How do I modify the ball on a beam system to accommodate for Friction?

I was following the tutorial within this link: https://ctms.engin.umich.edu/CTMS/index.php?example=BallBeam&section=SystemModeling

However, for my studies, I wish to modify the system equation and subsetquent MATLAB script to account for friction. Unfortunately I do not know how to do this.

Any help would be appreciated.

Thank-you for your time and support

Answer 1

[Sam Chak]

Hi @Kez

If Fr is the friction, and you know the friction formula, then you inject it in this differential equation.

I don't have a MATLAB script. If you already have the MATLAB code for describing the frictionless motion

then paste the code here so that I can show you where to insert Fr.

[Kez]

Hi Sam,

In the exercise regarding a frictionless surface, the Laplace transformation would have to be taken for the initial equation, resulting in the equation becoming reprisented as the following script, in which the input values may be altered based on preference.

%% Input Values %%

m = 0.111;

R = 0.015;

g = -9.8;

L = 1.0;

d = 0.03;

J = 9.99e-6;

%% Calculation %%

s = tf('s');

P_ball = -m*g*d/L/(J/R^2+m)/s^2

My understanding is that friction may be modelled as Mu*N, in which Mu is the co-efficient of friction and N is the normal force (mass of the ball x gravity).

so the inputs would accomodate for this as follows, with an aditional calculation for MuN:

%% Input Values %%

m = 0.111;

R = 0.015;

g = -9.8;

L = 1.0;

d = 0.03;

J = 9.99e-6;

Mu = 0.3

%% Calculation %%

Fr = Mu*m*g

I cannot deduce how to perform the laplace of an equation including the Fr as my brain is struggling to compute it.

Thank-you for taking the time to help, it is really appreciated.

[Sam Chak]

Oops! @Kez, I forgot that the surface-based friction formula can be obtained by the formula . It seems that friction does not depend on the system states such as r, , α, and . Therefore, it can be treated as a second external input to the system.

% Parameters
m  = 0.111;
R  = 0.015;
g  = -9.8;
L  = 1.0;
d  = 0.03;
J  = 9.99e-6;
Mu = 0.3;
Fr = - Mu*m*g;  % <-- I think the sign is 'negative', since it is stopping the motion
H  = - m*g/(J/(R^2) + m);

% Matrices
A = [0 1 0 0;
     0 0 H 0;
     0 0 0 1;
     0 0 0 0];
B = [0 0;
     0 1/(J/(R^2) + m); % <-- Fr is injected in this channel
     0 0;
     1 0];
C = [1 0 0 0];
D = 0*C*B;


% Linearized state-space
ball_Fr = ss(A, B, C, D)

ball_Fr =
 
  A = 
       x1  x2  x3  x4
   x1   0   1   0   0
   x2   0   0   7   0
   x3   0   0   0   1
   x4   0   0   0   0
 
  B = 
          u1     u2
   x1      0      0
   x2      0  6.435
   x3      0      0
   x4      1      0
 
  C = 
       x1  x2  x3  x4
   y1   1   0   0   0
 
  D = 
       u1  u2
   y1   0   0
 
Continuous-time state-space model.

[Kez]

Hi Sam,

thank-you so much. I just have one question, it might be me but I cant see the friction effecting the ss model in any way?

[Kez]

God bless you Sam, you are a lifesaver.

[Sam Chak]

Hi @Kez, The state equation looks like this:

where the friction (a constant, assuming μ is also constant throughout the entire surface) is an external force.

[Kez]

Hi sam,

So the expansion of the B4x2 matrix against tau Fr, would it look like this?

0 0

0 (1/(J/(R^2) + m))-Fr

0 0

1 0

[Sam Chak]

Hi @Kez

From here

,

the whole equation is divided by the coefficient of the term

The 2nd column of the input matrix B is actually the coefficient of the friction .

[Kez]

Hi Sam,

Thank-you for all of the help and support you have given me.

I really appreciate it and I wish you well. Thank-you for taking the time to answer all of my questions.

I hope you have a great week