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Question Number 81968 by mr W last updated on 17/Feb/20

Commented by mr W last updated on 17/Feb/20

$[d e r i v e d f r o m Q 81804]$ $g i v e n :$ $s t r i n g l e n g t h = s (> 0.5 L)$ $f r i c t i o n c o e f f i c i e n t w i t h g r o u n d = μ$ $t o f i n d :$ $t h e m i n i m a l a n d m a x i m a l p o s s i b l e$ $i n c l i n a t i o n o f t h e r o d a n d t h e$ $c o r r e s p o n d i n g t e n s i o n i n t h e s t r i n g .$
	Answered by mr W last updated on 27/Feb/20

	Commented by mr W last updated on 27/Feb/20

	$C a s e I : m a x i m a l i n c l i n a t i o n$ $w i t h f r i c t i o n f o r c e a c t i n g r i g h t w a r d s$ $φ = \tan^{- 1} μ$ $b o t h r e a c t i o n s f r o m g r o u n d a n d f r o m$ $p u l l e y m e e t m g a t t h e s a m e p o i n t O .$ $β = \frac{π}{2} - (θ + α - ϕ)$ $γ = \frac{π}{2} - (θ - α + ϕ)$ $\frac{B C}{\sin γ} = \frac{C D}{\sin β} = \frac{B D}{\sin 2 θ}$ $\Rightarrow B C = \frac{L \cos (θ - α + ϕ)}{2 \sin 2 θ}$ $\Rightarrow C D = \frac{L \cos (θ + α - ϕ)}{2 \sin 2 θ}$ $B C + C D = s$ $\Rightarrow \frac{L \cos (θ - α + ϕ)}{2 \sin 2 θ} + \frac{L \cos (θ + α - ϕ)}{2 \sin 2 θ} = s$ $\Rightarrow L \cos θ \cos (α - ϕ) = s \sin 2 θ$ $\Rightarrow \sin θ = \frac{L \cos (α - ϕ)}{2 s}$ $l e t Θ = α - ϕ$ $\Rightarrow \sin θ = \frac{L \cos Θ}{2 s} . . . (i)$ $h = \frac{L \sin α}{2} + B C \times \cos (θ - ϕ)$ $h = \frac{L \sin α}{2} + \frac{L \cos (θ - α + ϕ) \cos (θ - ϕ)}{2 \sin 2 θ}$ $\frac{2 h}{L} = \sin α + \frac{\cos (θ - α + ϕ) \cos (θ - ϕ)}{\sin 2 θ}$ $\frac{2 h}{L} = \sin α + \frac{[\cos θ \cos (α - ϕ) + \sin θ \sin (α - ϕ)] (\cos θ \cos ϕ + \sin θ \sin ϕ)}{2 \sin θ \cos θ}$ $\Rightarrow \frac{4 h}{L} = 2 \sin α + [\cos (α - ϕ) + \tan θ \sin (α - ϕ)] (\frac{\cos ϕ}{\tan θ} + \sin ϕ)$ $\Rightarrow \frac{4 h}{L} = 2 \sin α + (\cos Θ + \tan θ \sin Θ) [\frac{\cos (α - Θ)}{\tan θ} + \sin (α - Θ)] . . . (i i)$ $G C = B C \times \sin (θ - ϕ)$ $G C = \frac{L \cos (θ - α + ϕ) \sin (θ - ϕ)}{2 \sin 2 θ}$ $G C = \frac{L [\cos θ \cos (α - ϕ) + \sin θ \sin (α - ϕ)] (\sin θ \cos ϕ - \cos θ \sin ϕ)}{4 \sin θ \cos θ}$ $G C = \frac{L [\cos (α - ϕ) + \tan θ \sin (α - ϕ)] (\cos ϕ - \frac{\sin ϕ}{\tan θ})}{4}$ $O G = \frac{L \cos α}{2 \tan φ} - h$ $O G \times \tan ϕ = G C$ $(\frac{L \cos α}{2 \tan φ} - h) \tan ϕ = \frac{L [\cos (α - ϕ) + \tan θ \sin (α - ϕ)] (\cos ϕ - \frac{\sin ϕ}{\tan θ})}{4}$ $\Rightarrow 2 (\frac{\cos α}{μ} - \frac{2 h}{L}) \tan ϕ = [\cos (α - ϕ) + \tan θ \sin (α - ϕ)] (\cos ϕ - \frac{\sin ϕ}{\tan θ})$ $\Rightarrow 2 (\frac{\cos α}{μ} - \frac{2 h}{L}) \tan (α - Θ) = (\cos Θ + \tan θ \sin Θ) [\cos (α - Θ) - \frac{\sin (α - Θ)}{\tan θ}] . . . (i i i)$ $b y p u t t i n g (i) i n t o (i i) a n d (i i i) w e g e t$ $t w o e q u a t i o n s f o r α a n d Θ .$ $e x a m p l e :$ $L = 5 m, h = 3 m, s = 4 m, μ = 0.6$ $\Rightarrow α = 22.882 °$ $\Rightarrow Θ = α - ϕ = 7.0125 °$
	Commented by mr W last updated on 28/Feb/20

	$\frac{B C}{\sin (2 θ + β)} = \frac{C D}{\sin β} = \frac{L}{2 \sin 2 θ}$ $B C = \frac{L \sin (2 θ + β)}{2 \sin 2 θ}$ $C D = \frac{L \sin β}{2 \sin 2 θ}$ $s = \frac{L [\sin 2 θ \cos β + (1 + \cos 2 θ) \sin β]}{2 \sin 2 θ}$ $\Rightarrow \frac{2 s}{L} = \cos β + \frac{(1 + \cos 2 θ) \sin β}{\sin 2 θ}$ $\Rightarrow \frac{2 s}{L} \times \frac{\sin 2 θ}{\sqrt{2 (1 + \cos 2 θ)}} = \sin λ \cos β + \cos λ \sin β$ $\Rightarrow \frac{2 s}{L} \times \frac{\sin 2 θ}{\sqrt{2 (1 + \cos 2 θ)}} = \sin (λ + β)$ $\Rightarrow β = \sin^{- 1} [\frac{2 s}{L} \times \frac{\sin 2 θ}{\sqrt{2 (1 + \cos 2 θ)}}] - \tan^{- 1} \frac{\sin 2 θ}{1 + \cos 2 θ}$ $h = \frac{L \sin α}{2} + B C \times \sin (β + α)$ $\frac{2 h}{L} = \sin α + \frac{\sin (2 θ + β) \sin (β + α)}{\sin 2 θ}$ $\frac{2 h}{L} = \sin α + (\cos β + \frac{\sin β}{\tan 2 θ}) (\sin β \cos α + \cos β \sin α)$ $\frac{2 h}{L} = (1 + \cos^{2} β + \frac{\sin β \cos β}{\tan 2 θ}) \sin α + (\sin β \cos β + \frac{\sin^{2} β}{\tan 2 θ}) \cos α$ $\frac{2 h}{L} \times \frac{1}{\sqrt{{(1 + \cos^{2} β + \frac{\sin β \cos β}{\tan 2 θ})}^{2} + {(\sin β \cos β + \frac{\sin^{2} β}{\tan 2 θ})}^{2}}} = \cos δ \sin α + \sin δ \cos α$ $\frac{2 h}{L} \times \frac{1}{\sqrt{{(1 + \cos^{2} β + \frac{\sin β \cos β}{\tan 2 θ})}^{2} + {(\sin β \cos β + \frac{\sin^{2} β}{\tan 2 θ})}^{2}}} = \sin (α + δ)$ $\Rightarrow α = \sin^{- 1} [\frac{2 h}{L} \times \frac{1}{\sqrt{{(1 + \cos^{2} β + \frac{\sin β \cos β}{\tan 2 θ})}^{2} + {(\sin β \cos β + \frac{\sin^{2} β}{\tan 2 θ})}^{2}}}] - \tan^{- 1} [\frac{\sin β \cos β + \frac{\sin^{2} β}{\tan 2 θ}}{1 + \cos^{2} β + \frac{\sin β \cos β}{\tan 2 θ}}]$ $\Rightarrow 2 (\frac{\cos α}{μ} - \frac{2 h}{L}) \tan ϕ = [\cos (α - ϕ) + \tan θ \sin (α - ϕ)] (\cos ϕ - \frac{\sin ϕ}{\tan θ})$ $ϕ = θ + α + β - \frac{π}{2}$ $\Rightarrow 2 (\frac{\cos α}{μ} - \frac{2 h}{L}) \frac{1}{\tan (θ + α + β)} + [\sin (θ + β) + \tan θ \cos (θ + β)] [\sin (θ + α + β) + \frac{\cos (θ + α + β)}{\tan θ}] = 0$
	Commented by ajfour last updated on 29/Feb/20

	$Even with your solution its no$ $simpler sir . . .$
	Commented by mr W last updated on 29/Feb/20

	$n o s i r!$ $i t c a n n o t b e c o m e m o r e s i m p l e, i t h i n k .$ $b u t w e g e t a f i n a l e q u a t i o n f o r a s i n g l e$ $p a r a m e t e r θ .$
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