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Question Number 52312 by ajfour last updated on 06/Jan/19

Commented by ajfour last updated on 06/Jan/19

$I f l e n g t h s o f s t r i n g s l a n d L a r e$ $s u c h t h a t s h o w n g e o m e t r y r e s u l t s,$ $f i n d t e n s i o n i n b o t h s t r i n g s .$
Commented by tanmay.chaudhury50@gmail.com last updated on 06/Jan/19

Commented by tanmay.chaudhury50@gmail.com last updated on 06/Jan/19

$m g = T_{R} s i n θ_{1} + T_{B} s i n θ_{2}$ $T_{R} c o s θ_{1} = T_{B} c o s θ_{2}$ $M g = T_{R} {s i n}_{} θ_{3} + T_{B} s i n θ_{4}$ $T_{R} c o s θ_{3} = T_{B} c o s θ_{4}$ $n e e d t i m e a n d d i r e c t i o n t o r e a c h g o a l . . . s o w a i t$ $p l s . . .$
	Answered by mr W last updated on 06/Jan/19

	Commented by mr W last updated on 06/Jan/19

	$l e t μ = \frac{m}{M}, λ = \frac{L}{b}, κ = \frac{l}{b}$ $T_{1} \sin α - T_{2} \sin δ = 0$ $T_{1} \cos α + T_{2} \cos δ = M g$ $\Rightarrow T_{1} = \frac{M g \sin δ}{\sin (α + δ)}$ $\Rightarrow T_{2} = \frac{M g \sin α}{\sin (α + δ)}$ $T_{1} \sin β + T_{2} (\sin δ - \sin γ) = 0$ $T_{1} \cos β + T_{2} (\cos γ - \cos δ) = m g$ $\Rightarrow T_{1} = \frac{m g (\sin γ - \sin δ)}{\sin (β + γ) - \sin (β + δ)}$ $\Rightarrow T_{2} = \frac{m g \sin β}{\sin (β + γ) - \sin (β + δ)}$ $\Rightarrow \frac{M g \sin δ}{\sin (α + δ)} = \frac{m g (\sin γ - \sin δ)}{\sin (β + γ) - \sin (β + δ)}$ $\Rightarrow \frac{\sin δ}{\sin (α + δ)} = \frac{μ (\sin γ - \sin δ)}{\sin (β + γ) - \sin (β + δ)} . . . (i)$ $\Rightarrow \frac{M g \sin α}{\sin (α + δ)} = \frac{m g \sin β}{\sin (β + γ) - \sin (β + δ)}$ $\Rightarrow \frac{\sin α}{\sin (α + δ)} = \frac{μ \sin β}{\sin (β + γ) - \sin (β + δ)} . . . (i i)$ $\frac{O R}{\cos β} = \frac{O P}{\cos γ} = \frac{b}{\sin (β + γ)}$ $\Rightarrow O R = \frac{\cos β b}{\sin (β + γ)}$ $\Rightarrow O P = \frac{\cos γ b}{\sin (β + γ)}$ $∠ O P Q = \frac{π}{2} - α - (\frac{π}{2} - β) = β - α$ $\frac{O Q}{\sin (β - α)} = \frac{O P}{\sin (α + δ)} = \frac{P Q}{\sin (δ + β)}$ $\Rightarrow O Q = \frac{\sin (β - α) \times O P}{\sin (α + δ)} = \frac{\sin (β - α) \cos γ b}{\sin (α + δ) \sin (β + γ)}$ $\Rightarrow P Q = \frac{\sin (δ + β) \times O P}{\sin (α + δ)} = \frac{\sin (δ + β) \cos γ b}{\sin (α + δ) \sin (β + γ)}$ $O P + P Q = L$ $\frac{\cos γ b}{\sin (β + γ)} + \frac{\sin (δ + β) \cos γ b}{\sin (α + δ) \sin (β + γ)} = L$ $\Rightarrow \frac{\cos γ}{\sin (β + γ)} [1 + \frac{\sin (δ + β)}{\sin (α + δ)}] = λ . . . (i i i)$ $O R + O Q = l$ $\frac{\cos β b}{\sin (β + γ)} + \frac{\sin (β - α) \cos γ b}{\sin (α + δ) \sin (β + γ)} = l$ $\Rightarrow \frac{1}{\sin (β + γ)} [\cos β + \frac{\sin (β - α) \cos γ}{\sin (α + δ)}] = κ . . . (i v)$ $4 e q n . f o r 4 u n k n o w n s : α, β, γ, δ$ $. . . . . .$
	Commented by ajfour last updated on 06/Jan/19

	$T h a n k y o u S i r, i c o u l d g o n o f u r t h e r$ $t o o, t h e a n g l e s a r e i n t r i c a t e l y$ $r e l a t e d, d o n t e v e n p e r m i t u s t o r e d u c e$ $t h e m t o t w o e q s . i n t w o u n k n o w n s!$
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