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Channel: GetAClass - Физика в опытах и экспериментах

Галилео. Эксперимент. Маятник Фуко

Галилео. Эксперимент. Маятник Фуко

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физика

physics

science

наука

Центр тяжести

равновесие сил

параллелограмм сил

Торричелли

Ферма

Штейнер

гетакласс

getaclass

нгу

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today we will talk about how physics

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can unexpectedly help solve a

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mathematical problem;

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this problem was first proposed back in the 17th

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century by Pierre Fermat and then we solved

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Evangelista Torricelli and many

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mathematicians and problems; this is called the

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transport center problem and is

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formulated as follows: let

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plane there are three points, you can

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imagine that these are three cities on

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the map and you need to find the fourth point

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such that the sum of the distances to these

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given points

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is minimal. To solve this problem, we will

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use this mechanical model,

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we replaced the three points with three blocks and

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now we will hang 3 on these blocks

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loads of equal mass that are connected to

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each other by threads

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and now the loads swayed a little and the whole

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system came to a state of equilibrium, the

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loads in this system tend

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to fall down so that the center of gravity of the

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entire system takes on and the lowest possible

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position, well then it turns out that the

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sum of the lengths of all outer

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vertical segments threads takes a

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maximum value in the equilibrium position,

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and from this it follows that the sum of the lengths of the

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internal segments takes a

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minimum value, thereby the node

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connecting all three threads is

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now in the very transport center

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that we needed to find in this

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problem, and we also see that the nicks form between

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are equal angles and it is no coincidence that

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three forces of equal magnitude act on the node from the side of the loads

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and you are in equilibrium, the vector sum of these

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forces must be equal to zero,

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which means the sum of any two forces must

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balance the third for

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forces of equal magnitude, this is only possible

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when the angles between them are paired are equal,

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that is, they are 120 degrees each,

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now we know the correct answer to this problem,

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obtained by means of

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physics, but a real mathematician will be

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satisfied only when they justify

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this answer by means of geometry, and now

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I will tell you a beautiful proof

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that the German mathematician

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Hansberg Nier came up with when he was still a student at

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Knicks Birks University, we mark in the

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triangle a b c an arbitrary point p

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from which we draw a segment to all the

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vertices of the triangle and we need to find

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such a position of point p

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so that the sum of the lengths of these segments is the

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smallest, we will do an additional

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construction so that these segments become

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links of one polyline and

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for this we select the triangle bbc and

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Let's rotate it around point b by 60

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degrees so that it is outside the

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triangle a b c and point p

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will go to the point along stroke a. c to point

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c stroke and due to the fact that the rotation angle

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was 60 degrees we got 2

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equilateral triangles, firstly

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this triangle bcd stroke and

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note that the position of point c

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stroke does not depend in any way on where we

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chose point p

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secondly this is a triangle bpp stroke its

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side pp stroke is also equal to b small

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and we have a broken line apps 3 kHz and

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and stroke the length of which is exactly equal to a +

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b + c its ends are fixed at points a

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and c stroke and now we need to make this

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broken line as much as possible In short, by

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choosing a suitable point p,

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it is obvious that its length will be minimal

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if the links stretch along a

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straight line segment, prime and

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minimal. at zero should lie

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somewhere on this segment, how

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does it follow that the segments connecting the point at

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0 of a perfect

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square make angles of

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120 degrees to each other, let's figure this out, I

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erased everything unnecessary from the drawing, minimal. at

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0 and when rotated by 60 degrees it

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should also be on the segment from and the stroke

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so that an equilateral triangle is obtained

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bp zero at the zero stroke

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in an equilateral triangle all angles are

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60 degrees so on the

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segment from the stroke there is only one

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such .

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and thus we have shown that the solution to

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our problem is unique and the angle a at 0 b is

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equal to 120 degrees,

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but it is clear that we could make the initial

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rotation not around point b but, for example,

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around point c and then the angle op and 0 c

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will be equal to 120 degrees 120

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degrees remains at the third angle p p 0

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c and thus we substantiated the result

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obtained from considerations of the balance of forces, the

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groove at zero also says that from

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it all sides of the triangle bc are visible

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at different angles and knowing the Denov solution

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immediately follows a very simple way to

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construct the transport center of the

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triangle let's construct external equilateral triangles on all three

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sides of triangle bc,

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draw a segment

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fathers stroke, we know that the

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transport center must lie somewhere

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on this segment, draw a segment cd

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stroke the transport center lies on

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it too, which means it coincides with the

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intersection point of these segments and we

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constructed from here it immediately follows that the

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segment bh3 also passes through this

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point beautifully, isn’t it difficult

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to show that the solution found is suitable

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for triangles in which there is no obtuse

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angle exceeding 120 degrees, and if there is

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such an angle, then the transport center

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coincides with its top

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tire and now I want ask our

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traditional final question,

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let's consider not 3 but 4 cities and

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let them be at the vertices of the square

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and ask what the

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road network that connects

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these four cities will look like and at the same time has a

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minimum length, write your thoughts on

00:07:19

this

00:07:20

in the comments to our video on

00:07:23

youtube

00:07:25

[music]

Description:

При решении задачи об отыскании транспортного центра треугольника удобно прибегнуть к механической модели с тремя грузами, соединёнными нитями и переброшенными через блоки в вершинах треугольника. Ключевые слова: задача Штейнера о минимальном дереве, центр масс, центр тяжести, Steiner tree problem, Steiner network problem, rectilinear Steiner tree. Наш канал с дополнительными материалами https://t.me/getaclass_channel Новосибирский Государственный Университет https://www.nsu.ru/n/ Физический факультет НГУ

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