Find the acceleration of the body of mass `m_(2)` in the arrangement shown in figure. If the mass`m_(2)` is `eta` time great as the mass `m_(1)` and the angle that the inclined plane forms with the horizontal is equal to `theta`. The masses of the pulley and threads, as well as the friction, are assumed to be negligible.

Explanation

Let us write Newton's second law for both, bar 1 and body 2 in terms of projection having taken the positive direction of `x_1` and `x_2` as shown in the figure and assuming that body 2 starts sliding, say, upward along the incline
`T_1-m_1gsin alpha=m_1w_1` (1)
`m_2g-T_2=m_2w` (2)
For the pulley, moving in vertical direction from the equation `F_x=mw_x`
`2T_2-T_1=(m_P)w_1=0`
(as mass of the pulley `m_P=0`)
or `T_1=2T_2` (3)
As the length of the threads are constant, the kinematical relationship of accelerations becomes
`w=2w_1` (4)
Simultaneous solutions of all these equations yields:
`w=(2g(2m_2/m_1-sin alpha))/((4m_2/m_1+1))=(2g(2eta-sin alpha))/((4eta+1))`
As `etagt1`, w is directed vertically downward, and hence in vector from
`vecw=(2vec(g)(2eta-sin alpha))/(4eta+1)`