Rope between inclines with maximum length of hanging middle portion

[Meden Agan]

Homework Statement

A homogeneous cord rests on a pair of connected, rigid slopes with adjustable inclination angles, as shown in the figure. The coefficient of friction between the slopes and the cord is $\mu$. How should we choose the inclination angles $\alpha$ and $\beta$, and how should we deposit the cord to make the length of its hanging middle section maximal?

Relevant Equations

Newton's $2^{\text{nd}}$ Law of motion, some Calculus.

Let us take an infinitesimal piece of cord, long $\mathrm ds$, resting on a plane inclined at an angle $\theta$ to the horizontal. The weight of the elements is $\lambda g \, \mathrm ds$, where $\lambda$ is the linear density of the cord. The component parallel to the plane (directed downwards) is $\lambda g \sin \theta \, \mathrm ds$. The normal reaction is $\lambda g \cos \theta \, \mathrm ds$, so the maximum static friction available (directed upwards) is $\mu \, \lambda g \cos \theta \, \mathrm ds$. Writing the axial equilibrium (positive upward direction) according to Newton's $2^{\text{nd}}$ Law of motion: $$T(s+ \mathrm ds) - T(s) = \lambda g \, \mathrm ds \left(\mu \cos \theta - \sin \theta \right), \tag{1}$$ and taking the limit $\mathrm ds \to 0$ on the LHS, we have $T(s+ \mathrm ds) - T(s) \to \mathrm dT$; so, $(1)$ becomes $$\mathrm dT = \lambda g \, \mathrm ds \left(\mu \cos \theta - \sin \theta \right). \tag{2}$$ Integrating $(2)$ from the top end of the rope $(T_{\text{top}} = 0, \, s = 0)$ to the point where the rope detaches from the slope $(T_0, \, s)$ gives $$T_0 = \lambda g \, s \left(\sin \theta - \mu \cos \theta \right). \tag{3}$$

Now, let's work with two inclined planes (angles $\alpha$ and $\beta$ with respect to the horizontal plane).
If we call the two tensions $T_{\alpha}$ and $T_{\beta}$ for the values of $T_0$ at the two sides (they will not be equal) and $s_{\alpha}$ and $s_{\beta}$ as the lengths in contact with the slope at the two sides, then equation $(3)$ tells us the relationship between $T_{\alpha}$ and $s_{\alpha}$, and also between $T_{\beta}$ and $s_{\beta}$. We have
$$T_{\alpha} = \lambda g \, s_{\alpha} \left(\sin \alpha- \mu \cos \alpha \right) \tag{A}$$ and
$$T_{\beta} = \lambda g \, s_{\beta} \left(\sin \beta- \mu \cos \beta \right). \tag{B}$$
We have other constraints as well, because

- the net horizontal force on the rope must be zero, so we get $$T_{\alpha} \cos \alpha - T_{\beta} \cos \beta = 0 \iff T_{\alpha} \cos \alpha = T_{\beta} \cos \beta. \tag{C}$$

- if we suppose the length of the unsupported rope is $\ell$, then its weight is $\lambda g \, \ell$ has to be supported by the force at the two ends of the unsupported section, i.e. by the vertical component of the tension $T_0$. So, the total upwards force must equal the weight of the unsupported rope, and we get $$T_{\alpha} \sin \alpha + T_{\beta} \sin \beta = \lambda g \, \ell. \tag{D}$$

From $(\text{C})$, we have $T_{\beta} = T_{\alpha} \, \dfrac{\cos \alpha}{\cos \beta}$. Plugging the expression $(\text{A})$ for $T_{\alpha}$ into the previous equation, we get $$T_{\beta} = \lambda g \, s_{\alpha} \left(\sin \alpha- \mu \cos \alpha \right) \dfrac{\cos \alpha}{\cos \beta}. \tag{B1}$$
Plugging the expressions for $(\text{A})$ and $(\text{B1})$ into $(\text{D})$, we get $$\cancel{\lambda g} \, s_{\alpha} \left(\sin \alpha- \mu \cos \alpha \right) \sin \alpha + \cancel{\lambda g} \, s_{\alpha} \left(\sin \alpha- \mu \cos \alpha \right) \tan \beta \cos \alpha = \cancel{\lambda g} \, \ell,$$ or, equivalently, $$\boxed{\frac{\ell}{s_{\alpha}} = \left(\sin \alpha- \mu \cos \alpha \right) \left(\sin \alpha + \tan \beta \cos \alpha \right)}.$$

Here I am stuck, because I can't understand how to find the maximum angles $\alpha$ and $\beta$ that maximize $\ell$.

 

[JLT]

You might want to study:

https://en.wikipedia.org/wiki/Catenary

The trick for cables hanging under their own weight, the horizontal force is the same through the entire cable. The lowest point in cable has only horizontal forces. The angle of the cable is the amount of weight (vertical force). The angle gets steeper and steeper as more and more weight is held under it.

Pretend the cable is a paper chain of people - holding hands. If you were one of the people in the chain, which person would you like to be? Ans: the person at the lowest point in chain has it good, is held up by everyone above them. The person at the highest point has to hold up everyone under them - has both the weight, and that horizonal force too.

Read wiki about Caternary, and analyze the free-swinging piece of the rope.

 

[haruspex]

The question seems incomplete. Shouldn’t there be some limit on the total length of the rope? Or should we interpret it as asking for the fraction of unsupported rope to be maximised, $\frac l{s_\alpha+s_\beta}$?

I feel there’s a constraint you are not using, that the hanging rope is tangential to the slopes at point of contact. Maybe use the catenary equation.

 

[Meden Agan]

The question seems incomplete. Shouldn’t there be some limit on the total length of the rope? Or should we interpret it as asking for the fraction of unsupported rope to be maximised, $\frac l{s_\alpha+s_\beta}$?

What do you mean by "some limit on the total length of the rope"? If you mean that the rope must have a total length $L = \ell + s_\alpha + s_\beta$, well, I think so.

 

[kuruman]

Here I am stuck, because I can't understand how to find the maximum angles $\alpha$ and $\beta$ that maximize $\ell$.

Refer to your drawing in post #1. Suppose that you found a specific solution $\alpha = \alpha_1$ and $\beta = \beta_1$ that maximizes the hanging section. By symmetry, it follows that the choice $\alpha = \beta_1$ and $\beta = \alpha_1$ also does the job. What does this tell you about the relation between $\alpha$ and $\beta$ ? I suggest that you start from there.

 

[Meden Agan]

I feel there’s a constraint you are not using, that the hanging rope is tangential to the slopes at point of contact. Maybe use the catenary equation.

If $L = \ell + s_\alpha + s_\beta$, then there is no need to use the catenary equation, is there?
We can derive the expression for $s_\alpha$ and $s_\beta$ from my first post.

Expression for $s_\alpha$:

From $\frac{\ell}{s_{\alpha}} = \left(\sin \alpha- \mu \cos \alpha \right) \left(\sin \alpha + \tan \beta \cos \alpha \right)$, we have $$s_\alpha = \frac{\ell}{\left(\sin \alpha- \mu \cos \alpha \right) \left(\sin \alpha + \tan \beta \cos \alpha \right)}. \tag{Expression 1}$$

Expression for $s_\beta$:

From $(\text{B})$, we have $$T_{\beta} = \lambda g \, s_{\beta} \left(\sin \beta- \mu \cos \beta \right) \iff s_{\beta} = \frac{T_{\beta}}{\lambda g \, \left(\sin \beta- \mu \cos \beta \right)}. \tag{*}$$ Using $T_{\beta} = \lambda g \, s_{\alpha} \left(\sin \alpha- \mu \cos \alpha \right) \dfrac{\cos \alpha}{\cos \beta}$ as per $(\text{B1})$, we would have:
$$\begin{aligned}
s_{\beta} &= \frac{T_{\beta}}{\lambda g \, \left(\sin \beta- \mu \cos \beta \right)} \\
&= \frac{\lambda g \, s_{\alpha} \left(\sin \alpha- \mu \cos \alpha \right) \dfrac{\cos \alpha}{\cos \beta}}{\lambda g \, \left(\sin \beta- \mu \cos \beta \right)} \\
&= \frac{\cancel{\lambda g} \, s_{\alpha} \left(\sin \alpha- \mu \cos \alpha \right) \dfrac{\cos \alpha}{\cos \beta}}{\cancel{\lambda g} \, \left(\sin \beta- \mu \cos \beta \right)} \\
&= \frac{s_{\alpha} \left(\sin \alpha- \mu \cos \alpha \right) \cos \alpha}{\left(\sin \beta- \mu \cos \beta \right) \cos \beta} \\
&= \frac{\dfrac{\ell}{\left(\sin \alpha- \mu \cos \alpha \right) \left(\sin \alpha + \tan \beta \cos \alpha \right)} \left(\sin \alpha- \mu \cos \alpha \right) \cos \alpha}{\left(\sin \beta- \mu \cos \beta \right) \cos \beta} \\
&= \frac{\ell \left(\sin \alpha- \mu \cos \alpha \right) \cos \alpha}{\left(\sin \alpha- \mu \cos \alpha \right) \left(\sin \alpha + \tan \beta \cos \alpha \right) \left(\sin \beta- \mu \cos \beta \right) \cos \beta} \\
&= \frac{\ell \cancel{\left(\sin \alpha- \mu \cos \alpha \right)} \cos \alpha}{\cancel{\left(\sin \alpha- \mu \cos \alpha \right)} \left(\sin \alpha + \tan \beta \cos \alpha \right) \left(\sin \beta- \mu \cos \beta \right) \cos \beta} \\
&= \frac{\ell \cos \alpha}{\left(\sin \alpha + \tan \beta \cos \alpha \right) \left(\sin \beta - \mu \cos \beta \right) \cos \beta}.
\end{aligned}$$
So, we have $$s_{\beta} = \frac{\ell \cos \alpha}{\left(\sin \alpha + \tan \beta \cos \alpha \right) \left(\sin \beta - \mu \cos \beta \right) \cos \beta}. \tag{Expression 2}$$

Plugging $\text{(Expression 1)}$ and $\text{(Expression 2)}$ into the expression for $L$, we obtain
$$\begin{aligned}
L &= \ell + s_\alpha + s_\beta \\
&= \ell + \frac{\ell}{\left(\sin \alpha- \mu \cos \alpha \right) \left(\sin \alpha + \tan \beta \cos \alpha \right)} + \frac{\ell \cos \alpha}{\left(\sin \alpha + \tan \beta \cos \alpha \right) \left(\sin \beta - \mu \cos \beta \right) \cos \beta} \\
&= \ell \left[1+ \frac{1}{\left(\sin \alpha- \mu \cos \alpha \right) \left(\sin \alpha + \tan \beta \cos \alpha \right)} + \frac{\cos \alpha} {\left(\sin \alpha + \tan \beta \cos \alpha \right) \left(\sin \beta - \mu \cos \beta \right) \cos \beta}\right] \\
&= \ell \left[1 + \frac{1}{\left(\sin \alpha- \mu \cos \alpha \right) \left(\sin \alpha + \tan \beta \cos \alpha \right)} \left(1 + \frac{\cos \alpha}{\cos \beta}\right)\right],
\end{aligned}$$
from which $$\ell = \frac{L}{1 + \dfrac{1}{\left(\sin \alpha- \mu \cos \alpha \right) \left(\sin \alpha + \tan \beta \cos \alpha \right)} \left(1 + \dfrac{\cos \alpha}{\cos \beta}\right)}.$$

But here, we have a function with two variables $\alpha$ and $\beta$, so I don't know how to determine the maximum values of $\alpha$ and $\beta$ from the expression above.

 

[Meden Agan]

Refer to your drawing in post #1. Suppose that you found a specific solution $\alpha = \alpha_1$ and $\beta = \beta_1$ that maximizes the hanging section. By symmetry, it follows that the choice $\alpha = \beta_1$ and $\beta = \alpha_1$ also does the job. What does this tell you about the relation between $\alpha$ and $\beta$ ? I suggest that you start from there.

This is interesting, because it seems to get to the core of my trouble with angles.
Well, intuitively, I would say that the relationship between $\alpha$ and $\beta$ you ask is $\alpha = \beta$. But I cannot give a precise rationale for this. And it would not be clear to me why the two angles should be equal.

 

[haruspex]

But here, we have a function with two variables α and β, so I don't know how to determine the maximum values of α and β from the expression above.

I assume you mean maximise $l$ wrt $\alpha, \beta$.
What’s wrong with writing $\frac{\partial l}{\partial \alpha}=\frac{\partial l}{\partial \beta}=0$?
Assuming the angles are equal in the solution probably gives the right answer, but it is not guaranteed. Consider the function $((x-a)^2+(y-b)^2)((x-b)^2+(y-a)^2)$. Clearly it has minima at $(a,b), (b,a)$.

 

[Meden Agan]

I assume you mean maximise $l$ wrt $\alpha, \beta$.
What’s wrong with writing $\frac{\partial l}{\partial \alpha}=\frac{\partial l}{\partial \beta}=0$?
Assuming the angles are equal in the solution probably gives the right answer, but it is not guaranteed. Consider the function $((x-a)^2+(y-b)^2)((x-b)^2+(y-a)^2)$. Clearly it has minima at $(a,b), (b,a)$.

Setting $\dfrac{\partial \ell}{\partial \beta}=0$ gives me $\alpha = \arctan \mu$, and setting $\dfrac{\partial \ell}{\partial \alpha}=0$ gives me $\beta= - \arctan \mu$.
I can write down the maths, but it is a bit tedious. And these results are not so convincing.

 

[haruspex]

Your expressions 1 and 2 in post #6 cannot both be right. There is a symmetry here. If you swap $\alpha$ and $\beta$ over in one you should get the other, but that’s not what I see.
Btw, $\sin(\alpha)+\tan(\beta)\cos(\alpha)=\frac{\sin(\alpha+\beta)}{\cos(\beta)}$, which simplifies things a little.

 

[Meden Agan]

Your expressions 1 and 2 in post #6 cannot both be right. There is a symmetry here. If you swap $\alpha$ and $\beta$ over in one you should get the other, but that’s not what I see.
Btw, $\sin(\alpha)+\tan(\beta)\cos(\alpha)=\frac{\sin(\alpha+\beta)}{\cos(\beta)}$, which simplifies things a little.

If we swap $\alpha$ and $\beta$ over in $\text{(Expression 1)}$ for $s_\alpha$, we actually get $\text{(Expression 2)}$ for $s_\beta$.

Let’s start with $\text{(Expression 1)}$ for $s_\alpha$. We swap $\alpha$ and $\beta$.
We have $$s_\alpha = \frac{\ell}{\left(\sin \alpha- \mu \cos \alpha \right) \left(\sin \alpha + \tan \beta \cos \alpha \right)}.$$
Under the substitutions $\alpha \longmapsto \beta$ and $\beta \longmapsto \alpha$:
$$\begin{aligned}
s_\beta &= \frac{\ell}{\left(\sin \beta- \mu \cos \beta\right) \left(\sin \beta+ \tan \alpha \cos \beta \right)} \\
&= \frac{\ell}{\left(\sin \beta- \mu \cos \beta\right) \left(\sin \beta+ \dfrac{\sin \alpha}{\cos \alpha} \cos \beta \right)} \\
&= \frac{\ell \cos \alpha}{\left(\sin \beta- \mu \cos \beta\right) \left(\sin \beta \cos \alpha+ \sin \alpha \cos \beta \right)} \\
&= \frac{\ell \cos \alpha}{\left(\sin \beta- \mu \cos \beta\right) \left(\dfrac{\sin \beta \cos \alpha}{{\color{red}{\cos \beta}}}+ \dfrac{\sin \alpha \cos \beta}{{\color{red}{\cos \beta}}} \right) {\color{red}{\cos \beta}}} \\
&= \frac{\ell \cos \alpha}{\left(\sin \beta- \mu \cos \beta\right) \left(\dfrac{\sin \beta \cos \alpha}{\cos \beta}+ \dfrac{\sin \alpha \cancel{\cos \beta}}{\cancel{\cos \beta}} \right) \cos \beta} \\
&= \frac{\ell \cos \alpha}{\left(\sin \beta- \mu \cos \beta\right) \left(\tan \beta \cos \alpha+ \sin \alpha \right) \cos \beta}.
\end{aligned}$$
Thus, we have $$s_\beta =\frac{\ell \cos \alpha}{\left(\sin \beta- \mu \cos \beta\right) \left(\tan \beta \cos \alpha+ \sin \alpha \right) \cos \beta},$$ which is $\text{(Expression 2)}$.

Everything is consistent: by swapping $\alpha$ and $\beta$ over in $\text{(Expression 1)}$, we get $\text{(Expression 2)}$.

The questions in posts #6, #7 and #9 stand.

 

[Steve4Physics]

$$T_0 = \lambda g \, s \left(\sin \theta - \mu \cos \theta \right). \tag{3}$$

If you want $T_0$ to have a positive value this should be:
$T_0 = \lambda g \, s \left( \mu \cos \theta - \sin \theta \right)$
(E.g. consider the case with $\theta = 0$.)

It’s worth noting that the same result would be obtained if the internal tension is ignored. Consider a single object of weight $W (= \lambda g s)$ on a slope and subject to an external downhill force of $T_0$. Limiting friction is $\mu W \cos\theta$, the downhill component of weight is $W \sin\theta$ so we can immediately write $T_0 + W \sin \theta\ =\mu W \cos \theta$.

You seem to be looking for a solution without incorporating the fact that the hanging part (length $l$) of the cord is a catenary. IMO this is not possible because $l$ is partly determined by the fact that we have a catenary and not some other shape. We require that $s_{\alpha} + s_{\beta} + l =$ a constant (the total length). Other (non-catenary) curves can change the value of $l$ while still providing the required angles at the ends. Simply trying to maximise the weight that can be supported by the straight sections of the cord (which I think is what you trying to do) does not address this.

Edit - typo.

 

[Meden Agan]

You seem to be looking for a solution without incorporating the fact that the hanging part (length $l$) of the cord is a catenary. IMO this is not possible because $l$ is partly determined by the fact that we have a catenary and not some other shape. We require that $s_{\alpha} + s_{\beta} + l =$ a constant (the total length). Other (non-catenary) curves can change the value of $l$ while still providing the required angles at the ends. Simply trying to maximise the weight that can be supported by the straight sections of the cord (which I think is what you trying to do) does not address this.

I think I understand your POV. But, did you see post #6?
There, I derive the length $\ell$ of the hanging section as a function of $\alpha, \beta, \mu$ and the constant total length $L$ of the cord.
The trouble remains (please see posts #6, #7, #9).

 

[Steve4Physics]

I think I understand your POV. But, did you see post #6?
There, I derive the length $\ell$ of the hanging section as a function of $\alpha, \beta, \mu$ and the constant total length $L$ of the cord.
The trouble remains (please see posts #6, #7, #9).

In Post #6 you express $\ell$ as a function of $\alpha, \beta, \mu$ and $L$ only. I would argue that this (incorrectly) ignores the shape of the hanging section. For example if the shape of the hanging section were as shown in blue below (unphysical but OK geometrically), this would change the relative proportions $s_{\alpha} : s_{\beta} : \ell$.

The fact that the hanging part is a catenary and not some other shape should (AFAICS) affect the answer.

Edits.

 

[Meden Agan]

In Post #6 you express $\ell$ as a function of $\alpha, \beta, \mu$ and $L$ only. I would argue that this (incorrectly) ignores the shape of the hanging section. For example if the shape of the hanging section were as shown in blue below (unphysical but OK geometrically), this would change the relative proportions $s_{\alpha} : s_{\beta} : \ell$.
View attachment 361568
The fact that the hanging part is a catenary and not some other shape should (AFAICS) affect the answer.

Edits.

Good point! I started calculating $\ell$ by setting the constraint of the catenary equation.
I obtained $$\ell = L \frac{\tan \alpha + \tan \beta}{\dfrac{1}{\cos \alpha \left(\mu \cos \alpha - \sin \alpha \right)} + \tan \alpha + \dfrac{1}{\cos \beta \left(\mu \cos \beta- \sin \beta\right)} + \tan \beta}$$ (I can write down the maths if you want).

But, the trouble as illustrated in posts #7 and #9 stands as well. How do we calculate the maximum angles $\alpha$ and $\beta$? I suspect we need a preliminary relationship between $\alpha$ and $\beta$.

 

[Steve4Physics]

Good point! I started calculating $\ell$ by setting the constraint of the catenary equation.
I obtained $$\ell = L \frac{\tan \alpha + \tan \beta}{\dfrac{1}{\cos \alpha \left(\mu \cos \alpha - \sin \alpha \right)} + \tan \alpha + \dfrac{1}{\cos \beta \left(\mu \cos \beta- \sin \beta\right)} + \tan \beta}$$ (I can write down the maths if you want).

But, the trouble as illustrated in posts #7 and #9 stand as well. How do we calculate the maximum angles $\alpha$ and $\beta$? I suspect we need a preliminary relationship between $\alpha$ and $\beta$.

I don't know!

It may helpful (or at least comforting!) to try a simpler problem first.

Consider the fully symmetrical case: $\alpha = \beta, s_{\alpha} = s_{\beta}$. This is manageable as each straight section of cord supports exactly half the weight of the hanging part - so the shape of the hanging part (catenary or not) doesn’t matter in this case. You can express (for example) $s_{\alpha}$ as a function of $\alpha$ and then minimise $s_{\alpha}$. This may correspond to the general solution but I don’t see if/how that could be justified.

If I may ask, where does this problem come from? At what academic level is it targetted?

 

[Meden Agan]

Consider the fully symmetrical case: $\alpha = \beta, s_{\alpha} = s_{\beta}$. This is manageable as each straight section of cord supports exactly half the weight of the hanging part - so the shape of the hanging part (catenary or not) doesn’t matter in this case. You can express (for example) $s_{\alpha}$ as a function of $\alpha$ and then minimise $s_{\alpha}$. This may correspond to the general solution but I don’t see if/how that could be justified.

I understand the setup, but I cannot work out which expressions for $s_\alpha$ and $s_\beta$ I should consider. The statement I am confused by is "so the shape of the hanging part (catenary or not) doesn’t matter in this case".
Well, I am able to derive expressions in the case where the shape is a catenary. But in general, no, because $s_\alpha$ and $s_\beta$ depend on $T_\alpha$ and $T_\beta$, for which the expressions are different depending on the shape of the cord (?)

The final answer is $\alpha = \beta = \dfrac{\arctan \mu}{2}$. But, it is not at all obvious to me why $\alpha=\beta$ necessarily. Post #5 by @kuruman seems to go this way, but I am not sure.

If I may ask, where does this problem come from? At what academic level is it targetted?

This problem is a variant (of a classic problem) proposed by my teacher. The original source should be Problem 89 in Problem of the Week by David Morin.
In Morin's problem, however, the situation is fully symmetric, with $\alpha = \beta = \theta$. In this version, however, the two angles are not equal for assumption. It is neither obvious nor taken for granted that they should be (as would then turn out to be the case).
It should be labelled at “undergraduate level”.

 

[Steve4Physics]

I understand the setup, but I cannot work out which expressions for $s_\alpha$ and $s_\beta$ I should consider. The statement I am confused by is "so the shape of the hanging part (catenary or not) doesn’t matter in this case".

Sorry, I wasn't clear. The hanging part must be symmetrical - but any (symmetrical) shape is OK. We can then equate the vertical component of each tension ($T_0$) to the weight of half of the hanging part.

In Morin's problem, however, the situation is fully symmetric, with $\alpha = \beta = \theta$. In this version, however, the two angles are not equal for assumption. It is neither obvious nor taken for granted that they should be (as would then turn out to be the case).
It should be labelled at “undergraduate level”.

The symmetry of the solution to the general problem is not unexpected; but, like you, I don't see how to justify it.

I don't think I can add anything else.

 

[Meden Agan]

The symmetry of the solution to the general problem is not unexpected; but, like you, I don't see how to justify it.

Hope someone can help us find out!

 

[Lnewqban]

Isn't this a balance act between friction force and tension?
If so, it can't be but symmetrical.
What would happen at the extreme end of those angles?

As both angles tend to zero, it is all friction and no tension.
As both surfaces and rope halves tend to be vertical, it is all tension (induced by weight) and no friction (no normal forces induced by non-suspended weight).

 

[haruspex]

Consider the fully symmetrical case: $\alpha = \beta, s_{\alpha} = s_{\beta}$. This is manageable as each straight section of cord supports exactly half the weight of the hanging part - so the shape of the hanging part (catenary or not) doesn’t matter in this case. You can express (for example) $s_{\alpha}$ as a function of $\alpha$ and then minimise $s_{\alpha}$. This may correspond to the general solution but I don’t see if/how that could be justified.

How if we chop the problem in half? At the point where the rope is horizontal, replace the right half with a vertical wall to which the rope is anchored. If we find the $\alpha$ maximising the fraction of suspended rope, subject to the rope meeting the wall at a right angle, the full solution must work also for the right half (and provide matching horizontal tension at the join).

 

[Steve4Physics]

Isn't this a balance act between friction force and tension?
If so, it can't be but symmetrical.
What would happen at the extreme end of those angles?

As both angles tend to zero, it is all friction and no tension.
As both surfaces and rope halves tend to be vertical, it is all tension (induced by weight) and no friction (no normal forces induced by non-suspended weight).

Remember it’s not just the angles ($\alpha$ and $\beta$) that can vary independently, it’s also the contact-lengths ($s_{\alpha}$ and $s_{\beta})$. There are four variables to be juggled.

For a given value of $\mu$, the frictional force on each side depends on both the angle and contact-length on that side.

I’ll make up some values for illustration – they’re not to be taken literally.

Suppose we have the symmetrical case, where for some given value of $\mu$ the maximum length of the hanging-section occurs when:
$\alpha = 40^\circ, s_{\alpha} = 0.25L$
$\beta = 40^\circ, s_{\beta} = 0.25L$
The length of the hanging-section is then $L – 0.25L – 0.25L = 0.5L$

The problem is to rigorously prove that there isn’t another (asymmetrical) configuration which can make the hanging-section longer, for example:
$\alpha = 10^\circ, s_{\alpha} = 0.1L$
$\beta = 50^\circ, s_{\beta} = 0.3L$
In this case the maximum possible length of the hanging-section would be $L – 0.1L – 0.3L = 0.6L$.

Or maybe I’m missing some simple argument.

How if we chop the problem in half? At the point where the rope is horizontal, replace the right half with a vertical wall to which the rope is anchored. If we find the $\alpha$ maximising the fraction of suspended rope, subject to the rope meeting the wall at a right angle, the full solution must work also for the right half (and provide matching horizontal tension at the join).

In general, if the catenary is assymetric (one end higher than the other), then the point (P) where the rope is horizontal is not central -we don't know where it is without analysing the shape of the catenary. This sounds like a difficult case to deal with as we don't know the length of rope on each side of P.

Edits.

 

[Lnewqban]

The problem is to rigorously prove that there isn’t another (asymmetrical) configuration which can make the hanging-section longer,

Introducing the asymmetrical situation may not be needed, I believe.
It seems not to be a natural balanced situation to me.

For very small angles, the magnitude of the tension would tend to infinite.
The available static friction and the tension decrease at different ratios as the angles increase.

The condition of balance seems to happen when both values (tension and friction) coincide (interception of two curves perhaps).

 

[haruspex]

In general, if the catenary is assymetric (one end higher than the other), then the point (P) where the rope is horizontal is not central

I didn't assume it was. (I should have clarified I meant conceptually in half, not literally.)

 

[Meden Agan]

The problem is to rigorously prove that there isn’t another (asymmetrical) configuration which can make the hanging-section longer

That's exactly it. Again, post #5 by @kuruman suggests a preliminary relationship between $\alpha$ and $\beta$. But, I cannot understand their hint.

 

[kuruman]

The symmetry of the solution to the general problem is not unexpected; but, like you, I don't see how to justify it.

In this problem, there is nothing in the physical situation to distinguish left from right. The gravitational field is uniform and vertical. It can only distinguish "up" from "down". Suppose you solve this problem looking down at the diagram in post #1 and you find that the maximum requires the angle on the left labeled $\alpha$ in the diagram be smaller than the angle on the right labeled $\beta$. Should you be confident that you solved the problem correctly?

The answer is "No". If the correct solution necessitates the angle on the left to be less than the angle on the right, then an observer looking at the same diagram from behind the screen will disagree with you about which angle ought to be smaller because his left side is your right side. If there is nothing in the physical situation to distinguish left from right, then you are not, and should not be, in a position to do so.

Contrast this with the physical situation where one incline is more slippery than the other by having a smaller coefficient of static friction than the other. Aha! This removes the symmetry and you can draw conclusions such as "the angle with the smaller coefficient of static friction should be the smaller of the two."

 

[Lnewqban]

If we could anchor the ends of the homogeneous cord to the rigid slopes (which could be seen as having infinite available friction), the cord would form a beautiful catenary for any adjustment of the inclination angle we choose (symmetrical or not).

In that case, the length of the hanging middle section of the cord would be maximal and equal to its natural length.

The opposite situation would be true for a coefficient of friction between the slopes and the cord equal to zero: there would be no catenary or any length of hanging middle section of the cord.

We should deposit the minimal length of the cord on the rigid slopes (smaller angles induce bigger friction) in order to make the length of its hanging middle section maximal and pulling as little as possible (greater angles induce less pulling tension on the end length of the cord).

Although personally unable to follow the math to calculate it, I see a convergence toward a balancing angle, at least for the symmetrical case.

 

[haruspex]

In this problem, there is nothing in the physical situation to distinguish left from right. The gravitational field is uniform and vertical. It can only distinguish "up" from "down". Suppose you solve this problem looking down at the diagram in post #1 and you find that the maximum requires the angle on the left labeled $\alpha$ in the diagram be smaller than the angle on the right labeled $\beta$.

As I noted in post #8, it is not quite that clear. You could find there are two values for $\alpha$ and $\beta$, and it does not matter which is which as long as one takes one value and the other takes the other.
Is that likely? No. Is it possible? Yes.

 

[Meden Agan]

Read the posts, but still have doubts.
I calculated $\ell$ if the cord is a catenary, and for $\alpha = \beta$ the result is the same as in post #6 (not accounting for the catenary equation). You guys pointed out the problem only makes sense if $\alpha = \beta$. So, should I conclude that the cord shape doesn't affect the expression for $\ell$?

 

[Steve4Physics]

In this problem, there is nothing in the physical situation to distinguish left from right. The gravitational field is uniform and vertical. It can only distinguish "up" from "down".

Agree 100%.

Suppose you solve this problem looking down at the diagram in post #1 and you find that the maximum requires the angle on the left labeled $\alpha$ in the diagram be smaller than the angle on the right labeled $\beta$. Should you be confident that you solved the problem correctly?

The problem is the possibility of a pair of solutions such as these :

Solution-1:
$\alpha = 10^\circ, s_{\alpha} = 0.1L$
$\beta = 50^\circ, s_{\beta} = 0.3L$
and
Solution-2:
$\alpha = 50^\circ, s_{\alpha} = 0.3L$
$\beta =10^\circ, s_{\beta} = 0.1L$

These describe the same (assymetrical) configration - seen from the front and from behind. There is the possbility that the length of the catenary for each solution is the required maximum.

Edit - typo's

 

[Meden Agan]

And so guys, what shall I do to go on?

 

[Steve4Physics]

And so guys, what shall I do to go on?

You can already solve (I think) the symmetric case where $\alpha = \beta$ and $s_{\alpha} = s_{\beta}$.

What remains is to demonstrate that this symmetry is necessary to achieve the maximum catenary length. Is that what are you trying to achieve? I can't offer any suggestions on this.

 

[Meden Agan]

What remains is to demonstrate that this symmetry is necessary to achieve the maximum catenary length. Is that what are you trying to achieve?

Yes.

 

[Meden Agan]

I've been thinking about this: can we prove that $\alpha = \beta$ is the only possible solution maximizing the hanging section of the catenary through such a Theorem¹?

¹Theorem. If $\ell(\alpha, \beta)$ 1) is a continuously differentiable function of two variables $\alpha$ and $\beta$, 2) satisfies $\ell(\alpha, \beta) = \ell(\beta, \alpha)$ for all $(\alpha,\beta)$ (i.e., is symmetric), and 3) has a unique interior maximizer $(\alpha,\beta)$, then that maximizer must lie on the diagonal $\alpha = \beta$, i.e. $\alpha_{\max} = \beta_{\max}$.

I'm not at all sure about the validity of this theorem.

 

[haruspex]

I've been thinking about this: can we prove that $\alpha = \beta$ is the only possible solution maximizing the hanging section of the catenary through such a Theorem¹?

¹Theorem. If $\ell(\alpha, \beta)$ 1) is a continuously differentiable function of two variables $\alpha$ and $\beta$, 2) satisfies $\ell(\alpha, \beta) = \ell(\beta, \alpha)$ for all $(\alpha,\beta)$ (i.e., is symmetric), and 3) has a unique interior maximizer $(\alpha,\beta)$, then that maximizer must lie on the diagonal $\alpha = \beta$, i.e. $\alpha_{\max} = \beta_{\max}$.

I'm not at all sure about the validity of this theorem.

Even if you can, how would you show the third condition is satisfied?

What about my post #21? I haven’t seen a reaction to it since I clarified the idea in post #24.

To recap, just consider the $\alpha$ side, up to the point where the rope is horizontal, wherever that is. Let the suspended length that side be $r$. We can drop the $\alpha$ subscripts.

Write the vertical force balance equation and maximise $r/s$ wrt $\alpha$.
This allows us to maximise $r_\alpha/s_\alpha$ separately from $r_\beta/s_\beta$. It remains to show this maximises $(r_\alpha+r_\beta)/(s_\alpha+s_\beta)$.

The total normal force from the slope is $\lambda sg\cos(\alpha)$, and its vertical component is $\lambda sg\cos^2(\alpha)$.

The total upslope friction is $\mu\lambda sg\cos(\alpha)$, and its vertical component is $\mu\lambda sg\sin(\alpha)\cos(\alpha)$.

Adding, $\lambda sg(\cos^2(\alpha)+\mu\sin(\alpha)\cos(\alpha))=\lambda g(s+r)$.

$1+\frac rs=\frac 12(\mu\sin(2\alpha)+1+\cos(2\alpha))$

Maximising wrt $\alpha$: $\mu=\tan(2\alpha)$.

 

[Steve4Physics]

I've been thinking about this: can we prove that $\alpha = \beta$ is the only possible solution maximizing the hanging section of the catenary through such a Theorem¹?

¹Theorem. If $\ell(\alpha, \beta)$ 1) is a continuously differentiable function of two variables $\alpha$ and $\beta$, 2) satisfies $\ell(\alpha, \beta) = \ell(\beta, \alpha)$ for all $(\alpha,\beta)$ (i.e., is symmetric), and 3) has a unique interior maximizer $(\alpha,\beta)$, then that maximizer must lie on the diagonal $\alpha = \beta$, i.e. $\alpha_{\max} = \beta_{\max}$.

I'm not at all sure about the validity of this theorem.

I don't think it works. E.g. consider @haruspex 's (minimisation) example in Post #8:
$(x-a)^2+(y-b)^2)(x-b)^2+(y-a)^2$

Even when $a \ne b$ the minimum is obtained at two points, $(a,b)$ and $(b,a)$.

Also note that the original problem is a a four (not two) variable problem - all four variables ($\alpha, \beta, s_{\alpha}$ and $s_{\beta}$) must be considered.

EDIT. That's all rendered moot following @haruspex's Post #35 analysis.

 

[Meden Agan]

Even if you can, how would you show the third condition is satisfied?

You're spot on. We could have theoretically proved the third condition rigorously by taking the derivatives with respect to $\alpha$ and $\beta$ and showing that they are equal to zero at a single point $\alpha = \beta$. But, this doesn't happen, as I wrote in post #9.
I suspect this is because, in Multivariable Calculus, taking a partial derivative alone is not sufficient to conclude that we’ve found an extremum.
Maybe we can refer to the second-derivative test for functions of two variables (the Hessian test). However, I fear performing it analytically can be quite tedious.

Let $L_h$ be the length of the hanging section of the cord and $L$ be the total length of the cord. Then, $$\frac{L_h}{L} = \frac{\tan \alpha + \tan \beta}{\dfrac{1}{\cos \alpha \left(\mu \cos \alpha - \sin \alpha \right)} + \tan \alpha + \dfrac{1}{\cos \beta \left(\mu \cos \beta- \sin \beta\right)} + \tan \beta}, \tag{1}$$ as I said in post #15. I sketched an approximate graph following a logically grounded outline, through a heat map of $L_h/L$ vs. angles $\alpha, \beta$ for $\mu = 1$. Here it is the plot with the maximum marked with a white cross.

The function is symmetric (as per $(1)$) and the further the point away from the line of symmetry the lower the value of the function. We take an interval from point infinitely far away from the line to the diagonal itself and say that there is no local maximum. Same applies for other side, means the maximum is on the diagonal itself. So, the maximum is achieved when $\alpha = \beta$.
I fear a more involved analysis using the full three-dimensional slope tends to be cumbersome.

What about my post #21? I haven’t seen a reaction to it since I clarified the idea in post #24.

To recap, just consider the $\alpha$ side, up to the point where the rope is horizontal, wherever that is. Let the suspended length that side be $r$. We can drop the $\alpha$ subscripts.

Write the vertical force balance equation and maximise $r/s$ wrt $\alpha$.
This allows us to maximise $r_\alpha/s_\alpha$ separately from $r_\beta/s_\beta$.

At first, I didn't fully understand post #21. If you could make an approximate graphical sketch of your setup, I would write down the required maths :-)

It remains to show this maximises $(r_\alpha+r_\beta)/(s_\alpha+s_\beta)$.

How would you show that?

 

[Steve4Physics]

I’d like to throw this in.

In the diagram below we have P as the lowest point (vertex) of the catenary and lengths
$\text{AB} = s_{\alpha}, \text{arcBP} = r_{\alpha}, \text{arcPC} = r_{\beta}$ and $\text{CD} = s_{\beta}$.

In https://en.wikipedia.org/wiki/Catenary there is a reference to the 'Whewell equation' for the catenary. In the context of the current problem I interpret this as telling us (surprisingly) that arc length and tangential angle are proportional such that:$$\frac {r_{\alpha}}{r_{\beta}} = \frac {\tan \alpha} {\tan \beta}$$So this is an additional relationship which affects the catenary length.

I’m not sure if/how to use this and it’s my bedtime. No doubt I’ll have catenary dreams (or nightmares).

Edits as I can't preview to check LaTeX.
Edit to improve diagram (though it still looks like point P should be a bit more to the left).

 

[haruspex]

Edits as I can't preview to check LaTeX.

I've been doing it by first taking a copy of the post I'm about to make, just as text, then doing preview/refresh in some combination. This shows the preview correctly but it screws the post, so discard the post, make any corrections in the saved copy, then paste that back in and post it.

 

[haruspex]

We could have theoretically proved the third condition rigorously by taking the derivatives with respect to α and β and showing that they are equal to zero at a single point α=β. But, this doesn't happen, as I wrote in post #9.

But the result you got there, with opposite signs, cannot be right. As you agreed, swapping the angles around must convert Exp 1 into Exp 2, so swapping the angles in a solution should also be a solution.

taking a partial derivative alone is not sufficient to conclude that we’ve found an extremum

True, but it largely works. It’s just that instead of having only to discriminate maxima from minima you also have to worry about saddle points.

If you could make an approximate graphical sketch of your setup, I would write down the required maths :-)

@Steve4Physics has kindly done that in post #38, though I would have put point P a bit to the left; doesn’t seem to be at the bottom of the curve (or could be my astigmatism).

How would you show that?

Given an expression of the form $F=\frac{p+q}{r+s}$, and having shown that $\frac{p}{r}\leq t$ and $\frac{q}{s}\leq t$, all variables positive, what bound can you put on $F$?

 

[erobz]

Don't you have the constraint from arc length to somehow use yet?

$$ s_{\alpha} + s_{ \beta} - L = \int \sqrt{ 1 + \left( \frac{dy}{dx} \right)^2 } dx $$

 

[haruspex]

Don't you have the constraint from arc length to somehow use yet?

$$ s_{\alpha} + s_{ \beta} - L = \int \sqrt{ 1 + \left( \frac{dy}{dx} \right)^2 } dx $$

As in another recent thread on hanging ropes, it does not seem to matter whether the hanging section of the rope is uniform. There are enough equations without it, so it must be irrelevant.

 

[Meden Agan]

Given an expression of the form $F=\frac{p+q}{r+s}$, and having shown that $\frac{p}{r}\leq t$ and $\frac{q}{s}\leq t$, all variables positive, what bound can you put on $F$?

We have $$\frac{p}{r} \le t. \tag 1$$ Multiplying both sides of the inequality by $r > 0$, we have $p \le t r$. By adding $q > 0$ to both sides of the previous inequality, we obtain $$p + q \le tr + q. \tag{1.1}$$

Same way, we have $$\frac{q}{s} \le t. \tag 2$$ Multiplying both sides of the inequality by $s > 0$, we have $q \le t s$. By adding $tr > 0$ to both sides of the previous inequality, we obtain $$q + tr \le ts + tr. \tag{2.1}$$

Taking the two inequalities $(1.1)$ and $(2.1)$, we obtain the chain of inequalities $$p + q \le tr + q \le ts + tr.$$ Taking the first and the last one:
$$\begin{aligned}
p + q &\le ts + tr \\
&= t(s+r),
\end{aligned}$$
or, equivalently,
$$\boxed{\underbrace{\frac{p+q}{s+r}}_{\displaystyle F} \le t}.$$


But, I've a few doubts. Here $t = \left(\dfrac{r}{s}\right)_{\max}$, with $\left(\dfrac{r}{s}\right)_{\max}$ obtained by substituting $\alpha_\max$ for $\alpha$ into the expression for $r/s$. Correct?
If so, haven't we only shown that $\dfrac{r_\alpha +r_\beta}{s_\alpha + s_\beta}$ is maximized by $\left(\dfrac{r}{s}\right)_{\max}$? It would still remain to show explicitly that $\alpha_\max$ is the angle that also maximizes $\dfrac{r_\alpha +r_\beta}{s_\alpha + s_\beta}$.

 

[haruspex]

We have $$\frac{p}{r} \le t. \tag 1$$ Multiplying both sides of the inequality by $r > 0$, we have $p \le t r$. By adding $q > 0$ to both sides of the previous inequality, we obtain $$p + q \le tr + q. \tag{1.1}$$

Same way, we have $$\frac{q}{s} \le t. \tag 2$$ Multiplying both sides of the inequality by $s > 0$, we have $q \le t s$. By adding $tr > 0$ to both sides of the previous inequality, we obtain $$q + tr \le ts + tr. \tag{2.1}$$

Taking the two inequalities $(1.1)$ and $(2.1)$, we obtain the chain of inequalities $$p + q \le tr + q \le ts + tr.$$ Taking the first and the last one:
$$\begin{aligned}
p + q &\le ts + tr \\
&= t(s+r),
\end{aligned}$$
or, equivalently,
$$\boxed{\underbrace{\frac{p+q}{s+r}}_{\displaystyle F} \le t}.$$


But, I've a few doubts. Here $t = \left(\dfrac{r}{s}\right)_{\max}$, with $\left(\dfrac{r}{s}\right)_{\max}$ obtained by substituting $\alpha_\max$ for $\alpha$ into the expression for $r/s$. Correct?
If so, haven't we only shown that $\dfrac{r_\alpha +r_\beta}{s_\alpha + s_\beta}$ is maximized by $\left(\dfrac{r}{s}\right)_{\max}$? It would still remain to show explicitly that $\alpha_\max$ is the angle that also maximizes $\dfrac{r_\alpha +r_\beta}{s_\alpha + s_\beta}$.

As you found, $F\leq t$, and that does it.
You previously found that $t$ is the maximum value of each of $p/r$ and $q/s$, and you found an angle that achieves that maximum for each.
You also knew that $F$ achieves that value if you plug in that angle for both sides. What was missing was proof that no higher value could be achieved for $F$ by allowing different angles.

 

[Meden Agan]

To recap, just consider the $\alpha$ side, up to the point where the rope is horizontal, wherever that is. Let the suspended length that side be $r$. We can drop the $\alpha$ subscripts.

Write the vertical force balance equation and maximise $r/s$ wrt $\alpha$.
This allows us to maximise $r_\alpha/s_\alpha$ separately from $r_\beta/s_\beta$. It remains to show this maximises $(r_\alpha+r_\beta)/(s_\alpha+s_\beta)$.

Here below is my attempt. Please take a look at it and tell me if there are any flaws.

Let us take an infinitesimal piece of cord, long $\mathrm ds$, resting on a plane inclined at an angle $\alpha$ to the horizontal. The weight of the elements is $\lambda g \, \mathrm ds$, where $\lambda$ is the linear density of the cord. The component parallel to the plane (directed downwards) is $\lambda g \sin \alpha \, \mathrm ds$. The normal reaction is $\lambda g \cos \alpha \, \mathrm ds$, so the maximum static friction available (directed upwards) is $\mu \, \lambda g \cos \alpha \, \mathrm ds$. Writing the axial equilibrium (positive upward direction) according to Newton's $2^{\text{nd}}$ Law of motion: $$T(s+ \mathrm ds) - T(s) = \lambda g \, \mathrm ds \left(\mu \cos \alpha- \sin \alpha \right), \tag{1}$$ and taking the limit $\mathrm ds \to 0$ on the LHS, we have $T(s+ \mathrm ds) - T(s) \to \mathrm dT$; so, $(1)$ becomes $$\mathrm dT = \lambda g \, \mathrm ds \left(\mu \cos \alpha- \sin \alpha \right). \tag{2}$$ Integrating $(2)$ from the point where the rope detaches from the slope $(T= T_\alpha, \, s = s_\alpha)$ to the top end of the rope $(T_{\text{top}}=0, \, s=0)$ gives $$T_\alpha = \lambda g \, s_\alpha \left(\mu \cos \alpha- \sin \alpha \right). \tag{3}$$

Let the length of the hanging section on that side be $r_\alpha$. So, its weight is $\lambda g \, r_\alpha$.
This way, the total length $L_{\text{tot, left}}$ of the cord under consideration is equal to the length of the cord from its left top end to the point where the string detaches from the left slope $(s_\alpha)$, plus the length of the hanging section on the left side $(r_\alpha)$.
Thus, $L_{\text{tot, left}} = s_\alpha + r_\alpha$.

The weight of the left hanging section of the rope, $\lambda g \, r_\alpha$, must be balanced by the vertical component, $T_\alpha \sin \alpha$, of the tension at the point where it joins the part of the cord touching the left incline. We hence have $T_\alpha \sin \alpha = \lambda g \, r_\alpha$ or, equivalently, $$T_\alpha = \frac{\lambda g \, r_\alpha}{\sin \alpha}. \tag{4}$$

Equating the two expressions $(3)$ and $(4)$ for $T_\alpha$:
$$\begin{aligned}
\lambda g \, s_\alpha \left(\mu \cos \alpha - \sin \alpha \right) &= \frac{\lambda g \, r_\alpha}{\sin \alpha} \\
\cancel{\lambda g} \, s_\alpha \left(\mu \cos \alpha- \sin \alpha \right) &= \frac{\cancel{\lambda g} \, r_\alpha}{\sin \alpha} \\
s_\alpha \left(\mu \cos \alpha - \sin \alpha \right) &= \frac{r_\alpha}{\sin \alpha} \\
s_\alpha \sin \alpha \left(\mu \cos \alpha - \sin \alpha \right) &= r_\alpha,
\end{aligned}$$
which gives $$r_\alpha = s_\alpha \sin \alpha \left(\mu \cos \alpha - \sin \alpha \right). \tag{5}$$

The suspended fraction $f$ we are trying to maximize is $f = \dfrac{r_\alpha}{L_{\text{tot, left}}}$. So:
$$\begin{aligned}
f &= \frac{r_\alpha}{L_{\text{tot, left}}} \\
&= \frac{r_\alpha}{s_\alpha + r_\alpha} \\
&= \frac{s_\alpha \sin \alpha \left(\mu \cos \alpha - \sin \alpha \right)}{s_\alpha + s_\alpha \sin \alpha \left(\mu \cos \alpha - \sin \alpha \right)} \\
&= \frac{s_\alpha \sin \alpha \left(\mu \cos \alpha - \sin \alpha \right)}{s_\alpha \left(1+ \sin \alpha \left(\mu \cos \alpha - \sin \alpha \right)\right)} \\
&= \frac{\cancel{s_\alpha} \sin \alpha \left(\mu \cos \alpha - \sin \alpha \right)}{\cancel{s_\alpha} \left(1+ \sin \alpha \left(\mu \cos \alpha - \sin \alpha \right)\right)} \\
&= \frac{\sin \alpha \left(\mu \cos \alpha - \sin \alpha \right)}{1+ \sin \alpha \left(\mu \cos \alpha - \sin \alpha \right)},
\end{aligned}$$
thus $f=\dfrac{\sin \alpha \left(\mu \cos \alpha - \sin \alpha \right)}{1+ \sin \alpha \left(\mu \cos \alpha - \sin \alpha \right)}$.

Let $F(\alpha) = \sin \alpha \left(\mu \cos \alpha - \sin \alpha \right)$. Then: $$f = \frac{F(\alpha)}{1+ F(\alpha)} \qquad \text{where} \quad F(\alpha) = \sin \alpha \left(\mu \cos \alpha - \sin \alpha \right).$$
This expression for $f$ clearly is a monotonically increasing function of $F(\alpha)$. And since the suspended fraction increases as $F(\alpha)$ increases, it suffices to maximise $F(\alpha)$ in order to maximise $f$.
Differentiating $F(\alpha)$, we have $F'(\alpha) = \mu \cos(2 \alpha) - \sin(2 \alpha)$. It is zero when $\mu = \tan (2 \alpha_{\max})$. Then: $$\boxed{\alpha_{\max} = \frac{\arctan \mu}{2}},$$ as desired.


Analogous reasoning leads to $\beta_{\max} = \dfrac{\arctan \mu}{2}$, so $\alpha_{\max} = \beta_{\max}$.

 

[Steve4Physics]

Bu@Steve4Physics has kindly done that in post #38, though I would have put point P a bit to the left; doesn’t seem to be at the bottom of the curve (or could be my astigmatism).

Probably my astigmatism!

Point P edited (moved a bit left) in post #38.

 

[haruspex]

Here below is my attempt. Please take a look at it and tell me if there are any flaws.

Let us take an infinitesimal piece of cord, long $\mathrm ds$, resting on a plane inclined at an angle $\alpha$ to the horizontal. The weight of the elements is $\lambda g \, \mathrm ds$, where $\lambda$ is the linear density of the cord. The component parallel to the plane (directed downwards) is $\lambda g \sin \alpha \, \mathrm ds$. The normal reaction is $\lambda g \cos \alpha \, \mathrm ds$, so the maximum static friction available (directed upwards) is $\mu \, \lambda g \cos \alpha \, \mathrm ds$. Writing the axial equilibrium (positive upward direction) according to Newton's $2^{\text{nd}}$ Law of motion: $$T(s+ \mathrm ds) - T(s) = \lambda g \, \mathrm ds \left(\mu \cos \alpha- \sin \theta \alpha \right), \tag{1}$$ and taking the limit $\mathrm ds \to 0$ on the LHS, we have $T(s+ \mathrm ds) - T(s) \to \mathrm dT$; so, $(1)$ becomes $$\mathrm dT = \lambda g \, \mathrm ds \left(\mu \cos \alpha- \sin \alpha \right). \tag{2}$$ Integrating $(2)$ from the point where the rope detaches from the slope $(T= T_\alpha, \, s = s_\alpha)$ to the top end of the rope $(T_{\text{top}}=0, \, s=0)$ gives $$T_\alpha = \lambda g \, s_\alpha \left(\mu \cos \alpha- \sin \alpha \right). \tag{3}$$

Let the length of the hanging section on that side be $r_\alpha$. So, its weight is $\lambda g \, r_\alpha$.
This way, the total length $L_{\text{tot, left}}$ of the cord under consideration is equal to the length of the cord from its left top end to the point where the string detaches from the left slope $(s_\alpha)$, plus the length of the hanging section on the left side $(r_\alpha)$.
Thus, $L_{\text{tot, left}} = s_\alpha + r_\alpha$.

The weight of the left hanging section of the rope, $\lambda g \, r_\alpha$, must be balanced by the vertical component, $T_\alpha \sin \alpha$, of the tension at the point where it joins the part of the cord touching the left incline. We hence have $T_\alpha \sin \alpha = \lambda g \, r_\alpha$ or, equivalently, $$T_\alpha = \frac{\lambda g \, r_\alpha}{\sin \alpha}. \tag{4}$$

Equating the two expressions $(3)$ and $(4)$ for $T_\alpha$:
$$\begin{aligned}
\lambda g \, s_\alpha \left(\mu \cos \alpha - \sin \alpha \right) &= \frac{\lambda g \, r_\alpha}{\sin \alpha} \\
\cancel{\lambda g} \, s_\alpha \left(\mu \cos \alpha- \sin \alpha \right) &= \frac{\cancel{\lambda g} \, r_\alpha}{\sin \alpha} \\
s_\alpha \left(\mu \cos \alpha - \sin \alpha \right) &= \frac{r_\alpha}{\sin \alpha} \\
s_\alpha \sin \alpha \left(\mu \cos \alpha - \sin \alpha \right) &= r_\alpha,
\end{aligned}$$
which gives $$r_\alpha = s_\alpha \sin \alpha \left(\mu \cos \alpha - \sin \alpha \right). \tag{5}$$

The suspended fraction $f$ we are trying to maximize is $f = \dfrac{r_\alpha}{L_{\text{tot, left}}}$. So:
$$\begin{aligned}
f &= \frac{r_\alpha}{L_{\text{tot, left}}} \\
&= \frac{r_\alpha}{s_\alpha + r_\alpha} \\
&= \frac{s_\alpha \sin \alpha \left(\mu \cos \alpha - \sin \alpha \right)}{s_\alpha + s_\alpha \sin \alpha \left(\mu \cos \alpha - \sin \alpha \right)} \\
&= \frac{s_\alpha \sin \alpha \left(\mu \cos \alpha - \sin \alpha \right)}{s_\alpha \left(1+ \sin \alpha \left(\mu \cos \alpha - \sin \alpha \right)\right)} \\
&= \frac{\cancel{s_\alpha} \sin \alpha \left(\mu \cos \alpha - \sin \alpha \right)}{\cancel{s_\alpha} \left(1+ \sin \alpha \left(\mu \cos \alpha - \sin \alpha \right)\right)} \\
&= \frac{\sin \alpha \left(\mu \cos \alpha - \sin \alpha \right)}{1+ \sin \alpha \left(\mu \cos \alpha - \sin \alpha \right)},
\end{aligned}$$
thus $f=\dfrac{\sin \alpha \left(\mu \cos \alpha - \sin \alpha \right)}{1+ \sin \alpha \left(\mu \cos \alpha - \sin \alpha \right)}$.

Let $F(\alpha) = \sin \alpha \left(\mu \cos \alpha - \sin \alpha \right)$. Then: $$f = \frac{F(\alpha)}{1+ F(\alpha)} \qquad \text{where} \quad F(\alpha) = \sin \alpha \left(\mu \cos \alpha - \sin \alpha \right).$$
This expression for $f$ clearly is a monotonically increasing function of $F(\alpha)$. And since the suspended fraction increases as $F(\alpha)$ increases, it suffices to maximise $F(\alpha)$ in order to maximise $f$.
Differentiating $F(\alpha)$, we have $F'(\alpha) = \mu \cos(2 \alpha) - \sin(2 \alpha)$. It is zero when $\mu = \tan (2 \alpha_{\max})$. Then: $$\boxed{\alpha_{\max} = \frac{\arctan \mu}{2}},$$ as desired.


Analogous reasoning leads to $\beta_{\max} = \dfrac{\arctan \mu}{2}$, so $\alpha_{\max} = \beta_{\max}$.

All looks ok to me.