Solving equations of static equilibrium for all angles
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I have a system of equations representing equations of equilibirum. I am trying to solve for all values of an input which satisfy that the result of three equations are simultaneously zero. I have three additional variables which I defined in terms of only alpha (the thing I need to solve). Which method is recommended to solve: loop, solve function, or symbolic tool box? Thank you for your feedback
Defining alpa where start is 270 degrees and converting to radians
deg2rad = pi / 180 ;
alpha = (270:1:630)*deg2rad ;
Variables in terms of things I need to solve for
a_y = 500*cos(alpha) + 600*sin(alpha);
c_x = 300*sin(alpha);
c_y = 800*cos(alpha) - 600*sin(alpha);
Equations that state the necessary conditions for alpha to satisfy
f_x = c_x + 300*sin(alpha) == 0;
f_y = -a_y + c_y - 300*cos(alpha) == 0;
m_c = 0.150*a_y + -75*cos(alpha) + -90*sin(alpha) == 0;
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Answers (1)
SAI SRUJAN
on 12 Oct 2023
Hi Maggie Forest,
I understand that you are trying to figure out an appropriate method to use for obtaining solutions within a specific range by using loops or the “solve” function in MATLAB.
One approach is to employ loops, which involves checking at specific intervals within the range [l, r]. Alternatively, the "solve" function in MATLAB provides the advantage of obtaining solutions within a continuous interval between [l, r].
You can follow the below given example to resolve the issue.
syms z
deg2rad = pi / 180 ;
alpha = (270:1:630)*deg2rad ;
a_y = 500*cos(z) + 600*sin(z);
c_x = 300*sin(z);
c_y = 800*cos(z) - 600*sin(z);
f_x = c_x + 300*sin(z) == 0;
f_y = -a_y + c_y - 300*cos(z) == 0;
m_c = 0.150*a_y + -75*cos(z) + -90*sin(z) == 0;
%mention constraints
symanswer=solve([f_x f_y m_c z>=alpha(1) z<=alpha(end)],z)
You can refer to the below documentation to understand more about “solve” in MATLAB
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