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/sci/ - Science & Math

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9877614 No.9877614 [Reply] [Original]

How do you solve this?

>> No.9877619 [DELETED] 

Use half of the cable to find height as a function of length and then double it. Use the top of one pole as a starting point and the dip as another

>> No.9877638

>>9877614
Can't i just use Pythagoras? You've got two triangles with a missing side of 56.6, so double that.

>> No.9877640

>>9877614
The image is deceptive. If the poles are 50m, and half the cable is 40 meters, then the cable would never dip below 10 meters. If the poles were right next to each other, the cable would be 10m from the ground. The poles are 0m away from each other.

>> No.9877646

>>9877614
Typical problem in advanced mechanics using calculus of variations (Google calculus or variations and this problem will be one of the first examples to show up). Turns out to be a hyperbolic cosine, don't remember the exact form, but would parameterize the form by the given distances.

>> No.9877650

>>9877640
This is the "correct" answer, I saw it in a riddle book some time ago. Though I would be much more pleased someone knew analytical mechanics than realizing its a trick questions. Trick questions trick everyone, smart creative whatever, but interviewers think it shows "thinking outside the box" because they use pseudoscience and memes.

>> No.9877651

Wouldn't amazon be more concerned with asking questions about your knowledge of algorithms and systems? Also I too browse vee :^)

>> No.9877652

>>9877650
Yeah I would definitely have no idea how to solve it otherwise. But I always think in extremes, (What would it take for the rope to be taut? What if the poles were next to each other?) this is usually how I approach problems.

>> No.9877692

>>9877614
you need some differential equations, and a bit of differential geometry

>> No.9877721

>>9877614
>>9877652
How about this?:

Suppose the image is a 2D plane with the middle being x=0 and the ground being y=0. Assume the cable is a parabola, so ax^2+bx+c=f(x). At x=0 we have 10 so c=10. At f(d/2) and f(-d/2) we have 50. Plugging this in gives us a=160/d^2 and b=0.

Now we have the cable as f(x)=160/d^2*x^2 +10. Finding the arc length of this curve gives us 80, so we just need to solve arclength(160/d^2*x^2 +10)=80 from -d/2 to d/2 for d.

The arclength formula gives us some ugly shit that I'm too lazy to solve, but this would eventually lead to d=0.

>> No.9877728

Ropes hang in the shape of a catenary if you want an exact calculation.

>> No.9877736

>>9877721
>assume the cable is a parabola
it's a hyperbola dipshit

>> No.9877748

>>9877646
>calculus of variations
but you dont need calculus of variations to derive the catenary eqn.

https://www.youtube.com/watch?v=N1LRgHYxMKo

>> No.9877751

>>9877736
no it isn't. its the hyperbolic cosine. there's a difference.

>> No.9877755

>>9877638
You can do a ballpark estimation using the pythagorian theorum. Not solve, but estimate.

>> No.9877758

80m/2 +10m + d = 50m
40m + 10m + d = 50m
d = 0

>> No.9877790

>>9877614
Couldn't you just describe the cable as a catenary and solve it?

>> No.9877818

>>9877614
Cables under uniform gravity fixed at two points follow a catenary (hyperbolic cosine) curve. cosh(0) = 1, but we are given a minimum of 10 m. The general formula for the catenary of a hanging wire is f(x) = a*cosh(x/a), so the formula for the position of the cable is 10*cosh(x/10).

10*cosh(x_f /10) = 50

x_f = 10*arccosh(5)

d = 2*x_f is the distance between the poles

d = 20*arccosh(5) = 45.8486 m

>> No.9877825

I just did it, it is impossible you fucking troll. What a waste of time.

>> No.9877830

>>9877818
>>9877650
I don’t know if your method is wrong, but your answer is wrong

>> No.9877832

>>9877818
>>9877830
Meant to link here>>9877640

>> No.9877837
File: 91 KB, 645x729, hyuuuuhhhhfhffhff.png [View same] [iqdb] [saucenao] [google]
9877837

>>9877736

>> No.9877849

Guys, it's impossible. If you look at the extreme case (where the two poles are at 0m from each others) you get that the height of the rope is 80m/2 = 40m. 40m + 10m gives you the right height for the poles (50m). So the solution is that the two poles are at the same location which is impossible.

>> No.9877853

>>9877830
oh it's even gayer than working for Amazon, it's a trick question about triangles, didn't think about the distances

>> No.9877862

>>9877818
The problem would be that the arclenght of your function is not 80. So a contradiction.

>> No.9877863

>>9877650
>I would be much more pleased someone knew analytical mechanics than realizing its a trick questions. Trick questions trick everyone, smart creative whatever
You mean you'd be happier if someone blindly plugged figures into a formula they knew by rote without taking note of basic aspects?

The point isn't for it to be a trick question, or to test your memory of math that you could easily look up in a real-world situation, but for you to approach it in a logical manner and demonstrate your ability to reason. Plenty of people will durdle around with half-remembered math, sweating bullets and producing no insights, like they're struggling with an exam question.

A practical thinker will first notice that the distance has to be less than 80m, then start thinking about what other constraints they can place on it. A reasonable first step is to subtract 10m from the poles, and cut the cable in half, so you can tighten that upper bound starting with a simply right-angle triangle. Then you're looking at 40m and 40m, and now it should be obvious that either the poles can't be any distance apart, or the cable has stretched longer than 80m.

You shouldn't just be able to give the "trick" answer of 0m, you should also offer the possibilities of the cable stretching or the measurements being in error. In the working world, you can't just robotically take what information you're given and process it as instructed, you've got to notice when something is fishy and dig into why, or it could cost the company a fortune.

>> No.9877873

>>9877849
They're not in the same location, there is 0 m between them. When measuring my distance from the ground (commonly referred to as height) I don't measure the distance from my center to the center of the planet you fucking goon.

>> No.9877877

>>9877818
Brainlet engineer mindlessly using formulas

>> No.9877882

>>9877873
Well, he’s kinda right. If half the rope is 40m it will only be 10m above the ground if it falls straight down. But the ropes are joined, so so they will have a slight curve at the bottom even if their wholes are touching. Instead of the last millimeters of the rope reaching straight down, they’re being pulled laterally due to the connection with the other half of the rope, which would make the rope barely above 10m

>> No.9877883

>>9877873
You are fucking moron, if a particule Z is at x = 4 and another particule K is at K(Z) = Z + 0m , the two particules are both at x = 4

>> No.9877885

>>9877877
trick questions are the lowest form of intellectual tests, like riddles with an answer based on wordplay

>> No.9877887

This is clearly in hyperbolic geometry. Therefore, this is a rectangle and the distance is 80 meters as well.

>> No.9877889

>>9877885
What are some high form intellectual tests?

>> No.9877891

>>9877889
Testing mastery of actual physics, mathematics, and scientific knowledge? Or just a quality riddle.

>> No.9877892

>>9877883
These aren't mathematical points dipshit, they're poles.

>> No.9877900

>>9877892
They're poles with no definite dimension. You can't assume they have width dumbass.

>> No.9877903

>>9877891
Yet you still got the problem wrong when you tried to use “actual mathematics” lmao.

>> No.9877904

For those who are interested,
https://www.youtube.com/watch?v=l_ffdarcJiQ

>> No.9877916

>>9877900
An academic can't. Someone who lives in reality, where points are simply a mathematical abstraction and every actual object has length, width, and height can.

>> No.9877918

>>9877916
Fuck you, poles have no dimension

>> No.9877923

>>9877891

> cant solve shit with his higher math
> gets outskilled by a person using common sense and basic arithmetic
> whines that this is possible

pathetic

>> No.9877930

>>9877863
Its not math memory in any way or form. Except you remeber all the details of how to solve that problem, you need to be able to do a proper of the problem, either by tension analisis or by variational methods. This shows comprehension of difficult concepts and ability to solve problems efficiently with many tools. Obviously if you remember "Oh it's a hyperbolic cosine" then yes it's just plug and chug, and because it's an interesting fact that a lot of people could potentially know, it's not a good question on comprehension, so the best would be to show a proper model that then you can derive the equations from it. Your procedures is just a more direct approach shows a poor understanding of the problem, even if you can use it to arrive to a correct answer, because you don't know at what points will the curve strech to, and so you will get a distance that's not really insightful. But if you found a clever way to get the answer, the point here is that anyone that has done these problems, understands that dimensions are not really well represented in picturea, but degenerate casea are almost never considered, and in many cases is betrer to find general expresions, and numerical valua are just used to confirm your answer. Saying "practical thinker" or whatever has no real basis besides gut feelings fomented by retarded company heads and interview experts. You shouldn't just show your curriculum, but the questions shouls be on how capable are you in applying your knowledge in whatever shit the company needs you. That's why in proper software developer interviews (which are many becuase CEOs don't medel with techy shit), you are asked to solve certain concrete problems, that could be tedious or "tricky", but that anyone with proper training in the field should be able to answer. And if I have a guy that can properly explain a method of solving the "chain under gravity problem", usually shows he understands his shit

>> No.9877938

>>9877918
> citation needed

>> No.9877939

>>9877923
>hyperbolic cosine is higher math than triangle inequality
okay, trick questions are still gay

>> No.9877951

>>9877891
Hardly a riddle and more basic math and practical logic. trying to make something more complicated and still being incorrect is a sign of a shitty engineer

>> No.9877953

>>9877650
What riddle book are you talking about? I actually find fun trying to figure out why math can't work in those riddles outside just saying is bullshit.

>> No.9877965

>>9877953
>why math can't work in those riddles
>can’t work
?

>> No.9877976

>>9877638
>>9877652
>>9877692
>>9877721
>>9877736
>>9877825
>>9877853
>>9877887
>>9877755
Do none of you retards know what a catenary is?

>> No.9877979

>>9877939
Your mom's gay

>> No.9877982

>>9877916
This is a math problem in which some properties of the real world are being used but it still is a math problem, not real world.

>> No.9877988

>>9877965
For example, in OP riddle if you solve it using the catenary function, you will find the correct value of x but when you calculate the arc length it shows the cable should be way longer than 80m. That's what I look for, a proper explanation of why math doesn't solve the particular riddle. Is just fun for me, that's why I asked for the book.

>> No.9877992

>>9877976
Why would I know that? I'm not an engineer

>> No.9877995

>>9877982
It's more than a math problem - it's actually a physics problem.

>> No.9877996

>>9877992
Interestingly, for suspension bridges the path is actually a parabola.

>> No.9878000

>>9877992
I'm a physicist and I derived catenary function in my calculus lectures. Sure, six years later I forgot how to do it but I still remember the system.

>> No.9878002

>>9877996
They look similar if the catenary is open enough. I suppose a parabaola is an easier model to work with.

>> No.9878004

>>9877988
But you didn’t properly explain why the math doesn’t work. You just noted that it doesn’t work and contrasted the mathematical answer with the real answer

>> No.9878024

>>9877614
Use Euler-Lagrange equations to minimize the potential energy of the configuration of the cable.

>> No.9878030

>>9878004
using
>>9877818

arc length s=a*sinh(x/a)

s=10*sinh(22.924/10)

s=48.988m

2*s = 97.976 m

97.976 m > 80 m, contradiction

>> No.9878039

>>9878030
You haven’t explained anything. You show that the math doesn’t lead to the correct answer, but what about this particular problem causes this?

>> No.9878042

>>9878039
it's not actually a catenary curve, so assuming it is a catenary curve leads to a contradiction, because the cable is essentially two straight lines

>> No.9878057

>>9878042
And that wraps up the thread

>> No.9878068 [DELETED] 

160 / pi

>> No.9878075

I would have failed this IQ quiz.

>> No.9878086
File: 638 KB, 1194x748, ez.png [View same] [iqdb] [saucenao] [google]
9878086

Now where's my job?

>> No.9878090

>>9877995
Damn you are dumb. Physics uses the tool that is mathematics; so physics problems are math problems, but math problem aren't always physics problems.

>> No.9878101

>>9878086
Where are my fries?

>> No.9878103

>>9878090
How do you mathematically prove that the configuration the string aquires minimizes it's energy, and that at those dimensions one can approximate the gravitational force as constant?

>> No.9878108

Is that a parabola

>> No.9878141

>>9878103
You don't prove things in physics, you simply observe and translate the information into different languages : one of which is mathematics.

>> No.9878192

>>9878141
Exactly, retard.

>> No.9878217

>>9878090
So essentially what you are saying is physics is one level of abstraction above mathematics and also that you are a retarded drooling wretch. Makes sense.

>> No.9878220

>>9878192
My name is nirthgurd IV

>> No.9878272

>>9877638
>Can't i just use Pythagoras?
HYPERBOLIC Pythagoras.

>> No.9878277

Assuming this is a parabola, first we bring it down 10 meters, then we get a system of equations

kx^2 = 40;
integral sqrt(1+2kx) dx = (2 (40/x^2) x + 1)^(3/2)/(3 (40/h^2)) = 40;

Solving gives a solution of about 22 and doubling that gives a solution of about 44.

>> No.9878293

>>9878277
h^2 is x^2 typo

>> No.9878337

>>9878277
The dimensions of the (post height - cable minimum) and cable length (hypotenuse) don't satisfy the triangle inequality A+B>C for lengths of sides A B C of an arbitrary triangle, or rather they satisfy the equality A+B=C, which is the degenerate case of two straight lines 40m each running parallel. The posts are d=0 apart.

>> No.9878349

Bunch of fucking brainlets around this place, I'll tell you.

Divide all numbers by 10.
Catenary means y=cosh(x)
arccosh(5)=x
x=ln(5+√(25-1))=ln(5+2√6)
Multiply by two for full span, and then by ten to restore scale.

EXACT ANSWER is width = 20ln(5+2√6)

>> No.9878370
File: 1.15 MB, 3264x2448, F4196FF0-CDB2-4DCB-8098-AE040D02BAD4.jpg [View same] [iqdb] [saucenao] [google]
9878370

Handwriting is messy but here is my answer.

>> No.9878378

>>9878370
the horizontal sides aren't 40, nice bait

>> No.9878395

>>9878378
The rope is 80m long so if you pulled the poles apart so that the rope is straight the distance would be 80m. Thats where I got 40 from.

>> No.9878399

>>9878395
but that's not the question

>> No.9878715

>>9877614
If you try using Pythagoras' Theorem to get an estimate, both the hypotenuse and the vertical side of the triangle will be 40m. So the distance between the poles will be ~0m.

>> No.9878737

>>9877614
i don't even need to calculate anything!
question don't say i have to define "?", so the answer is "?".
that was easy

>> No.9878808

are you all brainlets? it's 80

the rope is 80 m and goes from one pole to the other

>> No.9878963
File: 23 KB, 1152x648, Untitled.png [View same] [iqdb] [saucenao] [google]
9878963

I have a method to solve the problem but my calc 2 knowledge went out the window. Imagine the cable as a parabolic function with the form f(x)=ax^2 where "a" is the coefficient that determines the shape of the parabola, we do no need the +b... portion (the shifting up part) as we take the derivative of the original function.
Let us now define p as 2*a where p comes from f'(x)=(2*a)x. We can then set the midpoint of the parabola at the origin of the cartesian plane such that the arclength of the parabola from 0 to L is half the length of the chain (40 m here).
Thus in the picture the problem is lasyed out. Now what I'm having trouble with is that I belive the integral is solved with trig sub (nameley using x=tan(x) or something like that) Would anybody care to take up what I have done? Also please leave your thoughts on the validity of the assumbtion that the chain can be modeled using a parabola and not a caternary.

>> No.9878965

>>9878963
I forgot to add this but it is probably self explanatory that 2B is the distance between the pillars.

>> No.9878988

>>9878808
Either lame troll or you are clinically retarded.

>> No.9878996
File: 34 KB, 1152x648, Untitled2.png [View same] [iqdb] [saucenao] [google]
9878996

>>9878965
Same poster here, I figured out th trig sub part but what is missing is the definition of p=2*a which then intuitively needs the definition of "a". If "a" was given then the problem would be solvable, but it seems the obly way to determine "a" is to empirically test varying values in symbolab to approximate the general shape of the curve.

>> No.9879000

>>9877996
>>9878002
It's still a catenary (hyperbolic cosine) but with the right ratio of weight to horizontal tension, several of the 4th and higher order Taylor series terms vanish and the whole thing is approximately parabolic to a rather low amount of error.

>> No.9879017

>>9877976
>knowing arbitrary names or formulas for special curves makes you smart
>not just being able to solve exactly the problem by analytical thinking

>> No.9879021

>>9877640
>If the poles are 50m, and half the cable is 40 meters, then the cable would never dip below 10 meters
Why?

>> No.9879034

>>9879021
Imagine the 2 ends of the cable attached to one pole, where does the bottom dip down to now? Well the cable is folded perfectly in half, and because its 80m long, it is going straight down 40m. Therefore it dips down to a point of 50m - 40m = 10m.

Now try spreading the two ends, do you think the middle, lowest point, will go down or up? Yes, it will go up the further you stretch it to a maximum height of 50m (if the two poles are 80m apart). Therefore the only time it actually "reaches" the 10m distance is when the two poles are right next to each other, the picture is wrong.

>> No.9879036

>>9877614
Trick question: There is NO cable!

>> No.9879049

>>9877614

the poles are touching each other.

>> No.9879050

>>9879049
Lewd

>> No.9879057

>>9879049
moar

>> No.9879068
File: 520 KB, 960x906, Natalie.png [View same] [iqdb] [saucenao] [google]
9879068

>>9879049

for the cable to be 10 above the ground and pole 50 high then half cable must be 40 meters from top. Therefore poles must be touching (assuming poles are straight up at 90 degrees to ground)

>> No.9879127

>>9877640
I second this

>> No.9879131

>>9879127
Shut the fuck up, nigger

>> No.9879150

>>9878349
I solved this shit last night using hyperbolic trig. Why the fuck are we still debating this?

>> No.9879164

>>9877748
Good to see it done without. Didn't say calculus of variations was required, just a typical problem.

>> No.9879169

>>9879150
Because the right answer is 0 you absolute moron.

>> No.9879181

>>9879034
Thanks. I quite misunderstood. You cleared it up for me

>> No.9879260

I had an Amazon job interview and the problems I had to solve were a lot harder.

This is the worst example of what an Amazon interview is like.

>> No.9879268

A uniform cable like that forms an arc that can be described by a hyperbolic cosine function. Just use vectors and the function to find the x value that corresponds to 10m and you've found half the unknown length. OF course this is kind of a bullshit question

>> No.9879277

>>9877614
This is an engineering question. You would need gravity and properties of the cable.

>> No.9879325

>>9879260

this

>> No.9879340

>>9877923
Still can't solve variation when lowest point is 20m high instead of 10m.

>> No.9879341

Jeff Bezos himself wouldn't even be able to solve those problems.

>> No.9879353

>>9879260
Examples?

>> No.9879356

>>9878349
>brainlets
>first divide event by 10
kek

>> No.9879361

>>9879150
Hello brainlet

>> No.9879363

>>9878349
Brainlet

>> No.9879366

Guys, throughout all of this we are forgetting the elasticity of the cable and thermal expansion. We have to include those for the exact answer.

>> No.9879373

>>9879260
Post examples

>> No.9879386

>>9879373
An apple costs 40 cents, a banana costs 60 cents and a grapefruit costs 80 cents. How much does a pear cost?

>> No.9879388

>>9879386
100?

>> No.9879393

>>9879388
We'll let you know

>> No.9879419
File: 12 KB, 1220x633, intermediate.png [View same] [iqdb] [saucenao] [google]
9879419

55mm stainless steel cable
66 km/h northern wind
different angle in pic related

>> No.9879420

>>9879386
What kind? Bartlett?

>> No.9879421

>>9877976
>All these responses covering for their lack of knowledge or why they shouldn't have to know.
If you read this and don't know it's hyperbolic trig functions. Something you go over in calc 2.

>> No.9879431

>>9879421
I know what hyperbolic trig functions are. I don't know what shape cables form when hanged. Why would this be mentioned in calc 2? This isn't math, it's physics/engineering

>> No.9879479

>>9879431
Because if your prof isn't shit and just spewing out the information you need to just pass the class he should be giving you the real life situations and examples for the things you're learning.

>> No.9879485

>>9879431
If you don't know the shape, fine. If you cannot figure out the shape, then you do not truly understand the mathematics if you cannot even use the mathematics.

>> No.9879503

Lol, way easier than you guys think. Just make them triangles. turns out answer is 0. Also, just for concept, 50m pole, run the cable down 40m and back up 40m, cable is 80m long. :D

>> No.9879512

>>9877996
>Interestingly, for suspension bridges the path is actually a parabola.

That's because it's forced to be a parabola in order to maintain tension in the bridge. hence why suspension bridges are so damn strong. if it was an unsupported span, it would be a catenary.

>> No.9879516

>>9877992
>2nd semester economics and computer science
>literally had this in "introduction to analysis and linear algebra" in our first semester
is this amerifat education?

>> No.9879517

>>9878030


You're literally solving for the wrong variable.

The other anon didn't do it correctly either.

the arc length is a constraint. solving for x. Go.

>> No.9879518

>>9877614
y = c + a*cosh(x/a)

c = 10

Then here’s my method for converging:
Once you get A you can tell the distance by finding where f(x)=50 and that is easily done with basic math as long as you know that the inverse cosh is a function.

Ok so here we go: if the distance is within the limits of 1 or 2, then we use another midpoint calculation then begin narrowing down our length and getting a closer approximation of A by selecting halves. This is a semester one algorithm for any programming student.

Then we just need a method if the length isn’t within the range of 1 2, and to do that I either multiply each side by two or divide it by two.

The number by which you divide/multiply by should be the upper limit - the lower limit.

Or you can just brute force linearly starting at a small number and then moving up. But you won’t get the job.

>> No.9879521

>>9879516
>9879485

No. most amerifats learn this high school/ first year of uni.

>> No.9879530

>>9878086

At McDonalds.

>> No.9879531

>>9877640
>The image is deceptive. If the poles are 50m, and half the cable is 40 meters, then the cable would never dip below 10 meters. If the poles were right next to each other, the cable would be 10m from the ground. The poles are 0m away from each other.

yep.

>> No.9879545

>>9879521
sure amerifat
you should watch some videos of foreigners taking SATs lol

>> No.9879685

>>9879517
Re-read the conversation. The point was to show that assuming the wire is a catenary curve leads to a contradiction, namely that the given arc length cannot be equal to the arc length required for it to be a real catenary.

>> No.9879708

>>9879386
40 cents. It’s 20 cents per vowel

>> No.9879961

>>9877614
One way is to recreate it experimentally on a smaller scale and then simply multiply the result by the scale factor.

>> No.9879968

So much fucking autism in this thread. Answer is fucking obvious, idiots. Should take no more than 5 seconds WITHOUT any shitty formulas.
Are you people all trolling or what?

>> No.9880082

>>9879968
>>9879369
No, /sci/ is just genuinely retarded. Anyone who pretends to be this retarded is beyond trolling.

>> No.9880353

There is a calculus problem relating to the sag of a cable, I have to look it up.

Point is this is a nontrivial problem.

>> No.9881670 [DELETED] 

>>9877614
x<80

>> No.9881673

>>9880353
No, jesus christ.

>> No.9881678

HINT ;?<1000m

>> No.9881708

>>9877891
>Testing mastery of actual physics, mathematics, and scientific knowledge?
Using physics, mathematics, and scientific knowledge, you should first ask yourself "does this problem even make sense?" And you failed.

>> No.9881719

>>9877640
i juat realized how fucking stupid i am

>> No.9881864

>>9879708
Are you retarded? Why would fruit costs be related to vowels?

>> No.9881894

>>9879485
How are you supposed to figure out the shape with just math? You can't, it requires physics