I am confused about a method used in the following problem. There is an arrangement as shown below. The surface is smooth, and the pulleys are light. We have to find the acceleration $a_0$ of $m_1$.
The method I used to solve it was to consider the pulley B and masses $m_2$ and $m_3$ as a single system that goes down with the same acceleration as that of $m_1$. If this acceleration be $a_0$, then the equations of motion give $$a_0=\frac {m_2+m_3}{m_1+m_2+m_3}g$$
However, the textbook solution treats motions of all objects individually, where $m_1$ has an acceleration $a_0$, $m_2$ has an acceleration $a_0-a$ and $m_3$ has an acceleration $a_0+a$, all from the lab frame(inertial). The answer calculated thus does not match with mine. The texbook gives $$a_0=\frac {g}{1+ \frac {m_1(m_2+m_3)}{4m_2m_3}}$$
The question is, what is the problem with considering the pulley B and the masses $m_2$ and $m_3$ as a single system of mass $(m_2+m_3)$? Or do we have to take some precautions, when the system is accelerated? (The textbook solution is perfectly alright and I understood it too, but what is the problem with mine?)
Best Answer
The vertically moving object is an Atwood machine and the two masses have their own accelerations that are in different directions. The acceleration of $m_2$ and $m_3$ (separate from the total system) is given by $$ a=\frac{m_3-m_2}{m_2+m_3}g\tag{1} $$
Mass $m_2$ is accelerating upwards, hence the acceleration in your case of $a_0-a$; likewise mass $m_3$ is accelerating downwards with an acceleration of $a_0+a$.
Newton's 2nd law says that the sum of the forces is equal to $ma$, so you should be using all the forces in the set up.