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Question Number 40928 by rahul 19 last updated on 29/Jul/18

	Answered by MrW3 last updated on 29/Jul/18

	$l e n g t h o f r o p e l = 2 π r = π R$ $l e t λ = \frac{W}{l} = \frac{W}{π R}$ $\cos θ = \frac{r}{R} = \frac{1}{2}$ $\Rightarrow θ = \frac{π}{3} = 60 °$ $p = p r e s s u r e b e t w e e n r o p e a n d s p h e r e$ $i n t h e p l a n e o f r o p e$ $p = \frac{λ}{\tan θ}$ $T = t e n s i o n i n r o p e$ $T = r p = \frac{r λ}{\tan θ} = \frac{R W}{2 \tan θ π R} = \frac{W}{2 π \sqrt{3}} = \frac{W}{π \sqrt{12}} = \frac{W}{π \sqrt{K}}$ $\Rightarrow K = 12$ $\Rightarrow \frac{K}{6} = 2$
	Commented by MrW3 last updated on 29/Jul/18

	Commented by MrW3 last updated on 30/Jul/18

	$a t e a c h c o n t a c t p o i n t b e t w e e n r o p e a n d$ $s p h e r e t w o e x t e r n f o r c e s a r e a c t i n g :$ $w e i g h t o f r o p e : λ [N / m]$ $n o r m a l f o r c e : n [N / m]$ $t h e r e s u l t a n t o f b o t h i s p [N / m] w h i c h$ $p r e s s e s t h e r o p e o u t w a r d s a n d c a u s e s$ $a t e n s i o n f o r c e i n t h e r o p e s i n c e t h e$ $r o p e i s a c l o s e d c i r c l e .$
	Commented by rahul 19 last updated on 30/Jul/18
	thank u so much sir��
	Commented by MrW3 last updated on 30/Jul/18

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