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Question Number 29198 by ajfour last updated on 05/Feb/18

Commented by ajfour last updated on 05/Feb/18

$u n i f o r m r o d .$
	Answered by mrW2 last updated on 05/Feb/18

	Commented by mrW2 last updated on 05/Feb/18

	$N_{1} = f_{2}$ $N_{2} = m g - f_{1}$ $N_{2} l \cos θ - m g \frac{l}{2} \cos θ - N_{1} l \sin θ = 0$ $c a s e 1 : f_{1} = μ_{1} N_{1}, f_{2} ⩽ μ_{2} N_{2} = μ_{2} (m g - μ_{1} f_{2})$ $f_{2} ⩽ \frac{μ_{2} m g}{1 + μ_{1} μ_{2}}$ $(m g - μ_{1} f_{2}) l \cos θ - m g \frac{l}{2} \cos θ - f_{2} l \sin θ = 0$ $f_{2} (μ_{1} + \tan θ) = \frac{m g}{2}$ $\Rightarrow f_{2} = \frac{m g}{2 (μ_{1} + \tan θ)} ⩽ \frac{μ_{2} m g}{1 + μ_{1} μ_{2}}$ $\Rightarrow \frac{1}{μ_{1} + \tan θ} ⩽ \frac{2 μ_{2}}{1 + μ_{1} μ_{2}}$ $\Rightarrow \tan θ ⩾ \frac{1 - μ_{1} μ_{2}}{2 μ_{2}}$ $\Rightarrow θ ⩾ \tan^{- 1} (\frac{1 - μ_{1} μ_{2}}{2 μ_{2}})$ $c a s e 2 : f_{2} = μ_{2} N_{2}, f_{1} ⩽ μ_{1} N_{1} = μ_{1} μ_{2} (m g - f_{1})$ $\Rightarrow f_{1} ⩽ \frac{μ_{1} μ_{2} m g}{1 + μ_{1} μ_{2}}$ $(m g - f_{1}) l \cos θ - m g \frac{l}{2} \cos θ - μ_{2} (m g - f_{1}) l \sin θ = 0$ $- f_{1} \cos θ + \frac{m g}{2} \cos θ - μ_{2} m g \sin θ + μ_{2} f_{1} \sin θ = 0$ $(1 - μ_{2} \tan θ) f_{1} = \frac{m g}{2} (1 - 2 μ_{2} \tan θ)$ $\Rightarrow f_{1} = \frac{m g (1 - 2 μ_{2} \tan θ)}{2 (1 - μ_{2} \tan θ)}$ $\Rightarrow f_{1} = \frac{m g (- \frac{1}{2} + 1 - μ_{2} \tan θ)}{1 - μ_{2} \tan θ}$ $\Rightarrow f_{1} = m g [1 - \frac{1}{1 - μ_{2} \tan θ}] ⩽ \frac{m g μ_{1} μ_{2}}{1 + μ_{1} μ_{2}}$ $\frac{1}{1 + μ_{1} μ_{2}} ⩽ \frac{1}{1 - μ_{2} \tan θ}$ $\tan θ ⩾ - μ_{1} \Rightarrow a l w a y s t r u e$ $\Rightarrow t h e r o d s l i p s w h e n θ ⩽ \tan^{- 1} (\frac{1 - μ_{1} μ_{2}}{2 μ_{2}})$
	Commented by ajfour last updated on 05/Feb/18

	$G r e a t! y o u h a v e b e e n t o o c a r e f u l$ $S i r, i s i m p l y t o o k f r i c t i o n a t$ $e a c h s u r f a c e a t i t s m a x i m u m,$ $s h a l l t r y t o u n d e r s t a n d y o u r$ $c a r e f u l n e s s S i r .$
	Commented by mrW2 last updated on 05/Feb/18

	$Y o u r r i g h t . T h e r e i s n o n e e d t o b e$ $s u c h c a r e f u l . I f o n e e n d s l i p s, t h e$ $o t h e r e n d s l i p s t o o . W e c a n p u t m a x .$ $f r i c t i o n o n b o t h e n d s a t t h e s a m e t i m e .$
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