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Question Number 82873 by mr W last updated on 25/Feb/20

Commented by mr W last updated on 25/Feb/20

$A r o p e w i t h l e n g t h l a n d m a s s m i s$ $f i x e d a t o n e e n d o n t h e t o p o f a r o l l e r$ $w i t h d i a m e t e r h a s s h o w n . F i n d t h e$ $m a x i m a l p o s s i b l e h o r i z o n t a l d i s t a n c e$ $b f r o m t h e f r e e e n d t o t h e f i x e d e n d,$ $i f t h e f r i c t i o n c o e f f i c i e n t b e t w e e n$ $t h e r o p e a n d t h e g r o u n d i s μ .$
Commented by jagoll last updated on 25/Feb/20

$waw phisic$
	Answered by mr W last updated on 25/Feb/20

	Commented by mr W last updated on 27/Feb/20
	$let ρ=(m/l) the free end B is right most when the max. friction between rope and ground is activated. tension in rope at poind D is T_0 with T_0 =μρdg let a=(T_0 /(ρg))=μd the rope is divided into 3 parts: arc AC=((hθ)/2) free hanging segment CD=s segment on ground DB=d l=((hθ)/2)+s+d ⇒s=l−((hθ)/2)−d in the xy−coordinate system the shape of the hanging segment is a catenary with equation y=a cosh (x/a) y′=sinh (x/a) at point C: x_C =c=b−d−((h sin θ)/2) ...(i) y_C =a+(h/2)(1+cos θ)=a cosh (c/a) ...(ii) y_D ′=tan θ=sinh (c/a) ...(iii) s=l−d−((hθ)/2)=a sinh (c/a) ...(iv) from (iii) and (iv): l−d−((hθ)/2)=μd tan θ ⇒d=((2l−hθ)/(2(1+μ tan θ))) (ii)^2 −(iii×a)^2 : a^2 +ah(1+cos θ)+((h^2 (1+cos θ)^2 )/4)−a^2 tan^2 θ=a^2 tan^2 θ a^2 −h(1+cos θ)a−((h^2 (1+cos θ)^2 )/4)=0 ⇒a=μd=((h cos θ (1+cos θ))/(2(1−cos θ))) ⇒((μ(2l−hθ))/(cos θ+μ sin θ))=((h(1+cos θ))/(1−cos θ)) let η=(l/h) ⇒((μ(2η−θ))/(cos θ+μ sin θ))=((1+cos θ)/(1−cos θ)) ...(I) from(iii): c=a sinh^(−1) (tan θ)=μd sinh^(−1) (tan θ) from (i): b=c+d+((h sin θ)/2) b=((h cos θ (1+cos θ)[1+μ sinh^(−1) (tan θ)])/(2μ(1−cos θ)))+((h sin θ)/2) ⇒((2b)/h)=((cos θ (1+cos θ)[1+μ sinh^(−1) (tan θ)])/(μ(1−cos θ)))+sin θ ...(II) we get θ from (I) and then b from (II). example: l=10m,h=2m, η=(l/h)=((10)/2)=5, μ=0.5 ⇒θ≈0.8840 (50.65°) ⇒((2b)/h)≈9.3479 ⇒b≈9.3479 m (h/l)=(2/(θ+(((1+cos θ)(cos θ+μ sin θ))/(μ(1−cos θ))))) (b/l)=(h/(2l)){((cos θ (1+cos θ)[1+μ sinh^(−1) (tan θ)])/(μ(1−cos θ)))+sin θ} using θ as parameter we can get the relation between (b/l) and (h/l), see diagram.$
	$l e t ρ = \frac{m}{l}$ $t h e f r e e e n d B i s r i g h t m o s t w h e n$ $t h e m a x . f r i c t i o n b e t w e e n r o p e a n d$ $g r o u n d i s a c t i v a t e d .$ $t e n s i o n i n r o p e a t p o i n d D i s T_{0} w i t h$ $T_{0} = μ ρ d g$ $l e t a = \frac{T_{0}}{ρ g} = μ d$ $t h e r o p e i s d i v i d e d i n t o 3 p a r t s :$ $a r c A C = \frac{h θ}{2}$ $f r e e h a n g i n g s e g m e n t C D = s$ $s e g m e n t o n g r o u n d D B = d$ $l = \frac{h θ}{2} + s + d \Rightarrow s = l - \frac{h θ}{2} - d$ $i n t h e x y - c o o r d i n a t e s y s t e m t h e$ $s h a p e o f t h e h a n g i n g s e g m e n t i s a$ $c a t e n a r y w i t h e q u a t i o n$ $y = a \cosh \frac{x}{a}$ $y^{'} = \sinh \frac{x}{a}$ $a t p o i n t C :$ $x_{C} = c = b - d - \frac{h \sin θ}{2} . . . (i)$ $y_{C} = a + \frac{h}{2} (1 + \cos θ) = a \cosh \frac{c}{a} . . . (i i)$ $y_{D}^{'} = \tan θ = \sinh \frac{c}{a} . . . (i i i)$ $s = l - d - \frac{h θ}{2} = a \sinh \frac{c}{a} . . . (i v)$ $f r o m (i i i) a n d (i v) :$ $l - d - \frac{h θ}{2} = μ d \tan θ$ $\Rightarrow d = \frac{2 l - h θ}{2 (1 + μ \tan θ)}$ ${(i i)}^{2} - {(i i i \times a)}^{2} :$ $a^{2} + a h (1 + \cos θ) + \frac{h^{2} {(1 + \cos θ)}^{2}}{4} - a^{2} \tan^{2} θ = a^{2}$ $\tan^{2} θ a^{2} - h (1 + \cos θ) a - \frac{h^{2} {(1 + \cos θ)}^{2}}{4} = 0$ $\Rightarrow a = μ d = \frac{h \cos θ (1 + \cos θ)}{2 (1 - \cos θ)}$ $\Rightarrow \frac{μ (2 l - h θ)}{\cos θ + μ \sin θ} = \frac{h (1 + \cos θ)}{1 - \cos θ}$ $l e t η = \frac{l}{h}$ $\Rightarrow \frac{μ (2 η - θ)}{\cos θ + μ \sin θ} = \frac{1 + \cos θ}{1 - \cos θ} . . . (I)$ $f r o m (i i i) :$ $c = a \sinh^{- 1} (\tan θ) = μ d \sinh^{- 1} (\tan θ)$ $f r o m (i) :$ $b = c + d + \frac{h \sin θ}{2}$ $b = \frac{h \cos θ (1 + \cos θ) [1 + μ \sinh^{- 1} (\tan θ)]}{2 μ (1 - \cos θ)} + \frac{h \sin θ}{2}$ $\Rightarrow \frac{2 b}{h} = \frac{\cos θ (1 + \cos θ) [1 + μ \sinh^{- 1} (\tan θ)]}{μ (1 - \cos θ)} + \sin θ . . . (I I)$ $w e g e t θ f r o m (I) a n d t h e n b f r o m (I I) .$ $e x a m p l e :$ $l = 10 m, h = 2 m, η = \frac{l}{h} = \frac{10}{2} = 5, μ = 0.5$ $\Rightarrow θ \approx 0.8840 (50.65 °)$ $\Rightarrow \frac{2 b}{h} \approx 9.3479 \Rightarrow b \approx 9.3479 m$ $\frac{h}{l} = \frac{2}{θ + \frac{(1 + \cos θ) (\cos θ + μ \sin θ)}{μ (1 - \cos θ)}}$ $\frac{b}{l} = \frac{h}{2 l} {\frac{\cos θ (1 + \cos θ) [1 + μ \sinh^{- 1} (\tan θ)]}{μ (1 - \cos θ)} + \sin θ}$ $u s i n g θ a s p a r a m e t e r w e c a n g e t t h e$ $r e l a t i o n b e t w e e n \frac{b}{l} a n d \frac{h}{l}, s e e$ $d i a g r a m .$
	Commented by mr W last updated on 26/Feb/20

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