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Question Number 98461 by mr W last updated on 14/Jun/20

Commented by mr W last updated on 14/Jun/20

$A l o n g r o p e w i t h m a s s m a n d l e n g t h$ $L i s f i x e d a t o n e e n d o n t h e c e i l i n g$ $a t p o i n t A . A t i t s o t h e r e n d a m a s s M$ $i s c o n n e c t e d . T h e l o n g r o p e i s h a n g e d$ $o n t h e c e i l i n g t h r o u g h a n o t h e r t h i n,$ $s h o r t a n d m a s s l e s s r o p e a t p o i n t B .$ $F i n d t h e t e n s i o n i n t h e s h o r t r o p e .$
	Answered by 1549442205 last updated on 14/Jun/20

	Commented by 1549442205 last updated on 15/Jun/20

	$Suppose that the rope L is tied at the end C$ $so that AC = c . Then from {Herong}^{'} s formula$ $we have h_{c} = \frac{2}{b} \sqrt{p (p - a) (p - b) (p - c)}, where$ $p = \frac{a + b + c}{2} . From that \cos \hat{MCH} = \cos \hat{BCD} = \frac{h_{c}}{a}$ $i) Consider the case that a^{2} + c^{2} = b^{2} we have$ $\cos \hat{MCH} = \cos \hat{BAC} = \frac{c}{b} = \frac{\sqrt{a^{2} - b^{2}}}{b} . Henxe,$ $The tension of the rope BC is$ $T = CH = CM . \cos \hat{MCH} = (M + m) g . \frac{\sqrt{b^{2} - a^{2}}}{b} (m - the mass of the rope L)$ $ii) The case L ⩾ c ⩾ a^{2} + b^{2} then T = (M + m) g because CB ⊥ AB$ $]$
	Commented by mr W last updated on 14/Jun/20

	$t h a n k y o u s i r .$ $b u t {i t}^{'} s n o t c o r r e c t .$ $t h e l o n g r o p e h a s m a s s, t h e r e f o r e$ $A C i s n o t a s t r a i g h t, b u t a c u r v e .$
	Commented by mr W last updated on 14/Jun/20

	$e v e n i f t h e l o n g r o p e w e r e m a s s l e s s,$ $b o t h r o p e s a r e n o t t i e d a t p o i n t C,$ $b u t c o n n e c t e d w i t h a s m a l l p u l l e y,$ $t h a t m e a n s l e n g t h c i s n o t g i v e n, b u t$ $t o d e t e r m i n e f r o m t h e c o n d i t i o n t h a t$ $t h e t e n s i o n i n A C i s e q u a l t o t h e$ $t e n s i o n i n C M .$
	Answered by mr W last updated on 15/Jun/20

	Commented by mr W last updated on 15/Jun/20

	$c a s e m = 0, i . e . l o n g r o p e m a s s l e s s$ $2 β + \frac{π}{2} - α = π$ $\Rightarrow α = 2 β - \frac{π}{2}$ $\frac{a}{\sin α} = \frac{b}{\sin (α + \frac{π}{2} - β)}$ $- \frac{a}{\cos 2 β} = \frac{b}{\sin β}$ $- \frac{a}{1 - 2 \sin^{2} β} = \frac{b}{\sin β}$ $2 \sin^{2} β - \frac{a}{b} \sin β - 1 = 0$ $\Rightarrow \sin β = \frac{1}{4} [\frac{a}{b} + \sqrt{{(\frac{a}{b})}^{2} + 8}]$ $T_{C} = M g$ $T = 2 T_{C} \cos β$ $\Rightarrow T = 2 M g \sqrt{1 - \frac{1}{16} {[\frac{a}{b} + \sqrt{{(\frac{a}{b})}^{2} + 8}]}^{2}}$
	Answered by mr W last updated on 15/Jun/20

	Commented by mr W last updated on 15/Jun/20

	$c a s e m \neq 0$
	Commented by mr W last updated on 16/Jun/20

	$i n t h i s c a s e t h e r o p e s e c t i o n A C i s$ $a p a r t o f t h e c a t e n a r y .$ $y = h \cosh \frac{x}{h}$ $w i t h h = \frac{T_{0}}{ρ g} = \frac{T_{C} \cos α}{ρ g} a n d ρ = \frac{m}{L}$ $α = \frac{π}{2} - 2 β$ $\tan α = y^{'} = \sinh \frac{e}{h} = \frac{1}{\tan 2 β}$ $\Rightarrow e = h \sinh^{- 1} \frac{1}{\tan 2 β}$ $f = h \cosh \frac{e}{h} = h \cosh (\sinh^{- 1} \frac{1}{\tan 2 β})$ $c = b - a \sin β - e = b - a \sin β - h \sinh^{- 1} \frac{1}{\tan 2 β}$ $r o p e l e n g t h \overset{⌢}{A E} = s_{1}$ $s_{1} = h \sinh \frac{c}{h} = h \sinh (\frac{b - a \sin β}{h} - \sinh^{- 1} \frac{1}{\tan 2 β})$ $r o p e l e n g t h \overset{⌢}{E C} = s_{2}$ $s_{2} = h \sinh \frac{e}{h} = \frac{h}{\tan 2 β}$ $r o p e l e n g t h C D = s_{3}$ $s_{3} = L - s_{1} - s_{2}$ $s_{3} = L - h \sinh (\frac{b - a \sin β}{h} - \sinh^{- 1} \frac{1}{\tan 2 β}) - \frac{h}{\tan 2 β}$ $T_{C} = M g + ρ {g s}_{3}$ $T_{C} = M g + ρ g [L - h \sinh (\frac{b - a \sin β}{h} - \sinh^{- 1} \frac{1}{\tan 2 β}) - \frac{h}{\tan 2 β}]$ $T_{C} = (M + m) g - h ρ g [\sinh (\frac{b - a \sin β}{h} - \sinh^{- 1} \frac{1}{\tan 2 β}) + \frac{1}{\tan 2 β}]$ $\frac{h ρ g}{\sin 2 β} = (M + m) g - h ρ g [\sinh (\frac{b - a \sin β}{h} - \sinh^{- 1} \frac{1}{\tan 2 β}) + \frac{1}{\tan 2 β}]$ $\frac{1 + \cos 2 β}{\sin 2 β} = \frac{(M + m) L}{m h} - \sinh (\frac{b - a \sin β}{h} - \sinh^{- 1} \frac{1}{\tan 2 β})$ $\Rightarrow \sinh (\frac{b - a \sin β}{h} - \sinh^{- 1} \frac{1}{\tan 2 β}) = \frac{(M + m) L}{m h} - \frac{1 + \cos 2 β}{\sin 2 β} . . . (i)$ $d = f + a \cos β$ $h \cosh \frac{c}{h} = h \cosh (\sinh^{- 1} \frac{1}{\tan 2 β}) + a \cos β$ $\Rightarrow \cosh (\frac{b - a \sin β}{h} - \sinh^{- 1} \frac{1}{\tan 2 β}) = \cosh (\sinh^{- 1} \frac{1}{\tan 2 β}) + \frac{a}{h} \cos β . . . (i i)$ $\Rightarrow {[\cosh (\sinh^{- 1} \frac{1}{\tan 2 β}) + \frac{a}{h} \cos β]}^{2} - {[\frac{(M + m) L}{m h} - \frac{1 + \cos 2 β}{\sin 2 β}]}^{2} = 1$ $\Rightarrow {[\cosh (\sinh^{- 1} \frac{1}{\tan 2 β}) h + a \cos β]}^{2} - {[\frac{(M + m) L}{m} - (\frac{1 + \cos 2 β}{\sin 2 β}) h]}^{2} = h^{2}$ $\Rightarrow {[\cosh (\sinh^{- 1} \frac{1}{\tan 2 β})]}^{2} h^{2} + a^{2} \cos^{2} β + 2 a \cos β \cosh (\sinh^{- 1} \frac{1}{\tan 2 β}) h - {[\frac{(M + m) L}{m}]}^{2} - {(\frac{1 + \cos 2 β}{\sin 2 β})}^{2} h^{2} + 2 (\frac{1 + \cos 2 β}{\sin 2 β}) \frac{(M + m) L}{m} h = h^{2}$ $\Rightarrow (\frac{1 + 2 \cos 2 β}{\sin^{2} 2 β}) h^{2} - \frac{1}{\sin β} [a \sqrt{1 + \cos^{2} 2 β} + (\frac{1 + \cos 2 β}{\cos β}) (\frac{M}{m} + 1) L] h + {[\frac{(M + m) L}{m}]}^{2} - a^{2} \cos^{2} β = 0 . . . (i i i)$ $f r o m (i i i) w e g e t h i n t e r m s o f β .$ $p u t t h i s i n t o (i) o r (i i) w e g e t β .$ $T_{C} = \frac{ρ g h}{\sin 2 β}$ $\Rightarrow T = 2 T_{C} \cos β = \frac{ρ g h}{\sin β}$
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