Problem with two pulleys and three masses

[erobz]

A different approach, is based on the idea of effective mass and effective acceleration. When mass $m_1$ has constant acceleration $a_1$ it doesn't matter whether there is another Atwood machine or an effective mass $m_{\text{eff.}}$ attached to the other nd of the string.
Step 1: Setup.
We assume that $m_1$ accelerates down ($m_1>m_{\text{eff.}}$) Using the standard Atwood machine formulas, the acceleration and the tension are $$a_1=\frac{m_1-m_{\text{eff.}}}{m_1+ m_{\text{eff.}}}g~;~~T_1=\frac{2m_1 m_{\text{eff.}}}{m_1+ m_{\text{eff.}}}g.$$Step 2: Find the effective mass.
On the other side of the rope we have an Atwood machine that is accelerating up with acceleration $a_1$. The situation is equivalent to an Atwood machine attached to a fixed support in an environment where the effective acceleration of gravity is $g_{\text{eff.}}=g+a_1$. Then tension $T_2$ in the second Atwood machine is $$T_2=\frac{2m_2 m_3}{m_2+ m_3}(g+a_1)=\frac{2m_2 m_3}{m_2+ m_3}\left({1+\frac{m_1-m_{\text{eff.}}}{m_1+ m_{\text{eff.}}}}\right)g.$$(I replaced $M$ in the original problem with $m_3$ for parallel structure.)
But $T_2=\frac{1}{2}T_1$ which gives $$\frac{2m_2 m_3}{m_2+ m_3}\left({1+\frac{m_1-m_{\text{eff.}}}{m_1+ m_{\text{eff.}}}}\right)g=\frac{m_1 m_{\text{eff.}}}{m_1+ m_{\text{eff.}}}g\implies m_{\text{eff.}}=\frac{4m_2m_3}{m_2+m_3}.$$
Step 3: Find the acceleration of $m_1.$
$$a_1=\frac{m_1-m_{\text{eff.}}}{m_1+ m_{\text{eff.}}}g=\frac{m_1-\frac{4m_2m_3}{m_2+m_3}}{m_1+ \frac{4m_2m_3}{m_2+m_3}}g=\frac{m_1(m_2+m_3)-4m_2m_3}{m_1(m_2+m_3)+4m_2m_3}g.$$Step 4: Find the tensions. $$\begin{align} & T_1=\frac{2m_1 m_{\text{eff.}}}{m_1+ m_{\text{eff.}}}g=\frac{2m_1 \frac{4m_2m_3}{m_2+m_3}}{m_1 +\frac{4m_2m_3}{m_2+m_3}}g=\frac{8m_1m_2m_3}{m_1(m_2+m_3)+4m_2m_3}g \nonumber \\ & T_2=\frac{1}{2}T_1=\frac{4m_1m_2m_3}{m_1(m_2+m_3)+4m_2m_3}g. \nonumber \end{align}$$
Step 5: Find the acceleration of $m_2$.
Assuming that $a_2$ accelerates down, $T_2-m_2g=-m_2 a_2\implies a_2=g-\dfrac{T_2}{m_2}.$ Then, $$a_2=g-\frac{4m_1m_3}{m_1(m_2+m_3)+4m_2m_3}g=\frac{m_1(m_2+m_3)+4m_3(m_2-m_1)}{m_1(m_2+m_3)+4m_2m_3}g.$$

Just for clarification, our formulas don't match (and that is fine, but could lead to confusion).

That is because I'm measuring $a_2$ w.r.t the accelerating pulley ( $m_1 \to 0, a_2 \to 0g$) . You are measuring $a_2$ w.r.t the inertial frame ( $m_1 \to 0, a_2 \to g$ ).

Had me scared for a minute!

 

[kuruman]

Your expression, $$a_2 = \frac{ 2m_1(m_2 - M) }{m_1M + m_1 m_2 + 4 M m_2}g \tag{3}$$must be the acceleration relative to movable pulley. That's because when $m_2=M$, it predicts that the acceleration is zero. That cannot be correct because in this case one has an Atwood machine with mass $m_1$ on one side and mass $2M$ on the other. The magnitude of the acceleration ought to be $$a_2=\frac{|m_1-2M|}{m_1+2M}g.$$ Now look at my expression in post #99. $$a_2=\frac{m_1(m_2+m_3)+4m_3(m_2-m_1)}{m_1(m_2+m_3)+4m_2m_3}g.$$When $m_2=m_3=M$, $$a_2=\frac{m_1(2M)+4M(M-m_1)}{m_1(2M)+4M^2}g=\frac{m_1+2(M-m_1)}{m_1+M}g=\frac{2M-m_1}{m_1+2M}g$$ The acceleration of mass $m_1$ is $$a_1=\frac{m_1(m_2+m_3)-4m_2m_3}{m_1(m_2+m_3)+4m_2m_3}g=\frac{m_1(2M)-4M^2}{m_1(2M)+4M^2}g=\frac{m_1-2M}{m_1+2M}=-a_2.$$Now these are magnitudes of accelerations and ought to be positive. However, if you follow my derivation, in Step 3 I assume that $a_1$ is down and in Step 5 that $a_2$ is also. This cannot be when the $m_2=m_3$ and the second pulley has zero angular acceleration. The result that one acceleration is the negative of the other validates the equations in that if one uses them and puts in all the numbers for the masses, a negative $a_1$ or $a_2$ means that the actual acceleration must be up instead of down.

On edit: Thanks for the confirmation that your $a_2$ is the non-inertial frame expression.

 

[kuruman]

Just for clarification, our formulas don't match (and that is fine, but could lead to confusion).

That is because I'm measuring $a_2$ w.r.t the accelerating pulley ( $m_1 \to 0, a_2 \to 0g$) . You are measuring $a_2$ w.r.t the inertial frame ( $m_1 \to 0, a_2 \to g$ ).

Had me scared for a minute!

That's right. See post #102. Sorry for the scare.