Problem with two pulleys and three masses

In summary, the conversation revolved around understanding the tension and weight distribution in a system involving masses and a pulley. It was concluded that in order for mass m1 to have no acceleration, the tension T1 in the string attached to m1 must be equal to m1*g. However, when mass M is zero, the tension in the string is also zero, and m1 will accelerate downwards at g. The conversation also touched upon Newton's law and the use of free body diagrams to understand the forces acting on each mass. Finally, an alternative approach was suggested involving finding the tension T(M,m2) in the string connecting the pulley to the ceiling and using this to solve for M when m1 is at rest.
  • #71
erobz said:
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))
 
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  • #72
kuruman said:
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?
 
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  • #73
Erdi said:
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.
Erdi said:
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?
 
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  • #74
kuruman said:
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!
 
  • #75
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?
 
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  • #76
Erdi said:
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 )
 
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  • #77
Yes a couple People on here have mentioned LaTex. So i think i Will look into it.
 
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  • #78
erobz said:
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.
 
  • #79
Erdi said:
Okai, i Will try to solve that little later today.
No rush, its currently challenging my understanding too (open mouth-insert foot)!
 
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  • #80
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
 
  • #81
What should we get at ##m_1=0## as a sanity check? I think ##a_2## should be ##g \downarrow ##?
 
  • #82
hutchphd said:
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?
 
  • #83
That seems reasonable to me. Am I screwing something up?
 
  • #84
hutchphd said:
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##.
 
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  • #85
Yeah you"re correct thanks. Let me do it again. I'll correct it in a bit.
 
  • #86
hutchphd said:
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.
 
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  • #87
1665404669890.png


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?
 
  • #88
erobz said:
[ 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 !
 
  • #89
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?
 
  • #90
erobz said:
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)
 
  • #91
hutchphd said:
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)
1665483306870.png

erobz said:
$$\mathcal{L} = T - U $$

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

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

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

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

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

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

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 ).
 
  • #92
hutchphd said:
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.
 
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  • #93
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!
 
  • #94
hutchphd said:
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.
 
  • #95
hutchphd said:
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?
 
  • #96
$$ T_1 - 2mg = -2m a_1$$Where does this come from?
 
  • #97
hutchphd said:
$$ T_1 - 2mg = -2m a_1$$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?
 
  • #98
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.
 
  • #99
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, $$a_2=g-\frac{4m_1m_3}{m_1(m_2+m_3)+4m_2m_3}g=\frac{m_1(m_2+m_3)+4m_3(m_2-m_1)}{m_1(m_2+m_3)+4m_2m_3}g.$$
 
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  • #100
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 :-p, 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...)
 
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  • #101
kuruman said:
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!
 
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  • #102
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.
 
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  • #103
erobz said:
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.
 
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