Double pulley Atwood machine (with 3 masses)

[haruspex]

Using my convention, the relationship that I got is now

a = -a1-a2 / 2 ; So I can use this without changing my equations right?

a, in the earlier algebra, was the acceleration of the pulley connecting m1 with m2. If you mean a3, what convention are you using? 2a3=-a1-a2 if all are positive upwards.

 

[Sho Kano]

Yes, my original convention was that downward was positive, so the acceleration of the pulley should be negative, which checks out.
The equations from the free body diagrams...what conventions are they based on?

 

[haruspex]

Yes, my original convention was that downward was positive, so the acceleration of the pulley should be negative, which checks out.
The equations from the free body diagrams...what conventions are they based on?

Are you referring to your equations in post #1? If you mean in general, the convention is up to you to choose, and is not fixed by the FBD unless you write arrows in it to show which directions are taken as positive.
Please state explicitly what conventions you wish to use in this problem and post the correponding equations you get. Then I will be able to check the signs for you.

 

[Sho Kano]

Sorry for the late response, here is my diagram,

 

[Sho Kano]

I think I made an error in the diagram, the negative sign is supposed to be on the opposite side (the left side) in equation 5. So the acceleration of the pulley should be -a pointing upwards.

 

[haruspex]

Sorry for the late response, here is my diagram,
View attachment 98828

In the first two equations, you have the same acceleration for m1 and m2.

It is still not clear what your sign conventions are. The diagram has an up arrow marked -a and a downarrow marked +a, implying down is positive. But you have another up arrow marked a, which says the opposite.

Tensions are not quite the same as forces. Rather, a tension of a straight massless string is an equal and opposite pair of forces. Best is to consider them as forces exerted on the masses. With positive down, the mg terms all act positively, so the sum of forces on each mass should be of the form T+mg, not -T+mg or T-mg etc. Likewise, the resulting accelerations should all be of the form ma, not -ma.

In short, the basic form of Newton's law is ΣF=ma. Note there are no minus signs in there. You only need to introduce minus signs into it if you are using an inconsistent convention, such as gravity positive down, other accelerations positive up.

If you are not sure you have the signs right in an equation, think through how each term affects others. E.g. in your first equation, suppose the acceleration will turnout to have a positive value. That means (with your positive down convention) the acceleration will be downwards. Would increasing the value of g increase the downward acceleration or reduce it? Which way does your equation say it will work?

 

[Sho Kano]

In the first two equations, you have the same acceleration for m1 and m2.

It is still not clear what your sign conventions are. The diagram has an up arrow marked -a and a downarrow marked +a, implying down is positive. But you have another up arrow marked a, which says the opposite.

Tensions are not quite the same as forces. Rather, a tension of a straight massless string is an equal and opposite pair of forces. Best is to consider them as forces exerted on the masses. With positive down, the mg terms all act positively, so the sum of forces on each mass should be of the form T+mg, not -T+mg or T-mg etc. Likewise, the resulting accelerations should all be of the form ma, not -ma.

In short, the basic form of Newton's law is ΣF=ma. Note there are no minus signs in there. You only need to introduce minus signs into it if you are using an inconsistent convention, such as gravity positive down, other accelerations positive up.

If you are not sure you have the signs right in an equation, think through how each term affects others. E.g. in your first equation, suppose the acceleration will turnout to have a positive value. That means (with your positive down convention) the acceleration will be downwards. Would increasing the value of g increase the downward acceleration or reduce it? Which way does your equation say it will work?

-My intended sign conventions are negative upwards, and positive downwards

-Do you mean that tensions are not quite the same as forces in the way like a normal force is to a force? I'm not sure what you mean by all masses should be of the form T+mg, if that was the case, then how would I distinguish one from the other? If mg is positive, then T is opposite- meaning it's -T.

-According to the first equation, I'll get a positive acceleration, which is inconsistent with my convention- so that's wrong.
Does this new equation for mass 1 look good?
m1g - t12 = m1a
I'll be getting a negative acceleration this time

 

[Sho Kano]

Using what I think would be an ordinary way of doing it, I get
T12 - m1g = m1a
and
m2g - T12 = m2a
Both ways seem to get a positive acceleration, but this is only the magnitude of the acceleration right? You are then free to choose which way is positive or negative? ie you are free to choose the convention after you solve for a.

 

[haruspex]

-Do you mean that tensions are not quite the same as forces in the way like a normal force is to a force?

If a string is under tension, it exerts a force at each end, so it is a pair of equal forces, each acting on a different body, rather than a single force. Because of pulleys etc. the directions need not be parallel or opposite.

I'm not sure what you mean by all masses should be of the form T+mg, if that was the case, then how would I distinguish one from the other? If mg is positive, then T is opposite- meaning it's -T.

Not all masses, the forces on all masses.
The forces on m1 are m1g and T. If we take positive down for all forces and accelerations then the net force on m1 is m1g+T. g will have a positive value and T a negative one, because it actually acts upwards.
In most problems it is evident which way each force acts, and most students tend to take each force variable as positive in that direction. Thus, g would be positive down and T positive up. With that convention, the net downward force is m1g-T, and both variables will turn out to have positive values. But this cannot always be done because sometimes you cannot guess in advance which way a force will really act.
So I recommend getting used to using the same convention for the positive direction of all forces on each body and therefore writing the net force as simply their sum.
Likewise, if accelerations are also positive down then the other side of the equation is simply ma: ##\Sigma F=ma##, no minus signs anywhere.

It can get even more confusing with g. The commonest vertical convention is positive up for all forces, displacements, velocities and acceleration. So what to do with g then? The logical thing to do is to still write the gravitational force as mg, but ascribe a negative value to the symbol g. However, in practice, most students like g to be positive so write the force as -mg.

-According to the first equation, I'll get a positive acceleration, which is inconsistent with my convention- so that's wrong.
Does this new equation for mass 1 look good?
m1g - t12 = m1a
I'll be getting a negative acceleration this time

No, for the reasons given, it should be m1g+ T12 = m1 a1. T12 will turn out negative. If it exceeds m1g in magnitude then that will make a1 negative, i.e. an upward acceleration, which would be correct.

 

[Sho Kano]

If a string is under tension, it exerts a force at each end, so it is a pair of equal forces, each acting on a different body, rather than a single force. Because of pulleys etc. the directions need not be parallel or opposite.

Not all masses, the forces on all masses.
The forces on m1 are m1g and T. If we take positive down for all forces and accelerations then the net force on m1 is m1g+T. g will have a positive value and T a negative one, because it actually acts upwards.
In most problems it is evident which way each force acts, and most students tend to take each force variable as positive in that direction. Thus, g would be positive down and T positive up. With that convention, the net downward force is m1g-T, and both variables will turn out to have positive values. But this cannot always be done because sometimes you cannot guess in advance which way a force will really act.
So I recommend getting used to using the same convention for the positive direction of all forces on each body and therefore writing the net force as simply their sum.
Likewise, if accelerations are also positive down then the other side of the equation is simply ma: ##\Sigma F=ma##, no minus signs anywhere.

It can get even more confusing with g. The commonest vertical convention is positive up for all forces, displacements, velocities and acceleration. So what to do with g then? The logical thing to do is to still write the gravitational force as mg, but ascribe a negative value to the symbol g. However, in practice, most students like g to be positive so write the force as -mg.

No, for the reasons given, it should be m1g+ T12 = m1 a1. T12 will turn out negative. If it exceeds m1g in magnitude then that will make a1 negative, i.e. an upward acceleration, which would be correct.

-Tensions exert a force on each end, meaning each ends of m1 and m2 in this case?
-I think it will be clearer if we do this,
What conventions are these equations holding?
T12 - m1g = m1a
m2g - T12 = m2a
T3 - 2T12 = 0
m3g - T3 = m3a
2*a = a1+a2

There are no conventions right? You are free to choose since acceleration in this case is a magnitude?

 

[haruspex]

-Tensions exert a force on each end, meaning each ends of m1 and m2 in this case?
-I think it will be clearer if we do this,
What conventions are these equations holding?
T12 - m1g = m1a
m2g - T12 = m2a
T3 - 2T12 = 0
m3g - T3 = m3a
2*a = a1+a2

There are no conventions right? You are free to choose since acceleration in this case is a magnitude?

As soon as you write that the net force on m1 is T12-m1g you have adopted the convention that the positive direction for the one force is the negative direction for the other. If you stick to the same sign convention for all forces the net force is the SUM of the forces, not a mix of sums and differences.

 

[Sho Kano]

As soon as you write that the net force on m1 is T12-m1g you have adopted the convention that the positive direction for the one force is the negative direction for the other. If you stick to the same sign convention for all forces the net force is the SUM of the forces, not a mix of sums and differences.

I think I know what you're saying. On a separate note, are these equations are wrong? Isn't acceleration (a) from those equations an absolute value of the actual acceleration?

 

[haruspex]

I think I know what you're saying. On a separate note, are these equations are wrong? Isn't acceleration (a) from those equations an absolute value of the actual acceleration?

If you get the equations right, it should have the correct sign.
By the way, despite my comments, you keep writing m1a and m2a instead of m1a1 and m2a2. Do you think the accelerations are the same, or are you just being a bit lazy?

 

[Sho Kano]

If you get the equations right, it should have the correct sign.
By the way, despite my comments, you keep writing m1a and m2a instead of m1a1 and m2a2. Do you think the accelerations are the same, or are you just being a bit lazy?

Sorry, you sort of did imply loosely that I am supposed to be putting subscripts for the accelerations, but I didn't notice it.
So you mean that if I work out the accelerations from those equations, and if they have the right signs according to my convention, then they are correct?

 

[Sho Kano]

How about this?
T12 - m1g = m1*|a12|
m2g - T12 = m2*|a12|
T3 - 2T12 = 0
m3g - T3 = m3|a3|
2*|a| =|a1|+|a2|

 

[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.

[itex]{ 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 } } [/itex]
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?
[itex]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 } } [/itex]