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How to solve this simple system of 2 equation in MATLAB

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Abdulaziz Abutunis
Abdulaziz Abutunis on 4 May 2018
Commented: John BG on 10 May 2018
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

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Accepted Answer

John BG
John BG on 6 May 2018
Edited: John BG on 6 May 2018
Hi Abdulaziz
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

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More Answers (2)

Stephan
Stephan on 4 May 2018
Edited: Stephan on 4 May 2018
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

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Jan
Jan on 7 May 2018
@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
John BG on 7 May 2018
Abdulaziz may not need the anonymous function, just the expressions of the results, have a look at my answer.
Abdulaziz Abutunis
Abdulaziz Abutunis on 7 May 2018
That what I meant John. Sorry, Stephan if my question statement was not clear enough.

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Abdulaziz Abutunis
Abdulaziz Abutunis on 7 May 2018
Thank you John and all of you who have valuable suggestions. John, that was exactly what I want

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John BG
John BG on 10 May 2018
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

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