How to solve this simple system of 2 equation in MATLAB

Hi All;

I wonder if Matlab can solve this two equations to find the Fx and Fy as a function in the other constants. I know I can do it by hand just want to validate. Please see the initial code below

syms D L COST SINT Fx Fy
Fx * COST + Fy *SINT - D =0; 
Fx * SINT + Fy * COST -L =0;
solve ( text: the two equations for Fx and Fy)

Thank you for your valuable suggestion

Aziz

Answer 1

[John BG]

Hi Abdulaziz

this is John BG <mailto:[e-mail] [e-mail]>

syms v Fx Fy D L
  b=[D ;L]
  b =
   D
   L
  A=[(1-v^2)^.5 v;v (1-v^2)^.5]
  A =
  [ (1 - v^2)^(1/2),               v]
  [               v, (1 - v^2)^(1/2)]
  F=A\b
   =
   (D*v^2 - D + L*v*(1 - v^2)^(1/2))/((1 - v^2)^(1/2)*(2*v^2 - 1))
                             (D*v - L*(1 - v^2)^(1/2))/(2*v^2 - 1)

.

these are the expressions

.

    Fx=F(1)
     =
    (D*v^2 - D + L*v*(1 - v^2)^(1/2))/((1 - v^2)^(1/2)*(2*v^2 - 1))
     Fy=F(2)
    Fy =
    (D*v - L*(1 - v^2)^(1/2))/(2*v^2 - 1)

.

with

    v=sin(t)

.

if you find this answer useful would you please be so kind to consider marking my answer as Accepted Answer?

To any other reader, if you find this answer useful please consider clicking on the thumbs-up vote link

thanks in advance for time and attention

John BG

<mailto:[e-mail] [e-mail]>

Answer 2

[Stephan]

Hi,

you could do so:

% declare syms
syms D L t;

% Coefficient Matrix
A = [cos(t) sin(t); sin(t) cos(t)];

% RHS
b = [D ; L];

% Unknown: Fx, Fy
F = A\b;

% create Matlab function
fun = matlabFunction(F);

% Test for D = 0, L = -1 and t = pi()
D = 0;
L = -1;
t = pi();
[F] = fun(D, L, t)

this gives you a vector F containing Fx and Fy:

F =

      0.0000
      1.0000

or you do the same in a live script with symbolic toolbox and get the same result but nice:

Best regards

Stephan

[Abdulaziz Abutunis]

Thank you, Stephan, I meant I want to show Fx and Fy as a function of the other variables and not the final results.

[Jan]

@Abdulaziz Abutunis: The shown output for F is the x and y component already in dependence to the other variables. So what do you call "not the final result"?

[John BG]

Abdulaziz may not need the anonymous function, just the expressions of the results, have a look at my answer.

[Abdulaziz Abutunis]

That what I meant John. Sorry, Stephan if my question statement was not clear enough.

Answer 3

[Abdulaziz Abutunis]

Thank you John and all of you who have valuable suggestions. John, that was exactly what I want

[John BG]

Thanks Abdulaziz

feel free to ask me about having a look at any other particular question that you may consider I would be able to assist with.

Regards

John BG

<mailto:[e-mail] [e-mail]>