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Question Number 187675 by mr W last updated on 20/Feb/23

Commented by mr W last updated on 20/Feb/23

$a n u n i f o r m r o p e o f l e n g t h L a n d$ $m a s s M i s f i x e d o n b o t h e n d s a t a$ $d i s t a n c e d a s s h o w n . a s m a l l o b j e c t$ $o f m a s s m m o v e s v e r y s l o w l y a l o n g$ $t h e r o p e f r o m o n e e n d t o t h e o t h e r .$ $f i n d t h e l o c u s o f t h e s m a l l o b j e c t$ $a n d t h e t o t a l d i s t a n c e i t m o v e d .$
	Answered by mr W last updated on 22/Feb/23

	Commented by mr W last updated on 22/Feb/23

	$\underset{―}{r o p e w i t h o u t s m a l l o b j e c t}$ $y = a \cosh \frac{x}{a}$ $a t x = \frac{d}{2} :$ $\frac{L}{2} = a \sinh \frac{d}{2 a}$ $\Rightarrow \frac{\sinh \frac{d}{2 a}}{\frac{d}{2 a}} = \frac{L}{d}$ $w e c a n s o l v e f o r \frac{d}{2 a} a n d g e t a .$
	Commented by mr W last updated on 24/Feb/23
	$small object at distance d_1 to left end and deflection f T_1 cos θ_1 =T_2 cos θ_2 =T_0 T_2 sin θ_2 −T_1 sin θ_1 =mg T_0 (tan θ_2 −tan θ_1 )=mg T_0 =((mg)/(tan θ_2 −tan θ_1 )) a=(T_0 /(ρg))=((mgL)/(Mg(tan θ_2 −tan θ_1 )))=((μL)/(tan θ_2 −tan θ_1 )) with μ=(m/M) for rope L_1 : y=a cosh (x/a) at point C: tan θ_1 =sinh (x_C /a) ⇒(x_C /a)=sinh^(−1) (tan θ_1 ) y_C =a cosh (x_C /a)=a (√(1+tan^2 θ_1 ))=(a/(cos θ_1 )) at point A: y_C +f=a cosh ((x_C −d_1 )/a) ⇒(f/a)=cosh (sinh^(−1) (tan θ_1 )−(d_1 /a))−(1/(cos θ_1 )) L_1 =a(sinh (x_C /a)−sinh ((x_C −d_1 )/a)) ⇒(L_1 /a)=tan θ_1 −sinh (sinh^(−1) (tan θ_1 )−(d_1 /a)) for rope L_2 : y=a cosh (x/a) at point C: tan θ_2 =sinh (x_C /a) ⇒(x_C /a)=sinh^(−1) (tan θ_2 ) y_C =a cosh (x_C /a)=a (√(1+tan^2 θ_2 ))=(a/(cos θ_2 )) at point B: y_C +f=a cosh ((x_C +d_2 )/a) ⇒(f/a)=cosh (sinh^(−1) (tan θ_2 )+(d_2 /a))−(1/(cos θ_2 )) L_2 =a(sinh ((x_C +d_2 )/a)−sinh (x_C /a)) ⇒(L_2 /a)=sinh (sinh^(−1) (tan θ_2 )+(d_2 /a))−tan θ_2 d_2 =d−d_1 L_1 +L_2 =L cosh [sinh^(−1) (tan θ_1 )−(((tan θ_2 −tan θ_1 )d_1 )/(μL))]−(1/(cos θ_1 ))=cosh [sinh^(−1) (tan θ_2 )+(((tan θ_2 −tan θ_1 )(d−d_1 ))/(μL))]−(1/(cos θ_2 )) sinh [sinh^(−1) (tan θ_1 )−(((tan θ_2 −tan θ_1 )d_1 )/(μL))]−sinh [sinh^(−1) (tan θ_2 )+(((tan θ_2 −tan θ_1 )(d−d_1 ))/(μL))]+(1+(1/μ))(tan θ_2 −tan θ_1 )=0 (f/a)=cosh (sinh^(−1) (tan θ_1 )−(d_1 /a))−(1/(cos θ_1 )) with ξ=(d_1 /d), η=(f/d), λ=((μL)/d), μ=(m/M) cosh [sinh^(−1) (tan θ_1 )−(((tan θ_2 −tan θ_1 )ξ)/λ)]−(1/(cos θ_1 ))=cosh [sinh^(−1) (tan θ_2 )+(((tan θ_2 −tan θ_1 )(1−ξ))/λ)]−(1/(cos θ_2 )) ...(i) sinh [sinh^(−1) (tan θ_1 )−(((tan θ_2 −tan θ_1 )ξ)/λ)]−sinh [sinh^(−1) (tan θ_2 )+(((tan θ_2 −tan θ_1 )(1−ξ))/λ)]+(1+(1/μ))(tan θ_2 −tan θ_1 )=0 ...(ii) η=(λ/(tan θ_2 −tan θ_1 )){cosh [sinh^(−1) (tan θ_1 )−(((tan θ_2 −tan θ_1 )ξ)/λ)]−(1/(cos θ_1 ))} ...(iii) for given ξ∈[0,1] we can solve (i) and (ii) for θ_1 and θ_2 and then get η from (iii). η=η(ξ) is the locus of the small object.$
	$\underset{―}{s m a l l o b j e c t a t d i s t a n c e d_{1} t o l e f t}$ $\underset{―}{e n d a n d d e f l e c t i o n f}$ $T_{1} \cos θ_{1} = T_{2} \cos θ_{2} = T_{0}$ $T_{2} \sin θ_{2} - T_{1} \sin θ_{1} = m g$ $T_{0} (\tan θ_{2} - \tan θ_{1}) = m g$ $T_{0} = \frac{m g}{\tan θ_{2} - \tan θ_{1}}$ $a = \frac{T_{0}}{ρ g} = \frac{m g L}{M g (\tan θ_{2} - \tan θ_{1})} = \frac{μ L}{\tan θ_{2} - \tan θ_{1}}$ $w i t h μ = \frac{m}{M}$ $\underset{―}{f o r r o p e L_{1} :}$ $y = a \cosh \frac{x}{a}$ $a t p o i n t C :$ $\tan θ_{1} = \sinh \frac{x_{C}}{a}$ $\Rightarrow \frac{x_{C}}{a} = \sinh^{- 1} (\tan θ_{1})$ $y_{C} = a \cosh \frac{x_{C}}{a} = a \sqrt{1 + \tan^{2} θ_{1}} = \frac{a}{\cos θ_{1}}$ $a t p o i n t A :$ $y_{C} + f = a \cosh \frac{x_{C} - d_{1}}{a}$ $\Rightarrow \frac{f}{a} = \cosh (\sinh^{- 1} (\tan θ_{1}) - \frac{d_{1}}{a}) - \frac{1}{\cos θ_{1}}$ $L_{1} = a (\sinh \frac{x_{C}}{a} - \sinh \frac{x_{C} - d_{1}}{a})$ $\Rightarrow \frac{L_{1}}{a} = \tan θ_{1} - \sinh (\sinh^{- 1} (\tan θ_{1}) - \frac{d_{1}}{a})$ $\underset{―}{f o r r o p e L_{2} :}$ $y = a \cosh \frac{x}{a}$ $a t p o i n t C :$ $\tan θ_{2} = \sinh \frac{x_{C}}{a}$ $\Rightarrow \frac{x_{C}}{a} = \sinh^{- 1} (\tan θ_{2})$ $y_{C} = a \cosh \frac{x_{C}}{a} = a \sqrt{1 + \tan^{2} θ_{2}} = \frac{a}{\cos θ_{2}}$ $a t p o i n t B :$ $y_{C} + f = a \cosh \frac{x_{C} + d_{2}}{a}$ $\Rightarrow \frac{f}{a} = \cosh (\sinh^{- 1} (\tan θ_{2}) + \frac{d_{2}}{a}) - \frac{1}{\cos θ_{2}}$ $L_{2} = a (\sinh \frac{x_{C} + d_{2}}{a} - \sinh \frac{x_{C}}{a})$ $\Rightarrow \frac{L_{2}}{a} = \sinh (\sinh^{- 1} (\tan θ_{2}) + \frac{d_{2}}{a}) - \tan θ_{2}$ $d_{2} = d - d_{1}$ $L_{1} + L_{2} = L$ $\cosh [\sinh^{- 1} (\tan θ_{1}) - \frac{(\tan θ_{2} - \tan θ_{1}) d_{1}}{μ L}] - \frac{1}{\cos θ_{1}} = \cosh [\sinh^{- 1} (\tan θ_{2}) + \frac{(\tan θ_{2} - \tan θ_{1}) (d - d_{1})}{μ L}] - \frac{1}{\cos θ_{2}}$ $\sinh [\sinh^{- 1} (\tan θ_{1}) - \frac{(\tan θ_{2} - \tan θ_{1}) d_{1}}{μ L}] - \sinh [\sinh^{- 1} (\tan θ_{2}) + \frac{(\tan θ_{2} - \tan θ_{1}) (d - d_{1})}{μ L}] + (1 + \frac{1}{μ}) (\tan θ_{2} - \tan θ_{1}) = 0$ $\frac{f}{a} = \cosh (\sinh^{- 1} (\tan θ_{1}) - \frac{d_{1}}{a}) - \frac{1}{\cos θ_{1}}$ $w i t h ξ = \frac{d_{1}}{d}, η = \frac{f}{d}, λ = \frac{μ L}{d}, μ = \frac{m}{M}$ $\cosh [\sinh^{- 1} (\tan θ_{1}) - \frac{(\tan θ_{2} - \tan θ_{1}) ξ}{λ}] - \frac{1}{\cos θ_{1}} = \cosh [\sinh^{- 1} (\tan θ_{2}) + \frac{(\tan θ_{2} - \tan θ_{1}) (1 - ξ)}{λ}] - \frac{1}{\cos θ_{2}} . . . (i)$ $\sinh [\sinh^{- 1} (\tan θ_{1}) - \frac{(\tan θ_{2} - \tan θ_{1}) ξ}{λ}] - \sinh [\sinh^{- 1} (\tan θ_{2}) + \frac{(\tan θ_{2} - \tan θ_{1}) (1 - ξ)}{λ}] + (1 + \frac{1}{μ}) (\tan θ_{2} - \tan θ_{1}) = 0 . . . (i i)$ $η = \frac{λ}{\tan θ_{2} - \tan θ_{1}} {\cosh [\sinh^{- 1} (\tan θ_{1}) - \frac{(\tan θ_{2} - \tan θ_{1}) ξ}{λ}] - \frac{1}{\cos θ_{1}}} . . . (i i i)$ $f o r g i v e n ξ \in [0, 1] w e c a n s o l v e (i) a n d$ $(i i) f o r θ_{1} a n d θ_{2} a n d t h e n g e t η$ $f r o m (i i i) . η = η (ξ) i s t h e l o c u s o f t h e$ $s m a l l o b j e c t .$
	Commented by mr W last updated on 24/Feb/23

	Commented by mr W last updated on 24/Feb/23

	Commented by mr W last updated on 24/Feb/23

	Commented by mr W last updated on 24/Feb/23

	Commented by mr W last updated on 24/Feb/23

	Commented by mr W last updated on 24/Feb/23

	Commented by mr W last updated on 25/Feb/23

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