Find the acceleration of a block (some sort of double Atwood machine?)

[michelp]

Homework Statement

Three objects m0, m1, m2

Relevant Equations

The files are attached

I hope the sketch is good enough

My attempt:

anything with k means it's the pulley's. ak means the pulley's acceleration, xk means the pulley's rope length.
So my idea here was to use the constraint a1 = 2ak - a2, find the eq for ak and a2 in terms of a1, group them and find a1, but instead I got that a1 = -9.8 which I don't think is the correct answer. Please help me, it'd be nice to point out where I got it wrong. I am also new to this website

 

[jbriggs444]

I had to zoom way in to read the handwriting.

Your first assertion is that $a_1 + a_2 = 2a_k$. This is obviously correct. The acceleration of the upper block is the average of the accelerations of the lower two blocks.

Your next two assertions looked strange to me. You have $T - m_1g = m_1a_1$ and $m_2g - T = m_2a_2$. It seems that you have adopted a sign convention where $a_1$ is positive upward and $a_2$ is positive downward.

But this means that your earlier assertion is now incorrect. It depended on a sign convention where $a_1$ and $a_2$ were both positive downward.

 

[BvU]

Homework Statement: Three objects m0, m1, m2
Relevant Equations: The files are attached

I am also new to this website

​

Homework Statement: Three objects m0, m1, m2
Relevant Equations: The files are attached

I hope the sketch is good enough

The picture is fine, but the hieroglyphs are more difficult.:

is hard to decipher ...

In PF you often get more questions than answers:

I suppose the problem statement says no friction between surface and $m_0$ ?

How can you write $\ T-m_1g=m_1a_1 \$ and $\ m_2g - T = m_2a_2 \$ without getting confused with the directions of these $a_i$ ? the picture suggests $\ T - m_2g= m_2a_2 \$ if up is positive !

[edit]: ah ! JB was faster -- fortunately we agree !

$\$

 

[kuruman]

Did you notice that your expression cannot possibly be correct because its numerator is the negative of the denominator and simplifies to $a_1=-g$? This is free fall!

Note that mass $m_0$ doesn't "know" (or care) whether there is a mass or an Atwood machine at the other end of the string. It will accelerate just the same. So the pulley of an Atwood machine can be replaced by an equivalent hanging mass, a step which greatly simplifies such problems. To find out how this is done, read this article.

In this case, replacing the Atwood machine with an equivalent mass allows you to find the acceleration of the Atwood pulley $a_0$. Then all you have to do is find the acceleration of the masses in an Atwood machine hanging from the ceiling of an elevator that accelerates down with acceleration $a_0$.

 

[berkeman]

Homework Statement: Three objects m0, m1, m2
Relevant Equations: The files are attached

My attempt:

I sent you a DM with tips for posting math here at PF using our LaTeX engine.

 

[haruspex]

@michelp: Further to @jbriggs444 response in post #2, it helps readers greatly if you define your variables up front, particularly in regard to sign conventions. It will help you too.

 

[michelp]

​

The picture is fine, but the hieroglyphs are more difficult.: View attachment 369124 is hard to decipher ...

In PF you often get more questions than answers:

I suppose the problem statement says no friction between surface and $m_0$ ?

How can you write $\ T-m_1g=m_1a_1 \$ and $\ m_2g - T = m_2a_2 \$ without getting confused with the directions of these $a_i$ ? the picture suggests $\ T - m_2g= m_2a_2 \$ if up is positive !

[edit]: ah ! JB was faster -- fortunately we agree !

$\$

Ahhh im so sorry for my handwriting T_T

 

[michelp]

​

The picture is fine, but the hieroglyphs are more difficult.: View attachment 369124 is hard to decipher ...

In PF you often get more questions than answers:

I suppose the problem statement says no friction between surface and $m_0$ ?

How can you write $\ T-m_1g=m_1a_1 \$ and $\ m_2g - T = m_2a_2 \$ without getting confused with the directions of these $a_i$ ? the picture suggests $\ T - m_2g= m_2a_2 \$ if up is positive !

[edit]: ah ! JB was faster -- fortunately we agree !

$\$

Sorry to reply twice, but I did the early steps with my teacherand thats what he wrote. I assume that's incorrect?

 

[michelp]

Did you notice that your expression cannot possibly be correct because its numerator is the negative of the denominator and simplifies to $a_1=-g$? This is free fall!

Note that mass $m_0$ doesn't "know" (or care) whether there is a mass or an Atwood machine at the other end of the string. It will accelerate just the same. So the pulley of an Atwood machine can be replaced by an equivalent hanging mass, a step which greatly simplifies such problems. To find out how this is done, read this article.

In this case, replacing the Atwood machine with an equivalent mass allows you to find the acceleration of the Atwood pulley $a_0$. Then all you have to do is find the acceleration of the masses in an Atwood machine hanging from the ceiling of an elevator that accelerates down with acceleration $a_0$.

Thank you so much!

 

[Steve4Physics]

@michelp, this is a bit late in the day, but it might explain the misunderstanding...

Your next two assertions looked strange to me. You have $T - m_1g = m_1a_1$ and $m_2g - T = m_2a_2$. It seems that you have adopted a sign convention where $a_1$ is positive upward and $a_2$ is positive downward.

... I did the early steps with my teacherand thats what he wrote. I assume that's incorrect?

In the case of a simple Atwood machine (with $m_2 > m_1$ say), it is not uncommon to find equations written in the form:
$T – m_1 g = m_1 a$ and
$m_2 g – T = m_2 a$

For example: https://en.wikipedia.org/wiki/Atwood_machine.

I’d guess this is how your teacher taught it.

In effect, $T, a$ and $g$ are magnitudes and directional signs are allocated ‘manually’. This is equivalent to treating the system as ‘straightened-out’ with the positive direction to be the direction of the larger mass’s acceleration:

This works because the pulley's only function is to change the direction of the tension. For more complicated systems this approach is unsuitable, as noted in previous posts.

 

[kuruman]

This works because the pulley's only function is to change the direction of the tension.

In which case, if the system is taken to be the two masses plus string, one can immediately write $$F_{\text{net}}=(m_2-m_1)g\implies a=\frac{F_{\text{net}}}{(m_1+m_2)}=\left(\frac{m_2-m_1}{m_1+m_2}\right) g$$without bothering about the tension.

 

[wrobel]

Just a nice block problem

 

[jbriggs444]

In the case of a simple Atwood machine (with $m_2 > m_1$ say), it is not uncommon to find equations written in the form:
$T – m_1 g = m_1 a$ and
$m_2 g – T = m_2 a$

For example: https://en.wikipedia.org/wiki/Atwood_machine.

I’d guess this is how your teacher taught it.

In effect, $T, a$ and $g$ are magnitudes and directional signs are allocated ‘manually’. This is equivalent to treating the system as ‘straightened-out’ with the positive direction to be the direction of the larger mass’s acceleration:
View attachment 369187

This is fine in isolation. One often picks a sign convention to try to keep all quantities positive. If one guesses wrong, the worst thing that happens is that upon solving the algebra some quantity turns out to be negative instead.

This is a common occurrence in circuit analysis. We may not immediately know which way the current will flow through some circuit element. So we pick a direction and call it positive.

As pointed out in #2 the problem here that the 'manual' assignment of sign convention in writing equations for the lower blocks is not the same 'manual' assignment that went into writing an equation for the upper block. We end up with the fallacy of equivocation: two different meanings for a single variable.

 

[Steve4Physics]

This is fine in isolation. One often picks a sign convention to try to keep all quantities positive. If one guesses wrong, the worst thing that happens is that upon solving the algebra some quantity turns out to be negative instead.
...
As pointed out in #2 the problem here that the 'manual' assignment of sign convention in writing equations for the lower blocks is not the same 'manual' assignment that went into writing an equation for the upper block. We end up with the fallacy of equivocation: two different meanings for a single variable.

Yes. Agree 100%. In case there's any confusion, I certainly wasn't (in Post #10) promoting the OP's (Post #1) approach.

But in Post #8, they wrote:

but I did the early steps with my teacherand thats what he wrote. I assume that's incorrect?

No one had specifically replied to this. My Post #10 was an attempt to do this in order to help the OP better understand what was going on.

 

[haruspex]

If one guesses wrong, the worst thing that happens is that upon solving the algebra some quantity turns out to be negative instead.

Mostly, yes. An awkward exception is kinetic friction. When writing, say, $\mu mg\sin(\theta)$ in an equation, the sign matters and might not be apparent.

 

[michelp]

@michelp, this is a bit late in the day, but it might explain the misunderstanding...



In the case of a simple Atwood machine (with $m_2 > m_1$ say), it is not uncommon to find equations written in the form:
$T – m_1 g = m_1 a$ and
$m_2 g – T = m_2 a$

For example: https://en.wikipedia.org/wiki/Atwood_machine.

I’d guess this is how your teacher taught it.

In effect, $T, a$ and $g$ are magnitudes and directional signs are allocated ‘manually’. This is equivalent to treating the system as ‘straightened-out’ with the positive direction to be the direction of the larger mass’s acceleration:
View attachment 369187
This works because the pulley's only function is to change the direction of the tension. For more complicated systems this approach is unsuitable, as noted in previous posts.

This makes so much more sense, thank you so much. But if I had used the pulley as frame of reference and applied pseudo forces would
$T – m_1 g = m_1 a$ and
$m_2 g – T = m_2 a$
still be usable?

 

[jbriggs444]

This makes so much more sense, thank you so much. But if I had used the pulley as frame of reference and applied pseudo forces would
$T – m_1 g = m_1 a$ and
$m_2 g – T = m_2 a$
still be usable?

If you are using an accelerating frame and pseudo-forces then those pseudo-forces should appear in your force balance equations.

If, for instance, the pulley is accelerating downward with some acceleration ($a_k$) then the pseudo-force will act upward -- opposite to gravity. The net apparent force of gravity would be reduced. To $g-a_k$. That is the gravitational acceleration that should then appear in your two equations above.

But you would need to keep in mind that the two accelerations you get from these equations are frame-relative accelerations. You would need to add $a_k$ back in to get accelerations relative to an inertial frame.

 

[michelp]

Turns out the problem really was the sign of convention being inconsistent. I hope I got it right this time. I'm still a little confused with the LaTeX so I'll be using my handwriting for now.

 

[jbriggs444]

View attachment 369247Turns out the problem really was the sign of convention being inconsistent. I hope I got it right this time. I'm still a little confused with the LaTeX so I'll be using my handwriting for now.

In line $\LaTeX$:: ##a_1 = \frac{4 m_2 m_1 + m_0 ( m_1 - m_2 )} {4 m_2 m_1 + m_0 (m_1 + m_2)} g##

Which should render as $a_1 = \frac{4 m_2 m_1 + m_0 ( m_1 - m_2 )} {4 m_2 m_1 + m_0 (m_1 + m_2)} g$

I tossed in some white space in the unrendered LaTeX for what I think is better readability of the code. Rendering removes that. If you wanted white space in the rendered version, you would use a backslash escape for the spaces: "\ "

To put it in a separate line: \$$a_1 = \frac{4 m_2 m_1 + m_0 ( m_1 - m_2 )} {4 m_2 m_1 + m_0 (m_1 + m_2)} g\$$

Which should render as $$a_1 = \frac{4 m_2 m_1 + m_0 ( m_1 - m_2 )} {4 m_2 m_1 + m_0 (m_1 + m_2)} g$$

In order to display back to back pound signs and dollar signs and not render the LaTeX I had to use some different trickery.

 

[Steve4Physics]

View attachment 369247

I get the same equation as you - but don’t forget that you haven’t completely answered the question. You are given values for the masses, so you will need to calculate a value for $a_1$’s magnitude and state its direction.

Turns out the problem really was the sign of convention being inconsistent. I hope I got it right this time.

We commonly take upwards as positive and to-the-right as positive. Here, the pulley accelerates downwards so we’d get a negative value for its acceleration. But $m_0$ accelerates to the right giving a (conflicting) positive value for its acceleration.

An appropriate minus-sign would resolve the problen but so would a suitable sign-convention (which is what I used):
horizontally: positive = to-the-right
vertically: positive = downwards

The horizontal and vertical accelerations are then consistent. So the 4 equations describing the system are then:
$m_1g – T = m_1a_1$
$m_2g – T = m_2a_2$
$2T = m_0a_0$ (prefer to use $a_o$ rather than $a_k$)
$a_1 + a_2 = 2a_0$
That may not be to everybody’s taste though!

Minor edits.