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Question Number 31792 by ajfour last updated on 14/Mar/18

Commented by ajfour last updated on 14/Mar/18

Commented by ajfour last updated on 15/Mar/18

$T (\cos β - \cos α) + N_{1} \sin α - N_{2} \sin β = M A$ $. . . . . . (i)$ $m g \sin α + m A \cos α - T = m a$ $T + m A \cos β - m g \sin β = m a$ $\Rightarrow 2 T = m g (\sin α + \sin β) + m A (\cos α - \cos β)$ $. . . . . . (i i)$ $N_{1} = m g \cos α - m A \sin α . . . (i i i)$ $N_{2} = m g \cos β + m A \sin β . . . (i v)$ $u s i n g (i i), (i i i), a n d (i v) i n (i) :$ $[m g (\sin α + \sin β) + m A (\cos α - \cos β)] [\cos β - \cos α]$ $+ 2 (m g \cos α - m A \sin α) \sin α$ $- 2 (m g \cos β + m A \sin β) \sin β = 2 M A$ $A = \frac{m g [(\sin α + \sin β) (\cos β - \cos α) - 2 \sin α \cos α + 2 \sin β \cos β]}{2 M + m [{(\cos β - \cos α)}^{2} + 2 (\sin^{2} α + \sin^{2} β)]}$ $A = \frac{m g s i n (α - β) [1 - 3 c o s (α + β)]}{2 M + m [4 - {(c o s α + c o s β)}^{2}]} .$
Commented by ajfour last updated on 15/Mar/18

$S i r, k i n d l y s o l v e a n d h e l p c h e c k i n g$ $m y s o l u t i o n .$
Commented by mrW2 last updated on 15/Mar/18

$n i c e a n d r i g h t s o l u t i o n s i r!$
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