Question

Find the acceleration $w$ of body $2$ in the arrangement shown in figure above, if its mass is $\eta$ times as great as the mass of bar $1$ and the angle that the inclined plane forms with the horizontal is equal to $\alpha$. The masses of the pulleys and the threads, as well as the friction, are assumed to be negligible. Look into possible cases.

Answer 1

Let us write Newton's second law for both, bar 1 and body 2 in terms of projection having take the positive direction of $x_1$ and $x_2$ as shown in figure below and assuming that body 2 starts sliding, say, upward along the incline
$T_1-m_{1}g\sin \alpha =m_1{w}_1$ (1)
$m_{2}g-T_2=m_{2}w$ (2)
For the pulley, moving in vertical direction from the equation $F_x=m{w}_x$
$2T_2-T_1=\left(m_p\right)w_1=0$
(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:
$w=\frac{2g(2\frac{m_2}{m_1}-\sin \alpha )}{(4\frac{m_2}{m_1}+1)}=\frac{2g(2\eta -\sin \alpha )}{(4\eta +1)}$
As $\eta >1$, $w$ is directed vertically downward, and hence in vector form
$\overrightarrow{w}=\frac{2\overrightarrow{g}(2\eta -\sin \alpha )}{4\eta +1}$