Thanks for your words and your feedback. I have been asked this question a few times, so I'll pin this comment to the top. Eqn 23 is correct as written. What I did in a single step (without adequate explanation) is: 1. subtracted λ/c1 from both sides, 2. divided both sides by mg, and 3. absorbed the mg into the λ (which is okay because λ is simply a constant - I could have called it a new constant λ' or λ2 or something else, but I just kept it as λ).

@@Freeball99 this is incorrect because when after you subtract the lambda/c1 you later have to multiply both sides by c1 anyway in order to isolate the y, which cancels out the c1, so absorbing that into a new arbitrary constant is not valid (because it’s not really arbitrary). Then dividing both sides by mg yields lambda/mg, as the original commenter correctly stated

@@dababyonfire6868 Yes, the λ should not be divided by c1. The mg, however, can be absorbed into the λ without a problem.

@@Freeball99 oh I see what you’re saying. I forgot that lambda itself was an arbitrary constant. Thank you!

Sir, THANK YOU! You literally saved my worksheet I have to explain on friday

I absorbed the mg into the λ since λ is just a constant (I should have mentioned that). I have pinned a comment to the top of the discussion section which explains this in more detail.

Most welcome!

Glad you enjoy it!

Interestingly, the shape of the catenary minimizes the tensile load in the chain. As a result, arches that are built in the shape of a catenary are the strongest as they minimize the compressive loads.

@Amit Correct. The parameter mg/c_1 sets a common scale for horizontal and vertical distances. In units of mg/c_1, the shape will follow the cosh function; see Equation 23.

@@Freeball99 See, e.g., Sagrada Famiglia en.wikipedia.org/wiki/Sagrada_Família

You have 3 equations and 3 unknowns so this is solvable though it will likely require a numerical solver in all but perhaps the simplest cases. I have shown an example in a different video of how to treat this for a pendulum problem where I substituted the constraint equation into the other two equations of motion. kzfaq.info/get/bejne/rbabZJl4nNPPio0.html

This really requires a mathematical derivation of the method to explain it. The basic idea is that by adding λG in this form (where G=0) does not affect the necessary condition for finding an extremal of the modified functional. Unfortunately I just don't have the ability to explain it using text only, so will have to save it for a video.

The shape of the two are the same (ie the cosh function). In our case, we need to fix the cosh function curve to the starting and ending points PLUS we need to generalize it for chains of different length. This is why we end up with the constants c1, c2 and λ.

There are numerous root-finding algorithms available including the two that you mentioned.

Brilliant explanation, I agree. However, I think it’s important to note that the solution is independent of mass and energy. This can be easily implemented by replacing lambda with m*g*lambda_2 in Formula (4) at minute 6:10. Then, you can divide by m*g and it disappears, since the other side equals zero. Am I right? 🙂

​@@breitbandfunker4332 I'm pretty sure that works out the same. You can see the final solution in 21:53, if you replace λ with mg λ_2, then c_1 will always be divided by mg. Substituting c_3 = c_1/mg, you will not see mg in the final solution. Although, you are right, explicit independence relations are always good in physics.

@@Zxymr well, I think if you absorb mg in the beginning, it vanishes completely in λ_2. Then you can determine the constants only with starting and endpoint. It should be completely independent from mass and g-factor, which means, any chain has the same shape - even on alien planets!

Originally a Durban accent, now with a California influence... Interesting question. I'll need to give it some more thought, but fundamentally, the discrete mass produces a discontinuity and would thus produce a boundary condition. Thus we would have to treat the rope on each side of the mass separately (if it is symmetric, we could just solve for one half). Also, it is not immediately clear to me whether this problem would have an analytical solution, but certainly we could set up the governing equations. It's a difficult problem to explain in these comments; really requires a video. I had planned to make a video demonstrating the case of a beam with a discrete mass. This will be fairly similar to the problem you have posed in that it was demonstrate how to treat discrete masses - after all, a rope is just a beam without any flexural rigidity (EI = 0). Perhaps I'll have to promote to the top of my To Do List.

"elastic catenary" might be thought of a chain where each link is a spring with a finite spring rate. A realistic problem in practical engineering, e.g., suspension bridge.

The app is called "Paper" by WeTransfer. Running on an iPad Pro 13 inch and using an Apple pencil. Video is edited on my Mac using iMovie.

@@Freeball99 Very detailed, thank you!

Since lambda is just a constant, I have absorbed the mg into lambda since a constant divided by a constant is still a constant. What I have done in a single step is to define a new constant, say, lambda2 = lambda / mg and then recognized that lambda2 is a dummy variable so I could just call it lambda.

@@Freeball99 Thank you for the reply, however I still don't understand why the "new lambda" is shown being divided by C1 in equation 23.

@@Freeball99 It is inaccurate to have it written as you have it in the video though since mg is not equal to C1.

Since lambda is just a constant, I absorbed the mg (which is also a constant) into the lambda. It's a different lambda from the line before (I didn't want to call it something different) but it's just a constant all the same and will come from solving the given conditions.

@@Freeball99 Ah, fair enough. you have lambda/C1 though, wouldn't it be more efficient to absorb that too? or is this a sort of canonical form and these specific constants have meaning behind this equation? And wouldn't it be better to specify C1/mg as another constant, then it takes the form y=a*cosh((x+b)/a)-c, which imo looks much cleaner?

Yes, what you've written is exactly correct.

Not exactly sure what you have in mind. Can you give an example?

Do you mean using a multiplier on a conserved quantity found through a symmetry?

This is a very good question and really requires a video deriving each method to get a proper understanding of why this is. I haven't made the video yet because it will be very mathematical, dry and likely not very popular. In short, however, these two problems you mentioned are not the same type of problem due to the different nature of the constraints. This problem (the hanging chain problem) is known as a CONSTRAINED OPTIMIZATION problem because the constraint is written as an integral and serves as an overall constraint of the system. On the other hand, the pendulum problem is known as a CONSTRAINED COORDINATES problem since the constraints involve the coordinates themselves (could also include derivatives) and apply to every point within the domain, rather than being satisfied in an integral sense like for the hanging chain. While both problems are solved using the Lagrange multiplier method, the reason that we do not subtract L from the constraint in this video is due to the difference in the types of constraints. So the methods used in each of these two videos are correct, i.e. in the case of an integral constraints, we DO NOT write G in the form G=0 while in the pendulum problem, we DO write G in the form G=0. Hope this makes sense.

Thank you for the comprehensive reply! 🙏 Your solid understanding of the topic is truly motivating. Much appreciated! 😊@@Freeball99

Not sure what this is called. If the chain is massless, then perhaps a mass on a string. If the chain has mass, I'd probably call it a compound pendulum

@@Freeball99 yes its probablx a compound pendulum with infinitely small pendula with the weight cancelling out any (?) chaos.

What I did in a single step here was: 1. Multiply each of the terms by mg/c1 2. Recognize that the 2nd term is mg/c1 * c2, which is just a constant so I absorbed it into c2 - I probably should have made this more clear in the video. I could have instead named it c3, but I didn’t want to start adding new unknowns so it was easiest just to include it as part of c2. This is a commonly used technique when multiplying some unknown constant by additional constants and I use it in some of my other videos (I might have explained it in one of those).

@@Freeball99 I see, I did think about that but didn't know it was something which happend more often, thanks!

Yes, this is correct. I probably should have mentioned it in the video.

No other currently. But probably my next video will contain Lagrange multipliers.

@@Freeball99 I will be waiting for it 😌

See the comment pinned to the top of this section.

The c1 should not be there. The mg, however, I absorbed into the λ which is fine because λ is just a constant.

I explained immediately after that m = ρA is the mass per unit length. So the equation for I is correct as written.

Sorry about that, I misheard, only paid attention to formulas and the usual meaning of notations + used to that mistake in some old calculus classes

How many meters of wire are needed for a 1000m electrical transmission line, given pole height, spacing and ground clearance?

Catenaries appear in many different applications. Clearly in the design of suspension bridges and in designing power lines. They are used in architecture when designing arches - arches designed in the shape of an (inverted) catenary are the strongest arches you can build (since the catenary shape minimizes the compressive load). Catenaries also used in ship building for optimizing the design of anchor chains.

​@@Freeball99compressive loads (unless stability related) are not a problem in archs though. The real reason why a catenary shape for load carrying beam structures is desirable, is because we only want the structure to be loaded by membrane forces and not by bending moments. This is because membrane stresses are distributed uniformly over the cross section and therefore make use of "all the material" to carry the load.

Correct m = ρA is the mass per unit length.

I have described it in the video above. It is the form that the Euler-Lagrange equation reduces to if the functional, F, is not explicitly dependent on x. The mathematics behind deriving the Euler-Lagrange equation are described in the following video (kzfaq.info/get/bejne/jKl4eaaJzL_Ipo0.html).

Lambda is just a constant, so I absorbed the mg into this. I really should have called it lambda_prime but I thought that would be confusing.

@@Freeball99 ok the mg absorption makes sense now thanks, but since there is a c1 multiplying the cosh and nothing dividing y, i really don't see why there still is a c1 dividing lambda

Sorry for the delayed response, but I somehow missed this until now...in this case you have added an extra unknown (the length of the chain) so we need an extra constraint equation. This comes from using the relationship between the length of the chain and the height, y (this comes from the shape of the catenary which you already have an equation for).

why do you need to justify that

s(b) and s(a) are two different locations/values for the path. The difference between the two is literally the definition of the path length, L. Similarly, x(b) - x(b) would be the distance along the path in the x-direction.

Do you have suggestion where I can learn more about the definitions is it curvilinear coordinate system? Its confusing . Thanks for everything

Now I understand. But how is h=yds. It is the potential energy but having a chain length multiplied.

In this problem, m = ρA is the linear density and not the mass of the chain. I mention this at 2:05 in the video. So, I am integrating ρAgy with respect to x. At the 10:17 mark, the "F" that I am using does not stand for "force", but rather it is the "functional" - that's why the F. In this problem, the functional has units of energy per unit length, but in general could have various other units depending on the quantity that we are trying to minimize.

realest comment

This video takes only gravity and the (uniform) mass of the chain into consideration. Any effects caused by the movement of earth are ignored.

We cannot grasp anything on It's fullest, we only haver models, and preferably useful ones. So ignoring a lot of negligible influences makes the solution easier - sometimes, it makes them computationally or even analitically viable...

Yes; I understand, the work is done in portions or puzzles. The bigger picture being too complex for the human mind. But, if we never work on the big picture; we will never find the solutions. The ignored portions considered nuisance become missing links.

@@biancabonet You are right, and sometimes the details make all the difference. This is very evident in dynamical systems, for example: the tiniest detail can make a huge difference over time. In the example of the video, however, it could probably be shown that the effects you cited have a very negligible and controlled effect overall, so that the solution is still imensely similar to the one this model already gives.

@@raphaelreichmannrolim25 Thank you. I tend to focus on purity and accuracy and conservation of the big picture. While focusing on retainment of the CMB, I guess you may zoom in...but zoom in to see plank sizes...you would still have to zoom out and see the macro-universe to begin to understand--to see the big picture..