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2 blocks & 3 mass

ValeryOchkov
24-Ruby IV

2 blocks & 3 mass

Help please to solve this ODEs task. The first try (Prime 3.1) in attach

3-mass-2-block.png

1 ACCEPTED SOLUTION

Accepted Solutions

F.M. написал(а):

Beautiful, .... but the woman ... the one of the photo, is even more ...

Woman and Flowers!

tulpan.png

You can change L, masses, start point and get new woman, flowers etc!

View solution in original post

108 REPLIES 108
vlehner
13-Aquamarine
(To:ValeryOchkov)

Wouldn't it be better, to study the geometrical relationships before starting calculation?

Volker
-MFra-
21-Topaz II
(To:vlehner)

Hi Valery,

Can we describe the problem the following way?

The material system consists of five bodies: the masses M0, M1, M2 and the fixed disks' masses μ1 and μ2, of equal rays r. The wire is inextensible and, by chance, it does not creeping.

Determine the system motion and the tension in the wire.

Can we choose the location of the reference system's origin, at our discretion? For example so:

tre masse.jpg


or is it fixed at L/2?

Thank you

F. M.

vlehner
13-Aquamarine
(To:-MFra-)

Hi F.M.

I guess it's fixed at L/2 and y0.

After releasing the weight m0 at this position, the system finds its equilibrium.

Putting the center of the first disk to the origin of the coordinate-system would be advantageous for solving with Lagrangesche Formalism.

The equations should be more simple than.

Viel Erfolg!

Volker

Volker
-MFra-
21-Topaz II
(To:vlehner)

Salve V. L., in fact I will use the Lagrange equations.

We usually choose the origin of the axes in such a way as to make the equations as simple as possible. Valery put it in (L / 2, 0) not in (L / 2, y0). This last is your choice.

Viel Spaß

F. M.

vlehner
13-Aquamarine
(To:-MFra-)

You're right.

I think too it's the best way to choose the origin like you suggested.

Now I try the formalism.

Cappucino now

Espresso Lagrangesche

tanto divertimento

Volker

Volker
vlehner
13-Aquamarine
(To:-MFra-)

Tried to use the Lagrangesche Method.

creating the differential EQN for the x-direction is ready, but it's very large. (it took a time of 30 minutes!)

Now there is to do the same for the y-direction.

see attachment please.

I'm not sure if both equations will work in a solve block.

What generalized koordinates can we choose for the Lagrangesche, that the equations become more comfortable and smaller?

Maybe with angles?

I really don't know and it seems to me that a solution here with Newton is necessary.

Viel Erfolg!

Volker

Volker

Thanks, Volker!

But es ist zu komplizirt for me.

I think we must begin with the static task with r=0 (warum nicht?).

It will be a key to solve the dynamic task.

Static-1.png

Static-2.png

Static-3.png

You will laugh, but it seems to me that I solved the problem. We have an attractor at air resistance and one perpetuum mobile without it.
Dynamic-1.png
Dynamic-2.png
Dynamic-3.png
Dynamic-4.png

In oil not in air

Dynamic-oil.png

Strictly speaking you haven't implemented the damping correctly here!  If the damping force is proportional to velocity squared (or velocity*|velocity|) then you need to calculate the damping force using the resultant velocity and then resolve that along the x and y axes.  This will not be the same as calculating x and y damping forces using the separate x and y components of velocity, in general.

Alan

AlanStevens написал(а):

Strictly speaking you haven't implemented the damping correctly here!  If the damping force is proportional to velocity squared (or velocity*|velocity|) then you need to calculate the damping force using the resultant velocity and then resolve that along the x and y axes.  This will not be the same as calculating x and y damping forces using the separate x and y components of velocity, in general.

Alan

Thanks, Allan!

I know it but we use one simple model!

Help please here Sheets Mathcad 15 and Prime - one bug

vlehner
13-Aquamarine
(To:AlanStevens)

Alan,

if I understood you right, you meant this?

Volker

Volker, Yes.  We should have Fwx = Fw*cos(theta) and Fwy = Fw*sin(theta)    where Fw is as you've defined it and theta = tan^-1(vy/vx). 

However, this would only be of importance in a real problem - it doesn't matter too much in a toy problem.  It's just the pedant in me pointing it out!!

Alan

vlehner
13-Aquamarine
(To:AlanStevens)

Thank you!

Volker

AlanStevens написал(а):

Volker, Yes.  We should have Fwx = Fw*cos(theta) and Fwy = Fw*sin(theta)    where Fw is as you've defined it and theta = tan^-1(vy/vx).

However, this would only be of importance in a real problem - it doesn't matter too much in a toy problem.  It's just the pedant in me pointing it out!!

Alan

One compromise:

3-mass-2-block-dynamic-picture-1.png

3-mass-2-block-dynamic-picture-2.png

3-mass-2-block-dynamic-picture-3.png

3-mass-2-block-dynamic-picture-4.png

Ok. I might as well show my solution using damping in the form I mentioned previously.  Note that my worksheet is M15, and first does a static equilibrium calculation, then does the dynamics (only the dynamics part pictured below).

ThreeMass1.PNG

ThreeMass2.PNG

ThreeMass3.PNG

Alan

Thanks, Alan!

But a solution without an animation is not solution.

A solution with one animation is not solution too.

Tree animations are good!

Video Link : 7859

Video Link : 7860

Video Link : 7861

Valery Ochkov wrote:

Thanks, Alan!

But a solution without an animation is not solution.

Ok. Here is a (not very exciting!) animation:

Video Link : 7871

Alan

Sorry, Alan!

I do not see an animation.

We must convert avi to mov and than input it here.

Attach please the avi file - I will convert it and put here!

Ok Valery, thanks.  avi file attached.

Alan

AlanStevens написал(а):

Ok Valery, thanks.  avi file attached.

Alan

Video Link : 7872

Sorry. The solution with animation but without trajectory is not full solution

Animating a simpler task in Smath Studio

http://en.smath.info/forum/yaf_postst973_Animation-double-pendulum-and-a-pendulum-on-a-spring.aspx

Fluctuations of three loads .The system consists of two extreme loads mass m1 and medium load mass m2 (2m1> m2).If the average load deviate from the equilibrium position, the system begins to oscillate.

Kolebaniya.gif

Fridel Selitsky написал(а):

Animating a simpler task in Smath Studio

http://en.smath.info/forum/yaf_postst973_Animation-double-pendulum-and-a-pendulum-on-a-spring.aspx

Fluctuations of three loads .The system consists of two extreme loads mass m1 and medium load mass m2 (2m1> m2).If the average load deviate from the equilibrium position, the system begins to oscillate.

Kolebaniya.gif

Sorry. The solution with animation but without not simple, interest, fine trajectory is not full solution!

tulpan.png

Pendulum-123-1.png

vlehner
13-Aquamarine
(To:ValeryOchkov)

Valery Ochkov schrieb:

One compromise:

Maybe we don't need a compromise:

Volker

Zu compliziert for me

Sergey Dovlatov about one compromise:

"Дело происходило в газете "Новый американец". Рубин и Меттер страшно враждовали. Рубин обвинял Меттера в профнепригодности. (Не без основания). Я пытался быть миротворцем. Я внушал Рубину: — Женя! Необходим компромисс. То есть система взаимных уступок ради общего дела. Рубин отвечал: — Я знаю, что такое компромисс. Мой компромисс таков. Меттер приползает на коленях из Джерси-Сити. Моет в редакции полы. Выносит мусор. Бегает за кофе. Тогда я его, может быть, и прощу."

It happened in the newspaper "New American". Rubin and Metter were terribly hostile. Rubin accused Metter of being unfit for work. (Not without reason). I tried to be a peacemaker. I suggested to Rubin: "Eugene!" A compromise is needed. That is, a system of mutual concessions for the sake of the common cause. Rubin answered: - I know what a compromise is. My compromise is this. Metter crawls on his lap from Jersey City. Wash floors in the office. He takes out the garbage. Running for coffee. Then I, perhaps, will forgive him.

vlehner
13-Aquamarine
(To:ValeryOchkov)

Did you try it?

You cant say its zu kompliziert for me without trying it!

Volker

Volker Lehner написал(а):

Did you try it?

You cant say its zu kompliziert for me without trying it!

You are right. 

vlehner
13-Aquamarine
(To:ValeryOchkov)

Thanks, Volker!

But es ist zu komplizirt for me.

I think we must begin with the static task with r=0 (warum nicht?).

It will be a key to solve the dynamic task.

And meanwhile, i think so too.

I try to solve it now with Newton by myself.

Let's see what F.M. makes with the Lagrangesche.

Volker

I do not show the picture with solution!

You can see the attach or not see - try to create yourself and compare than!

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