This 3 mass atwood problem is confusing me

[Thai777]

Homework Statement

In the illustration that follows, a mass of $m_3=7kg$, drag the mass $m_1=2kg$ and $m_2=4kg$. The rope and pulley masses are negligible. What are the masses's acceleration modulus and the Tension of each ropes.

Relevant Equations

$$\sum \vec F = m\vec a$$

It has been a while since I did a atwood problem and this one has stumped me. I am not sure if this is the right place (please tell me if so). I want to see if anyone has insight.

Using the y-axis, I have defined and each forces on each masses. Knowing that $a_1=-a_3$ and each Tension $T$ are the same (massless pulley), I have written three equations for each masses.
1) $$-m_3 a = T-m_3 g$$
2) $$m_1 a = T - m_1 g$$
3) $$-m_2 a = T - m_2 g - T = -m_2 g$$
Using eq 1) and 2) it's easy to prove that
4) $$ a = \frac{g(m_3-m_1)}{m_3+m_1} $$
This can easily give the acceleration as $a=5.44m/s^2$ and $T=30.5N.$
However that 3rd equation centered in mass 2 is not consistent. it shows that a=g for that mass all the time. My thought would be that my understanding of the tension for that mass is wrong or it somehow is not accelerating at the same speed as the other two masses. The first one being a way better reasoning, changing one of the tension (the one of the rope under $m_2$) to the opposite direction yield
5) $$a=\frac{2T-m_2 g}{m_2}$$
which gives $5.45m/s^2$ for $T=30.5N$
I however do not understand the reasoning for why the tension should be going down.

Furthermore, eq 4) seems nonsensical to me. Even though I wrote in a general form, it seems paradoxical in some situation.
If $m_1=m_3$, the acceleration is therefore 0 leaving the two mass unmoving.
In that scenario, the force diagram of $m_1$ and $m_3$ (eq1 and eq2) both gives that $T=m_1 g$ since the acceleration is 0 and doesn't depend on $m_2$. Therefore $m_1$ and $m_3$ are free parameters (as long as both are equal they can be anything and there should be no movement).
However eq 5) show that for $m_2$ to not be in motion, we require that $m_2 = 2T/g$.

$m_1$ and $m_3$ are suddenly not independent as they create the tension which also link back to $m_2$. This confuses me greatly, the equation tells me that as long as $m_1$ and $m_3$ are equal, no motion should happen, $m_2$ be damned. But $m_2$'s equation rely on $T$ to not move.

To make it even worse, I have tried using Lagrangian Mechanics (rusty) to see what I get.
Taking into account that the lenght of the rope $\alpha$ is constant. We can write that
$$\alpha = h-y_1+\pi r + h + \pi r + h + \pi r + h - y_3$$
or
$\alpha = C - y_1 - y_3$ where $C = 4h+3\pi r$
Using that relation we can write $y_3 = C - y_1 - \alpha$ and $\dot y_3^2 = \dot y_1^2$
Therefore the Lagrangian is simply
$$L = T - U$$
$$L = \frac{1}{2}(m_1+m_3)\dot y_1^2 + \frac{1}{2}m_2\dot y_2^2 - m_1 g y_1 - m_3g(C-y_1-\alpha)$$
Using the Lagrangian equation (\frac{d}{dt}\frac{\partial L}{\partial \dot y} - \frac{\partial L}{\partial y}) with $y_1$ and $y_2$ it yield me these two equations
6) $$\ddot y_1 = \frac{g(m_3-m_1)}{m_1+m_3}$$
7) $$\ddot y_2 = -g$$
which are the same equations as 4) and 3).

I am just left confused and unsure what I have missed or misunderstood.
(Preview doesn't seem to show Latex form and this is my first post so sorry if it looks ugly).

 

[Lnewqban]

It seems to me that you are excluding the inertial effect of m₂ in the overall acceleration of the 3-mass system.

My calculated value for that acceleration is 6.79 m/s²

 

[kuruman]

If the masses are accelerating, which they are, not all pieces of rope have the same tension that you labeled $T$. Specifically, mass $m_3$ is shown being pulled up by tension $T$ and also down by the same tension. The two tensions would cancel each other out and the only force left acting on $m_3$ would be gravity which means that this mass is in free fall as your equation 3) shows.

I recommend that you label the tension in the piece of rope that goes over the top left pulley $T_1$ and the tension is the other piece $T_2$ and rewrite the equations.

If you want to go the Lagrangian route, note that the pulleys are "ideal". This means that all they do is change the direction of the tension but not its magnitude. In this approximation you can ignore their radius. I would measure position relative to the horizontal line at height $h$. THis should make writing the constraint equation easier. Also, the potential energy of the $i$th mass would be $V_i=-mgy_i.$

P.S. This is the right place for posting this question.

 

[Thai777]

It seems to me that you are excluding the inertial effect of m₂ in the overall acceleration of the 3-mass system.

My calculated value for that acceleration is 6.79 m/s²

This is actually on point! I have just redone it and I have made a grave mistake, I took it as one full long string/rope instead of two separate one with their own tension!

Doing so I get
$$a = \frac{-g(m_3+m_2-m_1)}{m_1+m_2+m_3}$$
which truly gives 6.79 $m/s^2$ with the value given.
You also get back the $m_2$ dependance on the other masses. This was truly a blunder and I can't believe I found it an hour after I wrote this.

I have also used the Lagrangian method considering two rope of fixed lenght $\alpha_1$ and $\alpha_2$ and I recover the exact same result for the acceleration.

Thanks for the help!

 

[kuruman]

To make it even worse, I have tried using Lagrangian Mechanics (rusty) to see what I get.

To help you shake off the rust, I am posting a solution using mechanical energy conservation which some have dubbed "the Lagrangian's poor relative."

Say the assembly starts from rest and the potential energy is defined to be zero at that configuration. This means that initial mechanical energy is zero. The system is now released. After some time three masses have accquired common speed $v$. Also mass $m_3$ has moved down by $y$, $m_2$ has also moved down by $y$ but $m_1$ has moved up by $y$. Since mechanical energy is conserved, $$\begin{align} & \frac{1}{2}(m_1+m_2+m_3)v^2+(m_1-m_2-m_3)gy=0 \nonumber \\
& \implies \frac{1}{2}v^2=\frac{(-m_1+m_2+m_3)}{(m_1+m_2+m_3) }gy.
\end{align} $$ Now $$a=\frac{dv}{dt}=\frac{dv}{dy}\frac{dy}{dt}=v\frac{dv}{dy}=\frac{d}{dy}\left(\frac{1}{2}v^2\right).$$ Taking the derivative of equation (1) with respect to $y$, we find $$a=\frac{(-m_1+m_2+m_3)}{(m_1+m_2+m_3) }g.$$ This is the same expression as yours except for an overall negative sign. You followed the convention "down" is negative. Here, I have assumed that "down" is positive and since the speed increases with increasing $y$, the quantity $v\dfrac{dv}{dy}$ is positive.