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Question Number 44872 by ajfour last updated on 05/Oct/18

Commented by ajfour last updated on 05/Oct/18

$F i n d e q u a t i o n o f t h e r o p e c u r v e,$ $b o t t o m e n d o f r o p e i s a t t h e f o o t$ $o f t h e i n c l i n e a n d a l e n g t h b o f i t$ $i s i n c o n t a c t w i t h t h e f r i c t i o n l e s s$ $i n c l i n e . H e i g h t o f c e i l i n g i s h$ $a b o v e t h e g r o u n d .$
	Answered by MrW3 last updated on 06/Oct/18

	Commented by MrW3 last updated on 08/Oct/18
	$trying to solve the problem by using catenary equation directly. the hanging part of the rope AB is a part of the catenary curve ACD. let ρ=(m/l) tension in rope at point B is T=bρg sin α the horizontal component of tension in the rope AB is T_0 =T cos α=bρg sin α cos α=((bρg sin 2α)/2) the equation of rope ACD in the coordinate system x′y′ is a catenary (we use here u,v instead ofx′,y′ to avoid confusion) v=a cosh (u/a) with a=(T_0 /(ρg))=((bρg sin 2α)/(2ρg))=(1/2)b sin 2α length of rope upon point C is s=a sinh (u/a) inclination of rope is (dv/du)=sinh (u/a) at point B: tan α=(dv/du)=sinh (x_1 /a) ⇒x_1 =a sinh^(−1) (tan α) ⇒y_1 =a cosh (x_1 /a)=a cosh {sinh^(−1) (tan α)} CB^(⌢) =s_1 =a sinh (x_1 /a)=a tan α CA^(⌢) =s_2 =a sinh (x_2 /a) s_2 −s_1 =AB^(⌢) =l−b a sinh (x_2 /a)−a tan α=l−b sinh (x_2 /a)=tan α+((l−b)/a) ⇒x_2 =a sinh^(−1) (tan α+((l−b)/a)) ⇒y_2 =a cosh (x_2 /a)=a cosh {sinh^(−1) (tan α+((l−b)/a))} ⇒ΔH=x_2 −x_1 =a {sinh^(−1) (tan α+((l−b)/a))−sinh^(−1) (tan α)} ⇒ΔV=y_2 −y_1 =a[cosh {sinh^(−1) (tan α+((l−b)/a))}−cosh {sinh^(−1) (tan α)}] eqn. of rope AB in the requested coordinate system xy: u=x_2 −y=a sinh^(−1) (tan α+((l−b)/a))−y v=y_2 −x=a cosh {sinh^(−1) (tan α+((l−b)/a))}−x putting this into v=a cosh (u/a) we can get following eqn. for rope curve AB: a cosh {sinh^(−1) (tan α+((l−b)/a))}−x=a cosh ((a sinh^(−1) (tan α+((l−b)/a))−y)/a) or (x/a)=cosh {sinh^(−1) (tan α+((l−b)/a))}−cosh {sinh^(−1) (tan α+((l−b)/a))−(y/a)} or (y/a)=sinh^(−1) (tan α+((l−b)/a))−cosh^(−1) [cosh {sinh^(−1) (tan α+((l−b)/a))}−(x/a)] with a=(1/2)b sin 2α and it′s valid for 0≤x≤ΔV and 0≤y≤ΔH$
	$t r y i n g t o s o l v e t h e p r o b l e m b y u s i n g$ $c a t e n a r y e q u a t i o n d i r e c t l y .$ $t h e h a n g i n g p a r t o f t h e r o p e A B i s a$ $p a r t o f t h e c a t e n a r y c u r v e A C D .$ $l e t ρ = \frac{m}{l}$ $t e n s i o n i n r o p e a t p o i n t B i s$ $T = b ρ g \sin α$ $t h e h o r i z o n t a l c o m p o n e n t o f t e n s i o n$ $i n t h e r o p e A B i s$ $T_{0} = T \cos α = b ρ g \sin α \cos α = \frac{b ρ g \sin 2 α}{2}$ $t h e e q u a t i o n o f r o p e A C D i n t h e$ $c o o r d i n a t e s y s t e m x^{'} y^{'} i s a c a t e n a r y$ $(w e u s e h e r e u, v i n s t e a d {o f x}^{'}, y^{'} t o a v o i d$ $c o n f u s i o n)$ $v = a \cosh \frac{u}{a} w i t h$ $a = \frac{T_{0}}{ρ g} = \frac{b ρ g \sin 2 α}{2 ρ g} = \frac{1}{2} b \sin 2 α$ $l e n g t h o f r o p e u p o n p o i n t C i s$ $s = a \sinh \frac{u}{a}$ $i n c l i n a t i o n o f r o p e i s$ $\frac{d v}{d u} = \sinh \frac{u}{a}$ $a t p o i n t B :$ $\tan α = \frac{d v}{d u} = \sinh \frac{x_{1}}{a}$ $\Rightarrow x_{1} = a \sinh^{- 1} (\tan α)$ $\Rightarrow y_{1} = a \cosh \frac{x_{1}}{a} = a \cosh {\sinh^{- 1} (\tan α)}$ $\overset{⌢}{C B} = s_{1} = a \sinh \frac{x_{1}}{a} = a \tan α$ $\overset{⌢}{C A} = s_{2} = a \sinh \frac{x_{2}}{a}$ $s_{2} - s_{1} = \overset{⌢}{A B} = l - b$ $a \sinh \frac{x_{2}}{a} - a \tan α = l - b$ $\sinh \frac{x_{2}}{a} = \tan α + \frac{l - b}{a}$ $\Rightarrow x_{2} = a \sinh^{- 1} (\tan α + \frac{l - b}{a})$ $\Rightarrow y_{2} = a \cosh \frac{x_{2}}{a} = a \cosh {\sinh^{- 1} (\tan α + \frac{l - b}{a})}$ $\Rightarrow Δ H = x_{2} - x_{1} = a {\sinh^{- 1} (\tan α + \frac{l - b}{a}) - \sinh^{- 1} (\tan α)}$ $\Rightarrow Δ V = y_{2} - y_{1} = a [\cosh {\sinh^{- 1} (\tan α + \frac{l - b}{a})} - \cosh {\sinh^{- 1} (\tan α)}]$ $e q n . o f r o p e A B i n t h e r e q u e s t e d$ $c o o r d i n a t e s y s t e m x y :$ $u = x_{2} - y = a \sinh^{- 1} (\tan α + \frac{l - b}{a}) - y$ $v = y_{2} - x = a \cosh {\sinh^{- 1} (\tan α + \frac{l - b}{a})} - x$ $p u t t i n g t h i s i n t o v = a \cosh \frac{u}{a} w e c a n g e t$ $f o l l o w i n g e q n . f o r r o p e c u r v e A B :$ $a \cosh {\sinh^{- 1} (\tan α + \frac{l - b}{a})} - x = a \cosh \frac{a \sinh^{- 1} (\tan α + \frac{l - b}{a}) - y}{a}$ $o r$ $\frac{x}{a} = \cosh {\sinh^{- 1} (\tan α + \frac{l - b}{a})} - \cosh {\sinh^{- 1} (\tan α + \frac{l - b}{a}) - \frac{y}{a}}$ $o r$ $\frac{y}{a} = \sinh^{- 1} (\tan α + \frac{l - b}{a}) - \cosh^{- 1} [\cosh {\sinh^{- 1} (\tan α + \frac{l - b}{a})} - \frac{x}{a}]$ $w i t h a = \frac{1}{2} b \sin 2 α a n d {i t}^{'} s v a l i d$ $f o r 0 ⩽ x ⩽ Δ V a n d 0 ⩽ y ⩽ Δ H$
	Commented by MrW3 last updated on 06/Oct/18

	Commented by MrW3 last updated on 06/Oct/18

	Commented by ajfour last updated on 08/Oct/18

	$T h a n k y o u S i r, p l e a s e c h e c k m y$ $s o l u t i o n S i r, i f i t s c o r r e c t i t i s$ $s i m p l e r . .$
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