How Do You Find The Shape of Hanging Rope? Classic Physics Problem

Рет қаралды 43,020

2 жыл бұрын

A rope that's hung up between two points forms a shape called a catenary. I'll show you how to derive it from start to finish. Get the notes for free here: courses.physicswithelliot.com...
When you hang up a rope between two points under the weight of gravity, it makes a distinctive arc-like shape that anyone can instinctively recognize. In this video, I'll show you how to actually derive the shape---and find the mathematical function that describes it---using F = ma and a little calculus.
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About the classic physics problem series:
In these intro-to-intermediate-level physics videos, I'll discuss classic physics "challenge" problems that you might meet in your introductory mechanics and electromagnetism classes. They might be based on simple concepts, but these problems can still get pretty tough!
About me:
I’m Dr. Elliot Schneider. I love physics, and I want to help others learn (and learn to love) physics, too. Whether you’re a beginner just starting out with your physics studies, a more advanced student, or a lifelong learner, I hope you’ll find resources here that enable you to deepen your understanding of the laws of nature. For more cool physics stuff, visit me at www.physicswithelliot.com.

Пікірлер: 120

@alphalunamare 2 жыл бұрын

Many years ago in my first job before going to college I was asked to work out how much concrete a bridge needed. The shape was a catenary so I figured I'd work out the length. It took me 2 days and when I gave the answer, the structural engineer put a piece of string along the drawing and said Yes you're right. I felt so dumb. I had used integration with substitution a few times and there were square roots everywhere but the answer was correct. Then a graduate asked me why I hadn't used hyperbolic functions? What are they I asked? So anyway I looked them up and just couldn't believe how easy they made everything. I quit my job and went to college :-)

@PhysicswithElliot 2 жыл бұрын

Ha! That's one way to do it!

@AkamiChannel 6 ай бұрын

I saw your comment on another video about hyperbolic trig! Tell us more of your story! How did college go, what did you study and what did you do after? Do you have any regrets?

@Salien1999 2 жыл бұрын

Excellent. I'm especially happy that you didn't just get to the integral part and say "if we look at an integration table, we find that this is the integral for cosh." That always seemed like a non-explanation for me.

@fitofight8540 2 жыл бұрын

Coz it is

@jacobharris3002 6 ай бұрын

True but most of the time it's an integral you've encountered in one of your calculus classes before, so you should be able to work it out yourself if you want to. I do get annoyed when the answer is a special function or something I'm unfamilar with though.

@dukenukem9770 2 жыл бұрын

What a fun little problem! I can wait to do it with my son. Somehow, I made it all the way through a Ph.D. in physics without seeing this problem. Thanks for posting!

@PhysicswithElliot 2 жыл бұрын

Ha! Good thing it wasn't on your qualifying exam!!

@ozzymandius666 2 жыл бұрын

@@PhysicswithElliot This is an approximation in the limit of long and heavy and floppy ropes. Ropes need a bit of force to bend. I'd be interested how the stiffness of the rope could be taken into account, as in real life, the force per degree of deflection over an infinitesimal length tends towards infinity, but in the approximation, it tends towards zero. No doubt the complexity of a solution would be intractable, as if the rope is stiff enough, you could make any curve of length l that fits between the endpoints, which would also happen if g was set to zero.

@utuberaj60 Жыл бұрын

You are really honest, though you are a doctor of Physics. Appreciate that and also your zeal. Btw, though I am an engineer, we never did this basic derivation stuff- which is really more of calculus skills, in our course, and just learnt (and forgotten) that the hanging chain/rope is called a catenary !

@aditsaini5094 Жыл бұрын

This problem came in my tests. Well a junior version of this. I couldnt do it

@dukenukem9770 Жыл бұрын

@@aditsaini5094 Deriving this result without any guidance would probably be pretty tough. Following along with a well-produced tutorial like this one is undoubtedly much easier than doing it on a test! Don’t feel badly about it…

@kingbeauregard 2 жыл бұрын

I really like the moment when you say (I paraphrase), "well, that's all the physics, now it's just math". That's a good way to look at these problems. I seem to recall there's a way to derive the catenary by using the Lagrangian, though if you do it wrong, the result you get is that the rope just falls to the ground.

@PhysicswithElliot 2 жыл бұрын

Yep you can certainly do it with the Lagrangian!

@aswathik4709 2 жыл бұрын

@@PhysicswithElliot i tried using lagrangian. but after integrating i got a Logarithmic equation, which isn't right

@Salien1999 2 жыл бұрын

@@aswathik4709 I've seen a video on it and apparently you have to use a "Lagrange multiplier" to get it done.

@nerdsgalore5223 2 жыл бұрын

@@Salien1999 I don't know if this counts as the "Lagrangian" method that you're all referring to, but I considered the gravitational potential energy of a dm and integrate over the whole curve to get the total gpe, then used the Euler Lagrange equation to minimize that and got a catenary.

@utuberaj60 Жыл бұрын

Mr. Elliot- you made this derivation look so simple. It was really interesting and from 1st principles. I appreciate your videos on ALL Physics topics you have made, some are quite advanced. Why don't you make a video of the derivation of the famous curves of the 17th century - the Brachistochrone and the Tautochrone- that took the brilliance of the Bernoullis and Newton and Leinlbniz. The Tautochrone was discovered by another genius Huygens.

@moonwatcher2001 11 ай бұрын

Really nice!!! Calm talking pace, clear and interesting explanation, rigorous but entertaining stuff. Gorgeous video, mate. Thanks

@unknown546b 5 күн бұрын

wow what a solution man , thanks a lot, love from india

@adrianf.5847 Ай бұрын

To make the change of variables u = iv rigorous, I think one has to consider contour integrals. As an alternative, I recently solved a similar integral (int sqrt(x^2 + 1)) using the identity sinh^2 + 1 = cosh^2. Please note that T(x) should point "in the same direction as" T(x + dx). In total, you managed with only one mistake, which makes this frankly the best physics video I ever watched, including many by distinguished professors.

@johnchristian5027 Жыл бұрын

This is a great video. Amazing how something so simple as a hanging rope has such deep physics and math.

@hrperformance 2 жыл бұрын

This is the first time I've seen this. Great explanation. I had to pause and think a few times but I was able to follow pretty well. Super enjoyable! Many thanks for posting 😁

@PhysicswithElliot 2 жыл бұрын

Great to hear!

@ccdavis94303 2 жыл бұрын

3 in a row, all very well done. Top ranked reference/learning resource. subbed

@PhysicswithElliot 2 жыл бұрын

Thanks Charlie!

@SuhasbharadwajChintada-kv8so 9 ай бұрын

As usual nailed it 👍 thank you

@omrinitzan2312 Жыл бұрын

Great explanation, thanks a lot!

@dav0625 2 жыл бұрын

What brilliant explanations, thanks.

@brendanfan3245 2 жыл бұрын

amazing explaination!

@sphakamisozondi 2 жыл бұрын

Im glad i came across your channel

@mohammedal-haddad2652 2 жыл бұрын

I read this proof in a book and it was very hard to follow, you make it so easy and fun. Thank you very much.

@PhysicswithElliot 2 жыл бұрын

Appreciate it Mohammed!

@darkol93king34 2 жыл бұрын

Great video as usual. I learn , again 🤗, something new and that makes me very happy

@PhysicswithElliot 2 жыл бұрын

Glad to hear it JH!

@arnoldkotlyarevsky383 2 жыл бұрын

This problem has bothered me since undergrad. I went to gradschool for physics and while I got better at math and answering test questions this never really connected for me. Im now in my thirties, and I have periodically tried to revisit this to see if I could make it feel more intuitive. Thank you for putting this out.

@PhysicswithElliot 2 жыл бұрын

Glad it helped, Arnold!

@agrajyadav2951 2 ай бұрын

Thanks a lot, sir!

@mohammedpatel3051 Жыл бұрын

Excellent lesson well explained

@TIO540S1 2 жыл бұрын

Excellent. It’s been a very long time since did this problem. I’d have had a very difficult time working through it myself.

@prostatecancergaming9531 Жыл бұрын

Im not joking when I tell you I’ve lost hours of sleep contemplating this EXACT problem in bed. I never knew it was such a famous problem, seems reasonable considering the elegant solution. Thanks!!!

@jimgarrett7209 2 жыл бұрын

One of my favorite centenary are power lines across a canyon. They are marked on aeronautical charts. Why? You can see them from the air (so you can know where you are) and more importantly so you don't run into them.

@alexkong93 2 жыл бұрын

You are amazing :)

@michaelmcgee335 3 ай бұрын

A little above my head, gleaned a little, good video.

@Nekuzir 2 жыл бұрын

Hey I was right, its the hyperbolic trig stuff. I haven't done the actual maths with it but its nice being able to recognize it when it shows up.

@chandlerketelsen8702 2 жыл бұрын

Fantastic! I am taking modern physics, thermodynamics, and a few other courses right now for my physics major at Washington State University. Seeing these videos makes me enthusiastic to put effort into my classes which will help me with my degree and beyond. I appreciate this!

@PhysicswithElliot 2 жыл бұрын

Glad you liked it!

@juliencasel6024 4 ай бұрын

I got asked this on an oral interrogation back when I was an undergrad student. I knew at the time about the "trick" of applying Newton's law to a small piece of the rope, but I was still completely helpless... 25 years later I finally grabbed a piece of paper and tried to figure it out... and it was way harder than I thought. I actually spent an shameful amount of time solving the differential equation, actually I don't think I resolved it directly but I used some tricks to simplify it. In the end I got the cosh but it was reaaally painful.

@samarthtandale9121 Жыл бұрын

Beautiful 🤠

@LatainHater 2 жыл бұрын

Amazing Video! Thank you

@PhilFogle 2 жыл бұрын

Excellent delivery, I had to pause a few times, but very clear. I have to create a video like this to teach power factor correction to electrical apprentices. Can I ask what software you use?

@PhysicswithElliot 2 жыл бұрын

Procreate!

@alo1236546 2 жыл бұрын

Can u make a video about elastica curve

@Amine-gz7gq 11 ай бұрын

Nice, e is everywhere !

@MostlyWrong-td6cx Жыл бұрын

Made it all the way through this, I'm really impressed because I thought you had to go into variational calculus to do this.

@aswathik4709 2 жыл бұрын

can you pl. show this using lagrange's eqn

@informalchipmunk5775 8 ай бұрын

7:57 alternatively, we could substitute: u = tanθ => du = sec²θ•dθ and also, 1 + u² = 1 + tan²θ = sec²θ hence we have: ∫(sec²θ/√(sec²θ)•dθ = ∫secθ•dθ Now some of you may already know the result to that integral, if not then we can multiply and divide by cosθ to get: ∫(cosθ/(1 - sin²θ))•dθ Now let sinθ = v => cosθ•dθ = dv Now we have ∫(1/(1 - v²))•dv Which is a pretty simple partial fraction integral -which I'm too lazy to type- which is left as a challenge to the reader. We finally arrive at ln(|secθ + tanθ|) + c Now just sub θ = tan⁻¹u And as I finish writing this I've realised that your method is far superior 😂.

@kub8675 Ай бұрын

Absolutely wonderful video! Can anyone explain why the tension forces at 4:20 have to be tangential to the curve? I don’t understand completely.

@markkennedy9767 4 ай бұрын

Your explanations are very clear. Would those three equations help me understand why when a rope (of uniform mass per unit length) hangs between two posts of the same height that the midpoint of that rope will be the lowest point and will be horizontal. I know this is "obvious" physically but would those three equations show it for certain?

@jamesraymond1158 8 ай бұрын

Excellent. I had forgotten how difficult this problem is. What if the end points are 1000 miles apart, so that direction of gravity changes along the length of the rope?

@billmangamer1199 10 ай бұрын

You could also do u=sinh(h)

@qcard76 25 күн бұрын

I followed the logic up to defining the horizontal tension components at x and x+dx as constant: maybe I missed something, but the intuition behind that part of the tension explanation seemed sort of hand wavy, at least compared to the treatment of the vertical components.

@DaleCapewell 9 ай бұрын

This is absolutely incredible. What if the rope is thicker and not so bendy, like a pool noodle. Does this model account for that?

@ssym2 2 жыл бұрын

Thanks Dr. Elliot! Would you give thought to doing subject by subject videos according to college level physics course?

@PhysicswithElliot 2 жыл бұрын

I'm working on creating dedicated courses, make sure you're subscribed to my email list to hear when they're available!

@ssym2 2 жыл бұрын

@@PhysicswithElliot Great! Thanks!

@Darthvanger 11 ай бұрын

huh, so much trickery :)

@CosineKitty Жыл бұрын

This was a great explanation and I found it very helpful. The only part that bothered me was saying that sqrt(cos(x)^2) = cos(x). Shouldn't it be |cos(x)| -- the absolute value? Since the cosine can be negative, you can't just cancel out cos(x) / |cos(x)|... you get +1 or -1 depending on the value of x. I'm sure I'm missing something...

@galus_anonimus Жыл бұрын

i have the same problem, maybe theta is defined to be within such interval that cos(theta) ≥ 0, but i can't figure it out.

@droxyklo Жыл бұрын

In case you're still wondering, v goes from -1 to 1 (otherwise 1 - v² would be negative and the square root would be undefined) so sin(theta) also goes from -1 to 1, meaning that theta varies from -pi/2 to pi/2, interval on which the cosine function is positive, so we're good

@galus_anonimus Жыл бұрын

@@droxyklo well, i don't think that's really the case here, the only value of v for which the intergal would be undefined is 1 because there would be 0 in the denominator. However when v < 1 it's just a complex number and I don't think there would be any reason to assume that that every part of the expression should be only real. What's more, the next expression (in notes) usses imaginary numbers to get to this integral that further proves my point. Nevertheless, thank for your reply.

@galus_anonimus Жыл бұрын

I have reached out to my mathematics teacher about solving such intergal, and from what I understood solving it that way isn't really the way to go here. Instead you should just consider a function y = arcsin(x) ( or sin^-1(x) in different notation), then find it's derivative by using the derivative of an inverse function theorem (my own translation from polish "twierdzenie o pochodnej funckji odwrotnej" so it may be known under a bit different name), which would be 1/√(1-x²) which is our integrant. Therefore integral of 1/√(1-x²) dx = arcsin(x) + C I hope it helps

@CosineKitty Жыл бұрын

@@droxyklo This explanation doesn't work because v is an imaginary number, not a real one; he defines v = u/i, and u is the slope of the catenary, which can have any real value from negative infinity to positive infinity. Not only is it begging the question to disallow a negative radicand, v can't have a real value, so we certainly can't constrain it to a real range range like [-1,+1].

@ervishal21 2 жыл бұрын

Plz tell me the best books on mechanics

@multienergy3684 6 ай бұрын

Is there a way to solve this problem assuming that the gravitational acceleration g changes when the height changes? Say, postulate g(y)=g(y(x))?

@929Luc 2 жыл бұрын

Couldn't you solve that integral by making use of the identity: sec²(x)-tan²(x)=1? Great video, did one simplified exercise where I had to solve that exact differential on my linear differential equation class this semester, I don't remember much of the process tough, and I did not know the physics behind it till I saw this video.

@PhysicswithElliot 2 жыл бұрын

The most straightforward way would be to substitute u = \sinh(\theta) and then use the fact that \cosh^2(\theta) - \sinh^2(\theta) = 1, but I didn't want to assume folks were already familiar with hyperbolic functions

@Anistuffs 2 жыл бұрын

@@PhysicswithElliot Isn't the most straightforward way is to substitute u=tan(theta) since you're already assuming the viewers are familiar with trigonometric functions?

@declanwk1 2 жыл бұрын

i thought that historically, solving the problem of the catenary led to the development of the calculus of variations. But Physics of Elliot has shown here that the problem can be solved using less sophisticated math, without using the calculus of variations

@Xayuap 2 жыл бұрын

I always wanted to learn the catenary 'cause I miss this lecture, I always though it gonna be prettier

@stuartl7761 2 жыл бұрын

Interesting stuff. I learnt that sinh was pronounced shine. I suppose yours is a little closer to it's spelling.

@suzimurphy1904 6 ай бұрын

Why does the tension vector of the rope HAVE to point tangentially to the mass element?

@penguinshin 5 ай бұрын

Why does ty/tx need to be equal to dy/dx?

@Sonu-hv4ef Жыл бұрын

🔥

@keypo790 Жыл бұрын

9:50 When the dopamine just hit right...🤯

@Snoopies622 2 жыл бұрын

I'm wondering if this same problem can be solved by instead assuming that the total potential energy of the rope is minimized. Perhaps using the Euler Lagrange equation?

@PhysicswithElliot 2 жыл бұрын

Sure can!

@erichaag5229 2 жыл бұрын

Let's see a Bessel function problem.

@silversisask2328 2 жыл бұрын

Would anything change if the rope is made of rubber and stretches significantly under tension?

@PhysicswithElliot 2 жыл бұрын

Probably; for one thing the length wouldn't be fixed anymore and the mass density could vary along the rope

@antormosabbir4750 2 жыл бұрын

I had been trying this for 2 years! No one answered this

@user-uc8nz8io2k 2 ай бұрын

E=mc^2 Schwarzschild radius

@anilkumarsharma8901 9 ай бұрын

What actually change?x or something else???

@gytoser801 2 жыл бұрын

a * cosh(x/a)

@lumpi806 2 жыл бұрын

Two things are bothering me : 1) In Physics : I don’t understand why T(x) is toward the left, and T(x+dx) is toward the right. Your scheme means the tension T is NOT continuous. There are some confusions here. It is NOT T(x) but -T(x) which is really represented in your scheme. 2) In Maths : when you have u’ = k square root of (1 + u²), you just square that : u’² = k² ( 1 + u²), then you derive that : 2 u’ u’’ = 2 k² u u’, which leads to u’’ = k² u , which an easy LINEAR differential equation.

@PhysicswithElliot 2 жыл бұрын

There are two tension forces, T_R(x+dx) acting on the right and T_L(x) on the left. If v denotes the unit tangent vector to the curve, you can write them as T_R(x+dx) = T(x+dx) v and T_L(x) = -T(x) v. The x component of the tension is T(x+dx) times the x component of v to the right and -T(x) times the x component of v to the left. Yes that's a nice way of solving the differential equation! The general solution to u'' = k^2 u is a superposition of e^{kx} and e^{-kx}, or equivalently A \sinh(kx + B). But since you manipulated the original equation to get there you need to plug this guess back in to the first equation to check that it's actually a solution, which will require A = 1.

@edweinb 2 жыл бұрын

Seems like horizontal T would be a function of slope so not constant. Guess I need a little more explanation.

@PhysicswithElliot 2 жыл бұрын

The tension is the only force with a horizontal component, so it has to be constant for the force pulling to the right on one side to cancel the force pulling to the left on the other. Otherwise there would be a net horizontal force and the rope wouldn't be at rest

@edweinb 2 жыл бұрын

@@PhysicswithElliot Thanks. Got it.

@eamon_concannon 2 жыл бұрын

@@PhysicswithElliot I think it's also worth considering that at (x,y), the left end of each element, T_x, the x component of T(x), always has the same sign (negative) and by considering neighbouring elements using newton's 3rd law, T_x = c can be thought of as a constant vector at left end of every element. This means we don't have to worry about sign of c changing at different points of rope. Also, T_y =c dy/dx at every left point as tension vector is parallel (or anti-parallel) to element and when T_y is positive(negative), then dy/dx is negative(positive).

@fictionisfake4280 2 жыл бұрын

9:00 "The imaginary factor doesn't cancel, and we have shown that ropes don't exist" ~*The End*~

@chair547 2 жыл бұрын

I think that introducing complex numbers that don't represent anything physical and then letting them disappear it's just kind of a gross way to solve a physics integral. Is there any way to do this without complex numbers?

@PhysicswithElliot 2 жыл бұрын

Yes you can just make a substitution like u = sinh(\theta) to evaluate the integral without any i's. But I didn't want to assume that people had learned about cosh and sinh before so I did it using sines and cosines.

@Xayuap 2 жыл бұрын

Oh, who would think that female sinuses were in fact catenoids

@anilkumarsharma8901 Жыл бұрын

How much it's relationship with parabola and relative strength💪 of material👠💍👜💵 so we become a new better solution👌 for whole universe🌌🌌🌌🌌 As the 🕉️ universe🌌 itself curve 〰〰 likes an elliptical 👣path👣

@grantofat6438 Жыл бұрын

How I would find the shape of a hanging rope? By hanging a rope of course.

@DeepLyricist 2 жыл бұрын

Is there a way for me to do the calculation without using "i"? I did my own research and I don't believe "i" exists and won't be using it in my calculations..

@PhysicswithElliot 2 жыл бұрын

😂

@LatainHater 2 жыл бұрын

Lmao

@soulsilencer1864 2 жыл бұрын

I did my own research and I don't believe i exists. -Sun Tzu, Art of War

@alphalunamare 2 жыл бұрын

Yes you can. Just use the usual trigonometric substitutions and some ingenuity. It is ugly but doable. I did it in 1971 as a teenager. 'i' just makes life a whole lot easier. Real Analysis is fractured pottery compared to the Beautiful Vase of Complex Analysis. This was 50 years ago so obviously the details are a bit foggy but using multiple stages of the standard Integration by Parts utilising sin. tan & cos substitutions led to an answer ridden with Square Roots , so ugly but true. :-)

@alphalunamare 2 жыл бұрын

As for whether 'i' exists or not then that is a quite ridiculous and unnecessary question to ask. 'i' is just a means to an end that makes life far easier than it might otherwise be. You can define 'i' without even mentioning '-1' and having existential doubts. For example, in C++ merely define a class called Complex with elements of the form(a,b) where a and b are Real Numbers. Overload '+' so that (a,b) + (c,d) = (a+c,b+d) and Overload '*' by (a,b)*(c,d) = (ac-bd, ad+bc). ('/' is left as an exercise :-) ) You will then have a class mathematically equivalent to Complex Numbers without ever having mentioned 'i'. As you will see, 'i' is just a tool. (hint: multiply top and bottom by the conjugate of the denominator)