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Question Number 81804 by ajfour last updated on 15/Feb/20

Commented by ajfour last updated on 15/Feb/20

$I f t e n s i o n i n s t r i n g, T = \frac{m g}{5}$ $a n d r o d r e s t s i n e q u i l i b r i u m,$ $f i n d l e n g t h o f s t r i n g .$
	Answered by mr W last updated on 15/Feb/20

	Commented by mr W last updated on 16/Feb/20

	$β + θ = \frac{π}{2} - α$ $\Rightarrow β = \frac{π}{2} - (θ + α)$ $γ + θ = \frac{π}{2} + α$ $\Rightarrow γ = \frac{π}{2} - (θ - α)$ $\frac{B D}{\sin γ} = \frac{D C}{\sin β} = \frac{B C}{\sin 2 θ}$ $B D = \frac{L \sin γ}{2 \sin 2 θ} = \frac{L \cos (θ - α)}{2 \sin 2 θ}$ $D C = \frac{L \sin β}{2 \sin 2 θ} = \frac{L \cos (θ + α)}{2 \sin 2 θ}$ $l = B D + D C = \frac{L [\cos (θ - α) + \cos (θ + α)]}{2 \sin 2 θ}$ $\Rightarrow l = \frac{L \cos α}{2 \sin θ} . . . (i i i)$ $B D \sin θ = e \cos α$ $\frac{L \cos (θ - α) \sin θ}{2 \sin 2 θ} = e \cos α$ $e = \frac{L \cos (θ - α)}{4 \cos θ \cos α} = \frac{L}{4} (1 + \tan θ \tan α)$ $l e t T = \frac{m g}{τ} w i t h τ = 5 g i v e n$ $F = 2 T \cos θ$ $F (\frac{L}{2} + e) = m g \times \frac{L}{2}$ $\Rightarrow 3 \cos θ + \sin θ \tan α = τ . . . (i)$ $\frac{L \sin α}{2} + B D \times \cos θ = h$ $2 \sin α + \frac{\cos (θ - α)}{\sin θ} = \frac{4 h}{L}$ $w i t h η = \frac{h}{L}$ $\Rightarrow 3 \sin α + \frac{\cos α}{\tan θ} = 4 η . . . (i i)$ $f r o m (i) :$ $\frac{3}{\sqrt{9 + \tan^{2} α}} \cos θ + \sin θ \frac{\tan α}{\sqrt{9 + \tan^{2} α}} = \frac{τ}{\sqrt{9 + \tan^{2} α}}$ $\cos (θ - ϕ) = \frac{τ}{\sqrt{9 + \tan^{2} α}} w i t h \tan ϕ = \frac{\tan α}{3}$ $\Rightarrow θ = \tan^{- 1} (\frac{\tan α}{3}) + \cos^{- 1} (\frac{τ}{\sqrt{9 + \tan^{2} α}})$ $p u t t h i s i n t o (i i) w e c a n g e t α .$ $w i t h α a n d θ, w e g e t t h e l e n g t h o f$ $t h e s t r i n g a c c . t o (i i i) : l = \frac{L \cos α}{2 \sin θ} .$ $e x a m p l e : τ = \frac{m g}{T} = 3, η = \frac{h}{L} = \frac{4}{5} = 0.80$ $\Rightarrow α = 39.6988 °$ $\Rightarrow θ = 30.9364 °$ $\Rightarrow l = 3.7416 m$ $===================$ $i f s t r i n g l e n g t h l i s g i v e n, f i n d T = ?$ $f r o m (i i i) w e g e t$ $\frac{\cos α}{\sin θ} = \frac{2 l}{L} = λ$ $\Rightarrow \sin θ = \frac{\cos α}{λ}$ $p u t t h i s i n t o (i i) :$ $3 \sin α + \sqrt{λ^{2} - \cos^{2} α} = 4 η$ $\sqrt{λ^{2} - 1 + \sin^{2} α} = 4 η - 3 \sin α$ $8 \sin^{2} α - 24 η \sin α + 16 η^{2} + 1 - λ^{2} = 0$ $\Rightarrow \sin α = \frac{6 η - \sqrt{2 (2 η^{2} + λ^{2} - 1)}}{4}$ $w i t h α a n d θ w e c a n g e t f r o m (i) :$ $τ = 3 \cos θ + \sin θ \tan α$ $\Rightarrow τ = \frac{1}{λ} (3 \sqrt{λ^{2} - 1 + \sin^{2} α} + \sin α)$ $e x a m p l e : L = 4 m, h = 3 m, l = 3 m$ $η = \frac{h}{L} = \frac{3}{4} = 0.75, λ = \frac{2 l}{L} = \frac{6}{4} = 1.5$ $\sin α = \frac{6 η - \sqrt{2 (2 η^{2} + λ^{2} - 1)}}{4} = 0.580137$ $\Rightarrow τ = 2.905932$ $\Rightarrow T = \frac{m g}{τ} = 0.344 m g$
	Commented by ajfour last updated on 15/Feb/20

	$g r e a t w a y S i r, i t s r e a l l y n o t s o$ $c o m p l e x, t h a n k y o u, i t h i n k$ $m y s o l u t i o n a n s w e r s w e l l t o o .$
	Commented by mr W last updated on 15/Feb/20

	${t h a t}^{'} s s u r e s i r! y o u r s o l u t i o n s o l v e s$ $t h e p r o b l e m a l s o v e r y w e l l . m a n y$ $w a y s l e a d t o t h e s a m e g o a l!$
	Commented by mr W last updated on 16/Feb/20

	$f o r t h e c a s e : l i s g i v e n, f i n d T = ?$ $a n e x a c t s o l u t i o n i s p o s s i b l e, s e e a b o v e .$
	Answered by ajfour last updated on 15/Feb/20

	Commented by ajfour last updated on 15/Feb/20
	$I am enlisting just the eqs. that i have: h−((L )/2)sin θ=pcos θ ....(i) h−Lsin θ=qcos θ ...(ii) pcos α=qcos β psin α+qsin β=(L/2) cos α+2cos β=5cos θ α−β = 2θ (one of them is is dependent) cos α+((2pcos α)/q)=5cos θ ⇒ cos α=((5cos θ)/((1+2p/q))) cos β=((5(p/q)cos θ)/((1+2p/q))) p(√(1−{((5cos θ)/((1+2p/q)))}^2 ))+ q(√(1−{((5(p/q)cos θ)/((1+2p/q)))}^2 )) =(L/2) while we have p(θ) and q(θ), from first two eqs. ________________________$
	$I a m e n l i s t i n g j u s t t h e e q s .$ $t h a t i h a v e :$ $h - \frac{L}{2} \sin θ = p \cos θ . . . . (i)$ $h - L \sin θ = q \cos θ . . . (i i)$ $p \cos α = q \cos β$ $p \sin α + q \sin β = \frac{L}{2}$ $\cos α + 2 \cos β = 5 \cos θ$ $α - β = 2 θ$ $(o n e o f t h e m i s i s d e p e n d e n t)$ $\cos α + \frac{2 p \cos α}{q} = 5 \cos θ$ $\Rightarrow \cos α = \frac{5 \cos θ}{(1 + 2 p / q)}$ $\cos β = \frac{5 (p / q) \cos θ}{(1 + 2 p / q)}$ $p \sqrt{1 - {\frac{5 \cos θ}{(1 + 2 p / q)}}^{2}} +$ $q \sqrt{1 - {\frac{5 (p / q) \cos θ}{(1 + 2 p / q)}}^{2}} = \frac{L}{2}$ $w h i l e w e h a v e p (θ) a n d q (θ),$ $f r o m f i r s t t w o e q s .$ $________________________$
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