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Question Number 110467 by ajfour last updated on 29/Aug/20

Commented by ajfour last updated on 29/Aug/20

$F i n d m a x i m u m F s u c h t h a t t h e$ $c y l i n d e r r e s t i n g a g a i n s t t h e g r o u n d$ $a n d w a l l, n e i t h e r t u r n s n o r i t m o v e s .$
	Answered by mr W last updated on 29/Aug/20

	Commented by mr W last updated on 29/Aug/20

	$f_{1} = μ N_{1}$ $N_{2} = f_{1} = μ N_{1}$ $f_{2} = μ N_{2} = μ^{2} N_{1}$ $F - M g + N_{1} + f_{2} = 0$ $F - M g + (1 + μ^{2}) N_{1} = 0 . . . (i)$ $F R = (f_{1} + f_{2}) R$ $F = μ (1 + μ) N_{1} . . . (i i)$ $F [1 + \frac{1 + μ^{2}}{μ (1 + μ)}] = M g$ $\Rightarrow F = \frac{μ (1 + μ)}{2 μ^{2} + μ + 1} M g$
	Commented by ajfour last updated on 29/Aug/20

	$C o r r e c t a n d N i c e S i r, t h a n k s!$
	Answered by mr W last updated on 29/Aug/20

	Commented by mr W last updated on 29/Aug/20

	$\tan ϕ = μ$ $(\frac{e}{\tan ϕ} - R) \frac{1}{\tan ϕ} + e = R$ $\Rightarrow e = \frac{μ (1 + μ) R}{1 + μ^{2}}$ $F (R + e) = M g e$ $\Rightarrow F = \frac{e}{e + R} M g = \frac{\frac{μ (1 + μ) R}{1 + μ^{2}}}{\frac{μ (1 + μ) R}{1 + μ^{2}} + R} M g$ $\Rightarrow F = \frac{μ (1 + μ)}{2 μ^{2} + μ + 1} M g$
	Commented by ajfour last updated on 29/Aug/20

	$A w e s o m e! S i r .$
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