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Question Number 126080 by ajfour last updated on 17/Dec/20

Commented by ajfour last updated on 18/Dec/20

$C y l i n d e r a n d s p h e r e h a v e t h e$ $s a m e r a d i u s . A n e q u i l a t e r a l$ $t r i a n g u l a r p l a t e r e s t s o n s p h e r e$ $w i t h t w o v e r t i c e s a g a i n s t c u r v e d$ $s u r f a c e o f c y l i n d e r a n d t o p v e r t e x$ $o n r i m o f c y l i n d e r . I t i s i n c o n t a c t$ $w i t h s p h e r e a t p o i n t P w h i c h i s a t$ $a h e i g h t h f r o m b a s e o f c y l i n d e r .$ $(P Q = h) . F i n d h e i g h t o f c y l i n d e r .$
Commented by mr W last updated on 17/Dec/20

$i f t h e t o p v e r t e x h a s c o n t a c t f o r c e$ $w i t h t h e r i m o f c y l i n d e r, n o u n i q u e$ $s o l u t i o n e x i s t s . i f w e a s s u m e t h a t$ $t h e t o p v e r t e x i s j u s t o n t h e r i m, b u t$ $w i t h o u t r e a l c o n t a c t, t h e n i t h i n k$ $t h e r e i s a n u n i q u e s o l u t i o n . p l e a s e$ $c h e c k s i r .$
Commented by ajfour last updated on 18/Dec/20

$t r i a n g l e i s o r i e n t e d s y m m e t r i c a l l y$ $l e f t a n d r i g h t, a n d t o p v e r t e x j u s t$ $r e a c h i n g t h e r i m, n o c o n t a c t f o r c e .$ $(I g e t y o u r p o i n t, S i r, l e t s a s s u m e$ $f r i c t i o n l e s s s u r f a c e s o f s p h e r e,$ $t r i a n g u l a r p l a t e a n d c y l i n d e r) .$
	Answered by mr W last updated on 17/Dec/20

	Commented by mr W last updated on 17/Dec/20

	$s = s i d e l e n g t h o f e q u i l a t e r a l t r i a n g l e$ $B^{'} = m i d p o i n t o f B C$ $D = c e n t e r o f m a s s o f p l a t e$ ${A B}^{'} = \frac{\sqrt{3} s}{2}$ $A D = \frac{2}{3} \times \frac{\sqrt{3} s}{2} = \frac{\sqrt{3} s}{3}$ ${D B}^{'} = \frac{1}{3} \times \frac{\sqrt{3} s}{2} = \frac{\sqrt{3} s}{6}$ $h = r + r \sin θ$ $\Rightarrow \sin θ = \frac{h}{r} - 1$ $2 r = \frac{\sqrt{3} s}{2} \times \sin θ + r - \sqrt{r^{2} - \frac{s^{2}}{4}}$ $\Rightarrow \frac{\sqrt{3} s}{2} \times (\frac{h}{r} - 1) - r = \sqrt{r^{2} - \frac{s^{2}}{4}}$ $\Rightarrow [3 {(\frac{h}{r} - 1)}^{2} + 1] s = 4 \sqrt{3} (\frac{h}{r} - 1) r$ $\Rightarrow s = \frac{4 \sqrt{3} (\frac{h}{r} - 1) r}{3 {(\frac{h}{r} - 1)}^{2} + 1}$ $P D = S D \times \cos θ = {D B}^{'} \times \cos^{2} θ$ $= \frac{\sqrt{3} s}{6} [1 - {(\frac{h}{r} - 1)}^{2}]$ $A P = A D + P D = \frac{\sqrt{3} s}{3} + \frac{\sqrt{3} s}{6} [1 - {(\frac{h}{r} - 1)}^{2}]$ $= \frac{\sqrt{3} s}{6} [3 - {(\frac{h}{r} - 1)}^{2}]$ $h e i g h t o f c y l i n d e r = H$ $H = h + A P \times \cos θ$ $\Rightarrow H = h + \frac{\sqrt{3} s}{6} [3 - {(\frac{h}{r} - 1)}^{2}] \sqrt{1 - {(\frac{h}{r} - 1)}^{2}}$
	Commented by ajfour last updated on 18/Dec/20

	$V e r y s m a r t (i n t e l l i g e n t) d i a g r a m$ $a n d s o l u t i o n S i r, l o v e d f o l l o w i n g$ $i t; TAL (t h a n k s a l o t)!$
	Commented by mr W last updated on 18/Dec/20

	$i t h i n k i n 3 D, b u t c a n o n l y i l l u s t r a t e$ $w i t h 2 D - d i a g r a m s . i l i k e y o u r 3 D -$ $d i a g r a m s!$
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