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Question Number 28940 by ajfour last updated on 01/Feb/18

Commented by ajfour last updated on 01/Feb/18

$(i) F i n d t e n s i o n i n t h e t h r e e$ $t h r e a d s, e a c h o f l e n g t h l .$ $(i i) F i n d t e n s i o n f o r a = b = c$ $(T h e b l u e t r i a n g l e i s a p o r t i o n o f$ $t h e r o o f) .$
	Answered by mrW2 last updated on 02/Feb/18

	Commented by mrW2 last updated on 02/Feb/18

	$t h e m a s s m u s t h a n g u n d e r t h e$ $c i r c u m c e n t e r O o f t h e t r i a n g l e .$ $l e t R = r a d i u s o f c i r c u m c i r c l e$ $O A = O B = O C = R$ $a n g l e m a d e b y t h r e a d s a n d h o r i z o n t a l$ $i s θ w i t h \cos θ = \frac{R}{L} o r θ = \cos^{- 1} \frac{R}{L} .$ ${l e t}^{'} s s e e t h e v e r t i c a l c o m p o n e n t o f$ $t e n s i o n T_{2} :$ $t h e s u m o f t o r q u e a b o u t l i n e B C = 0,$ $T_{2} \sin θ \times h_{A} = m g e_{A}$ $\Rightarrow T_{2} = \frac{m g}{\sin θ \times \frac{h_{A}}{e_{A}}}$ $h_{A} = c \sin B$ $e_{A} = R \cos ∠ B O D = R \cos \frac{∠ B O C}{2} = R \cos A$ $R = \frac{a}{2 \sin A} = \frac{b}{2 \sin B} = \frac{c}{2 \sin C}$ $\Rightarrow c = 2 R \sin C$ $\Rightarrow h_{A} = 2 R \sin C \sin B$ $\Rightarrow \frac{h_{A}}{e_{A}} = \frac{2 \sin B \sin C}{\cos A}$ $\Rightarrow= \frac{- \cos (B + C) + \cos (B - C)}{\cos A}$ $\Rightarrow= \frac{\cos A + \cos (B - C)}{\cos A}$ $\Rightarrow= 1 + \frac{\cos (B - C)}{\cos A}$ $\Rightarrow T_{2} = \frac{m g}{\sin θ (\frac{2 \sin B \sin C}{\cos A})}$ $o r$ $\Rightarrow T_{2} = \frac{m g}{\sin θ [1 + \frac{\cos (B - C)}{\cos A}]}$ $T_{1} a n d T_{3} s i m i l a r l y .$ $I n c a s e o f a = b = c,$ $R = \frac{2}{3} \times \frac{\sqrt{3}}{2} a = \frac{\sqrt{3} a}{3}$ $\cos θ = \frac{\sqrt{3} a}{3 L}$ $\sin θ = \sqrt{1 - \frac{1}{3} {(\frac{a}{L})}^{2}}$ $A = B = C = 60 °$ $\Rightarrow T_{1} = T_{2} = T_{3} = \frac{m g}{3 \sqrt{1 - \frac{1}{3} {(\frac{a}{L})}^{2}}}$
	Commented by ajfour last updated on 02/Feb/18

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