Problem with two pulleys and three masses

[erobz]

What if you require a1 to be zero.? Do Eq 2and 3 look correct?...I don'tthink so. Just make positive up and use one inertial frame and you will get @SammyS result I believe. So what does your does the Lagrangian method give you for a1? I have a result for a1 but am tired of writing down wrong stuff! (It is not particularly pretty but seems correct)

$$\mathcal{L} = T - U $$

$$ \mathcal{L} = \frac{1}{2} m_1 { \dot l_1 }^2+ \frac{1}{2} m_2 \left( { \dot l_2}- {\dot l_1 } \right)^2 + \frac{1}{2} M \left( {\dot l_1}+ {\dot l_2} \right) ^2 + m_1 g l_1 + m_2 g ( L - l_1 + l_2) + M g ( L - l_1 + S - l_2 )$$

$$ \frac{ \partial \mathcal{L} }{ \partial l_1 } = \frac{d}{dt} \frac{ \partial \mathcal{L} }{ \partial \dot{l_1} }$$

$$ ( m_1 - m_2 -M ) g = \left ( m_1 + m_2+M \right) \ddot {l_1} + \left( M - m_2\right) \ddot {l_2}$$

$$ \ddot{l_1} = \frac{(m_1 -m_2 -M)g - (M-m_2) \ddot{l_2}}{m_1 + m_2 + M} \tag{1}$$

$$ \frac{ \partial \mathcal{L} }{ \partial l_2 } = \frac{d}{dt} \frac{ \partial \mathcal{L} }{ \partial \dot{l_2} }$$

$$( m_2 -M ) g = \left( M - m_2 \right) \ddot{l_1} + \left( M+m_2 \right) \ddot{l_2} \tag{2}$$

Substitute $(1) \to (2)$ and reduce:

$$\ddot{l_2} = \frac{ 2m_1(m_2 - M) }{m_1M + m_1 m_2 + 2 M m_2}g \tag{3}$$

There it is ( sub $(3) \to (1)$ ). Not particularly pretty indeed. For $m_1 = 0$ , it reduces to $\ddot {l_2} = 0g$ , and $\ddot{l_1} = -g$

I haven't checked other end cases...I was too exhausted when I had finished all the algebra (for the 3rd time ).

 

[erobz]

What if you require a1 to be zero.? Do Eq 2and 3 look correct?...I don'tthink so.

If you plug in $a_1 =0$ we do indeed get the solution to the OP's problem from the resulting system. So it doesn't seem like that criterion disqualifies it.

That being said, if you plug in $a_1 = 0$ to @SammyS system it reduces to the system that solves the OP's questions well.

There must be something wrong with how I've set up my relative accelerations.

 

[hutchphd]

I think I see it. Your suspicions about eqn 4 are correct, and they follow from not using the same inertial frame for the two "halves" of the upper pulley. In particular consider the case where all the masses are equal. So one frame per problem!

 

[erobz]

I think I see it. Your suspicions about eqn 4 are correct, and they follow from not using the same inertial frame for the two "halves" of the upper pulley. In particular consider the case where all the masses are equal. So one frame per problem!

Unfortunately, I have to go grocery shopping. My suspicion about $(4)$ were that we seem to have a pulley accelerating under no net force. I'm presently unsure if that is basically the issue you are referring to or not? Hopefully I can figure out what you mean later on today.

 

[erobz]

I think I see it. Your suspicions about eqn 4 are correct, and they follow from not using the same inertial frame for the two "halves" of the upper pulley. In particular consider the case where all the masses are equal. So one frame per problem!

Let $m_2 = M \equiv m$

This implies $a_2 = 0$

It follows that $(2)$ and $(3)$ both reduce to:

$$T_2 - mg = -ma_1 \implies T_2 = m \left( g - a_1\right)$$

Looking at the top pulley I would expect:

$$ T_1 - 2mg = -2m a_1 \implies T_1 = 2m(g-a_1)$$

It follows that:

$$ T_1 - 2T_2 = 2m(g-a_1) - 2[ m (g - a_1) ] = 0 $$

I'm not finding any obvious contradiction there. What am I missing?

 

[hutchphd]

$$ T_1 - 2mg = -2m a_1$$Where does this come from?

 

[erobz]

$$ T_1 - 2mg = -2m a_1$$Where does this come from?

In the case where $m_2 = M$

The tension in the rope on the right-hand side of the fixed pulley $T_1$ is resisting accelerating both hanging masses $2m$ at $-a_1$ ...is it not?

 

[erobz]

Don't hate me, but I just did re-solved everything, and I am now disgustingly close to the same result I got in the other thread using the Lagrangian. Its too close for coincidence...I must have made another algebra mistake that I'm going to waste all day finding. The algebra is absolutely atrocious.

 

[kuruman]

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, $$a_2=g-\frac{4m_1m_3}{m_1(m_2+m_3)+4m_2m_3}g=\frac{m_1(m_2+m_3)+4m_3(m_2-m_1)}{m_1(m_2+m_3)+4m_2m_3}g.$$

 

[erobz]

Ok! The error was in the math of my Largrangian approach on my last simplification in the denominator... (I'm not saying I didn't trust @Dale , I just didn't trust myself). Sorry if I drove everyone bonkers.

The system that I wrote in #87 does properly solve the EoM.

$$ a_2 = \frac{ 2m_1(m_2 - M) }{m_1M + m_1 m_2 + 4 M m_2}g \tag{3}$$

That result has now been obtained independently with both methods. I wasn't crazy after all on this one, just absolutely terrible at algebra.

This was a much more difficult but seemingly innocuous follow up question than anticipated (37 full sheets of paper until I ended up answering my own question...)

 

[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.