The Relationship Between Masses and Angles in a Pulley System

[Juanda]

Homework Statement

3 masses are hanging from 2 pullies as shown in the picture.
Assume $l_1$ and $l_4$ to be sufficiently large to allow for equilibrium to be reached if possible.
There is no friction. The pullies are small enough to be considered a point. Constant gravity is at play.

1) Knowing $m_1$, $m_2$ and $m_3$ find $\alpha$, $\beta$, $l_2$ and $l_3$.
2) If $m_1$ and $m_3$ are known, find the limiting values of $m_2$ that make equilibrium possible. Apply "question 1" to those 2 limiting scenarios.

Relevant Equations

Newton.
Vector summation.

The problem is based on a similar thread. In fact, the first question is extremely similar. However, the second question is the one I consider more interesting but I posted the first one too for context.

If this was just 1 pulley and two masses, then equilibrium is only possible if both masses have the same weight. If we have something as shown, then the possibilities grow. If $m_2$ is too light then it's just as the version with only 2 masses and 1 pulley. On the other hand, if it's too heavy then it'll drag both with it. There is a whole region in the middle of those two scenarios that make equilibrium possible where $m_2$ will hang lower or higher depending on how big it is compared with $m_1$ and $m_3$.

I could find the limits numerically using a graphic calculator but I'd like to know if it's possible to find them analytically and get more insight about the origin of the solution. If possible, while using just Newton. I'm sure it must be possible to set up the problem so the potential energy is to be minimized and I'd like to solve it that way too eventually but not at the moment.

I'll start with question 1. To find that, it's necessary to apply equilibrium to $m_2$.
$$\sum F_x=0 \rightarrow -m_1g\sin\alpha + m_3gsin\beta = 0\rightarrow m_1\sin\alpha = m_3\sin\beta \tag{1}$$
$$\sum F_y=0 \rightarrow m_1g\cos\alpha -m_2g+ m_3gcos\beta = 0\rightarrow m_1\cos\alpha + m_3\cos\beta = m_2 \tag{2}$$

From equations $1$ and $2$ it is already possible to find $\alpha$ and $\beta$. It's not pretty because it's a trigonometric system of equations so I couldn't completely clear the variables. This is what I got.
From $1$.
$$\alpha=\arcsin(\frac{m_3}{m_1}\sin\beta) \tag{3}$$
Plugging that into $2$ results in.
$$m_1\cos(\arcsin(\frac{m_3}{m_1}\sin\beta) )+ m_3\cos\beta = m_2 \tag{4}$$
I believe $\beta$ cannot be cleared from $4$ so it's only possible to find numerical solutions. Still, from the numerical solution of $4$, it's possible to plug in the value in $3$ and then we'd have both $\alpha$ and $\beta$ solved. Once that is known, it's now possible to find the values of $l_2$ and $l_3$.

The sum of the purple vectors must be 0. This gives two equations. Only $l_2$ and $l_3$ are unknown so it's solvable.
$$\sum x=0 \rightarrow l_2 \cos(3\pi/2+\alpha)+l_3\cos(\pi/2-\beta)-b=0 \tag{5}$$
$$\sum y=0 \rightarrow l_2 \sin(3\pi/2+\alpha)+l_3\sin(\pi/2-\beta)-a=0 \tag{6}$$
With that, the first question should be completely solved with the inconvenience of not being able to clear eqn $4$ to get a nice expression for the angles.

Now for question 2, I could only find the limiting values using a graphic calculator and I'd like to know if there is a better method that'd give some insight. I considered eqn $4$ to be a function. To deal with the online graphic calculator I changed $\beta$ with $x$.
$$m_1\cos(\arcsin(\frac{m_3}{m_1}\sin\beta) )+ m_3\cos\beta = m_2$$
$$\downarrow$$
$$f(x)=m_1\cos(\arcsin(\frac{m_3}{m_1}\sin x) )+ m_3\cos x - m_2 \tag{7}$$
Fixing $m_1$ and $m_3$ it's possible to change the value of $m_2$ and see if there's any $x$ ($x=\beta$) for which $f(x)=0$. As a first shot, the studied region will be within $0\leq\beta\leq \arctan(a/b)$. The angle will be 0 when $m_2$ is too heavy and the angle will be parallel to the line between the supports if $m_2=0$. In the following link, you can see the function.
https://www.desmos.com/calculator/8ddibiq3v0
Lowest possible $m_2$ that makes the function $f(x)=0$ within the shown range.

Greatest possible $m_2$ that makes the function $f(x)=0$ within the shown range.

The limit case where $m_2$ is too heavy is somewhat intuitive. From playing with the function it seems that if $m_2 = m_1+m_3$ then the angles need to be 0 to support it. This can be seen in eqn $2$ although it feels weird both angles approach 0 "at the same speed" as $m_2$ goes lower and lower.
For the smallest $m_2$ though... I have no intuition about where the limiting value comes from. Let me know if you have other point of view or resolution method that could make this a little clearer.

 

[Juanda]

I forgot to add the second part of the second question.

2) If $m_1$ and $m_3$ are known, find the limiting values of $m_2$ that make equilibrium possible. Apply "question 1" to those 2 limiting scenarios.

In the limiting case where $m_2=m_1+m_3$ → the angles $\alpha$ and $\beta$ will approach 0 to make equilibrium possible, and the sections of the ropes $l_2$ and $l_3$ will approach $\infty$ so that they can make those angles possible.

In the limiting case where $m_2$ is too light, I can only give a numerical solution so I need to assign values to all the variables (using SI for all of them). With $m_1=1$, $m_3=3$ and using the graph shown then the minimum value for $m_2$ will be $m_2 \approx 2.85$ which will cause $\beta \approx 0.3396$. Once $\beta$ is known, $\alpha$ can be known:
$$\alpha=\arcsin(\frac{m_3}{m_1}\sin 0.3396) =1.5341 \approx \pi/2$$
That suggests the limit for the lowest value of $m_2$ happens when the rope attached to the lower pulley becomes horizontal but I'm not sure why that is. This seems to get me closer to the insight I needed. Finding $l_2$ and $l_3$ seems trivial at this point so I'll omit it. EDIT: Actually it isn't. I want to check if $m_2$ always falls in the middle of the two.

 

[Juanda]

That suggests the limit for the lowest value of $m_2$ happens when the rope attached to the lower pulley becomes horizontal but I'm not sure why that is. This seems to get me closer to the insight I needed.

I believe it happens because if that rope goes above horizontality, then $m_3$ would need to increase so much to equilibrate the forces that it'd be too heavy and it'd drag the rest of the masses with it but I couldn't prove it so far.

 

[Juanda]

That suggests the limit for the lowest value of $m_2$ happens when the rope attached to the lower pulley becomes horizontal but I'm not sure why that is. This seems to get me closer to the insight I needed. Finding $l_2$ and $l_3$ seems trivial at this point so I'll omit it. EDIT: Actually it isn't. I want to check if $m_2$ always falls in the middle of the two.

I think it won't if there is not enough symmetry in the problem.
The mass $m_2$ will always fall in the middle if
$$l_2\sin\alpha=l_3\sin\beta \tag{8}$$
no matter what are the initial conditions (if equilibrium is possible).
From $(5)$.
$$\sum x=0 \rightarrow l_2 \cos(3\pi/2+\alpha)+l_3\cos(\pi/2-\beta)-b=0 \tag{5}$$
$$\downarrow$$
$$l_2 = \frac{b-l_3\cos(\pi/2-\beta)}{\cos(3\pi/2+\alpha)} \tag{9}$$
Plugging that into $(6)$.
$$\sum y=0 \rightarrow l_2 \sin(3\pi/2+\alpha)+l_3\sin(\pi/2-\beta)-a=0 \tag{6}$$
$$\downarrow$$
$$\frac{b-l_3\cos(\pi/2-\beta)}{\cos(3\pi/2+\alpha)} \sin(3\pi/2+\alpha)+l_3\sin(\pi/2-\beta)-a=0$$
$$\downarrow$$
$$\frac{b\sin(3\pi/2+\alpha)}{\cos(3\pi/2+\alpha)} - \frac{l_3\cos(\pi/2-\beta)\sin(3\pi/2+\alpha)}{\cos(3\pi/2+\alpha)}+l_3\sin(\pi/2-\beta)-a=0$$
$$\downarrow$$
$$l_3 = \frac{a-\frac{b\sin(3\pi/2+\alpha)}{\cos(3\pi/2+\alpha)}}{\sin(\pi/2-\beta)-\frac{\cos(\pi/2-\beta)\sin(3\pi/2+\alpha)}{\cos(3\pi/2+\alpha)}} \tag{10}$$

I doesn't look like $(9$ and $(10)$ simplify to make $(8)$ true. As a quick check, I'll find the resulting lengths for the smallest $m_2$ from post #2. For $a=5$; $b=6$; $m_1=1$; $m_2=2.85$; $m_3=3$; $\alpha=1.5341$; $\beta=0.3396$. Plugging all that into $(9)$ gives:
https://www.desmos.com/calculator/1ghqgwlocn
$$l_3=5.4655$$.
$$l_2=4.18$$.
$$l_{2x}=\sin\alpha=4.17$$
$$l_{3x}=\sin\beta=1.82$$
As a quick check, it can be seen how $l_{2x}+l_{3x}=b$ so it'd be an indicator of the calculations being correct.

So it's been checked that $m_2$ is not restricted horizontally to the middle of the pulleys. The following is still the pending point to better understand this problem.

$$\alpha=\arcsin(\frac{m_3}{m_1}\sin 0.3396) =1.5341 \approx \pi/2$$
That suggests the limit for the lowest value of $m_2$ happens when the rope attached to the lower pulley becomes horizontal but I'm not sure why that is. This seems to get me closer to the insight I needed.
I believe it happens because if that rope goes above horizontality, then $m_3$ would need to increase so much to equilibrate the forces that it'd be too heavy and it'd drag the rest of the masses with it but I couldn't prove it so far.

 

[erobz]

We have three equations Newtons Second ( dividing out $g$):

$$ m_1 \cos \alpha + m_3 \cos \beta = m_2 \tag{1}$$

$$ m_3 \sin \beta - m_1 \sin \alpha = 0 \tag{2} $$

and finally the Law of Cosines:

$$ a^2 + b^2 = l_2^2 + l_3^2 - 2 l_2 l_3 \cos ( \alpha + \beta ) \tag{3}$$

I believe you can use (3) and some trig identities to solve for $\sin \beta = f ( \sin \alpha )$ or visa-versa ( It isn't "light work" to do so, but in the end you'll be solving a quadratic in $\sin \beta$ ). Then you can solve (2) with that result for $\sin \alpha = \text{mess of constants}$ and with some more trig identities sub into 1 to get everything in terms constants. It would certainly be a monstrosity, then you would have to manipulate that to find whatever you are after.

Maybe you don't have to do all of it depending on what you are after, and what is specified. I'm just trying to point out an algebraic path to eliminate the trig functions.

 

[Steve4Physics]

Hint for Q1. Draw the equilibrium force triangle for $m_3$ and use the cosine rule. You can easily find $\alpha$ and $\beta$ as functions of $m_1, m_2$ and $m_3$.

Finding $l_2$ and $l_3$ in terms of $a, b, \alpha$ and $\beta$ is then just (probably messy) trig' and algebra.

 

[Juanda]

and finally the Law of Cosines:

$$ a^2 + b^2 = l_2^2 + l_3^2 - 2 l_2 l_3 \cos ( \alpha + \beta ) \tag{3}$$

I believe you can use (3) and some trig identities to solve for $\sin \beta = f ( \sin \alpha )$ or visa-versa ( It isn't "light work" to do so, but in the end you'll be solving a quadratic in $\sin \beta$ ). Then you can solve (2) with that result for $\sin \alpha = \text{mess of constants}$ and with some more trig identities sub into 1 to get everything in terms constants. It would certainly be a monstrosity, then you would have to manipulate that to find whatever you are after.

Did you mean $\cos$? And I don't know how to do that. It might be possible with trig identities as you said but I doubt it. It'd be convenient to avoid implicit functions but I guess it's OK if there is no other way around it.

Maybe you don't have to do all of it depending on what you are after, and what is specified. I'm just trying to point out an algebraic path to eliminate the trig functions.

I'd say right now the main topic now is:
The limit for the lowest value of $m_2$ seems to happen when the rope attached to the lower pulley becomes horizontal but I'm not sure why that is. The realization of the horizontal limit gets me closer to the insight I needed.
I believe it happens because if that rope (the left one in this case) goes above horizontality, then $m_3$ would need to increase so much to equilibrate the forces that it'd be too heavy and it'd drag the rest of the masses with it but I couldn't prove it so far.

Hint for Q1. Draw the equilibrium force triangle for $m_3$ and use the cosine rule. You can easily find $\alpha$ and $\beta$ as functions of $m_1, m_2$ and $m_3$.

Finding $l_2$ and $l_3$ in terms of $a, b, \alpha$ and $\beta$ is then just (probably messy) trig' and algebra.

I'd say Q1 is solved although not using the cosine rule. In fact, the cosine rule only gives 1 equation as @erobz showed on #5 but there are two unknowns ($l_2$ and $l_3$) so for that method I feel there's still something missing.
Even Q2 is solved numerically. I am now most interested in the text shown in italics above.

 

[haruspex]

What happens if you square both sides in eqns 1 and 2 and add them together?

 

[Lnewqban]

... I'd say right now the main topic now is:
The limit for the lowest value of $m_2$ seems to happen when the rope attached to the lower pulley becomes horizontal but I'm not sure why that is. The realization of the horizontal limit gets me closer to the insight I needed.
I believe it happens because if that rope (the left one in this case) goes above horizontality, then $m_3$ would need to increase so much to equilibrate the forces that it'd be too heavy and it'd drag the rest of the masses with it but I couldn't prove it so far.

Please, see:
https://www.engineeringtoolbox.com/wire-rope-slings-d_1613.html

Any weight of $m_2$ that is greater than 0 will create tremedous tension if the ropes are horizontal.
Reason for which, the location of $m_2$ will naturally go down in order to reduce those tensions and find a point of balance among horizontal and vertical components.

The relative vertical positions of the pulleys has no influence in the angles or position of equilibrium adopted by $m_2$.

 

[haruspex]

I believe it happens because if that rope goes above horizontality, then $m_3$ would need to increase so much to equilibrate the forces that it'd be too heavy and it'd drag the rest of the masses with it but I couldn't prove it so far.

Suppose we get that rope section close to horizontal, anchor $m_1$ in place, then pull $m_3$ a bit lower. It must be possible to get the rope above horizontal in a static arrangement. Whatever the tensions are at that point, we can get the same by suitable choice of $m_1$ and $m_3$. Whether that is stable equilibrium is another matter.

 

[haruspex]

Any weight of $m_2$ that is greater than 0 will create tremedous tension if the ropes are horizontal.

If both are, yes, but in general a>0.

 

[Juanda]

What happens if you square both sides in eqns 1 and 2 and add them together?

I don't see where you're trying to arrive but this is what I got.
$$\sum F_x=0 \rightarrow -m_1g\sin\alpha + m_3gsin\beta = 0\rightarrow m_1\sin\alpha = m_3\sin\beta \tag{1}$$
So squared it'll be:
$$m_1^2\sin^2\alpha = m_3^2\sin^2\beta$$
On the other hand:
$$\sum F_y=0 \rightarrow m_1g\cos\alpha -m_2g+ m_3gcos\beta = 0\rightarrow m_1\cos\alpha + m_3\cos\beta = m_2 \tag{2}$$
So squared it'll be:
$$m_1^2\cos^2\alpha + m_3^2\cos^2\beta + 2(m_1\cos\alpha)(m_3\cos\beta) = m_2^2$$
Summing up both expressions:
$$m_1^2\cos^2\alpha + m_3^2\cos^2\beta + 2(m_1\cos\alpha)(m_3\cos\beta)+m_1^2\sin^2\alpha - m_3^2\sin^2\beta= m_2^2$$
$$\downarrow$$
$$m_1^2+m_3^2(\cos^2\beta-\sin^2\beta)+2(m_1\cos\alpha)(m_3\cos\beta)=m_2^2$$
I don't know if there's something else to see in that. It's an equation with 2 unknowns ($\alpha$ and $\beta$ if all the masses are considered to be known).

 

[Juanda]

Please, see:
https://www.engineeringtoolbox.com/wire-rope-slings-d_1613.html

Any weight of $m_2$ that is greater than 0 will create tremedous tension if the ropes are horizontal.
Reason for which, the location of $m_2$ will naturally go down in order to reduce those tensions and find a point of balance among horizontal and vertical components.

The relative vertical positions of the pulleys has no influence in the angles or position of equilibrium adopted by $m_2$.

In this particular case, the tension in the ropes is set and it cannot grow to infinity. I know what you mean but it's a different case.
In this case, the system will try to find a new equilibrium position where the available tension is capable of keeping everything in place.
The point now is to understand the range of possible $m_2$ that make equilibrium possible. Especially, to achieve a better understanding of the case for the smallest $m_2$.

 

[Steve4Physics]

I'd say Q1 is solved

I wanted to point out an alternative approach. Often a first line of attack is to set up equations. But a more 'geometrical' approach sometimes helps. To be explicit...

EDIT. The following is wrong as I had $m_2$ and $m_3$ the wrong way round. The approach is correct though. For the correct version, see @Juanda's Post #20.

In Q1, if we construct $m_3$’s force triangle we can use the cosine law. Cancelling the '$g$'s we can immediately write:
$m_2^2 = m_1^2 +m_3^2 - 2m_1m_3 \cos \alpha$
so
$\alpha = \cos^{-1}(\frac {m_1^2 + m_3^2 - m_2^2}{2m_1m_3})$
and similarly for $\beta$.

 

[Lnewqban]

If both are, yes, but in general a>0.

Even in that case, $\alpha+\beta=2\pi$ only when the component of $m_2$ that is perpendicular to both taut ropes, equals zero or is very small, in practical terms.

Please, see:
https://www.wired.com/2016/12/pull-car-ditch-super-strength-physics/

 

[Juanda]

I wanted to point out an alternative approach. Often a first line of attack is to set up equations. But a more 'geometrical' approach sometimes helps. To be explicit...

In Q1, if we construct $m_3$’s force triangle we can use the cosine law. Cancelling the '$g$'s we can immediately write:
$m_2^2 = m_1^2 +m_3^2 - 2m_1m_3 \cos \alpha$
so
$\alpha = \cos^{-1}(\frac {m_1^2 + m_3^2 - m_2^2}{2m_1m_3})$
and similarly for $\beta$.

Ok, I think now I'm getting where you're getting at. Also, the use of the cosine law differs from the one in post #5 because you did it on the triangle of forces instead of using it on the actual geometry. That seems to be very useful because it allows us to get explicit expressions for $\alpha$ and $\beta$ in terms of the masses.
I'll try to draw the triangles of forces you used to get to that expression to better understand its origin and get back here to confirm if it's the same or equivalent to what you did.

 

[erobz]

And I don't know how to do that. It might be possible with trig identities as you said but I doubt it.

$$ cos( \alpha + \beta ) = \frac{l_2^2+ l_3^2 -( a^2 + b^2) }{2 l_2 l_3} = \lambda $$

$$ cos \alpha \cos \beta - \sin \alpha \sin \beta = \lambda $$

$$ \sqrt{1 - \sin^2 \alpha}\sqrt{1 - \sin^2 \beta} = \lambda + \sin \alpha \sin \beta $$

Square both sides, collect like terms:

$$ \sin^2 \alpha - 2 \lambda \sin \beta \sin \alpha + \lambda^2 + \sin^2 \beta - 1 = 0 $$

Solve the quadratic:

$A = 1$
$B = -2 \lambda \sin \beta$
$C = \lambda^2 + \sin^2 \beta - 1$

You will get:

$\sin \alpha = f(\sin \beta )$

Plug that into (2) and try to solve for $\sin \alpha$?

 

[Juanda]

Solve the quadratic:

$A = 1$
$B = -2 \lambda \sin \beta$
$C = \lambda^2 + \sin^2 \beta - 1$

You will get:

$\sin \alpha = f(\sin \beta )$

Plug that into (2) and try to solve for $\sin \alpha$?

The point is that even if you have that expression $\sin \alpha = f(\sin \beta )$ when you plug it into (2), you won't be able to have an explicit expression for $\beta$ so I don't see the benefit of doing that.

I'm working on what @Steve4Physics suggested in post #14. Somehow he managed to use the cosine law without having both $\alpha$ and $\beta$ in the same equation so he'd actually get the angles as an explicit function of the masses although I still couldn't reproduce the result.
In fact, I suspect there might be something wrong. For the initial conditions described in #2 I got:

In the limiting case where $m_2$ is too light, I can only give a numerical solution so I need to assign values to all the variables (using SI for all of them). With $m_1=1$, $m_3=3$ and using the graph shown then the minimum value for $m_2$ will be $m_2 \approx 2.85$ which will cause $\beta \approx 0.3396$. Once $\beta$ is known, $\alpha$ can be known:
$$\alpha=\arcsin(\frac{m_3}{m_1}\sin 0.3396) =1.5341 \approx \pi/2$$
That suggests the limit for the lowest value of $m_2$ happens when the rope attached to the lower pulley becomes horizontal but I'm not sure why that is. This seems to get me closer to the insight I needed.

However, when I use the expression shown in #14 I get a very different result.

I wanted to point out an alternative approach. Often a first line of attack is to set up equations. But a more 'geometrical' approach sometimes helps. To be explicit...

In Q1, if we construct $m_3$’s force triangle we can use the cosine law. Cancelling the '$g$'s we can immediately write:
$m_2^2 = m_1^2 +m_3^2 - 2m_1m_3 \cos \alpha$
so
$\alpha = \cos^{-1}(\frac {m_1^2 + m_3^2 - m_2^2}{2m_1m_3})$
and similarly for $\beta$.

Here is the difference between the values of $\alpha$ depending on the method to obtain it. $\alpha$ is from the procedure I showed and $\alpha_2$ is using the equation suggested at #14.

 

[erobz]

The point is that even if you have that expression $\sin \alpha = f(\sin \beta )$ when you plug it into (2), you won't be able to have an explicit expression for $\beta$ so I don't see the benefit of doing that.

It at least looks solvable to me?

$$ m_3 \sin \beta = m_1\left( \lambda \sin \beta \pm \sqrt{\lambda^2 \sin ^2 \beta - \lambda^2 - \sin^2 \beta + 1 } \right) $$

$$ \sin \beta = \text{ a mess of constants} $$

Its reasonable to believe that there is no real benefit to it, but you said you doubt it could be done?

 

[Juanda]

I wanted to point out an alternative approach. Often a first line of attack is to set up equations. But a more 'geometrical' approach sometimes helps. To be explicit...

In Q1, if we construct $m_3$’s force triangle we can use the cosine law. Cancelling the '$g$'s we can immediately write:
$m_2^2 = m_1^2 +m_3^2 - 2m_1m_3 \cos \alpha$
so
$\alpha = \cos^{-1}(\frac {m_1^2 + m_3^2 - m_2^2}{2m_1m_3})$
and similarly for $\beta$.

Ok, I think now I'm getting where you're getting at. Also, the use of the cosine law differs from the one in post #5 because you did it on the triangle of forces instead of using it on the actual geometry. That seems to be very useful because it allows us to get explicit expressions for $\alpha$ and $\beta$ in terms of the masses.
I'll try to draw the triangles of forces you used to get to that expression to better understand its origin and get back here to confirm if it's the same or equivalent to what you did.

Here is what I got. This is the triangle of forces (scaled with a factor $g$)

Using the law of cosines:
$$\alpha = \arccos(\frac {m_1^2 + m_2^2 - m_3^2}{2m_1m_2})$$

It seems there's a small typo in the expression @Steve4Physics proposed but the concept is totally right and very convenient. Doing it like this, it's possible to obtain the expressions for the angles in terms of the masses and nothing else. Once the typo is corrected, I get the same results using both methods (the one I showed and the one using the law of cosines which I believe is much more convenient).

 

[Juanda]

It at least looks solvable to me?

$$ m_3 \sin \beta = m_1\left( \lambda \sin \beta + \sqrt{\lambda^2 \sin ^2 \beta - \lambda^2 - \sin^2 \beta + 1 } \right) $$

$$ \sin \beta = \text{ a mess of constants} $$

Its reasonable to believe that there is no real benefit to it, but you said you doubt it could be done?

Oh. I just realized we are using the same number for different equations. I now understand better what you're saying.
You're right. It looks like it'd be solvable. Tedious for sure. But solvable. However, I think Steve nailed it this time with his approach using the law of cosines on the triangle of forces instead of the physical triangle resulting from the geometry of the problem.

 

[Steve4Physics]

Here is what I got. This is the triangle of forces (scaled with a factor $g$)
View attachment 333496
Using the law of cosines:
$$\alpha = \arccos(\frac {m_1^2 + m_2^2 - m_3^2}{2m_1m_2})$$

Agreed..

It seems there's a small typo in the expression

Apologies for the mistake. I accidentally had $m_2$ and $m_3$ the wrong way round in a diagram I'd sketched. Carelessness and/or old-age!

 

[kuruman]

I would assume that the pulleys are ideal, i.e. that they change the direction of the the tension but not its magnitude. Then because the tensions in the strings supporting $m_3$ are equal to the hanging weights on each side. The equilibrium condition is the that the sum of all the forces be zero.
$\vec T_1+\vec T_3 +m_2\vec g=0$ which, as you have observed, forms a closed triangle (see my version below).

Now apply the law of sines on the force triangle on the right noting that $\sin[\pi-(\alpha+\beta)]=\sin(\alpha+\beta).$ $$\frac{m_3}{\sin\alpha}=\frac{m_1}{\sin\beta}=\frac{m_2}{\sin(\alpha+\beta)}.$$The first equation gives $$\sin\beta=\frac{m_1}{m_3}\sin\alpha~;~~\cos\beta=\sqrt{1-\left(\frac{m_1}{m_3}\sin\alpha\right)^2}.$$The second equation gives $$\sin\alpha=\frac{m_3}{m_2}\sin(\alpha+\beta)=\frac{m_3}{m_2}(\sin\alpha \cos\beta+\cos\alpha\sin\beta).$$ We now eliminate the $\beta$ trig functions and do some elementary algebra to get $$\cos\alpha=\frac{m_2^2+m_1^2-m_3^2}{2m_1m_2}.\tag{1}$$We find $\cos\beta$ by swapping indices 1 and 2, $$\cos\beta=\frac{m_2^2+m_3^2-m_1^2}{2m_2m_3}.\tag{2}$$These expressions answer the first part of the question.

As for the second part, the condition for equilibrium to exist is obvious. When there is equiibrium, the forces and hence the masses must form a closed triangle. In other words, the masses must obey the triangle inequality, $$m_1+m_2>m_3~; ~~m_3+m_1>m_2~;~~m_2+m_3>m_1.$$Note that equations (1) and (2) that give the squares of the sines are positive only if the triangle inequality is obeyed.

On edit: I just noticed that I called the middle mass $m_3$ instead of $m_2$ and the mass on the right $m_2$ instead of $m_3.$ I apologize for the confusion this might generate when comparing solutions.

On second edit: I swapped mass indices 2 and 3 in the diagram and equations to bring the labeling in line with OP's labels.

On third edit: Fixed algebraic errors in my original derivation which brought my original expressions in agreement with those of OP.

 

[Juanda]

As for the second part, the condition for equilibrium to exist is obvious. When there is equiibrium, the forces and hence the masses must form a closed triangle. In other words, the masses must obey the triangle inequality, $$m_1+m_2>m_3~; ~~m_3+m_1>m_2~;~~m_2+m_3>m_1.$$Note that equations (1) and (2) that give the squares of the sines are positive only if the triangle inequality is obeyed.

On edit: I just noticed that I called the middle mass $m_3$ instead of $m_2$ and the mass on the right $m_2$ instead of $m_3.$ I apologize for the confusion this might generate when comparing solutions.

That's very interesting and sheds valuable light on the question I got with this. I wouldn't say it's obvious though. At least not for me. Maybe now that I saw the connection it seems obvious but I'd not have done it by myself.
Since $m_2$ is the value I'm changing, I'll express the inequalities with that cleared (the change in names doesn't affect the inequalities but I'll keep using the original naming for the variables).
$$m_1+m_2>m_3 \rightarrow m_2>m_3-m_1$$
$$m_3+m_1>m_2 \rightarrow m_2<m_3+m_1$$
$$m_2+m_3>m_1 \rightarrow m_2>m_1-m_3$$
In this case, let's assume $m_3>m_1>0$. Therefore, the inequality $m_2>m_1-m_3$ can be ignored because the other one will be more restrictive. As a consequence, we have an upper and lower boundary for $m_2$.
$$m_3-m_1<m_2<m_3+m_1$$
The use of the law of cosines in the triangle of forces to obtain $\alpha$ and $\beta$ agrees with this. If I try to go beyond the limits, the functions are not defined.

$$\alpha=\arccos(\frac{m_{1}^{2}+m_{2}^{2}-m_{3}^{2}}{2m_{1}m_{2}})$$
$$\beta=\arccos(\frac{m_{2}^{2}+m_{3}^{2}-m_{1}^{2}}{2m_{2}m_{3}})$$

However, even before the lowest limit for $m_2$, something weird happens. In the following link, you can see how, for those initial/boundary conditions, we'd be able to make $2<m_2<4$ but for $m_2=2.2$ I'm getting negative lengths. Is there an additional more restrictive restriction we're not considering? I believe it must be that the ropes cannot fall on the line that connects the two pulleys.

There is a second weird thing going on although I suspect this is just a mathematical artifact due to the use of $\arcsin()$ in $f(x)$ and how $\arcsin()$ is defined. For $m_2=2.3$ we should have no problem and it seems like that's the case. Angles are defined and lengths are positive. But the function $f(x)$ (see post #1 for a better description of $f(x)$ and the link to interact with it changing $m_2$ or other variables as desired) describing the equilibrium of vertical forces shows there is no possible $\beta$ ($\beta=x$ to deal with the calculator interface) that makes equilibrium possible. How is equilibrium being reached then?

 

[haruspex]

$$\sum F_x=0 \rightarrow -m_1g\sin\alpha + m_3gsin\beta = 0\rightarrow m_1\sin\alpha = m_3\sin\beta \tag{1}$$
$$\sum F_y=0 \rightarrow m_1g\cos\alpha -m_2g+ m_3gcos\beta = 0\rightarrow m_1\cos\alpha + m_3\cos\beta = m_2 \tag{2}$$

$m_1\sin\alpha = m_3\sin\beta \tag{1}$
$m_1\cos\alpha + m_3\cos\beta = m_2 \tag{2}$
Squaring:
$m_1^2\sin^2\alpha = m_3^2\sin^2\beta$
$m_1^2\cos^2\alpha + m_3^2\cos^2\beta +2m_1m_3\cos(\alpha)\sin(\beta)= m_2^2$
Adding:
$m_1^2+m_3^2+2m_1m_3\cos(\alpha)\sin(\beta)= m_2^2$.
I just thought it looked useful.

 

[haruspex]

Even in that case, α+β=2π only when the component of m2 that is perpendicular to both taut ropes, equals zero or is very small, in practical terms.

Sure, but we were not discussing the two ropes being in a straight line; we were discussing the left rope being horizontal.

 

[kuruman]

As a consequence, we have an upper and lower boundary for $m_2$.
$$m_3-m_1<m_2<m_3+m_1$$

That's the triangle inequality. Take two sticks of different lengths. Call the longer one $m_1$ and the shorter one $m_3$. Consider the upper and lower limits the length $m_2$ that a third stick must have to form a triangle. See picture below.

 

[Juanda]

That's the triangle inequality. Take two sticks of different lengths. Call the longer one $m_1$ and the shorter one $m_3$. Consider the upper and lower limits the length $m_2$ that a third stick must have to form a triangle. See picture below.

View attachment 333507

Yes. That's clear now. I did a worse sketch on my notebook but I got it when you pointed at it on post #23. Seeing presented here like that helps though. Especially to future readers.
This problem is turning out to be an even better learning experience for me than I anticipated.

At the moment, the thing I'm trying to understand is this from post #24.

However, even before the lowest limit for $m_2$, something weird happens. In the following link, you can see how, for those initial/boundary conditions, we'd be able to make $2<m_2<4$ but for $m_2=2.2$ I'm getting negative lengths. Is there an additional more restrictive restriction we're not considering? I believe it must be that the ropes cannot fall on the line that connects the two pulleys.

There is a second weird thing going on although I suspect this is just a mathematical artifact due to the use of $\arcsin()$ in $f(x)$ and how $\arcsin()$ is defined. For $m_2=2.3$ we should have no problem and it seems like that's the case. Angles are defined and lengths are positive. But the function $f(x)$ (see post #1 for a better description of $f(x)$ and the link to interact with it changing $m_2$ or other variables as desired) describing the equilibrium of vertical forces shows there is no possible $\beta$ ($\beta=x$ to deal with the calculator interface) that makes equilibrium possible. How is equilibrium being reached then?

 

[Steve4Physics]

@Juanda, I suspect this is part of the problem...

The line ($L$) joining the pulleys makes an angle $\theta =tan^{-1} {\frac ab}$ with the horizontal.

The maximum possible value of $\alpha$ is $\alpha_{max} = \frac {\pi}2 + \theta$ when the upper rope-sections are in a straight line. (For $\alpha$ larger than this, $m_2$ would be above $L$.)

With the values of $a$ and $b$ used in your link, I get $\alpha_{max} = 2.266~rad~(=129.8^o)$.

But the link shows $\alpha = 2.372~rad~ (=135.9^o)$, which exceeds $\alpha_{max}$. So equilibrium is not possible for the values of mass used in the link.

 

[Juanda]

@Juanda, I suspect this is part of the problem...

The line ($L$) joining the pulleys makes an angle $\theta =tan^{-1} {\frac ab}$ with the horizontal.

The maximum possible value of $\alpha$ is $\alpha_{max} = \frac {\pi}2 + \theta$ when the upper rope-sections are in a straight line. (For $\alpha$ larger than this, $m_2$ would be above $L$.)

With the values of $a$ and $b$ used in your link, I get $\alpha_{max} = 2.266~rad~(=129.8^o)$.

But the link shows $\alpha = 2.372~rad~ (=135.9^o)$, which exceeds $\alpha_{max}$. So equilibrium is not possible for the values of mass used in the link.

I see it the same way. For example, if $m_2=0$ then the rope would be straight from pulley to pulley, and equilibrium would only be possible if the remaining masses are the same.

However, the central point of the argument is that we "know" the line cannot fall on top of the line linking the two pulleys derived from the intuition of the case where $m_2=0$ but I can't get a conclusion about why the limiting angle for the line is the same in the general case where $m_2 \neq 0$.

Secondly, for $m_2=2.3$ the result is also strange because according to $f(x)$ there is no real value for the angle that can make equilibrium possible. However, we know equilibrium does happen because when obtained from the cosine law the results are physically possible. (I believe the problem with $f(x)$ is related to how the function $\arcsin()$ is defined similar to how the function $x^{0.5}$ only returns positive values).