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Question Number 163293 by ajfour last updated on 05/Jan/22

Commented by ajfour last updated on 05/Jan/22

$I f t h e p l a n k r e s t s o n t w o f i x e d c y l i n d e r s$ $i n c o n t a c t w i t h e a c h o t h e r a n d$ $g r o u n d, a n d i f t h e b o t t o m e d g e o f$ $p l a n k i s j u s t b i t a b o v e g r o u n d, f i n d$ $f r i c t i o n c o e f f i c i e n t μ_{m i n} b e t w e e n$ $p l a n k a n d c y l i n d e r, s u c h t h a t t h e$ $p l a n k c a n b e i n s t a t i c e q u i l i b r i u m .$
	Answered by mr W last updated on 06/Jan/22

	Commented by mr W last updated on 06/Jan/22

	$\sin \frac{θ}{2} = \frac{b - a}{b + a}$ $\Rightarrow θ = 2 \sin^{- 1} \frac{b - a}{b + a}$ $μ_{m i n} = \tan θ$ $s u c h t h a t e q u i l i b r i u m i s a t t h e g i v e n$ $p o s i t i o n,$ $L = \frac{b + a}{\tan \frac{θ}{2}}$
	Commented by ajfour last updated on 07/Jan/22

	$T h a n k u s i r, i m b i t u n w e l l . .$
	Commented by mr W last updated on 07/Jan/22

	$g e t w e l l s o o n s i r!$
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