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Question Number 47342 by ajfour last updated on 08/Nov/18

Commented by ajfour last updated on 08/Nov/18

$F i n d t h e T e n s i o n i n r o p e (l i g h t) i f$ $a l l s u r f a c e s a r e f r i c t i o n e s s .$
	Answered by MrW3 last updated on 08/Nov/18

	$a_{1} = a c c e l e r a t i o n o f w e d g e (\to)$ $a_{2} = a c c e l e r a t i o n o f s m a l l b l o c k o n g r o u n d (\leftarrow)$ $a_{3} = a c c e l e r a t i o n o f s m a l l b l o c k o n w e d g e (↙)$ $a_{3} = a_{1} + a_{2}$ $T = m_{o} a_{2}$ $m g \sin β - T = m (a_{3} - a_{1} \cos β) = m [(1 - \cos β) a_{1} + a_{2}]$ $m g \cos β - N = {m a}_{1} \sin β$ $\Rightarrow N = m (g \cos β - a_{1} \sin β)$ $T - T \cos β + N \sin β = {M a}_{1}$ $T (1 - \cos β) + m (g \cos β - a_{1} \sin β) \sin β = {M a}_{1}$ $m_{0} a_{2} (1 - \cos β) + m (g \cos β - a_{1} \sin β) \sin β = {M a}_{1}$ $\Rightarrow (M + m \sin^{2} β) a_{1} - m_{0} (1 - \cos β) a_{2} = m g \cos β \sin β . . . (i)$ $\Rightarrow m (1 - \cos β) a_{1} + (m + m_{0}) a_{2} = m g \sin β . . . (i i)$ $\Rightarrow [(m + m_{0}) (M + m \sin^{2} β) + m_{0} m {(1 - \cos β)}^{2}] a_{2} = m g \sin β [M + m (1 - \cos β)]$ $\Rightarrow a_{2} = \frac{m g \sin β [M + m (1 - \cos β)]}{(m + m_{0}) (M + m \sin^{2} β) + m_{0} m {(1 - \cos β)}^{2}}$ $\Rightarrow T = \frac{m_{0} m [M + m (1 - \cos β)] g \sin β}{(m + m_{0}) (M + m \sin^{2} β) + m_{0} m {(1 - \cos β)}^{2}}$
	Commented by ajfour last updated on 09/Nov/18

	$I f m = 1 k g, m_{0} = 2 k g, M = 3 k g,$ $a n d β = \tan^{- 1} \frac{4}{3}, t h e n w i t h$ $g = 10 m / s^{2}; y o u r e q . y i e l d s$ $T = \frac{2 (3 + \frac{2}{5}) \times \frac{40}{5}}{3 (3 + \frac{16}{25}) + 2 \times \frac{4}{25}} = \frac{80 \times 17}{281} N .$
	Commented by ajfour last updated on 09/Nov/18

	$T h a n k y o u S i r .$
	Answered by ajfour last updated on 09/Nov/18

	$T = (M + m) A - m (A + a_{0}) \cos β$ $= m_{0} a_{0}$ $m g \sin β - T = m [(A + a_{0}) - A \cos β]$ $A d d i n g$ $\Rightarrow m g \sin β = {m a}_{0} (1 - \cos β) + M A$ $+ 2 m A (1 - \cos β)$ $\Rightarrow m g \sin β = {m a}_{0} (1 - \cos β)$ $+ A [M + 2 m (1 - \cos β)]$ $\Rightarrow m g \sin β = {m a}_{0} (1 - \cos β)$ $+ \frac{a_{0} (m_{0} + m \cos β) [M + 2 m (1 - \cos β)}{M + m (1 - \cos β)}$ $a_{0} = \frac{m g \sin β [M + m (1 - \cos β)]}{m (1 - \cos β) [M + m (1 - \cos β)] + (m_{0} + m \cos β) [M + 2 m (1 - \cos β)]}$ $T = \frac{{m m}_{0} g \sin β [M + m (1 - \cos β)]}{m (1 - \cos β) [M + m (1 - \cos β)] + (m_{0} + m \cos β) [M + 2 m (1 - \cos β)]}$ $I f m = 1 k g, m_{0} = 2 k g, M = 3 k g$ $a n d β = \tan^{- 1} \frac{4}{3}$ $T = \frac{20 \times \frac{4}{5} (3 + \frac{2}{5})}{\frac{2}{5} (3 + \frac{2}{5}) + (2 + \frac{3}{5}) (3 + \frac{4}{5})}$ $= \frac{80 \times 17}{34 + 13 \times 19} = \frac{80 \times 17}{281} N .$
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