# Two-way pulley system considering inertia of pulleys This is the following question.

[![This is how i tried, but I know it is wrong.]]

This is our attempt but We can't seem to solve it with inertia. • Can you edit out the contact information for a seemingly random person that is appearing in the bottom of the image? – hazzey May 22 '17 at 14:20
• Done. It's not there anymore. – Shahryar May 22 '17 at 22:02

The first step in a problem like this is to draw the relevant free body diagrams. You sort of did this in the top left diagram, but not really. What you want to do is to split each mass and inertia up into a separate free-body diagram. i.e. one FBD for mass 1, one FBD for pulley 1, one FBD for mass 2, one FBD for pulley 2. Then, for each FBD, write down the equations of motion: $\sum F = m a$, $\sum M = \Theta \alpha$. That would get you 8 equations (although 2 will be trivial as the 2 masses have no applied torque and no rotations, so the equation will be just 0=0)
The diagram you drew, with everything all together, is going to get you confused. For example, your second equation looks like $\sum F = T_1 - m_1 g +T_2 + T_3=0$. That's probably wrong, because $T_3$ does not act upon the 1st mass. If you draw out separate diagrams it will be easier to set up the correct equations.