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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

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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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

Jan
on 7 May 2018

John BG
on 7 May 2018

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Abdulaziz Abutunis
on 7 May 2018

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