Accelerations in a Three Mass Pulley System: How Are They Related?

[Mr Davis 97]

Homework Statement

The system of masses M₁, M₂, and M₃ in the sketch uses massless pulleys and ropes. The horizontal table is frictionless How are the accelerations related?

Homework Equations

The Attempt at a Solution

I feel like the solution will come from the constraint equation involving the string of the pulley system. However, I am not sure how to do this. It seems that since M₁ and M₂ are different masses, they do not have to accelerate at the same rate, which means that M₃ would have all kinds of funky motion, not just linear in the y-direction. How do I think about this problem?

 

[TomHart]

To think about these kinds of problems, I like to go to the extreme cases. One extreme case would be if M2 and M3 were equal and M1 was infinite. What would happen in that situation. What would the acceleration of M2 be relative to M3? Another case (which really isn't an extreme) would be if M1 and M2 were equal. What would the acceleration of M3 be relative to M1 and M2?

Also, are there any forces available to move M3 in a direction other than vertical?

 

[Mr Davis 97]

To think about these kinds of problems, I like to go to the extreme cases. One extreme case would be if M2 and M3 were equal and M1 was infinite. What would happen in that situation. What would the acceleration of M2 be relative to M3? Another case (which really isn't an extreme) would be if M1 and M2 were equal. What would the acceleration of M3 be relative to M1 and M2?

Also, are there any forces available to move M3 in a direction other than vertical?

So this is what I tried, where x₁ is the length of the string from the first mass to the first pulley, and x₂ is the length of the string from the first pulley to the middle pulley, and x₃ is the length of the string from the third pulley to the third mass. Then

$l = x_1 + \pi r_1 + x_3 + \pi r_2 + x_3 + \pi r_3 + x_2$
$0 = \ddot{x}_1 + \ddot{x}_2 + 2 \ddot{x}_3$
$-2 \ddot{x}_3 = \ddot{x}_1 + \ddot{x}_2$

Is this the correct relation between the accelerations of the masses?

 

[TomHart]

It looks right to me, but I haven't thought it all the way through. Here's how I see it in my own way of thinking. For a car driving down a road, the velocity of the point of the tire making contact with the road is 0. The midpoint of the tire/wheel is moving at the same velocity as the car. The top of the tire is moving twice as fast as the car. That is the situation we have here if M1 is infinite mass. The left side of the lower pulley is the road, the middle of the pulley is moving at some velocity and the right-most side of the pulley is moving twice the velocity of the center. So therefore your equation works in that case (because a₁ = 0).

For the case of M1 = M2, the velocity (and thus acceleration) of M1 and M2 has to be equal to M3 because there is no rotation of the lower pulley. It's as though the 2 sides of the rope going to the lower pulley are connected directly to M3. So once again, your equation works in that case.

 

[Mr Davis 97]

It looks right to me, but I haven't thought it all the way through. Here's how I see it in my own way of thinking. For a car driving down a road, the velocity of the point of the tire making contact with the road is 0. The midpoint of the tire/wheel is moving at the same velocity as the car. The top of the tire is moving twice as fast as the car. That is the situation we have here if M1 is infinite mass. The left side of the lower pulley is the road, the middle of the pulley is moving at some velocity and the right-most side of the pulley is moving twice the velocity of the center. So therefore your equation works in that case (because a₁ = 0).

For the case of M1 = M2, the velocity (and thus acceleration) of M1 and M2 has to be equal to M3 because there is no rotation of the lower pulley. It's as though the 2 sides of the rope going to the lower pulley are connected directly to M3. So once again, your equation works in that case.

Well that's good. Also, can you explain why there is a negative sign? What does this mean physically? Is the negative sign supposed to be there?

 

[Charles Link]

Well that's good. Also, can you explain why there is a negative sign? What does this mean physically? Is the negative sign supposed to be there?

To answer your question, the position of the mass $M_3$ in the y-direction is $y=x_3+b$ where $y$ is measured from the plane of the table and $b$ is some constant. Thereby $\ddot{y}=\ddot{x_3}$. If you want to call that a " $-y$" so that $-y=x+b$ and $\ddot{y}=-\ddot{x_3}$, you can do that also. The sign of $x_3$ is determined by the length of the string: As $x_1$ and $x_2$ get longer, $x_3$ must get shorter, i.e. $x_1+x_2+2 x_3=Constant$.

 

[haruspex]

So this is what I tried, where x₁ is the length of the string from the first mass to the first pulley, and x₂ is the length of the string from the first pulley to the middle pulley, and x₃ is the length of the string from the third pulley to the third mass. Then

$l = x_1 + \pi r_1 + x_3 + \pi r_2 + x_3 + \pi r_3 + x_2$
$0 = \ddot{x}_1 + \ddot{x}_2 + 2 \ddot{x}_3$
$-2 \ddot{x}_3 = \ddot{x}_1 + \ddot{x}_2$

Is this the correct relation between the accelerations of the masses?

You seem to have swapped the meanings of x₂ and x₃ at some point. According to your definitions, x₁+2x₂+x₃ should be constant.

Edit: looks like it was simply a typo in the definitions. x₃ is the drop height to M₃.

 

[Miles123K]

Im also working on this book and I'd like to post my solution though I cannot guarantee that I am correct:
The constraints are defined in standard x-y coordinates:
$(l_1-x_1)+(l_1+x_2)+\pi r_1+2l_2+\pi r_2-2x_3=K$
Differentiate twice in respect to $t$ to get:
$-\ddot x_1 + \ddot x_2 + \ddot 2x_3 = 0$
$\ddot 2x_3 = \ddot x_2 - \ddot x_1$
From pulley 3, we can see that
$2T = m_3 g$
and on mass 1 and mass 3
$m_1 \ddot x_1 = T$
and
$m_2 \ddot x_2 = -T$
substitute
$\ddot x_1 = \frac 1 2 \frac {m_3} {m_1} g$
and
$\ddot x_2 = - \frac 1 2 \frac {m_3} {m_2} g$
$\ddot x_3 = \frac 1 2 (\ddot x_1 - \ddot x_2) = \frac 1 4 ( - \frac {m_3} {m_2} g - \frac {m_3} {m_1} g）$

 

[haruspex]

Im also working on this book and I'd like to post my solution though I cannot guarantee that I am correct:
The constraints are defined in standard x-y coordinates:
$(l_1-x_1)+(l_1+x_2)+\pi r_1+2l_2+\pi r_2-2x_3=K$
Differentiate twice in respect to $t$ to get:
$-\ddot x_1 + \ddot x_2 + \ddot 2x_3 = 0$
$\ddot 2x_3 = \ddot x_2 - \ddot x_1$
From pulley 3, we can see that
$2T = m_3 g$
and on mass 1 and mass 3
$m_1 \ddot x_1 = T$
and
$m_2 \ddot x_2 = -T$
substitute
$\ddot x_1 = \frac 1 2 \frac {m_3} {m_1} g$
and
$\ddot x_2 = - \frac 1 2 \frac {m_3} {m_2} g$
$\ddot x_3 = \frac 1 2 (\ddot x_1 - \ddot x_2) = \frac 1 4 ( - \frac {m_3} {m_2} g - \frac {m_3} {m_1} g）$

how, exactly, are you defining x₁? Is it the same way MrDavis defined it?
What are l₁ and l₂?

 

[Miles123K]

how, exactly, are you defining x₁? Is it the same way MrDavis defined it?
What are l₁ and l₂?

$l_1$ and $l_2$ are different portions of length on the strings. I defined $x_1$ as the horizontal displacement of $m_1$ from its initial position, positive toward right, and the same thing for all the others. I don't believe they are the same. The original user who posted the thread practically ignored the signs in the math.

 

[haruspex]

$l_1$ and $l_2$ are different portions of length on the strings. I defined $x_1$ as the horizontal displacement of $m_1$ from its initial position, positive toward right, and the same thing for all the others. I don't believe they are the same. The original user who posted the thread practically ignored the signs in the math.

Ok, that helps.

2T=m3g

Isn't m₃ accelerating?

 

[Miles123K]

It is, and I did calculate it

 

[haruspex]

It is, and I did calculate it

But the equation I quoted says the forces on it are in balance.

 

[Miles123K]

But the equation I quoted says the forces on it are in balance.

Actually, in this case I defined $x_3$ as the displacement of the massless pulley right above $m_3$. The forces are in balance but there will still be acceleration since we have the mass of the pulley as zero. Hope that this explanation makes it intuitive.

 

[haruspex]

Actually, in this case I defined $x_3$ as the displacement of the massless pulley right above $m_3$. The forces are in balance but there will still be acceleration since we have the mass of the pulley as zero. Hope that this explanation makes it intuitive.

That doesn't resolve it. If your equation represents the forces on the pulley then the m₃g term is wrong. If m₃ is exerting m₃g down on the pulley then the pulley must be exerting m₃g up on the mass. That would make the forces on m₃ in balance, so it would not accelerate.

 

[Miles123K]

That doesn't resolve it. If your equation represents the forces on the pulley then the m₃g term is wrong. If m₃ is exerting m₃g down on the pulley then the pulley must be exerting m₃g up on the mass. That would make the forces on m₃ in balance, so it would not accelerate.

Umm I don't really know how to explain this but I have seen this working out as it should. My method is identical to the one used in this question:
https://www.physicsforums.com/threads/kleppner-mass-and-pulleys.743779/
and in fact this equation was used throughout this book.

 

[haruspex]

My method is identical to the one used in this question:

No it isn't. The equation there is
$M_Cg−T_2=M_C\ddot y_C$
Your equation is missing the acceleration of the hanging mass.

As I am sure you know, the general equation is ΣF=ma. Your equation "2T=m₃g" is saying ΣF=0, i.e. no acceleration of that mass.

 

[Miles123K]

No it isn't. The equation there is
$M_Cg−T_2=M_C\ddot y_C$
Your equation is missing the acceleration of the hanging mass.

As I am sure you know, the general equation is ΣF=ma. Your equation "2T=m₃g" is saying ΣF=0, i.e. no acceleration of that mass.

The way I did it is by defining $x_3$ as the position of the pulley instead of the mass. Since the pulley is massless, the total force acting on the pulley must also be zero, or the acceleration of pulley will be infinite. The displacement of $m_3$ must be consistent with that of the third pulley, so the thing calculated is actually the same.
Again, I might be wrong and there's no solution provided to this.

 

[haruspex]

the total force acting on the pulley must also be zero

Yes, but as I wrote in post #15, the force mass 3 exerts on the pulley is not m₃g.

 

[Miles123K]

Yes, but as I wrote in post #15, the force mass 3 exerts on the pulley is not m₃g.

Damn bro you are correct. I totally screwed up. I will correct my solutions right away.

 

[Miles123K]

First of all I would like to thank @haruspex for pointing out my mistake. I don't know what happened but I totally screwed up with calculating the accelerations. Here is the proper solution.
Def $T_2$ as the tension between $m_3$ and the pulley above. $T_1$ as the tension of the strings to $m_1$ and $m_2$
$m_3 \ddot x_3 = T_2 - m_3 g$
$m_1 \ddot x_1 = T_1$
$m_2 \ddot x_2 = -T_1$
Solve this system of equations and you should get:
$\ddot x_1 = g \frac {2m_2 m_3} {4m_1 m_2 + m_2 m_3 + m_1 m_3}$
$\ddot x_2 = - g \frac {2m_1 m_3} {4m_1 m_2 + m_2 m_3 + m_1 m_3}$
$\ddot x_3 = - g \frac {m_1 m_3+m_2 m_3} {4m_1 m_2 + m_2 m_3 + m_1 m_3}$
The definition of the coordinates are still the same as above, but ignore the acceleration calculations above. They are non-sense.

 

[haruspex]

First of all I would like to thank @haruspex for pointing out my mistake. I don't know what happened but I totally screwed up with calculating the accelerations. Here is the proper solution.
Def $T_2$ as the tension between $m_3$ and the pulley above. $T_1$ as the tension of the strings to $m_1$ and $m_2$
$m_3 \ddot x_3 = T_2 - m_3 g$
$m_1 \ddot x_1 = T_1$
$m_2 \ddot x_2 = -T_1$
Solve this system of equations and you should get:
$\ddot x_1 = g \frac {2m_2 m_3} {4m_1 m_2 + m_2 m_3 + m_1 m_3}$
$\ddot x_2 = - g \frac {2m_1 m_3} {4m_1 m_2 + m_2 m_3 + m_1 m_3}$
$\ddot x_3 = - g \frac {m_1 m_3+m_2 m_3} {4m_1 m_2 + m_2 m_3 + m_1 m_3}$
The definition of the coordinates are still the same as above, but ignore the acceleration calculations above. They are non-sense.

And, of course, T₁=T₂.

 

[Miles123K]

And, of course, T₁=T₂.

I think you meant $T_2 = 2T_1$ right?

 

[haruspex]

I think you meant $T_2 = 2T_1$ right?

No. It's all the same light string and the pulleys are massless, so the tension is the same everywhere. The force it exerts on m3 is 2T.

 

[Miles123K]

No. It's all the same light string and the pulleys are massless, so the tension is the same everywhere. The force it exerts on m3 is 2T.

Oh I defined $T_2$ as one of the forces on $m_3$. I see what you mean.