The ratio of acceleration of points A, B and C is [assume all surfaces are smooth, pulley and strings are light]


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Question Number 24772 by Tinkutara last updated on 25/Nov/17

$The ratio of acceleration of points A,$ $B and C is [assume all surfaces are$ $smooth, pulley and strings are light]$
Commented by Tinkutara last updated on 25/Nov/17

Commented by mrW1 last updated on 26/Nov/17

$I t h i n k B a n d C c a n m o v e i n d e p e n d e n t l y$ $f r o m A, t h e r e f o r e t h e r e i s n o c e r t a i n$ $r a t i o b e t w e e n a_{A} a n d a_{B} .$
Commented by Tinkutara last updated on 26/Nov/17

$Answer is wrong .$ $Why a_{B} = a_{C} = \frac{9 F}{m} ?$
Commented by mrW1 last updated on 26/Nov/17

$a_{A} = \frac{3 F}{m}$ $a_{F} = 3 a_{A} = \frac{9 F}{m} (a c c e l e r a t i o n o f t h e e n d o f s t r i n g)$ $a_{B} = a_{C} = 0$
Commented by Tinkutara last updated on 26/Nov/17

$B u t a n s w e r i s s t i l l w r o n g a s r a t i o i s$ $i n t e g e r .$
Commented by mrW1 last updated on 26/Nov/17

$T h i s i s I t h i n k t h e r i g h t a n s w e r .$
Commented by ajfour last updated on 26/Nov/17

$I s B a r o p e p o i n t o r c e n t r e o f$ $p u l l e y ?$
Commented by mrW1 last updated on 26/Nov/17

$B (a s w e l l a s C) i s c e n t e r o f p u l l e y .$
Commented by ajfour last updated on 26/Nov/17

$3 x + 2 l + 3 y = c o n s t a n t$ $3 (a_{C} - a_{A}) - 3 a_{B} + a_{F} = 0 . . . . (i)$ ${F v}_{F} = {m v}_{A} a_{A}$ $\Rightarrow {F a}_{F} = {m a}_{A}^{2} + 0$ $F u r t h e r 3 F = {m a}_{A}$ $a_{A} = \frac{3 F}{m}$ $\Rightarrow a_{F} = 3 a_{A} = \frac{9 F}{m} .$ $N o w f r o m (i) :$ $a_{C} = a_{B} = (w h a t e v e r, d e p e n d s o n$ $m a s s o f s m a l l p u l l e y s) .$
Commented by Tinkutara last updated on 26/Nov/17

$A n s w e r i s 3 : 1 : 1$
Commented by jota+ last updated on 26/Nov/17

$L a r e s u l t a n t e s o b r e e l s i s t e m a B$ $y C i s z e r o . 2 F - 2 F = 0$ $\Rightarrow a_{B} = a_{C} = 0$
Commented by mrW1 last updated on 26/Nov/17

$A n s w e r 3 : 1 : 1 i s w r o n g i n e a c h c a s e,$ $t h a t w u r d e m e a n a_{A} = 3 a_{B} = 3 a_{C} .$ $B u t t h i s i s w r o n g e v e n i f B a n d C$ $a r e t h e r o p e p o i n t s . I f B a n d C a r e$ $t h e p o i n t s o n t h e r o p e, t h e n t h e$ $a n s w e r s h o u l d b e 1 : 3 : 3 w i t h$ $a_{A} = \frac{3 F}{m}$ $a_{B} = a_{C} = 3 a_{A} = \frac{9 F}{m}$ $T h i s w a s m y o r i g i n a l a n s w e r .$ $I f B a n d C a r e t h e c e n t e r o f p u l l e y,$ $t h e n a_{A} : a_{B} : a_{C} = u n c e r t a i n .$
Commented by Tinkutara last updated on 26/Nov/17

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