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Question Number 51700 by mr W last updated on 29/Dec/18

Commented by mr W last updated on 29/Dec/18

$A r o p e w i t h u n i f o r m m a s s i s f i x e d$ $a t o n e e n d o n p o i n t A . A t p o i n t B t h e$ $r o p e i s l a i d o v e r a p u l l e y . A l l c o n t a c t$ $i s f r i c t i o n l e s s a n d t h e s i z e o f p u l l e y$ $i s n e g l i g i b l e .$ $F i n d t h e m i n i m u m l e n g t h o f t h e r o p e$ $s u c h t h a t t h e r o p e r e m a i n s i n$ $e q u i l i b r i u m a s s h o w n i n t h e p i c t u r e .$
	Answered by mr W last updated on 30/Dec/18

	Commented by mr W last updated on 30/Dec/18

	$r o p e l e n g t h A B = l$ $r o p e l e n g t h B C = h$ $t o t a l r o p e l e n g t h = L = l + h$ $t e n s i o n i n r o p e a t B = T$ $T = h ρ g$ $T_{0} = T \cos θ$ $a = \frac{T_{0}}{ρ g} = h \cos θ$ $\Rightarrow h = \frac{a}{\cos θ}$ $y = a \cosh \frac{x}{a}$ $y^{'} = \sinh \frac{x}{a}$ $s = a \sinh \frac{x}{a}$ $a t B :$ $y^{'} = \sinh \frac{b}{2 a} = \tan θ$ $\Rightarrow a = \frac{b}{2 \sinh^{- 1} (\tan θ)}$ $\frac{l}{2} = a \sinh \frac{b}{2 a} = a \tan θ$ $\Rightarrow l = 2 a \tan θ$ $\Rightarrow L = l + h = 2 a \tan θ + \frac{a}{\cos θ} = a (2 \tan θ + \frac{1}{\cos θ})$ $\Rightarrow L = \frac{b}{2 \sinh^{- 1} (\tan θ)} (2 \tan θ + \sqrt{1 + \tan^{2} θ})$ $l e t t = \tan θ$ $\Rightarrow \frac{L}{b} = f (t) = \frac{2 t + \sqrt{1 + t^{2}}}{2 \sinh^{- 1} t} = \frac{2 t + \sqrt{1 + t^{2}}}{2 \ln (t + \sqrt{1 + t^{2}})}$ $f^{'} (t) = 0 \Rightarrow t = 1.0186 \Rightarrow θ = 45.528 °$ $\frac{L_{m i n}}{b} = m i n . f (t) \approx 1.9367$ $\Rightarrow L_{m i n} \approx 1.9367 b$
	Commented by ajfour last updated on 30/Dec/18

	$s u p e r t h a n k s f o r t h e h y p e r b o l i c w a y S i r!$
	Commented by ajfour last updated on 30/Dec/18
	$Tcos θ=T_0 =T_B cos α ⇒ dTcos θ = Tsin θdθ ⇒ dT = ((Tsin θdθ)/(cos θ)) ...(i) Also (T+dT)sin (θ+dθ)−Tsin θ=ρgdl ⇒ dTsin θ+Tcos θdθ = ρgdl using (i): ((Tsin^2 θdθ)/(cos θ))+Tcos θdθ = ρgdl ⇒ Tdθ = ρgcos θdl = ρgdx ...(ii) and T_B sin α = ρg((l/2)) ...(iii) ⇒ ((Tcos θdθ)/(cos θ)) = ρgdx or (((T_B cos α)sin α)/(cos θ.sin α))dθ = ρgdx using (iii) ((ρg((l/2))cos α)/(cos θ.sin α))dθ = ρgdx (l/2)((cos α)/(sin α))∫_0 ^( θ) sec θ = x ⇒ ln(sec θ+tan θ)= ((2xtan α)/l) (√(1+((dy/dx))^2 ))+(dy/dx) = e^((2xtan α)/l) squaring 1= e^((4xtan α)/l) −2((dy/dx))e^((2xtan α)/l) ⇒ (dy/dx) = (1/2)(e^((2xtan α)/l) −e^(−((2xtan α)/l)) ) and this at B (x=(b/2)), gives 2tan α = e^((btan α)/l) −e^(−((btan α)/l)) 2m= t−(1/t) (say) ⇒ t^2 −2mt −1 = 0 or (t−m)^2 = 1+m^2 t = e^((bm)/l) = m+(√(1+m^2 )) ⇒ ((btan α)/l) = ln( tan α+sec α) ...(I) further T_B = 𝛒gz = ((𝛒gl)/(2sin 𝛂)) or z = (l/(2sin α)) (z being the vertically hanging length) L = l+z = l(1+(1/(2sin α))) and from (I) (4 lines above) L = ((btan α)/(ln(tan α+sec α)))(((2sin α+1)/(2sin α))) L = ((b(2sin α+1))/(2cos α ln(tan α+sec α))) ...(A) ⇒ (b/(2L)) = ((cos α ln(tan α+sec α))/((2sin α+1))) (d/dα)((b/(2L))) = 0 ⇒ ((cos α)/((2sin α+1)))×sec α +(((2sin α+1)(−sin α)−2cos^2 α)/((2sin α+1)^2 )) ln(tan α+sec α)= 0 ___________________________ ⇒ ln(sec α+tan α)= ((2sin α+1)/(2+sin α)) ..(B) ⇒ { α = 45.5274° } ___________________________ utilizing this in (A) L = ((b(2sin α+1))/(2cos α))×(((2+sin α))/((2sin α+1))) ⇒ L = ((b(2+sin α))/(2cos α)) = 1.9367b .$
	$T \cos θ = T_{0} = T_{B} \cos α$ $\Rightarrow d T \cos θ = T \sin θ d θ$ $\Rightarrow d T = \frac{T \sin θ d θ}{\cos θ} . . . (i)$ $A l s o$ $(T + d T) \sin (θ + d θ) - T \sin θ = ρ g d l$ $\Rightarrow d T \sin θ + T \cos θ d θ = ρ g d l$ $u s i n g (i) :$ $\frac{T \sin^{2} θ d θ}{\cos θ} + T \cos θ d θ = ρ g d l$ $\Rightarrow T d θ = ρ g \cos θ d l = ρ g d x . . . (i i)$ $a n d T_{B} \sin α = ρ g (\frac{l}{2}) . . . (i i i)$ $\Rightarrow \frac{T \cos θ d θ}{\cos θ} = ρ g d x$ $o r \frac{(T_{B} \cos α) \sin α}{\cos θ . \sin α} d θ = ρ g d x$ $u s i n g (i i i)$ $\frac{ρ g (\frac{l}{2}) \cos α}{\cos θ . \sin α} d θ = ρ g d x$ $\frac{l}{2} \frac{\cos α}{\sin α} \int_{0}^{θ} \sec θ = x$ $\Rightarrow l n (\sec θ + \tan θ) = \frac{2 x \tan α}{l}$ $\sqrt{1 + {(\frac{d y}{d x})}^{2}} + \frac{d y}{d x} = e^{\frac{2 x \tan α}{l}}$ $s q u a r i n g$ $1 = e^{\frac{4 x \tan α}{l}} - 2 (\frac{d y}{d x}) e^{\frac{2 x \tan α}{l}}$ $\Rightarrow \frac{d y}{d x} = \frac{1}{2} (e^{\frac{2 x \tan α}{l}} - e^{- \frac{2 x \tan α}{l}})$ $a n d t h i s a t B (x = \frac{b}{2}), g i v e s$ $2 \tan α = e^{\frac{b \tan α}{l}} - e^{- \frac{b \tan α}{l}}$ $2 m = t - \frac{1}{t} (s a y)$ $\Rightarrow t^{2} - 2 m t - 1 = 0$ $o r {(t - m)}^{2} = 1 + m^{2}$ $t = e^{\frac{b m}{l}} = m + \sqrt{1 + m^{2}}$ $\Rightarrow \frac{b \tan α}{l} = l n (\tan α + \sec α) . . . (I)$ $f u r t h e r T_{B} = ρ g z = \frac{ρ g l}{2 \sin α}$ $o r z = \frac{l}{2 \sin α}$ $(z b e i n g t h e v e r t i c a l l y h a n g i n g l e n g t h)$ $L = l + z = l (1 + \frac{1}{2 \sin α})$ $a n d f r o m (I) (4 l i n e s a b o v e)$ $L = \frac{b \tan α}{l n (\tan α + \sec α)} (\frac{2 \sin α + 1}{2 \sin α})$ $L = \frac{b (2 \sin α + 1)}{2 \cos α l n (\tan α + \sec α)} . . . (A)$ $\Rightarrow \frac{b}{2 L} = \frac{\cos α l n (\tan α + \sec α)}{(2 \sin α + 1)}$ $\frac{d}{d α} (\frac{b}{2 L}) = 0 \Rightarrow$ $\frac{\cos α}{(2 \sin α + 1)} \times \sec α$ $+ \frac{(2 \sin α + 1) (- \sin α) - 2 \cos^{2} α}{{(2 \sin α + 1)}^{2}} l n (\tan α + \sec α) = 0$ $___________________________$ $\Rightarrow l n (\sec α + \tan α) = \frac{2 \sin α + 1}{2 + \sin α} . . (B)$ $\Rightarrow {α = 45.5274 °}$ $___________________________$ $u t i l i z i n g t h i s i n (A)$ $L = \frac{b (2 \sin α + 1)}{2 \cos α} \times \frac{(2 + \sin α)}{(2 \sin α + 1)}$ $\Rightarrow L = \frac{b (2 + \sin α)}{2 \cos α} = 1.9367 b .$
	Commented by mr W last updated on 30/Dec/18

	${i t}^{'} s a m a r v e l o u s j o b t o d e r i v e t h e e q n .$ $f r o m s c r a t c h! t h a n k s a l o t s i r!$
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