Question-31792

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

\dots \dots (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)

\dots \dots (i i)

N_{1} = m g \cos α - m A \sin α \dots (i i i)

N_{2} = m g \cos β + m A \sin β \dots (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!

Facebook Tweet Pin