How Do You Find The Shape of Hanging Rope? Classic Physics Problem
Vložit
- čas přidán 16. 09. 2021
- 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.
Get the animation: www.physicswithelliot.com/han...
Online tutoring inquiries: www.physicswithelliot.com/tut...
If you find the content I’m creating valuable and would like to help make it possible for me to continue sharing more, please consider supporting me! You can make a recurring contribution at / physicswithelliot , or make a one time contribution at www.physicswithelliot.com/sup.... Thank you so much!
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. - Věda a technologie
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 :-)
Ha! That's one way to do it!
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?
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.
Coz it is
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.
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!
Ha! Good thing it wasn't on your qualifying exam!!
@@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.
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 !
This problem came in my tests. Well a junior version of this. I couldnt do it
@@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…
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.
Yep you can certainly do it with the Lagrangian!
@@PhysicswithElliot i tried using lagrangian. but after integrating i got a Logarithmic equation, which isn't right
@@aswathik4709 I've seen a video on it and apparently you have to use a "Lagrange multiplier" to get it done.
@@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.
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.
Really nice!!! Calm talking pace, clear and interesting explanation, rigorous but entertaining stuff. Gorgeous video, mate. Thanks
This is a great video. Amazing how something so simple as a hanging rope has such deep physics and math.
As usual nailed it 👍 thank you
Great explanation, thanks a lot!
3 in a row, all very well done.
Top ranked reference/learning resource.
subbed
Thanks Charlie!
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 😁
Great to hear!
What brilliant explanations, thanks.
amazing explaination!
Im glad i came across your channel
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.
Appreciate it Mohammed!
Great video as usual. I learn , again 🤗, something new and that makes me very happy
Glad to hear it JH!
Excellent lesson well explained
Thanks a lot, sir!
A little above my head, gleaned a little, good video.
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.
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.
Glad it helped, Arnold!
Excellent. It’s been a very long time since did this problem. I’d have had a very difficult time working through it myself.
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.
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!!!
You are amazing :)
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.
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?
Procreate!
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?
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?
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!
Glad you liked it!
Beautiful 🤠
This is absolutely incredible. What if the rope is thicker and not so bendy, like a pool noodle. Does this model account for that?
Amazing Video! Thank you
Made it all the way through this, I'm really impressed because I thought you had to go into variational calculus to do this.
Can u make a video about elastica curve
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))?
You could also do u=sinh(h)
can you pl. show this using lagrange's eqn
Thanks Dr. Elliot!
Would you give thought to doing subject by subject videos according to college level physics course?
I'm working on creating dedicated courses, make sure you're subscribed to my email list to hear when they're available!
@@PhysicswithElliot Great! Thanks!
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...
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.
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
@@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.
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
@@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].
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.
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
@@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?
Nice, e is everywhere !
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 😂.
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
Interesting stuff. I learnt that sinh was pronounced shine. I suppose yours is a little closer to it's spelling.
I always wanted to learn the catenary 'cause I miss this lecture,
I always though it gonna be prettier
Plz tell me the best books on mechanics
Let's see a Bessel function problem.
Why does the tension vector of the rope HAVE to point tangentially to the mass element?
huh, so much trickery :)
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?
Sure can!
Why does ty/tx need to be equal to dy/dx?
Would anything change if the rope is made of rubber and stretches significantly under tension?
Probably; for one thing the length wouldn't be fixed anymore and the mass density could vary along the rope
I had been trying this for 2 years! No one answered this
🔥
9:50 When the dopamine just hit right...🤯
Seems like horizontal T would be a function of slope so not constant. Guess I need a little more explanation.
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
@@PhysicswithElliot Thanks. Got it.
@@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).
E=mc^2
Schwarzschild radius
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.
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.
What actually change?x or something else???
9:00
"The imaginary factor doesn't cancel, and we have shown that ropes don't exist"
~*The End*~
a * cosh(x/a)
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?
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.
Oh, who would think that female sinuses were in fact catenoids
How I would find the shape of a hanging rope? By hanging a rope of course.
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👣
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..
😂
Lmao
I did my own research and I don't believe i exists.
-Sun Tzu, Art of War
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. :-)
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)