Equações diferenciais de uma ponte de carregamento (simplificada)


9

Estou tendo problemas para calcular as equações diferenciais de uma ponte de carregamento simplificada.

O sistema é construído como mostrado na figura abaixo (apenas um esboço):

insira a descrição da imagem aqui

Se eu usar a abordagem de Newton, estou obtendo as seguintes equações negligenciando o atrito, a resistência do ar e as mudanças no comprimento da corda:

mkx¨k=FA+FSsin(φ)mGx¨G=FSsin(φ)mGz¨G=mGgFScos(φ)

When I look at the kinematic relationships from the gripper (the circle with the weight mG) I get the following equations.

xG=xk+lsin(φ)zG=lcos(φ)φ=ωt=φ˙t

I know the weights mk and mG and the length l but the values are not important right now.

The goal is to have two differential equations at the end. One equation shall show the relationship between the driving force FA and the path of the trolley xk (with derivations) The other equation shall show the relationship between driving force FA and angle of the rope φG.

After that I want to make the transfer functions (Laplace transformation etc.) but that is not the problem.

The problem is that I can not seem to find those equations. My best approach so far looks like this:

mkx¨k=FA+FSsin(φ)

So that means if

mGx¨G=FSsin(φ)FSsin(φ)=mGx¨G

I can say:

mkx¨k=FAmGx¨G

and if I derive xG like this:

xG=xk+lsin(φ)x˙G=x˙k+lφ˙cos(φ)x¨G=x¨k+l[φ¨cos(φ)φ˙2sin(φ)]

I am actually getting stuck here because I can not find a way to eliminate φ from the equations. The addition theorems are not helping me at all (or I'm using them correctly).

Does anyone have an idea of how I should continue at this point? I hope I don't need a complete solution. I am actually more interested at doing this myself and hope to get a push towards the right direction.

Respostas:


4

My guess is that you probably need another differential equation for the angular movement, that will involve the inertia, such as:

mGl2φ¨=mGglsin(φ)

which yields:

φ¨=glsin(φ)

You can then maybe use the small angles approximation:

sin(φ)φ

Check out the inverted pendulum example.


Especially the inverted pendulum is very helpful... thanks for that - i didn't think about that
tlp

5

Kinematics and dynamics

enter image description here

Those are the steps to solve problems of this nature.

  1. Analize the kinematics of the system.

orOP = orOR + orRP

orOP = orOR + R(φ)BrRP

orOP = (xkî+0j+0k) + (sin(φ)lî+0j+cos(φ)lk)

orOP = [(xk+sin(φ)l)î+0j+(cos(φ)l)k]

note: R(φ) is a rotation matrix and xG=xk+sin(φ)l.

Taking the time derivatives:

xG˙ = xk˙+cos(φ)φ˙l

xG¨ = xk¨+lcos(φ)φ¨lsin(φ)φ˙2

  1. Use Newton's equation:

mkxk¨=FAmGxG¨

Substitute xG:

mkxk¨=FAmG(xk¨+lcos(φ)φ¨lsin(φ)φ˙2)

(mk+mG)xk¨+mG(lcos(φ)φ¨)mG(lsin(φ)φ˙2)=FA

For the z axis:

FZ = mGgl(cos(φ)φ˙2+sin(φ)φ¨)

  1. Use Newton's second law for rotation:

Iφ¨ = FZlsin(φ)(mGxG¨)lcos(φ)

FZlsin(φ)=mGglsin(φ)l2(cos(φ)sin(φ)φ˙2+sin(φ)2φ¨)

(mGxG¨)lcos(φ)=mG(l2cos(φ)2φ¨)mG(l2cos(φ)sin(φ)φ˙2)+mGxK¨lcos(φ)

Using trigonometry identities:

(I+mGl2)φ¨ = mGglsin(φ)mklcos(φ)xk¨

  1. Done! Now you can rest... ¨
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