Working out the accelerations in a double Atwood pulley system

[hmparticle9]

Homework Statement

Find the acceleration of $m_1$.

Relevant Equations

$F = ma$

Above we have a diagram of the double atwood machine. There are LOADS of questions on here about the double Atwood, but none answer my question. They either go into using Lagrangian/Hamiltonian mechanics or use effective mass directly. For instance I can solve this problem in the following way (using effective mass). The equations are

$$T - m_1g = m_1a_1$$
$$T - 2T' = -\frac{4m_2m_3}{m_2 + m_3}a_1$$
$$T' - m_2g = m_2 a'$$
$$T' - m_3g = -m_3a'$$

We use the last two equations to solve for $T'$ and then put this into the first two equations to get

$$a_1 = \frac{4m_2 m_3 -m_1(m_2 + m_3)}{4m_2m_3 + m_1(m_2 + m_3)}g$$

My textbook now says, obtain the same result without using effective mass. Instead use relative accelerations. On thinking about it, these three equations hold

$$T - m_1g = m_1a_1$$
$$T' - m_2g = m_2 a'$$
$$T' - m_3g = -m_3a'$$

Also, since the pulley $p'$ is mass less I can say

$$T = 2T'$$

I am struggling to move forward. I think I am missing one equation. I don't want to just blindly apply formulas, I want to understand why. Can I have some help solving this problem using relative accelerations?. The textbook indicates that I should consider the acceleration relative to the ground. Surely the acceleration of $m_2$ relative to the ground is $a' - a_1$....

 

[haruspex]

Do the pulleys have mass? If not, what is $T-2T'$?
How are you defining $a'$ exactly? The acceleration of what and in what frame?

 

[hmparticle9]

$a'$ is the acceleration of $m_2$ if we were to sit on the pulley $p'$. The pulleys do not have mass. By $F = ma, T = 2T'$?

 

[haruspex]

$a'$ is the acceleration of $m_2$ if we were to sit on the pulley $p'$.

But $T'$ and $m_2g$ are the forces on $m_2$ in the rest frame, so the LHS of your equation gives its acceleration in the rest frame. If you want the equation in the frame of the pulley you need to include a virtual fictitious force to represent that frame's acceleration. Or get the same equation by working in the rest frame.

The pulleys do not have mass. By $F = ma, T = 2T'$?

Right. I do not know how you got your version from the effective mass.

 

[hmparticle9]

Thanks for your reply. I still don't have enough to go on... virtual force?

 

[haruspex]

Thanks for your reply. I still don't have enough to go on... virtual force?

Sorry, wrong terminology. I mean a fictitious or pseudo force. In this case, d'Alembert's inertial force. See https://en.wikipedia.org/wiki/Fictitious_force.
But you might as well just use an inertial frame.

 

[kuruman]

Can I have some help solving this problem using relative accelerations?.

The idea is simple. If B accelerates relative to A with acceleration a_BA (a one-dimensional signed vector) and A accelerates relative to the Earth with acceleration a_AE (also a one-dimensional signed vector), the acceleration a_BE of B relative to the Earth is given by

a_BE = a_BA + a_AE

If the Atwood machine in this problem were attached to the ceiling, the accelerations of the masses relative to the Earth and hence relative to the axle of the pulley from which they hang would be given by

a_2P = (m₃ - m₂) g / (m₃ + m₂)
a_3P = - (m₃ - m₂) g / (m₃ + m₂)

If the axle of the pulley is itself accelerating relative to the Earth with acceleration a_PE, what would be the expressions of the acceleration of each mass relative to the Earth?

Can you find an expression for a_PE? Remember that T = 2T' as you correctly surmised.

Note: I have used BBcode for equations instead of LaTeX because, according to @Greg Bernhardt, the core software was updated and that changed the editor and anything that modified the editor is now broken. He thinks that this is a serious problem and he's on it.

(Edited to fix subscript typo.)

 

[hmparticle9]

Surely the accelerations of mass 2 and 3 with respect to the earth would be:

$$a_{2E} = a_{2P} + a_{PE}$$
$$a_{3E} = a_{3P} + a_{PE}$$

I just can't make the leap to find an expression for $a_{PE}$. One thing I would also like to ask is, is my method using effective mass correct? I reaches the correct answer.

But back to this. I am really not doing great....

 

[kuruman]

I just can't make the leap to find an expression for $a_{PE}$.

How is a_PE related to the acceleration (relative to the Earth) a_1E of the top mass m₁? What equation do you get from the free body diagram of m₁?

One thing I would also like to ask is, is my method using effective mass correct?

It is, but you are asked not to use effective mass. Personally, I prefer using the term equivalent mass as laid out in this article, https://www.physicsforums.com/insights/how-to-solve-a-multi-atwood-machine-assembly/. Nevertheless, I think that finishing the problem using relative accelerations will consolidate your understanding. Getting the correct answer in two different ways will boost your confidence that you actually understand this stuff.

 

[hmparticle9]

The free body diagram for $m_1$ gives me:
$$T - m_1g = m_1a_1$$

$a_{PE}$ must be the same as $a_1$ but in the opposite direction.

 

[kuruman]

The free body diagram for $m_1$ gives me:
$$T - m_1g = m_1a_1$$

$a_{PE}$ must be the same as $a_1$ but in the opposite direction.

So far so good. Read all of the above posts. Can you put it together?

 

[hmparticle9]

$a_1$ is what we want, so I need an expression for $a_{PE}$. The only thing I can say is

$a_{2E} - a_{2P} = a_{PE}$

I know the acceleration of $m_2$ if we are sitting on the pulley $P'$, which is $a_{2P}$. But we don't know $a_{2E}$.

I'm sorry, I just don't get it.

 

[haruspex]

$a_1$ is what we want, so I need an expression for $a_{PE}$. The only thing I can say is

$a_{2E} - a_{2P} = a_{PE}$

I know the acceleration of $m_2$ if we are sitting on the pulley $P'$, which is $a_{2P}$. But we don't know $a_{2E}$.

I'm sorry, I just don't get it.

It's a bit hard without seeing all your a priori equations in one place.
You should be able to write down equations relating:
$a_1, a_{PE}$
$a_{2P}, a_{3P}$
$a_{2P}, a_2, a_{PE}$
$a_{3P}, a_3, a_{PE}$
$m_2, a_2, T'$
$m_3, a_3, T'$
$m_1, a_1, T$
$T, T'$
So 8 equations, 8 unknowns.

 

[kuruman]

I'm sorry, I just don't get it.

1. You hang mass m from the ceiling of an elevator at rest. The tension in the string will be T = mg.
2. The elevator is accelerating up with acceleration a. The tension in the string will be T = m(g + a).
3. The elevator is accelerating down with acceleration a. The tension in the string will be T = m(g - a).

If you are in the elevator, at rest with respect to the mass, in all three cases you see a hanging mass at rest. You add a spring scale between known mass m and the ceiling to measure the tension. You deduce that there is an effective acceleration of gravity that depends on the elevator's acceleration.

In case 1, g_eff. = g = 9.8 m/s².
In case 2, g_eff. = 9.8 m/s² + a.
In case 3, g_eff. = 9.8 m/s² - a.

Now suppose you hang a massless pulley from the ceiling next to hanging mass m. Masses m₃ and m₂ are connected with a string going over the pulley to form a simple Atwood machine. If you wanted to write Newton's second law equations for the two masses in a way that takes into account the elevator's acceleration, what would you use for the acceleration of gravity that would be consistent with the spring scale reading next to the Atwood setup? Having written these equations, you can find (a) expressions for the accelerations of the two masses which will have equal magnitudes and opposite directions, a₃ = - a₂ and (b) an expression for the tension in the connecting string.

Mini Quiz
What are a₃ and a₂ relative to?
1. The Earth.
2. The axle of the pulley.
3. I don't know.

Now imagine that the ceiling of the accelerating elevator becomes magically transparent. You look up and you see that the (massless) cable attached to the elevator goes over a pulley with mass m₁ attached to the other side of the cable.

How is the acceleration (relative to the Earth) of the elevator related to the acceleration of m₁?
How is the acceleration (relative to the Earth) of the pulley's axle inside the elevator related to the acceleration of m₁?

At this point, you can write another Newton's second law equation for m₁ noting that you know the tension in the upper cable connected to the elevator. What is it? What would you use for the acceleration of gravity in this equation, 9.8 m/s² or g_eff.?

I hope you get it after reading all this.

 

[hmparticle9]

Here are my 8 equations:
$$a_1 = -a_{PE}$$
$$a_2 = -a_3$$
$$a_{2E} = a_2 + a_{PE}$$
$$a_{3E} = a_3 + a_{PE}$$
$$T' - m_2g = m_2a_2$$
$$T' - m_3g = m_3a_3$$
$$T - m_1g = m_1a_1$$
$$2T' = T$$

I need to find $a_1$. So I am trying to get an expression for $a_{PE}$. Since I know neither $a_{2E}$ nor $a_{3E}$ there is no way for me to get an expression for $a_{PE}$

What is strange is that I can get an expression for $T'$. Then using $2T' = T$ I can put that into this equation

$$T - m_1g = m_1a_1$$

to get an expression for $a_1$, but it is incorrect. Why is it incorrect?

 

[kuruman]

Why is it incorrect?

Good question. My guess is that you made an algebraic mistake somewhere. Since you don't show your work, we cannot help you answer this question beyond guessing.

 

[hmparticle9]

But my 8 equations are correct?

 

[kuruman]

But my 8 equations are correct?

Yes.

 

[hmparticle9]

Here are my workings:

 

[haruspex]

Here are my workings:
View attachment 361256

You went wrong straight away. You took these two equations:

$$T' - m_2g = m_2a_2$$
$$T' - m_3g = m_3a_3$$

and replaced both lab frame accelerations by $\pm a'$, when clearly they are not simply equal and opposite (as they are in the frame of the pulley).

 

[JLT]

edit - either works:
T1 - m1 g = m1 a1,
T2 - m2 g = m2 a2,
T2 - m3 g = m3 a3,
T1 - (m2 + m3) g = m2 a2 + m3 a3,
a2 + a3 = - 2 a1

or use
T1 - m1 g = m1 a1,
T2 - m2 g = m2 a2,
T2 - m3 g = m3 a3,
T1 = 2 T2,
a2 + a3 = -2 a1

5 equations 5 unknowns: T1, T2, a1, a2, a3
test answer by plugging in different m's, make sure m1 accelerates up if it is smaller than the other two etc.

relative accelerations from lengths
a2+a3=-2a1
measured from ceiling (y2-Yp')+(y3-Yp') = Length of string connecting m2 and m3, take derivatives and you get the relative accelerations.

Plug in different m's
{m1,m2,m3}={50,25,25} = should get a=0
{m1,m2,m3}={50,0,50} = should get both 50's falling etc.

Challenge: what if the pulley does have some mass? T1 - 2 T2 = -mp a1 - can you get it to work if masses of pulley are accounted for?

 

[kuruman]

Challenge: what if the pulley does have some mass? T1 - 2 T2 = -mp a1 - can you get it to work if masses of pulley are accounted for?

If the pulley has mass, an additional equation is needed for the rotational acceleration of the pulley. It is unrealistic to say that the pulley has non-zero mass and then pretend that it has zero moment of inertia.

 

[hmparticle9]

For me, it was a problem with notation! I got it straight away when you told me I was wrong from the start.

Thanks :D