Double pulley Atwood machine (with 3 masses)

[haruspex]

T12 - m1g

No. Reread my posts #44 and #46.

a12

No. Reread the second para of my post #48 and the first line of my post #41.

|a12|

No. There should be no need to take absolute values. Reread the first line of my post #48.

We don't seem to be getting anywhere. You keep making the same errors, apparently ignoring all my corrections.

 

[Sho Kano]

Sorry I don't understand why I cannot use those equations, it was the way that I was taught in school, and solved pulley problems with without mistake. I've never encountered the methods you suggest me to approach this problem with, and am confused with a lot of your explanations. But I do realize that you are suggesting to just go with the SUM of all forces for all the objects- because we are not always sure which direction things go in... right?
So following your suggestions, you seem to imply these equations:
m1g+T12 = m1a
m2g+T12 = m2a
T12+T3 = 0
m3g+T3 = m3a
2a = a1+a1

On another note, I have an question:
a1 is equal to a2, because both of them are connected to each other with a taut string, can't they be considered as one moving system?

 

[haruspex]

T12+T3 = 0

Two problems there.
There is a sign problem. As discussed, you are taking the tensions as being forces acting on the masses, so they should all end up with negative values, i.e. they all act upwards. That equation insists that if one is negative the other must be positive.
There is also a coefficient error.
Both can be corrected by considering the free body diagram of the left hand pulley. The top pulley reflects the tension T3, so T3 still acts upwards on the left hand pulley, but T12 now acts downwards. Therefore you should put a minus sign on the T12 term since it is equal and opposite to the T12 that acts on m1 and m2.
How many forces act on the left hand pulley?

2a = a1+a1

See below.

a1 is equal to a2, because both of them are connected to each other with a taut string, can't they be considered as one moving system?

This is what we went through such a discussion to get right in post #26. No, a1 need not equal a2. The pulley connecting them can rotate. Consider the case where m2 and m3 are very heavy but m1 is light. Which way do you think m1 will accelerate? What about m2?

 

[haruspex]

I don't understand why I cannot use those equations, it was the way that I was taught in school,

Your equations with mixed signs on the sum of forces would be right if you were to adopt the naive convention that the positive direction for each force is assigned based on which way that particular force is expected to act. This is why I asked you very early on what convention you were using. If I don't know which convention, I cannot tell if your equations are right.

But note that if we were to go back to the beginning and take the equations you had there on the understanding of using the naive convention, there would still be some sign errors in some equations because you had not applied that convention consistently. So I feel that having got this far we should persist with down is always positive.

There are lessons here:
- State your conventions explicitly up front, both so that others can follow your working and to help yourself be consistent.
- To be able to cope with more complex problems as you progress, ditch the naive convention and pick a more rigid one, such as up is always positive (as I said, this is very common), and right is always positive (to make it consistent with Cartesian coordinates).
- Always check that your equations make sense in simple examples... if this force here becomes more positive, does it make sense that this acceleration over here will also be more positive?

 

[haruspex]

Before you start to solve a problem, you should make a figure, show the accelerations and forces and explain all your notations.

View attachment 98880]

The problem says that both pulleys rotate clockwise. The string around the upper pulley accelerates with A, the string round the lower pulley accelerates with B with respect to the pulley. The upward direction is taken positive. The tensions act all with positive forces on the blocks.
The acceleration with respect to the ground are
a₃ = -A,
acceleration of the moving pulley is a_p = A,
a₁=A+B, a₂=A-B.

Now the equations F=ma come.
m₃a₃=T₃-m₃g,
The moving pulley has zero mass, so
m_pa_p=T₃-2T₁₂=0
m₁a₁=T₁₂-m₁g --> m₁(A+B)=T₁₂-m₁g
m₂a₂=T₁₂-m₂g --> m₁(A-B)=T₁₂-m₂g

Can you go on?

Ehild, please!
I've had an enormous battle to get Sho Kano to settle on a convention and stick to it. Sho Kano opted for positive down everywhere, so now I'm trying to get the equations to conform to that. Now you have posted equations using what I refer to as the naive convention of assigning positive directions independently for each force. This is only going to confuse matters totally.

 

[ehild]

I deleted my previous post and changed to downward as positive.

The masses are connected to the strings, and the strings move on the rotating pulleys clockwise. A is the acceleration of the upper string, and B is the acceleration of the lower string with respect to the moving pulley, you can write the acceleration of the masses and that of the moving pulley in terms of A and B.
(a3=A, a_pulley=-A, a1=-(A+B), a2=B-A.The tensions act with upward forces on the masses, but T3 is upward and both T12 are downward on the moving pulley. Try to write your equations again.

 

[Sho Kano]

This is what we went through such a discussion to get right in post #26. No, a1 need not equal a2. The pulley connecting them can rotate. Consider the case where m2 and m3 are very heavy but m1 is light. Which way do you think m1 will accelerate? What about m2?

What I'm understanding now is that the acceleration of m1 and m2 relative to the pulley have the same magnitude, but the absolute accelerations of m1 and m2 (relative to the earth) are different. Is that right?

 

[haruspex]

What I'm understanding now is that the acceleration of m1 and m2 relative to the pulley have the same magnitude, but the absolute accelerations of m1 and m2 (relative to the earth) are different. Is that right?

Yes.

 

[Sho Kano]

Ehild, please!
I've had an enormous battle to get Sho Kano to settle on a convention and stick to it. Sho Kano opted for positive down everywhere, so now I'm trying to get the equations to conform to that. Now you have posted equations using what I refer to as the naive convention of assigning positive directions independently for each force. This is only going to confuse matters totally.

I see, so (so far) these are the ways of assigning conventions. One was assigning the positive direction to the way a predicted force is. This is a "naive" way of doing it because you do not always know which way the force actually acts. Therefore, you suggest not assigning directions at all, and put just a sum of forces, which offers no room for errors. Is this right?
BTW, ehild's post helped a lot. I think seeing more ways of doing a problem gives you a better understanding of it.

 

[haruspex]

you suggest not assigning directions at all, and put just a sum of forces, which offers no room for errors. Is this right?

Not quite. I suggest assigning all forces and accelerations as positive in the same direction. If that direction is down, it means you treat the pull of a rope on one of the masses as being down, but expect the value to turn out to be negative, i.e. it is really up.

Now, I would not have gone this route in the first place, but when I asked you what your convention was you said it was positive down everywhere, but it turns out that was not really true. Anyway, I think you will have gained something out of all this.

 

[Sho Kano]

Not quite. I suggest assigning all forces and accelerations as positive in the same direction. If that direction is down, it means you treat the pull of a rope on one of the masses as being down, but expect the value to turn out to be negative, i.e. it is really up.

Now, I would not have gone this route in the first place, but when I asked you what your convention was you said it was positive down everywhere, but it turns out that was not really true. Anyway, I think you will have gained something out of all this.

How would YOU do this problem convention wise?

 

[haruspex]

How would YOU do this problem convention wise?

It varies according to whim.

 

[Sho Kano]

It varies according to whim.

Can you give me an example of the non-naive way?

 

[Sho Kano]

I deleted my previous post and changed to downward as positive.

View attachment 98882

The masses are connected to the strings, and the strings move on the rotating pulleys clockwise. A is the acceleration of the upper string, and B is the acceleration of the lower string with respect to the moving pulley, you can write the acceleration of the masses and that of the moving pulley in terms of A and B.
(a3=A, a_pulley=-A, a1=-(A+B), a2=B-A.The tensions act with upward forces on the masses, but T3 is upward and both T12 are downward on the moving pulley. Try to write your equations again.

-T12 + m1g = m1a1
-T12 + m2g = m2a2
-T3 + 2T12 = 0

a1 = -A-B
a2 = -A+B
-2A = a1 + a2
A = -a3
2a3 = a1 + a2

-T3 + m3g = m3a3

 

[haruspex]

-T12 + m1g = m1a1
-T12 + m2g = m2a2
-T3 + 2T12 = 0

a1 = -A-B
a2 = -A+B
-2A = a1 + a2
A = -a3
2a3 = a1 + a2

-T3 + m3g = m3a3

Those equations are all correct if we take all accelerations as positive down but all tensions as positive up forces on the masses. So let's go with that.

Edit: sorry, there is still a sign error in the accelerations. I didn't notice, partly because it is in an equation that you previously had correct. See e.g. the last line of your post #45.

 

[Sho Kano]

Those equations are all correct if we take all accelerations as positive down but all tensions as positive up forces on the masses. So let's go with that.

{ a }_{ 2 }\quad =\quad \frac { { m }_{ 1 }{ m }_{ 3 }g\quad -\quad 4{ m }_{ 1 }{ m }_{ 2 }g\quad +\quad { m }_{ 2 }{ m }_{ 3 }g }{ { m }_{ 2 }{ m }_{ 3 }\quad +\quad { m }_{ 1 }{ m }_{ 3 }\quad -\quad 4{ m }_{ 1 }{ m }_{ 2 } }
The negative in the denominator again, but I can't find any error in my work. Can you verify if it's wrong?

 

[ehild]

Use A and B in the equations instead of a1, a2, a3. Eliminate the T-s and solve for A and B. You get the accelerations from them. There will be no negative term in the denominator.

 

[ehild]

-
A = -a3

No, a3=A

 

[Sho Kano]

No, a3=A

Oh that's right,
-A+a3=0

 

[Sho Kano]

Use A and B in the equations instead of a1, a2, a3. Eliminate the T-s and solve for A and B. You get the accelerations from them. There will be no negative term in the denominator.

Here are my results so far, are these right?
A\quad =\quad \frac { { m }_{ 1 }{ m }_{ 3 }g\quad +\quad { m }_{ 2 }{ m }_{ 3 }g\quad -\quad 4{ m }_{ 1 }{ m }_{ 2 }g }{ 4{ m }_{ 1 }{ m }_{ 2 }\quad +\quad { m }_{ 1 }{ m }_{ 3 }\quad +\quad { m }_{ 2 }{ m }_{ 3 } } \\ B\quad =\quad \frac { 2{ m }_{ 2 }{ m }_{ 3 }g\quad -\quad 2{ m }_{ 1 }{ m }_{ 3 }g }{ 4{ m }_{ 1 }{ m }_{ 2 }\quad +\quad { m }_{ 2 }{ m }_{ 3 }\quad +\quad { m }_{ 1 }{ m }_{ 3 } }

 

[ehild]

A is correct. Check the signs in the numerator for B. I got the opposite. I will check my derivation.

 

[Sho Kano]

A is correct. Check the signs in the numerator for B. I got the opposite. I will check my derivation.

I didn't find any error upon looking at my derivation- I'll check it once more

 

[ehild]

I didn't find any error upon looking at my derivation- I'll check it once more

It is OK, the mistake was at my side.

 

[Sho Kano]

It is OK, the mistake was at my side.

Got it. So now I just plug A and B into these equations and that is the answer right?
a1 = -A-B
a2 = -A+B
2a3 = a1+a2

 

[ehild]

Got it. So now I just plug A and B into these equations and that is the answer right?
a1 = -A-B
a2 = -A+B
2a3 = a1+a2

Remember, a3=A. See the picture in post #56

 

[Sho Kano]

Remember, a3=A. See the picture in post #56

Gotcha, won't believe how happy I was to see that the denominators of A and B were the same so I could easily put them together

 

[Sho Kano]

For the 3rd question, would this be the right answer?
m3 = m1 + m2

 

[ehild]

Gotcha, won't believe how happy I was to see that the denominators of A and B were the same so I could easily put them together

Congratulation! Now it is easy to answer the other questions.

 

[ehild]

For the 3rd question, would this be the right answer?
m3 = m1 + m2

No, why do you think it? You have the expression for a3=A. What follows from it?

 

[Sho Kano]

No, why do you think it? You have the expression for a3=A. What follows from it?

A = a3 = 0
thus, m1m3g+m2m3g−4m1m2g = 0
and solve for m3
is this right?

 

[ehild]

A = a3 = 0
thus, m1m3g+m2m3g−4m1m2g = 0

Yes, what does it mean for m3?

 

[Sho Kano]

Yes, what does it mean for m3?

After solving for m3 for that equation, that would be the answer for question 3.
Then for question 4, I'd just plug in that into the m3's of a1 and a2 right?

 

[ehild]

After solving for m3 for that equation, that would be the answer for question 3.
Then for question 4, I'd just plug in that into the m3's of a1 and a2 right?

Use that A=0, so a1 = -B and a2 = B. Plug in the expression for m3 into B.

 

[Sho Kano]

Use that A=0, so a1 = -B and a2 = B. Plug in the expression for m3 into B.

Gotcha, so it goes like this right?

3) m1m3g+m2m3g−4m1m2g = 0 ; solve for m3
4) a1 = -B ; a2 = +B ; plug in m3 from (3)

 

[ehild]

Gotcha, so it goes like this right?

3) m1m3g+m2m3g−4m1m2g = 0 ; solve for m3
4) a1 = -B ; a2 = +B ; plug in m3 from (3)

Yes, what do you get?

 

[Sho Kano]

Yes, what do you get?

m3 = 4m1m2/m1+m2

 

[ehild]

m3 = 4m1m2/m1+m2

It would be good if you had not forgot parentheses. m3 = 4m1m2/(m1+m2).

 

[Sho Kano]

B = 8m1m2m2g - 8m1m1m2g / (4m1m1 + 5m1m2 + m2m2 + 4m1m2m2 + 4m1m1m2)
a1 = -8m1m2m2g + 8m1m1m2g / (4m1m1 + 5m1m2 + m2m2 + 4m1m2m2 + 4m1m1m2)
a2 = 8m1m2m2g - 8m1m1m2g / (4m1m1 + 5m1m2 + m2m2 + 4m1m2m2 + 4m1m1m2)

edit: something went horribly wrong in my calculations. Here is my revised answer
B = m1m2m2g - m1m1m2g / (m1m1m2 + m1m2m2)
B = m2g - m1g / (m1 + m2)
a1 = -m2g + m1g / (m1 + m2)
a2 = m2g - m1g / (m1 + m2)

 

[ehild]

PARENTHESES!

 

[Sho Kano]

PARENTHESES!

B = (m2g - m1g) / (m1 + m2)
a1 = (-m2g + m1g) / (m1 + m2)
a2 = (m2g - m1g) / (m1 + m2)

 

[ehild]

Pull out g, it is nicer.
You get the result expected. m3 does not accelerate. It can be in rest, but then the hanging pulley is in rest, too, and your system is equivalent with a single pulley-two mass system. The "driving force" is the difference of the weights, the total mass is m1+m2.
I go to sleep...

 

[Sho Kano]

Thanks haruspex, and ehild!

 

[Sho Kano]

No, a3=A

On second thought, why is a3 = A?
a3 is down, while A is upwards

 

[haruspex]

On second thought, why is a3 = A?
a3 is down, while A is upwards

Depends on conventions! If all accelerations are positive up then a3 and A will have opposite signs, but if A is positive up and a3 is positive down they will have the same sign.

 

[Sho Kano]

Depends on conventions! If all accelerations are positive up then a3 and A will have opposite signs, but if A is positive up and a3 is positive down they will have the same sign.

So because originally, I specified downwards as positive, that's why a3 is positive.

 

[ehild]

On second thought, why is a3 = A?
a3 is down, while A is upwards

Look at the figure in Post # 56, 'A' was how the rope around the upper pulley accelerated along its length. It was said that the pulleys move clockwise. The rope moves with the pulley, so the right piece of it moves downward, and the left piece moves upward. The rope must keep its length! A is the magnitude of acceleration of all points of the rope.
The block m3 is attached to the right end of the rope, which moves downward, so the acceleration of m3 is equal to A. The hanging pulley is connected to the other end, so it accelerates upward, so its acceleration with respect to the ground is -A.
When ropes are involved in a problem, we can speak of the acceleration along their length .

 

[Anjum S Khan]

What are the answers ?

 

[Sho Kano]

What are the answers ?

See post 70, a1 = -A-B, a2 = B-A, a3 = A

These are the answers to part 1