ^{2} + 1} + \sqrt{{(μ + \sqrt{λ^{2} + 1})}^{2} - λ^{2}}}{\sqrt{λ^{2} + 1}})$ $= μ m$ $T_{0} = (μ + \sqrt{λ^{2} + 1}) \frac{M g}{2}$ $f o r L = 4, h = 3, \frac{ρ_{B}}{ρ} = 50$ $\frac{ρ_{B}}{ρ} = \frac{M g}{ρ g L} = 2 \times \frac{(M g / 2)}{ρ g h} \times \frac{h}{L} = \frac{2}{μ m}$ $\Rightarrow 50 = \frac{3 \times 2}{4 μ} \Rightarrow μ = \frac{3}{100} = 0.03$ $w i t h m = \frac{L}{h} = \frac{4}{3}$ $λ = 1.3664, T_{m a x} = T_{0} \approx 1.723 (\frac{M g}{2}) .$
Commented by MrW3 last updated on 07/Nov/18

$y o u r a n s w e r i s c o r r e c t s i r!$ $t h a n k y o u v e r y m u c h .$
	Answered by MrW3 last updated on 24/Oct/18

	Commented by MrW3 last updated on 25/Oct/18

	$T = t e n s i o n i n r o p e$ $T_{H} = h o r i z o n t a l c o m p o n e n t o f T = c o n s t$ $M = ρ_{B} l$ $p = \frac{2 l ρ}{M} = \frac{2 ρ}{ρ_{M}}, q = \frac{2 h ρ}{M} = \frac{2 h ρ}{l ρ_{B}}, λ = \tan α$ $a t p o i n t B :$ $T_{B} \sin α = \frac{M g}{2}$ $T_{H} = T_{B} \cos α = \frac{M g}{2 \tan α}$ $E q n . o f r o p e (c a t e n a r y) i n x y - s y s t e m :$ $y = a \cosh \frac{x}{a}$ $w i t h a = \frac{T_{H}}{ρ g} = \frac{M}{2 ρ \tan α}$ $\frac{d y}{d x} = \tan θ = \sinh \frac{x}{a}$ $a t p o i n t B :$ $\tan α = \sinh \frac{e}{a}$ $e = a \sinh^{- 1} (\tan α)$ $f = a \cosh \frac{e}{a} = a \cosh [\sinh^{- 1} (\tan α)]$ $a t p o i n t A :$ $f + h = a \cosh \frac{e + l}{a}$ $\Rightarrow a \cosh [\sinh^{- 1} (\tan α)] + h = a \cosh [\frac{l}{a} + \sinh^{- 1} (\tan α)]$ $\Rightarrow \cosh [\sinh^{- 1} (\tan α)] + \frac{h}{a} = \cosh [\frac{l}{a} + \sinh^{- 1} (\tan α)]$ $\Rightarrow \cosh [\sinh^{- 1} (\tan α)] + \frac{h}{a} = \cosh (\frac{l}{a}) \cosh [\sinh^{- 1} (\tan α)] + \sinh (\frac{l}{a}) \tan α$ $\Rightarrow [\cosh (\frac{l}{a}) - 1] \cosh [\sinh^{- 1} (\tan α)] + \sinh (\frac{l}{a}) \tan α = \frac{h}{a}$ $\Rightarrow [\cosh (\frac{2 l ρ \tan α}{M}) - 1] \sqrt{1 + \tan^{2} α} + \sinh (\frac{2 l ρ \tan α}{M}) \tan α = \frac{2 h ρ \tan α}{M}$ $\Rightarrow \sqrt{1 + λ^{2}} [\cosh (p λ) - 1] + λ [\sinh (p λ) - q] = 0 . . . (i)$ $\Rightarrow λ = f (p, q) (o n l y n u m e r i c a l l y)$ $m a x . T = T_{A} = \frac{T_{H}}{\cos θ_{A}} = \frac{M g}{2 λ} \sqrt{1 + \tan^{2} θ_{A}}$ $\tan θ_{A} = \sinh \frac{e + l}{a} = \sinh [\frac{2 l ρ \tan α}{M} + \sinh^{- 1} (\tan α)]$ $= \sinh (p λ + \sinh^{- 1} λ) = \sqrt{1 + λ^{2}} \sinh (p λ) + λ \cosh (p λ)$ $\tan^{2} θ_{A} = (1 + λ^{2}) \sinh^{2} (p λ) + λ^{2} \cosh^{2} (p λ) + 2 λ \sqrt{1 + λ^{2}} \sinh (p λ) \cosh (p λ)$ $= (\frac{1}{2} + λ^{2}) \cosh (2 p λ) + λ \sqrt{1 + λ^{2}} \sinh (2 p λ) - \frac{1}{2}$ $1 + \tan^{2} θ_{A} = \frac{1}{2} + (\frac{1}{2} + λ^{2}) \cosh (2 p λ) + λ \sqrt{1 + λ^{2}} \sinh (2 p λ)$ $\Rightarrow m a x . T = \frac{M g}{2 λ} \sqrt{\frac{1}{2} + (\frac{1}{2} + λ^{2}) \cosh (2 p λ) + λ \sqrt{1 + λ^{2}} \sinh (2 p λ)} . . . (i i)$ $e x a m p l e : h = 3 m, l = 4 m, \frac{ρ_{B}}{ρ} = 50$ $\Rightarrow λ = \tan α = 0.7318 f r o m (i)$ $\Rightarrow m a x . T = 1.7233 \times \frac{M g}{2} f r o m (i i)$ $i n c a s e t h a t r o p e i s m a s s l e s s, w e h a v e$ $λ = \tan α = \frac{3}{4} = 0.75$ $T = \frac{5}{3} \times \frac{M g}{2} = 1.6667 \times \frac{M g}{2}$
	Commented by ajfour last updated on 25/Oct/18

	$T h a n k y o u t o o, S i r .$ $V e r y v e r y n i c e q u e s t i o n!$
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