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Question Number 68487 by mr W last updated on 11/Sep/19

Commented by mr W last updated on 11/Sep/19

$r o d A B h a s l e n g t h b a n d m a s s M,$ $r o p e B C h a s l e n g t h l a n d m a s s m,$ $t h e f r i c t i o n b e t w e e n r o d a n d g r o u n d$ $i s s u f f i c i e n t, s u c h t h a t p o i n t A c a n$ $n o t s l i d e .$ $l + b > h$ $f i n d θ = ?$
Commented by mr W last updated on 13/Sep/19

$a j f o u r s i r :$ $c a n y o u t r y t o g e t a s i n g l e e q u a t i o n$ $t o d e t e r m i n e θ ?$
	Answered by mr W last updated on 12/Sep/19

	Commented by Prithwish sen last updated on 13/Sep/19

	$excellent sir!$
	Commented by mr W last updated on 13/Sep/19
	$mass of rope of unit length ρ=(m/l) T_0 =horizontal component of tension in rope let a=(T_0 /(ρg)) the rope BC is a part of a catenary with equation y=a cosh (x/a) in coordinate system as shown in the diagram above. T_B cos β=T_0 =aρg T_B b sin (θ+β)=((Mgb cos θ)/2) ((aρg)/(cos β))sin (θ+β)=((Mg cos θ)/2) ((am)/(l cos β))sin (θ+β)=((M cos θ)/2) ((a(sin θ cos β+cos θ sin β))/(cos β cos θ))=((Ml)/(2m)) ⇒tan θ+tan β=((μl)/a) with μ=(M/(2m)) at point B: k=a cosh (e/a) y′=tan β=sinh (e/a) at point C: f=b cos θ k+h−b sin θ=a cosh ((e+f)/a) a cosh (e/a)+h−b sin θ=a cosh ((e+f)/a) ⇒h−b sin θ=a(cosh ((e+f)/a)−cosh (e/a)) l=a(sinh ((e+f)/a)−sinh (e/a)) ⇒sinh ((e+f)/a)−sinh (e/a)=(l/a) ⇒tan θ+sinh (e/a)=((μl)/a) ⇒(e/a)=sinh^(−1) (((μl)/a)−tan θ) ⇒sinh ((e+f)/a)=((l(1+μ))/a)−tan θ ⇒((e+f)/a)=sinh^(−1) [((l(1+μ))/a)−tan θ] ⇒(e/a)+(f/a)=sinh^(−1) [((l(1+μ))/a)−tan θ] ⇒sinh^(−1) (((μl)/a)−tan θ)+((b cos θ)/a)=sinh^(−1) [((l(1+μ))/a)−tan θ] ⇒sinh^(−1) [((l(1+μ))/a)−tan θ]−sinh^(−1) (((μl)/a)−tan θ)=(l/a)×((b cos θ)/l) ⇒cosh {sinh^(−1) [((l(1+μ))/a)−tan θ]}−cosh {sinh^(−1) (((μl)/a)−tan θ)}=(l/a)×((h−b sin θ)/l) let λ=(l/a) ⇒sinh^(−1) [(1+μ)λ−tan θ]−sinh^(−1) (μλ−tan θ)=((λb cos θ)/l) ...(i) ⇒cosh {sinh^(−1) [(1+μ)λ−tan θ]}−cosh {sinh^(−1) (μλ−tan θ)}=((λ(h−b sin θ))/l) ...(ii) unknowns in (i) and (ii): λ and θ example: μ=(M/(2m))=2 l=2 b=3 h=4 ⇒θ=61.617° (θ=61.045° if rope were massless)$
	$m a s s o f r o p e o f u n i t l e n g t h ρ = \frac{m}{l}$ $T_{0} = h o r i z o n t a l c o m p o n e n t o f t e n s i o n$ $i n r o p e$ $l e t a = \frac{T_{0}}{ρ g}$ $t h e r o p e B C i s a p a r t o f a c a t e n a r y$ $w i t h e q u a t i o n$ $y = a \cosh \frac{x}{a} i n c o o r d i n a t e s y s t e m$ $a s s h o w n i n t h e d i a g r a m a b o v e .$ $T_{B} \cos β = T_{0} = a ρ g$ $T_{B} b \sin (θ + β) = \frac{M g b \cos θ}{2}$ $\frac{a ρ g}{\cos β} \sin (θ + β) = \frac{M g \cos θ}{2}$ $\frac{a m}{l \cos β} \sin (θ + β) = \frac{M \cos θ}{2}$ $\frac{a (\sin θ \cos β + \cos θ \sin β)}{\cos β \cos θ} = \frac{M l}{2 m}$ $\Rightarrow \tan θ + \tan β = \frac{μ l}{a} w i t h μ = \frac{M}{2 m}$ $a t p o i n t B :$ $k = a \cosh \frac{e}{a}$ $y^{'} = \tan β = \sinh \frac{e}{a}$ $a t p o i n t C :$ $f = b \cos θ$ $k + h - b \sin θ = a \cosh \frac{e + f}{a}$ $a \cosh \frac{e}{a} + h - b \sin θ = a \cosh \frac{e + f}{a}$ $\Rightarrow h - b \sin θ = a (\cosh \frac{e + f}{a} - \cosh \frac{e}{a})$ $l = a (\sinh \frac{e + f}{a} - \sinh \frac{e}{a})$ $\Rightarrow \sinh \frac{e + f}{a} - \sinh \frac{e}{a} = \frac{l}{a}$ $\Rightarrow \tan θ + \sinh \frac{e}{a} = \frac{μ l}{a}$ $\Rightarrow \frac{e}{a} = \sinh^{- 1} (\frac{μ l}{a} - \tan θ)$ $\Rightarrow \sinh \frac{e + f}{a} = \frac{l (1 + μ)}{a} - \tan θ$ $\Rightarrow \frac{e + f}{a} = \sinh^{- 1} [\frac{l (1 + μ)}{a} - \tan θ]$ $\Rightarrow \frac{e}{a} + \frac{f}{a} = \sinh^{- 1} [\frac{l (1 + μ)}{a} - \tan θ]$ $\Rightarrow \sinh^{- 1} (\frac{μ l}{a} - \tan θ) + \frac{b \cos θ}{a} = \sinh^{- 1} [\frac{l (1 + μ)}{a} - \tan θ]$ $\Rightarrow \sinh^{- 1} [\frac{l (1 + μ)}{a} - \tan θ] - \sinh^{- 1} (\frac{μ l}{a} - \tan θ) = \frac{l}{a} \times \frac{b \cos θ}{l}$ $\Rightarrow \cosh {\sinh^{- 1} [\frac{l (1 + μ)}{a} - \tan θ]} - \cosh {\sinh^{- 1} (\frac{μ l}{a} - \tan θ)} = \frac{l}{a} \times \frac{h - b \sin θ}{l}$ $l e t λ = \frac{l}{a}$ $\Rightarrow \sinh^{- 1} [(1 + μ) λ - \tan θ] - \sinh^{- 1} (μ λ - \tan θ) = \frac{λ b \cos θ}{l} . . . (i)$ $\Rightarrow \cosh {\sinh^{- 1} [(1 + μ) λ - \tan θ]} - \cosh {\sinh^{- 1} (μ λ - \tan θ)} = \frac{λ (h - b \sin θ)}{l} . . . (i i)$ $u n k n o w n s i n (i) a n d (i i) : λ a n d θ$ $e x a m p l e :$ $μ = \frac{M}{2 m} = 2$ $l = 2$ $b = 3$ $h = 4$ $\Rightarrow θ = 61.617 °$ $(θ = 61.045 ° i f r o p e w e r e m a s s l e s s)$
	Commented by Prithwish sen last updated on 13/Sep/19

	$but what if BC is a straight line .$
	Commented by mr W last updated on 13/Sep/19

	$i f B C i s s t r a i g h t l i n e, i . e . t h e r o p e i s$ $m a s s l e s s, t h e n i t i s a p u r e g e o m e t r i c$ $q u e s t i o n f o r a t r i a n g l e w i t h s i d e s$ $b, l a n d h .$ $\sin θ = \cos (\frac{π}{2} - θ) = \frac{b^{2} + h^{2} - l^{2}}{2 b h}$ $\Rightarrow θ = \sin^{- 1} (\frac{b^{2} + h^{2} - l^{2}}{2 b h})$
	Commented by Prithwish sen last updated on 13/Sep/19

	$Thank you sir .$
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