Relationship between accelerations of blocks 1 and 2

[vcsharp2003]

Homework Statement

In the following mass pulley system what would be the relationship between accelerations of blocks 1 and 2? Assume acceleration of block 1 is $a$ downwards, so that $a$ denotes the magnitude of acceleration of block 1.

Relevant Equations

None

Assume that the lengths of various sections of the string at an instant are as shown. So we can say that
$l_1 + l_2+l_3= k$ at an instant of time $t$, where $k$ is a constant.
Taking the second derivative of both sides w.r.t. $t$, we get $$\frac {d^2{l_1}} {dt^2} +\frac {d^2{l_2}} {dt^2} + \frac {d^2{l_3}} {dt^2} = 0$$.
After above step, I'm stuck. I know the second and third terms in LHS will be each equal to $a$.
$$\therefore \frac {d^2{l_3}} {dt^2} = -2a$$.

I need to now prove that LHS is negative of acceleration of block of 2. How will I prove that?

 

[Orodruin]

Homework Statement:: In the following mass pulley system what would be the relationship between accelerations of blocks 1 and 2? Assume acceleration of block 1 is $a$ downwards, so that $a$ denotes the magnitude of acceleration of block 1.
Relevant Equations:: None

I need to now prove that LHS is negative of acceleration of block of 2. How will I prove that?

What do you mean? This depends on which direction you state as positive for the acceleration of 2.

 

[anuttarasammyak]

As time goes by l_3 is shortened or decreasing with accelerated pace. Shortend means signature minus.

 

[vcsharp2003]

What do you mean? This depends on which direction you state as positive for the acceleration of 2.

The problem I'm having is how can block 2 have a negative acceleration. I cannot find any meaning in the acceleration of block 2 being negative. $a$ is positive since it's the magnitude of block 1 acceleration.

 

[vcsharp2003]

As time goes by l_3 is shortened or decreasing with accelerated pace. Shortend means signature minus.

$l_3$ is shortened with time and therefore the first derivative of it is negative. That I get. But now, how will I know whether second derivative of $l_3$ is positive or negative?

 

[Orodruin]

The problem I'm having is how can block 2 have a negative acceleration. I cannot find any meaning in the acceleration of block 2 being negative. $a$ is positive since it's the magnitude of block 1 acceleration.

Again, this depends on what direction you define as negative. If $a$ is positive and you define the positive direction for 2 as the direction of growing $\ell_3$, ie, 2 moving to the left, then the acceleration of 2 is negative. If you define it the other way around it is positive.

 

[vcsharp2003]

Again, this depends on what direction you define as negative. If $a$ is positive and you define the positive direction for 2 as the direction of growing $\ell_3$, ie, 2 moving to the left, then the acceleration of 2 is negative. If you define it the other way around it is positive.

I have taken $l_3$ as distance from the pulley to block 2. So, its as if I took the positive x-axis to be pointing leftwards with its origin at the pulley fixed to table. Since we know the block 2 will move towards the right i.e. in negative x-axis direction, so its velocity i.e. first derivative will be negative, and also acceleration will point in negative x-direction so its second derivative will also be negative. That is the meaning of second derivative coming out to be negative. Does that sound right?

 

[vcsharp2003]

Could we also say the following as an alternate method to get the acceleration relationships? Someone mentioned to me that this is a quick shortcut approach to getting the acceleration relationships.

Since block 1 experiences a tension of 2T if block 2 experiences a tension of T and also acceleration can be considered inversely proportional to the tension on the block, so block 2 has an acceleration that's half the acceleration of block 1.

 

[haruspex]

Since block 1 experiences a tension of 2T if block 2 experiences a tension of T and also acceleration can be considered inversely proportional to the tension on the block

Right so far, then

so block 2 has an acceleration that's half the acceleration of block 1.

Whoops!

 

[vcsharp2003]

Whoops!

But, it works in this situation and other pulley mass systems. I have tried it in other pulley mass systems and surprisingly it always gave the correct acceleration relationships. But, don't know why it works consistently across various pulley mass systems.

I applied it to two mass pulley systems and it worked giving the correct acceleration relationship. In first system $2T_1 = T_2$ so acceleration of $m_2$ is half the acceleration of $m1$.

 

[anuttarasammyak]

That I get. But now, how will I know whether second derivative of l3 is positive or negative?

As time goes by the length become shorter and shorter in enhanced pace.
For an example
Length sequence with time 100 96 91 84 74 ...
change -4 -5 -7 -10, ...
pace of change -1 -2 -3 ... negative.
Can you show a case that pace of change is positive and l_3 keep shortened more and more ?

 

[vcsharp2003]

Can you show a case that pace of change is positive and l_3 keep shortened more and more ?

You're right.

 

[haruspex]

But, it works in this situation and other pulley mass systems.

As I responded in post #9, your inversion principle is fine, but you did not apply it correctly.
You wrote "block 1 experiences a tension of 2T if block 2 experiences a tension of T". So does block 1 experience half the acceleration or twice the acceleration?

 

[vcsharp2003]

As I responded in post #9, your inversion principle is fine, but you did not apply it correctly.
You wrote "block 1 experiences a tension of 2T if block 2 experiences a tension of T". So does block 1 experience half the acceleration or twice the acceleration?

I said block 2 experiences twice the tension experienced by block 1 since $2T_1= T_2$. So block 2 acceleration will be half that of block 1

UPDATE 1: Sorry, you're right. I was thinking about the later post# 10 when I wrote the above para. In my orginal question diagram, block 1 would experience half the acceleration of block 2 since block 1 experiences twice the tension that block 2 experiences.

 

[vcsharp2003]

Right so far,

Why this quick shortcut works is still a mystery to me but it sure helps arriving at the relationship between accelerations very quickly.

 

[haruspex]

Why this quick shortcut works is still a mystery to me but it sure helps arriving at the relationship between accelerations very quickly.

It is effectively conservation of energy.

 

[erobz]

It seems like you just forgot to assume a convention for block 2.

let's say its this $\rightarrow^+$

The other bits are fine:

$$ \ddot{l} = \ddot{l_1} + \ddot{l_2} + \ddot{l_3} = 0 $$

$$ \ddot{l_3} = - \left( \ddot{l_1} + \ddot{l_2} \right) $$

$$ \ddot{l_1} = \ddot{l_2} \implies \ddot{l_3} = -2 \ddot{l_2} $$

Now you need to examine free body diagrams of each block under the assumed convention. If you correctly set those up the directions should remain aligned with the convention.

I get the following equations from the FBD's

$$ m_2 \, \ddot{l_3} = T $$

$$ m_1 \, \ddot{l_2} = m_1 g - 2T $$

 

[vcsharp2003]

It is effectively conservation of energy.

Ok, that's very interesting. Is it like equating the work done by the tensions? Equating in my orginal question, we would get $T \times s_1 = 2T \times s_2$, where $s_1$ and $s_2$ would be the displacements of masses $m_1$ and $m_2$ respectively. So, displacement of $m_1$ would be twice the displacement of $m_2$ in the same time interval, and therefore, we could say acceleration of $m_1$ is twice the acceleration of $m_2$, provided we assume that eac mass experiences a uniform acceleration that remains constant with time.
Is this what you meant?

UPDATE1: I think the net work done by tensions on masses should be zero. This is because the total energy, which will KE + GPE of the masses at two different instants of time remains the same since we are ignoring any friction resistance that may otherwise exist in a real situation. Thus individual work done by tensions on each mass must cancel out i.e. +ve and -ve tension work would cancel out.

 

[erobz]

$$ m_2 \, \ddot{l_3} = T $$

$$ m_1 \, \ddot{l_2} = m_1 g - 2T $$

I'm second guessing myself!

When I'm solving this I'm getting

$$ \ddot{l_2} = \frac{m_1}{m_1- 4 m_2} g $$

That can't be right!

I expect:

$$ \ddot{l_2} = \frac{m_1}{m_1+ 4 m_2} g $$

Seems like this is the system:

$$ m_2 \, \ddot{l_3} = -T $$

$$ m_1 \, \ddot{l_2} = m_1 g - 2T $$

It seems like we have to account for the direction of $\ddot{l_3}$ as negative to get there... for some reason its seeming like there is maybe some unnecessary steps in my approach? Because isn't it the case that we shouldn't have to correctly assume how this thing evolves, but rather be told how it evolves by the solution given our assumptions?

 

[vcsharp2003]

I'm second guessing myself!

When I'm solving this I'm getting

$$ \ddot{l_2} = \frac{m_1}{m_1- 4 m_2} g $$

That can't be right!

I expect:

$$ \ddot{l_2} = \frac{m_1}{m_1+ 4 m_2} g $$

Seems like this is the system:
It seems like we have to account for the direction of $\ddot{l_3}$ as negative to get there... for some reason its seeming like there is maybe some unnecessary steps in my approach? Because isn't it the case that we shouldn't have to correctly assume how this thing evolves, but rather be told how it evolves by the solution given our assumptions?

The mistake you committed was a very subtle mistake. With your axis system pointing right with origin at the table pulley, we know that the coordinate $l_3$ is negative and therefore the horizontal distance along table of string is $-l_3$ since distance must always be +ve. Now our initial equation becomes $-l_3 + l_2 + l_1= k$. This would then give you the correct values with correct directions.

Remember we must always take coordinates and not distances when we use an axis system in kinematics. I also committed the same mistake in my analysis of taking distances in my original post and that's why I was also very, very confused. It's good you pointed out this issue which helped clarify a key concept when using axis system in kinematics.

 

[haruspex]

Ok, that's very interesting. Is it like equating the work done by the tensions? Equating in my orginal question, we would get $T \times s_1 = 2T \times s_2$, where $s_1$ and $s_2$ would be the displacements of masses $m_1$ and $m_2$ respectively. So, displacement of $m_1$ would be twice the displacement of $m_2$ in the same time interval, and therefore, we could say acceleration of $m_1$ is twice the acceleration of $m_2$, provided we assume that eac mass experiences a uniform acceleration that remains constant with time.
Is this what you meant?

UPDATE1: I think the net work done by tensions on masses should be zero. This is because the total energy, which will KE + GPE of the masses at two different instants of time remains the same since we are ignoring any friction resistance that may otherwise exist in a real situation. Thus individual work done by tensions on each mass must cancel out i.e. +ve and -ve tension work would cancel out.

Yes.

 

[vcsharp2003]

Yes.

I am now wondering about it using the following energy equation that applies to any system for all cases of mechanical energy without any heat flow scenarios.
$$ W_{net} = \Delta {KE} + \Delta {GPE}$$
The question that comes up now is why tensions don't contribute to $W_{net}$. Is it because $W_{net}$ denotes the total work done by external forces and tension is not an external force but an internal force when we consider the mass pulley system as the target for above equation? If yes, then we could further say that net work done by internal forces in a pure mechanical energy scenario is always zero.

 

[vcsharp2003]

I am now wondering about it using the following energy equation that applies to any system for all cases of mechanical energy without any heat flow scenarios.
$$ W_{net} = \Delta {KE} + \Delta {GPE}$$
The question that comes up now is why tensions don't contribute to $W_{net}$. Is it because $W_{net}$ denotes the total work done by external forces and tension is not an external force but an internal force when we consider the mass pulley system as the target for above equation? If yes, then we could further say that net work done by internal forces in a pure mechanical energy scenario is always zero.

For a system of particles rather than just a particle the net work done is work done by external plus internal forces, when using the energy equation. So, my idea of internal forces not being considered is incorrect. They need to be considered.

 

[haruspex]

For a system of particles rather than just a particle the net work done is work done by external plus internal forces, when using the energy equation. So, my idea of internal forces not being considered is incorrect. They need to be considered.

It depends what you are counting in internal energy. A muscle produces mechanical energy from chemical energy, but the total energy does not change.

 

[vcsharp2003]

It depends what you are counting in internal energy. A muscle produces mechanical energy from chemical energy, but the total energy does not change.

But, does this mean we can disregard internal forces always for a system when using conservation of energy except when the system is a single particle when internal forces don't exist? Probably, not.

 

[vcsharp2003]

I am now wondering about it using the following energy equation that applies to any system for all cases of mechanical energy without any heat flow scenarios.
$$ W_{net} = \Delta {KE} + \Delta {GPE}$$
The question that comes up now is why tensions don't contribute to $W_{net}$. Is it because $W_{net}$ denotes the total work done by external forces and tension is not an external force but an internal force when we consider the mass pulley system as the target for above equation? If yes, then we could further say that net work done by internal forces in a pure mechanical energy scenario is always zero.

After much thinking I came to the following answer for why tension(s) in ideal mass pulley systems result in zero work by tension(s) even when its a complex system of masses and pulleys. To determine the work done by tension, we must focus on only those parts connected to the string that are moving (i.e. either in translational or rotational motion or both). If connected part to string is not moving then we already know that the string is doing no work on such parts as displacement is zero.

• Firstly, the string is incapable of storing energy which means it cannot do some +ve non-zero net work as it would then lose energy ( how can it lose energy when it cannot store energy) OR it cannot do some -ve non-zero net work as it would then gain energy ( how can it gain energy when it cannot store energy). The massless and inextensible string possesses no energy of its own and cannot store energy like a spring can.
• Secondly, the tension(s) for all moving pulley(s) that could be in pure rotation or rotation + translation will balance out to zero.
  • For translation motion of pulleys that has strings on both sides the net tension would be zero resulting in zero work and if it has string on one side only then tension work would balance out with equal non-zero work of opposite sign done on masses connected to the other end of the string
  • For rotational motion tension work will always be zero due to net torque on these ideal pulleys being zero as tensions are equal on both sides of the pulley.

@haruspex, could you please let me know if the above explanation sounds correct?

 

[Lnewqban]

Copied from
https://interestingengineering.com/machine-design-101-pulleys-and-counterweights

"An example of a compound pulley system that has the force equations labeled and written out. Closely examine all of the different equations to get an idea of how pulley systems are modeled mathematically."

"An example of many different pulleys and the forces that are involved in each design. You can see that as more pulleys are added, the less force that is needed to move the block the same displacement. However, the distance that force has to be applied over is also longer."

 

[haruspex]

After much thinking I came to the following answer for why tension(s) in ideal mass pulley systems result in zero work by tension(s) even when its a complex system of masses and pulleys. To determine the work done by tension, we must focus on only those parts connected to the string that are moving (i.e. either in translational or rotational motion or both). If connected part to string is not moving then we already know that the string is doing no work on such parts as displacement is zero.

• Firstly, the string is incapable of storing energy which means it cannot do some +ve non-zero net work as it would then lose energy ( how can it lose energy when it cannot store energy) OR it cannot do some -ve non-zero net work as it would then gain energy ( how can it gain energy when it cannot store energy). The massless and inextensible string possesses no energy of its own and cannot store energy like a spring can.
• Secondly, the tension(s) for all moving pulley(s) that could be in pure rotation or rotation + translation will balance out to zero.
  • For translation motion of pulleys that has strings on both sides the net tension would be zero resulting in zero work and if it has string on one side only then tension work would balance out with equal non-zero work of opposite sign done on masses connected to the other end of the string
  • For rotational motion tension work will always be zero due to net torque on these ideal pulleys being zero as tensions are equal on both sides of the pulley.

@haruspex, could you please let me know if the above explanation sounds correct?

You seem to be assuming massless pulleys, so you could simply note that as with the inextensible strings they can neither supply net work to nor absorb net work from the rest of the system.