D'Alembert, two blocks and an inclined plane
Ok, I am still confused about this. This time I considered all the forces and all the particles. However this is how I broke it down.
$\displaystyle F_{applied} = (\frac{m_1}{\sin \alpha}+m_2)g$
$\displaystyle F_{inertial} = (m_2 +\frac{m_1}{\sin \alpha})\ddot{s} $
However the solution misses out the sin(alpha) in the applied force. Are constraints always ignored when writing down applied forces?
