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Question Number 170601 by mr W last updated on 27/May/22

Commented by mr W last updated on 29/May/22

$a n u n i f o r m t h i n r o p e w i t h l e n g t h L$ $i s h a n g i n g a t o n e e n d a t a h e i g h t o f h$ $a b o v e a i n c l i n e d p l a n e a s s h o w n .$ $i f t h e f r i c t i o n c o e f f i c i e n t b e t w e e n$ $r o p e a n d t h e p l a n e i s μ, f i n d t h e$ $m i n i m u m a n d m a x i m u m l e n g t h o f$ $t h a t p a r t o f t h e r o p e w h i c h c a n l i e o n$ $t h e p l a n e .$ $(L > h)$
	Answered by aleks041103 last updated on 28/May/22

	Commented by aleks041103 last updated on 28/May/22

	$r e d c u r v e \to c a t e n a r y$ $y = a c o s h (\frac{x}{a})$ $l e n g t h f u n c t i o n :$ $\int \sqrt{1 + y^{' 2}} d x = \int \sqrt{1 + {s i n h}^{2} (\frac{x}{a})} d x =$ $= \int c o s h (\frac{x}{a}) d x = a s i n h (\frac{x}{a})$ $L e t l e n g t h o n s l o p e b e l, t h e n t h e r e i s$ $a r o p e o f l e n g t h L - l h a n g i n g o r$ $L - l = a (s i n h (\frac{x_{b}}{a}) - s i n h (\frac{x_{a}}{a}))$ $a l s o$ $y^{'} (x_{b}) = t a n θ = k$ $\Rightarrow s i n h (\frac{x_{b}}{a}) = k \Rightarrow x_{b} = a a r c s i n h (k)$ $a n d L - l = a (k - s i n h (\frac{x_{a}}{a}))$ $\Rightarrow s l o p e h a s e q n :$ $s (x) = k (x - a a r c s i n h (k)) + a \sqrt{1 + k^{2}}$ $\Rightarrow f o r x_{a} w e h a v e$ $a c o s h (\frac{x_{a}}{a}) - s (x_{a}) = h$ $\Rightarrow c o s h (\frac{x_{a}}{a}) - k \frac{x_{a}}{a} = \frac{h}{a} + \sqrt{1 + k^{2}} - k a r c s i n h (k)$ $A l s o, t h e h a n g i n g p a r t h a s t w o f o r c e s a c t i n g$ $T_{1} a t A a t a n g l e α t o x a n d T_{2} a t B a t a n g l e$ $θ t o x .$ $t a n α = s i n h (\frac{x_{a}}{a})$ $\Rightarrow s i n α = \frac{t a n α}{\sqrt{1 + {t a n}^{2} α}} = \frac{s i n h (x_{a} / a)}{c o s h (x_{a} / a)} = t a n h (\frac{x_{a}}{a})$ $c o s α = s e c h (\frac{x_{a}}{a})$ $\Rightarrow E q u i l i b r i u m r e q u i r e s :$ $T_{1} c o s α = T_{2} c o s θ$ $- T_{1} s i n α + T_{2} s i n θ = \frac{L - l}{L} m g$ $A l s o T_{2} ⩽ \frac{l}{L} μ m g c o s θ$ $\Rightarrow t g α ⩽ \frac{\frac{l}{L} μ m g s i n θ c o s θ + \frac{l - L}{L} m g}{\frac{l}{L} μ {m g c o s}^{2} θ} =$ $= t g θ - \frac{L - l}{μ l} {s e c}^{2} θ = - \frac{L - l}{μ l} (1 + k^{2}) + k ⩾ s i n h \frac{x_{a}}{a}$ $\Rightarrow L - l ⩽ a (\frac{L - l}{μ l} (1 + k^{2}))$ $\Rightarrow l ⩽ \frac{a (1 + k^{2})}{μ} \Rightarrow a ⩾ \frac{μ l}{1 + k^{2}}$ $. . .$ $A l o t o f i n e q u a l i t i e s l e f t t o d o$
	Commented by mr W last updated on 29/May/22

	$t h a n k s f o r t h e s o l u t i o n s i r!$
	Answered by mr W last updated on 29/May/22

	Commented by mr W last updated on 29/May/22

	${l e t}^{'} s h a v e a l o o k a t a n o b j e c t o n a n$ $i n c l i n e d p l a n e .$ $s u c h t h a t t h e o b j e c t i s i n e q u i l i b r i u m$ $o n s u c h a s l o p e, i . e . s u c h t h a t i t$ ${d o e s n}^{'} t s l i d e d o w n, a f r i c t i o n f o r c e f$ $m u s t a c t b e t w e e n t h e o b j e c t a n d t h e$ $s l o p e a s s h o w n . t h e f r i c t i o n c o e f f i c i e n t$ $b e t w e e n t h e m d e f i n e s a c t u a l l y t h e$ $m a x i m a l p o s s i b l e f r i c t i o n f o r c e$ $b e t w e e n t h e m .$ $f = μ N w i t h 0 ⩽ μ ⩽ μ_{n a x} .$ $t h o u g h w e s p e a k o f f r i c t i o n$ $c o e f f i c i e n t μ, b u t a c t u a l l y w e m e a n$ $t h e m a x i m a l p o s s i b l e f r i c t i o n$ $c o e f f i c i e n t μ_{m a x} . t h e a c t u a l f r i c t i o n$ $c o e f f i c i e n t μ m u s t f u l f u l l 0 ⩽ μ ⩽ μ_{m a x} .$ $i n f o l l o w i n g w i t h μ i^{'} l l d e n o t e t h e$ $a c t u a l f r i c t i o n c o e f f i c i e n t b e t w e e n$ $r o p e a n d s l o p e .$ $f = μ N$ $N = M g \cos θ$ $s u c h t h a t t h e o b j e c t {d o e s n}^{'} t s l i d e$ $d o w n,$ $f - M g \sin θ = 0$ $μ M g \cos θ - M g \sin θ = 0$ $μ \cos θ - \sin θ = 0$ $μ = \frac{\sin θ}{\cos θ} = \tan θ ⩽ μ_{m a x}$ $i f t h e f r i c t i o n c o e f f i c i e n t μ_{n a x} i s$ $s m a l l e r t h a n \tan θ, a n o b j e c t c a n n o t$ $b e i n e q u i l i b r i u m o n t h e s l o p e .$ $s i n c e i n o u r c a s e a p a r t o f t h e r o p e$ $c a n l i e o n t h e s l o p e, w e m u s t h a v e$ $μ_{m a x} ⩾ \tan θ .$
	Commented by mr W last updated on 29/May/22

	Commented by mr W last updated on 29/May/22

	$c a s e 1 : μ = \tan θ$ $w e s e e t h e r o p e c a n t a k e t h e s h a p e$ $a s s h o w n i n d i a g r a m . t h e l e n g t h o f$ $t h e p a r t w h i c h l i e s o n t h e s l o p e i s$ $b = L - h .$
	Commented by mr W last updated on 29/May/22

	Commented by mr W last updated on 29/May/22

	$c a s e 2 : μ > \tan θ$ $t h e f r i c t i o n f o r c e f i s l a r g e e n o u g h$ $s o t h a t i t c a n e v e n a l l o w a n$ $a d d i t i o n t e n s i o n f o r c e b e t w e e n t h e$ $t w o p a r t s o f t h e r o p e .$
	Commented by mr W last updated on 29/May/22

	Commented by mr W last updated on 29/May/22

	$s a y t h e l e n g t h o f t h e r o p e p a r t w h i c h$ $l i e s o n t h e i n c l i n e d p l a n e i s b .$ $ρ = \frac{m}{L}$ $t h e f r i c t i o n f o r c e b e t w e e n r o p e a n d$ $t h e s l o p e i s f .$ $f = μ ρ g b \cos θ w i t h \tan θ < μ ⩽ μ_{m a x}$ $t e n s i o n i n t h e r o p e a t p o i n t B :$ $T_{1} = f - ρ g b \sin θ = (μ \cos θ - \sin θ) ρ g b$ $l e t μ^{'} = μ \cos θ - \sin θ$ $T_{1} = μ^{'} ρ g b$ $T_{0} = T_{1} \cos θ = μ^{'} ρ g b \cos θ$ $d e f i n e a = \frac{T_{0}}{ρ g} = μ^{'} b \cos θ$ $t h e h a n g i n g r o p e f r o m p o i n t A t o B$ $i s p a r t o f a c a t e n a r y . w . r . t . t h e$ $c o o r d i n a t e s y s t e m a s s h o w n,$ $y = a \cosh \frac{x}{a}$ $y^{'} = \sinh \frac{x}{a}$ $s = a \sinh \frac{x}{a}$ $\tan θ = \sinh \frac{x_{B}}{a} \Rightarrow \frac{x_{B}}{a} = \sinh^{- 1} (\tan θ)$ $s_{1} = \overset{⌢}{C B} = a \sinh \frac{x_{B}}{a} = a \tan θ$ $\overset{⌢}{A C} = L - b - s_{1} = a \sinh \frac{x_{A}}{a}$ $L - b - a \tan θ = a \sinh \frac{x_{A}}{a}$ $\Rightarrow \frac{x_{A}}{a} = \sinh^{- 1} (\frac{L - b}{a} - \tan θ)$ $y_{A} - y_{B} = h - (x_{A} + x_{B}) \tan θ$ $a \cosh \frac{x_{A}}{a} - a \cosh \frac{x_{B}}{a} = h - (x_{A} + x_{B}) \tan θ$ $\cosh \frac{x_{A}}{a} - \cosh \frac{x_{B}}{a} = \frac{h}{a} - (\frac{x_{A}}{a} + \frac{x_{B}}{a}) \tan θ$ $\cosh [\sinh^{- 1} (\frac{L - b}{a} - \tan θ)] - \cosh [\sinh^{- 1} (\tan θ)] = \frac{h}{a} - [\sinh^{- 1} (\frac{L - b}{a} - \tan θ) + \sinh^{- 1} (\tan θ)] \tan θ$ $l e t \frac{b}{L} = \frac{1}{λ}$ $\cosh [\sinh^{- 1} (\frac{λ - 1}{μ^{'} \cos θ} - \tan θ)] - \cosh [\sinh^{- 1} (\tan θ)] = \frac{λ h}{μ^{'} L \cos θ} - [\sinh^{- 1} (\frac{λ - 1}{μ^{'} \cos θ} - \tan θ) + \sinh^{- 1} (\tan θ)] \tan θ$ $\Rightarrow [\sinh^{- 1} (\frac{λ - 1}{μ^{'} \cos θ} - \tan θ) + \sinh^{- 1} (\tan θ)] \sin θ = 1 + \frac{λ h}{μ^{'} L} - \sqrt{1 + {(\frac{λ - 1}{μ^{'}})}^{2} - 2 (\frac{λ - 1}{μ^{'}}) \sin θ}$ $t h i s e q u a t i o n s h o w s t h e r e l a t i o n s h i p$ $b e t w e e n λ a n d t h e a c t u a l f r i c t i o n$ $c o e f f i c i e n t μ (\tan θ < μ ⩽ μ_{m a x}) .$ $g e n e r a l l y t h i s r e l a t i o n s h i p l o o k s l i k e$ $a s s h o w n i n t h e d i a g r a m b e l o w .$ $e x a m p l e :$ $h = 3 m, L = 5 m, θ = 30 °, μ_{m a x} = 1.5$ $\Rightarrow \frac{b_{m i n}}{L} = 0.33202 a t μ = 1.5$ $\Rightarrow \frac{b_{m a x}}{L} = 0.43165 a t μ = 0.7$
	Commented by mr W last updated on 29/May/22

	Commented by mr W last updated on 29/May/22

	$p o i n t 1 :$ $t h i s i s t h e c a s e 1 w i t h$ $μ = \tan θ a n d \frac{b}{L} = \frac{L - h}{L} = 1 - \frac{h}{L} = 0.4$ $p o i n t 3 :$ $t h i s i s t h e s i t u a t i o n w h e n t h e f r i c t i o n$ $r e a c h e s i t s m a x i m u m a n d t h e p a r t o f$ $r o p e w h i c h l i e s o n t h e r o p e i s t h e$ $m i n i m u m .$ $p o i n t 2 :$ $t h i s i s t h e s i t u a t i o n w h e n t h e p a r t o f$ $r o p e o n t h e s l o p e h a s i t s m a x i m u m$ $l e n g t h . w e c a n n o t p u t m o r e r o p e o n$ $t h e s l o p e a n d s t i l l k e e p i t i n e q u i l i b r i u m .$
	Commented by mr W last updated on 29/May/22

	Commented by mr W last updated on 29/May/22

	Commented by mr W last updated on 29/May/22

	Commented by mr W last updated on 29/May/22

	Commented by mr W last updated on 29/May/22

	$c a s e 3 : μ < \tan θ$ $t h e f r i c t i o n f o r c e i s t o o s m a l l s o t h a t$ $t h e h a n g i n g p a r t o f t h e r o p e m u s t$ $h o l d t h e t h e p a r t l y i n g o n t h e s l o p e .$
	Commented by mr W last updated on 29/May/22

	Commented by mr W last updated on 29/May/22

	Commented by mr W last updated on 29/May/22

	$s a y t h e l e n g t h o f t h e r o p e p a r t w h i c h$ $l i e s o n t h e i n c l i n e d p l a n e i s b .$ $t h e f r i c t i o n f o r c e b e t w e e n r o p e a n d$ $t h e s l o p e i s f .$ $f = μ ρ g b \cos θ w i t h 0 ⩽ μ < \tan θ$ $t e n s i o n i n t h e r o p e a t p o i n t B :$ $T_{1} = ρ g b \sin θ - f = (\sin θ - μ \cos θ) ρ g b$ $l e t μ^{'} = μ \cos θ - \sin θ$ $T_{1} = - μ^{'} ρ g b$ $T_{0} = T_{1} \cos θ = - μ^{'} ρ g b \cos θ$ $d e f i n e a = \frac{T_{0}}{ρ g} = - μ^{'} b \cos θ$ $t h e h a n g i n g r o p e f r o m p o i n t A t o B$ $i s p a r t o f a c a t e n a r y . w . r . t . t h e$ $c o o r d i n a t e s y s t e m a s s h o w n,$ $y = a \cosh \frac{x}{a}$ $y^{'} = \sinh \frac{x}{a}$ $s = a \sinh \frac{x}{a}$ $\tan θ = \sinh \frac{x_{B}}{a} \Rightarrow \frac{x_{B}}{a} = \sinh^{- 1} (\tan θ)$ $s_{1} = \overset{⌢}{C B} = a \sinh \frac{x_{B}}{a} = a \tan θ$ $\overset{⌢}{A C} = L - b + s_{1} = a \sinh \frac{x_{A}}{a}$ $L - b + a \tan θ = a \sinh \frac{x_{A}}{a}$ $\Rightarrow \frac{x_{A}}{a} = \sinh^{- 1} (\frac{L - b}{a} + \tan θ)$ $y_{A} - y_{B} = h + (x_{A} - x_{B}) \tan θ$ $a \cosh \frac{x_{A}}{a} - a \cosh \frac{x_{B}}{a} = h + (x_{A} - x_{B}) \tan θ$ $\cosh \frac{x_{A}}{a} - \cosh \frac{x_{B}}{a} = \frac{h}{a} + (\frac{x_{A}}{a} - \frac{x_{B}}{a}) \tan θ$ $\cosh [\sinh^{- 1} (\frac{L - b}{a} + \tan θ)] - \cosh [\sinh^{- 1} (\tan θ)] = \frac{h}{a} + [\sinh^{- 1} (\frac{L - b}{a} + \tan θ) - \sinh^{- 1} (\tan θ)] \tan θ$ $l e t \frac{b}{L} = \frac{1}{λ}$ $\cosh [\sinh^{- 1} (- \frac{λ - 1}{μ^{'} \cos θ} + \tan θ)] - \cosh [\sinh^{- 1} (\tan θ)] = - \frac{λ h}{μ^{'} L \cos θ} + [\sinh^{- 1} (- \frac{λ - 1}{μ^{'} \cos θ} + \tan θ) - \sinh^{- 1} (\tan θ)] \tan θ$ $\Rightarrow [\sinh^{- 1} (\frac{λ - 1}{μ^{'} \cos θ} - \tan θ) + \sinh^{- 1} (\tan θ)] \sin θ = 1 - \frac{λ h}{μ^{'}} - \sqrt{1 + {(\frac{λ - 1}{μ^{'}})}^{2} - 2 (\frac{λ - 1}{μ^{'}}) \sin θ}$ $t h i s e q u a t i o n s h o w s t h e r e l a t i o n s h i p$ $b e t w e e n λ a n d t h e a c t u a l f r i c t i o n$ $c o e f f i c i e n t μ (0 ⩽ μ < \tan θ) .$ $g e n e r a l l y t h i s r e l a t i o n s h i p l o o k s l i k e$ $a s s h o w n i n t h e d i a g r a m b e l o w .$ $e x a m p l e :$ $h = 3 m, L = 5 m, θ = 30 °, μ_{m a x} = 1.5$ $\Rightarrow \frac{b_{m i n}}{L} = 0.22196 a t μ = 0 (p o i n t 4)$ $\Rightarrow \frac{b_{m a x}}{L} = 0.43165 a t μ = 0.7 (p o i n t 2)$
	Commented by mr W last updated on 30/May/22

	Commented by Tawa11 last updated on 06/Oct/22

	$Great sir$
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