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Question Number 48268 by ajfour last updated on 21/Nov/18

Commented by ajfour last updated on 21/Nov/18

$F i n d l e n g t h o f b o t h s e g m e n t s$ $o f s t r i n g A P = L_{1} a n d B P = L_{2},$ $a n d t h e t e n s i o n i n s t r i n g i f$ $r o d o f l e n g t h l, m a s s m h a n g s$ $i n e q u i l i b r i u m .$
Commented by mr W last updated on 21/Nov/18

$i s i t r e q u i r e d t h a t t h e t e n s i o n i n b o t h$ $s t r i n g s s h o u l d b e e q u a l ?$
Commented by ajfour last updated on 21/Nov/18

$o f c o u r s e i t s t h e s a m e s t r i n g,$ $p a s s e s t h r o u g h a s m o o t h r o l l e r$ $a t t h e t i p o f t h e r o d . .$
Commented by mr W last updated on 21/Nov/18

$I s e e .$
	Answered by mr W last updated on 21/Nov/18

	Commented by mr W last updated on 27/Nov/18

	${O P}^{'} = l \sin θ = l_{1}$ $∠ {A O P}^{'} = α$ $\frac{a}{\sin φ} = \frac{l_{1}}{\sin (α + φ)} . . . (i)$ $\frac{b}{\sin φ} = \frac{l_{1}}{\sin (\frac{π}{2} - α + φ)} = \frac{l_{1}}{\cos (α - φ)} . . . (i i)$ $\Rightarrow b \cos (α - φ) = a \sin (α + φ)$ $\Rightarrow b (\cos α \cos φ + \sin α \sin φ) = a (\sin α \cos φ + \cos α \sin φ)$ $\Rightarrow b (1 + \tan α \tan φ) = a (\tan α + \tan φ)$ $\Rightarrow (b \tan φ - a) \tan α = a \tan φ - b$ $\Rightarrow \tan α = \frac{b - a \tan φ}{a - b \tan φ} . . . (i i i)$ $\sin α = \frac{b - a \tan φ}{\sqrt{(a^{2} + b^{2}) (1 + \tan^{2} φ) - 4 a b \tan φ}}$ $\cos α = \frac{a - b \tan φ}{\sqrt{(a^{2} + b^{2}) (1 + \tan^{2} φ) - 4 a b \tan φ}}$ $f r o m (i) :$ $\frac{\sin α}{\tan φ} + \cos α = \frac{l_{1}}{a} = \frac{l \sin θ}{a}$ $\frac{1}{\sqrt{(a^{2} + b^{2}) (1 + \tan^{2} φ) - 4 a b \tan φ}} [\frac{b - a \tan φ}{\tan φ} + a - b \tan φ] = \frac{l \sin θ}{a}$ $\Rightarrow \frac{a b}{l \sqrt{(a^{2} + b^{2}) (1 + \tan^{2} φ) - 4 a b \tan φ}} (\frac{1}{\tan φ} - \tan φ) = \sin θ$ $w e g e t φ f r o m t h i s e q n .$ $t h e n w e g e t α f r o m (i i i) .$ ${A P}^{'} = \frac{a \sin α}{\sin φ}$ $\Rightarrow L_{1} = \sqrt{{(l \cos θ)}^{2} + {(\frac{a \sin α}{\sin φ})}^{2}}$ ${B P}^{'} = \frac{b \cos α}{\sin φ}$ $\Rightarrow L_{2} = \sqrt{{(l \cos θ)}^{2} + {(\frac{b \cos α}{\sin φ})}^{2}}$ $\Rightarrow T = \frac{m g}{2 [\frac{1}{\sqrt{1 + {(\frac{a \sin α}{l \cos θ \sin φ})}^{2}}} + \frac{1}{\sqrt{1 + {(\frac{b \cos α}{l \cos θ \sin φ})}^{2}}}]}$ $=================$ $(b \tan α - a) \tan φ = a \tan α - b$ $\Rightarrow \tan φ = \frac{a \tan α - b}{b \tan α - a}$ $\Rightarrow \sin α \frac{\frac{b}{a} \tan α - 1}{\tan α - \frac{b}{a}} + \cos α = \frac{l \sin θ}{a}$
	Commented by ajfour last updated on 21/Nov/18

	$VERY BRILLIANT s o l u t i o n$ $a n d q u i c k t o o .$ $S o i t a l l r e s t s o n$ ${(x - \frac{1}{x})}^{2} = k^{2} (1 + x^{2} - c x) .$ $T h a n k y o u S i r .$
	Commented by ajfour last updated on 21/Nov/18

	$U n d e r s t o o d e v e r y l i n e S i r .$
	Commented by mr W last updated on 21/Nov/18

	$t h a n k s f o r c h e c k i n g s i r!$
	Commented by mr W last updated on 22/Nov/18

	${i t}^{'} s m o r e d i f f i c u l t i f o n l y t h e l e n g t h$ $o f t h e s t r i n g L i s g i v e n, n o t t h e a n g l e$ $θ f o r t h e r o d .$
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