Problem with two pulleys and three masses

[erobz]

Well i think the formula is correct, so what you mean is:
T - mg = m*(-a)

Good. Please use capital $M$ though so you don't get confused about which hanging mass.

 

[Erdi]

Good. Please use capital $M$ though so you don't get confused about which hanging mass.

I know i edited it, guess it didnt go throug..

 

[erobz]

I know i edited it, guess it didnt go throug..

Also its $T_1$ you missed the subscript.

 

[erobz]

Now, of the forces on the LHS. Which one is supplied by the rope?

 

[Erdi]

Back to stage 1 again. That is m2g isn't it??

 

[erobz]

Back to stage 1 again. That is m2g isn't it??

The $m_2$ 's weight does not make $m_2$ go up. The other force i.e. $T_1$ (the tension) does. The tension is supplied by the rope.

I'm going to be honest here. It seems like you have some road ahead of you before you solve this problem. I got to get to bed, its midnight here. Don't lose hope, you'll figure it out. Someone will pick back up with you at some point. Goodnight.

In the meantime, it would be helpful if you learned to use Latex to format your equations. There is a guide on how to do this in the lower left corner of the new relpy box.

 

[Erdi]

Okay thanks for your help tho!

 

[Lnewqban]

I know that the tension from pulley B (T1) has to be equal the m*g of m1 for m1 to have acceleration = 0. But i can't figure how this works because the m2 is already heavier. And so the block(M) has to be negative weight?

Welcome, @Erdi !

I believe that that statement shows the misconception that is the root of your confusion.
Any mass is weightless in free fall (reason for which your guts feel funny during a roller coaster ride).

Please, see:
https://en.m.wikipedia.org/wiki/Weightlessness

The more you slowdown that free fall, the heavier the mass becomes.
If you are able to stop that mass completely (like Earth's surface and a bathroom scale do to your body), that mass "adquires" its natural or normally measurable weight when in repose.

If the same mass is suddely moved upwards (like your body in an elevator), its measurable weight increases, as weight is a force and an accelerated mass resists that acceleration showing a reaction force.

 

[Lnewqban]

Im sorry, I am lost. I really don't have a clue.

Just imagine that you are removing M and that you are grabing the rope while standing on the ground.
You keep a solid grip on that rope and m2, which is four times heavier that m1, starts moving down until m1 hits the top fixed pulley.
After that moment, your hand feels a force of m2g value pulling up.

If you slowly release your grip, the rope starts sliding up and m2 falls even lower.
Your hand starts feeling less force simultaneously as m2 is loosing weight.

If you fully release the rope, m2 becomes completely weightless and unable to exert any force on the movable pulley.
As a result, m1 is free to fall down, loosing weight at the same time.

There must be a balance point for which you keep some gripping force on the sliding rope in such a way that m2 continues falling, but m1 remains static.
At that point, your hand should feel a pulling upwards force, which value is the key to the solution of this problem.

 

[kuruman]

You might profit if you research "Atwood machine" on the web. Here is a compact treatment but there is plenty more including videos. Please read carefully and try to understand how one proceeds to solve such problems. The method is straightforward:

1. Choose a system, in this case one of the masses, say $M$.
2. Find the net force (sum of all the forces acting on it) $F_{\text{net,M}}.$
3. Set this sum equal to the mass of the system times its acceleration. This gives you one equation, $F_{\text{net,M}}=Ma$
4. Repeat for the other mass, $m_2$. Note that if $M$ accelerates up, $m_2$ must accelerate down. You should get a second equation $F_{\text{net},\text{m}_{2}}=-m_2a.$
5. Combine the two equations to find the acceleration.

See how this plan is executed in the link I provided. The link does not provide the tension, but you can find it easily with a little algebra from either $F_{\text{net,M}}=Ma$ or $F_{\text{net},\text{m}_{2}}=-m_2a$ if you have an expression for the acceleration $a$.

 

[erobz]

In addition to what the others have stated:

Might as well not waste what you have figured out so far. You are at a good point to examine the effects of $M$ on $T_1$ with a plot.

You have correctly identified the following set of equations necessary to solve for the tension in the lower rope $T_1$ as a function of $M$.

$$\begin{align} T_1 - m_2 g &= m_2 a \tag{1} \\ T_1 - Mg &= -Ma \tag{2} \end{align} $$

Solve this set of equations for $T_1$ and make a plot vs $M$. Let $M$ range from $0 \rm{kg}$ to $m_2$, study it, then let $M$ go to something arbitrarily large. Try to understand what it's telling you about the limits of the tension $T_1$.

Solving the system is a step toward completing the solution, so it's not an out of the way excursion. It should help to visually see this part of the solution.

 

[Erdi]

Thanks guys! I think i found an expression for T2 and i came to an answer for M.

 

[kuruman]

And what is your answer?

 

[erobz]

In which direction was the mass $M$ traveling?

 

[Orodruin]

You might profit if you research "Atwood machine" on the web. Here is a compact treatment but there is plenty more including videos. Please read carefully and try to understand how one proceeds to solve such problems. The method is straightforward:

1. Choose a system, in this case one of the masses, say $M$.
2. Find the net force (sum of all the forces acting on it) $F_{\text{net,M}}.$
3. Set this sum equal to the mass of the system times its acceleration. This gives you one equation, $F_{\text{net,M}}=Ma$
4. Repeat for the other mass, $m_2$. Note that if $M$ accelerates up, $m_2$ must accelerate down. You should get a second equation $F_{\text{net},\text{m}_{2}}=-m_2a.$
5. Combine the two equations to find the acceleration.

See how this plan is executed in the link I provided. The link does not provide the tension, but you can find it easily with a little algebra from either $F_{\text{net,M}}=Ma$ or $F_{\text{net},\text{m}_{2}}=-m_2a$ if you have an expression for the acceleration $a$.

It may seem, with the angst it can bring,
That an Atwood's machine's a harsh thing.
But you just need to say
That F is ma,
And use conservation of string!

https://www.physics.harvard.edu/undergrad/limericks

 

[Erdi]

I can quickly go through my math here, firstly i got:
m1: m1g -T1 = m1a
m2: m2g-T2 = m2a These are the downward "forces" (m*g) minus the upwardforces tension(T)
M: Mg - T2 = Ma
Pulley B: 2*T2 - T1, from here i can see out that T2 equals T1/2

Then i got to the acceleration relative to each other
a(m1) - a(pulleyB) = 0, a(m1) = -a(B)
a(M)-a(B)+a(m2)-a(B) = 0 => a(M)+a(m2) = 2*a(B) = -2*a(m1)

So i solved for
2*a(m1) = -(a(M)+a(m2)), and so i found expression for a(M,m2)
2*a(m1) =-(g-(T2/M) + g-(T2/m2))

So there was a lot of math between right there to my final answer, but i ended up with:
a(m1) = (g(m1*M+m2*m1-4*M*m2)) / (M*m1+m2*m1+4*M*m2) = 0
Solved for M:
M = (m2*m1)/(4*m2-g*m1) = 0,32kg

 

[Erdi]

In which direction was the mass $M$ traveling?

So M will be traveling upwards, if I am correct.

 

[erobz]

I can quickly go through my math here, firstly i got:
m1: m1g -T1 = m1a
m2: m2g-T2 = m2a These are the downward "forces" (m*g) minus the upwardforces tension(T)
M: Mg - T2 = Ma
Pulley B: 2*T2 - T1, from here i can see out that T2 equals T1/2

Then i got to the acceleration relative to each other
a(m1) - a(pulleyB) = 0, a(m1) = -a(B)
a(M)-a(B)+a(m2)-a(B) = 0 => a(M)+a(m2) = 2*a(B) = -2*a(m1)

So i solved for
2*a(m1) = -(a(M)+a(m2)), and so i found expression for a(M,m2)
2*a(m1) =-(g-(T2/M) + g-(T2/m2))

So there was a lot of math between right there to my final answer, but i ended up with:
a(m1) = (g(m1*M+m2*m1-4*M*m2)) / (M*m1+m2*m1+4*M*m2) = 0
Solved for M:
M = (m2*m1)/(4*m2-g*m1) = 0,32kg

$g$ shouldn't be in the final result. You also changed your variable names, and your coordinate direction from what we were working towards earlier... not making it easy to follow along.

It seems like your EoM for mass $M$ is not consistent with the coordinate direction you chose to describe the EoM of mass $m_2$?

Any chance you could try using Latex to format the math. It doesn't take much effort to learn it. It is so much easier to find\point out any errors.

LaTeX Guide

 

[erobz]

Hmm, well here's almost the rest of the math. Why shouldn't g be there?

Do the units make sense? You are subtracting a force from a mass in the denominator. Thats generally an indication something has gone haywire.

 

[kuruman]

M = (m2*m1)/(4*m2-g*m1) = 0,32kg

I got the same expression without the g for the reason already explained. The error is between the two steps below

So i solved for
2*a(m1) = -(a(M)+a(m2)), and so i found expression for a(M,m2)
2*a(m1) =-(g-(T2/M) + g-(T2/m2))

The right-hand side of the bottom equation has dimensions of force but the left hand side has dimensions of acceleration. check your substitution from the top to the bottom equation.

 

[Erdi]

LaTeX Guide

I could replace this with another expression. I see what you mean
So i solved for
2*a(m1) = -(a(M)+a(m2)), and so i found expression for a(M,m2)
2*a(m1) =-(g-(T2/M) + g-(T2/m2))

Replacement:
2*a(m1) =-(2g-(T1/2) * ((1/M) + (1/m2)), T1 = m1g - m1a
2*a(m1) = -(2*g - (m1/2) * (g-a(m1)) * ((1/M)+(1/m2))

 

[Erdi]

I got the same expression without the g for the reason already explained. The error is between the two steps below

The right-hand side of the bottom equation has dimensions of force but the left hand side has dimensions of acceleration. check your substitution from the top to the bottom equation.

I messed up on the solve for M part,
So i got M = 0.1333kg
ANy chance you solved for M with the values and got the same?

 

[kuruman]

2*a(m1) = -(2*g - (m1/2) * (g-a(m1)) * ((1/M)+(1/m2))

This looks wrong, again for dimensional reasons. Look at the term (g-a(m1)). You have an acceleration, g, from which you subtract a force , a(m1). If by a(m1) you mean "acceleration as a function of m1" then it is dimensionally correct, but the acceleration is not a function of m1 because m1 is fixed at 0.5 kg.

ANy chance you solved for M with the values and got the same?

I solved for for M and got $M=\dfrac{2}{15}~$kg. What do you get when you divide 2 by 15?

Maybe you got the answer but do you feel good about it?

 

[Erdi]

This looks wrong, again for dimensional reasons. Look at the term (g-a(m1)). You have an acceleration, g, from which you subtract a force , a(m1). If by a(m1) you mean "acceleration as a function of m1" then it is dimensionally correct, but the acceleration is not a function of m1 because m1 is fixed at 0.5 kg.

I solved for for M and got $M=\dfrac{2}{15}~$kg. What do you get when you divide 2 by 15?

Maybe you got the answer but do you feel good about it?

Hey! yeah i got 2/15 that is equal to 0.1333kg. Its my confusing writing on here that is the problem. I DO NOT mean function or force when i wrote a(m1). I don't know how to write a small the subscript. But i mean acceleration for m1 that is zero!

 

[erobz]

A good follow up to test your own understanding might be: What is the acceleration $a_2$ of mass $m_2$ as a function of $m_1,m_2,M$ with respect to a stationary frame?

 

[hutchphd]

Hey! yeah i got 2/15 that is equal to 0.1333kg. Its my confusing writing on here that is the problem. I DO NOT mean function or force when i wrote a(m1). I don't know how to write a small the subscript. But i mean acceleration for m1 that is zero!

If you wish coherent conversation on this venue learn LaTex. There is a built-in guide below. If you choose not to bother, we (I at least) will likely choose not to answer ! (Also it is a useful technical skill )

 

[Erdi]

Yes a couple People on here have mentioned LaTex. So i think i Will look into it.

 

[Erdi]

A good follow up to test your own understanding might be: What is the acceleration $a_2$ of mass $m_2$ as a function of $m_1,m_2,M$ with respect to a stationary frame?

Okai, i Will try to solve that little later today.

 

[erobz]

Okai, i Will try to solve that little later today.

No rush, its currently challenging my understanding too (open mouth-insert foot)!

 

[hutchphd]

I found it useful to define the following quantities $$\Delta=\frac {M-m_2} {M+m_2},~~~\mu=\frac {m_1} {M+m_2}$$ The result I get is $$a_1=-\frac {1+\Delta^2} \mu g$$ Seems to have the correct limits, but you should check it

 

[erobz]

What should we get at $m_1=0$ as a sanity check? I think $a_2$ should be $g \downarrow$?

 

[erobz]

I found it useful to define the following quantities $$\Delta=\frac {M-m_2} {M+m_2},~~~\mu=\frac {m_1} {M+m_2}$$ The result I get is $$a_1=-\frac {1+\Delta^2} \mu g$$ Seems to have the correct limits, but you should check it

the limit of $a_1$ as $m_1 \to 0$ seems to go to $\infty$. Am I interpreting that correctly?

 

[hutchphd]

That seems reasonable to me. Am I screwing something up?

 

[erobz]

That seems reasonable to me. Am I screwing something up?

I think the limit should be $g \uparrow$ for $a_1$. If $a_1$ was $\infty$ in the limit, then the hanging masses on the other side $M, m_2$ would be falling with infinite acceleration. They should only ever fall at $g$ at most.

In otherwords, if $m_1\to 0$ then the tension forces in ALL the ropes goes to $0$. The masses $M,m_2$ are just in freefall, accelerating at $g$.

 

[hutchphd]

Yeah you"re correct thanks. Let me do it again. I'll correct it in a bit.

 

[erobz]

Yeah you"re correct thanks. Let me do it again. I'll correct it in a bit.

I personally keep getting $m_1 = 0, a_2 = \frac{2}{3}g \downarrow$, . I'm presently stuck on finding a resolution.

 

[erobz]

Here is the system of equations I derive:

$$\begin{align} T_1 - m_1 g &= m_1 a_1 \tag{1} \\ T_2 - m_2g &= -m_2 \left( a_1 + a_2 \right) \tag{2} \\ T_2 - Mg &= M \left( a_2-a_1 \right) \tag{3} \end{align}$$

And finally, for the lower massless pulley (what I'm presently suspicious of as the culprit behind the error):

$$T_1 - 2T_2 = 0 \tag{4} $$

I get, after some tedious algebra:

$$ a_2 = \frac{ m_1 m_2 + M m_2 - M m_1 }{ M m_1 + 3 M m_2 + m1 m_2 } 2g $$

Which to my horror $m_1 \to 0, a_2 \to \frac{2}{3}g$

Where is the mistake?

 

[SammyS]

[ ATTACH type="full" width="226px" alt="1665404669890.png"]315357[/ATTACH]

Here is the system of equations I derive:

$$\begin{align} T_1 - m_1 g &= m_1 a_1 \tag{1} \\ T_2 - m_2g &= -m_2 \left( a_1 + a_2 \right) \tag{2} \\ T_2 - Mg &= M \left( a_2-a_1 \right) \tag{3} \end{align}$$
$$T_1 - 2T_2 = 0 \tag{4} $$
. . .

Where is the mistake?

The accelerations in equations (2) and (3) do not appear to be consistent with the figure nor with @Orodruin 's poem.

If $a_2$ is the acceleration of the block having mass, $m_2$, with 'up' being positive, then simply

$\displaystyle \quad T_2 - m_2 \, g = m_2 \,a_2$

and

$\displaystyle \quad T_2 - M \, g = M \,A \text{, where }A$ is the acceleration of the big block with mass, $M$.

The acceleration of the lower pulley is simply related to the acceleration of block 1. $\ \ a_P=-a_1$ .

The accelerations of block 2 and the big block, relative to the lower pulley are related by $A'=-a'_2$ so that $A'+a'_2=0$ . These are relative to the pulley. if we include it's acceleration we get the following

$\displaystyle \quad A'+a'_2=(A-a_P)+(a_2-a_P)=A+a_2+2a_1=0$

Solve for $A$ .

Equation (3) becomes: $\displaystyle \quad T_2 - M \, g = -M (2a_1+a_2)$ .

Equation (4) is correct !

 

[erobz]

I was treating $a_2$ as the acceleration of $m_2$ relative to the pulley. Which is different from how you are treating it?

That being said I found the solution via the Lagrangian in another thread I started, since this was an offshoot.

https://www.physicsforums.com/threads/system-of-masses-atwood-machine.1046304/

As $m_1 \to 0$, $a_2 \to 0$. I believe that is the result that is expected.

Mine above is going to $\frac{2}{3}g$, so it’s certainly wrong.

What do you get for the equations if you take $a_2$ as relative to the pulley as I intended?

 

[hutchphd]

I was treating a2 as the acceleration of m2 relative to the pulley. Which is different from how you are treating it?

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)

 

[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...)