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Question Number 71170 by mr W last updated on 12/Oct/19

Commented by mr W last updated on 12/Oct/19

$You can't use 'macro parameter character #' in math mode$ $i n a p a r a b o l o i d c u p, w h i c h i s a b s o l u t e l y$ $s m o o t h, a s t i c k r e m a i n s i n e q u i l i b r i u m$ $a s s h o w n . f i n d t h e m a x i m u m l e n g t h$ $t h e s t i c k m a y h a v e .$
	Answered by mr W last updated on 12/Oct/19

	Commented by mr W last updated on 01/Dec/19

	${l e t}^{'} s s a y t h e s t i c k t o u c h e s t h e i n s i d e$ $o f t h e c u p a t p o i n t A (- e, f) .$ $s u c h t h a t t h e s t i c k i s i n e q u i l i b r i u m,$ $t h e c o n t a c t f o r c e s F a n d N a s w e l l a s$ $t h e g r a v i t y f o r c e o f t h e s t i c k m u s t$ $i n t e r s e c t a t t h e s a m e p o i n t S .$ $l e t η = \frac{h}{R}, ϵ = \frac{e}{R}, λ = \frac{l}{R}$ $e q n . o f p a r a b o l a :$ $y = \frac{{h x}^{2}}{R^{2}} = \frac{η x^{2}}{R}$ $y^{'} = \frac{2 η x}{R}$ $a t A (- e, f) :$ $f = \frac{η e^{2}}{R} = η ϵ^{2} R$ $y^{'} = - \frac{1}{\tan φ} = - \frac{2 η e}{R} = - 2 η ϵ$ $\Rightarrow \tan φ = \frac{1}{2 η ϵ}$ $p o i n t B (R, h) :$ $\tan θ = \frac{h - f}{R + e} = \frac{h - η ϵ^{2} R}{R + e} = \frac{η (1 - ϵ^{2})}{1 + ϵ} = η (1 - ϵ)$ $e q n . o f l i n e A S :$ $\frac{y - f}{x + e} = \tan φ = \frac{1}{2 η ϵ}$ $\Rightarrow y = f + \frac{x + e}{2 η ϵ}$ $e q n . o f l i n e S B :$ $\frac{y - h}{x - R} = - \frac{1}{\tan θ} = - \frac{1}{η (1 - ϵ)}$ $\Rightarrow y = h - \frac{x - R}{η (1 - ϵ)}$ $p o i n t S (x_{S}, y_{S}) :$ $y_{S} = f + \frac{x_{S} + e}{2 η ϵ} = h - \frac{x_{S} - R}{η (1 - ϵ)}$ $η ϵ^{2} R + \frac{x_{S} + e}{2 η ϵ} = h - \frac{x_{S} - R}{η (1 - ϵ)}$ $η ϵ^{2} + \frac{\frac{x_{S}}{R} + ϵ}{2 η ϵ} = η - \frac{\frac{x_{S}}{R} - 1}{η (1 - ϵ)}$ $l e t s = \frac{x_{S}}{R}$ $η ϵ^{2} + \frac{s + ϵ}{2 η ϵ} = η - \frac{s - 1}{η (1 - ϵ)}$ $\frac{s + ϵ}{2 ϵ} + \frac{s - 1}{1 - ϵ} = η^{2} (1 - ϵ^{2})$ $\frac{s - ϵ}{2 ϵ (1 - ϵ)} = η^{2} (1 - ϵ)$ $\Rightarrow s = ϵ + 2 η^{2} ϵ {(1 - ϵ)}^{2}$ $x_{S} = x_{D} = - e + \frac{l \cos θ}{2} = - e + \frac{l}{2 \sqrt{1 + \tan^{2} θ}}$ $s = \frac{x_{S}}{R} = - ϵ + \frac{λ}{2 \sqrt{1 + η^{2} {(1 - ϵ)}^{2}}}$ $- ϵ + \frac{λ}{2 \sqrt{1 + η^{2} {(1 - ϵ)}^{2}}} = ϵ + 2 η^{2} ϵ {(1 - ϵ)}^{2}$ $\Rightarrow λ = 4 ϵ {[1 + η^{2} {(1 - ϵ)}^{2}]}^{\frac{3}{2}}$ $f o r t h e m a x i m u m o f l e n g t h l,$ $\frac{d λ}{d ϵ} = 4 {[1 + η^{2} {(1 - ϵ)}^{2}]}^{\frac{3}{2}} + 4 ϵ \frac{3}{2} {[1 + η^{2} {(1 - ϵ)}^{2}]}^{\frac{1}{2}} (- 2 ϵ η^{2}) = 0$ $1 + η^{2} {(1 - ϵ)}^{2} - 3 η^{2} ϵ (1 - ϵ) = 0$ $4 ϵ^{2} - 5 ϵ + 1 + \frac{1}{η^{2}} = 0$ $\Rightarrow ϵ = \frac{1}{8} (5 - \sqrt{9 - \frac{16}{η^{2}}})$ $\Rightarrow λ_{m a x} = \frac{1}{2} (5 - \sqrt{9 - \frac{16}{η^{2}}}) {[1 + \frac{η^{2}}{64} (3 + \sqrt{9 - \frac{16}{η^{2}}})]}^{\frac{3}{2}} ⩾ 4$ $f o r η ⩽ 1.5242, λ_{m a x} = 4$
	Commented by ajfour last updated on 12/Oct/19

	$d i d y o u n o t s o l v e i t t h e n, s i r ?$
	Commented by mr W last updated on 12/Oct/19

	Commented by mr W last updated on 12/Oct/19

	$s i r, i d e l e t e d t h e o l d p o s t b y a c c i d e n t,$ $b u t i w a n t e d t o k e e p t h e s o l u t i o n$ $s o m e h o w, t h e r e f o r e i r e p o s t e d i t h e r e .$
	Commented by mr W last updated on 12/Oct/19

	Commented by mr W last updated on 12/Oct/19

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