# What is the constraint acceleration equation of the system? [closed] In the question there is a hint given, $$y_A$$ is not constant. From this hint I began to look at the pulley that holds $$m_1$$ and $$m_2$$, and can see that it has the same tension as $$m_3$$. From this observation, I (think I) have a means to involve all the masses within a given equation.

My force equation for pulley A is $$T_B - (m_1 + m_2)g = (m_1 + m_2)a$$

and for $$m_3$$ $$T_B-m_3 g= m_3 a$$

Solving both equations for $$T_B$$ and setting equal to each other I get $$m_1 a + m_2 a + (m_1 + m_2)g = m_3 g + m_3 a$$

and then solving for a I get $$a= \frac{m_3 g - (m_1 + m_2)g}{m_1 + m_2 - m_3}$$

What I would like to know is if my assumption about pulley 2 is correct and my subsequent work.

Your procedure is not correct. This looks like a homework problem, so I just give few hints:

• write equations with respect to some inertial reference frame, such as ground
• each mass has its own acceleration - this is very important!
• since pulley is considered massless, tension in the cable from both sides has equal magnitude
• write equations of motion for the three masses separately; this will give you 3 equations with 5 unknowns ($$a_1$$, $$a_2$$, $$a_3$$, $$T_A$$, and $$T_B$$)
• to find unique solution you need 2 more equations: (i) equation of motion for the pulley A will give relationship between tensions $$T_A$$ and $$T_B$$, and (ii) displacements for the three masses will give relationship between accelerations $$a_1$$, $$a_2$$, and $$a_3$$.

Here are links to my answers to similar problems that can help you with setting up the equations:

What is positive rotation direction of a pulley in the Atwood machine?

How to find a condition for direction of motion in a system with hanging pulley?

• This procedure is the only sane way that I know of to approach problems like this. Mar 30 at 10:52