Equation of three variables but...

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Son Goku ssj4 2013-3-28

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Hello, I've got an equation which I don't know how to solve in Matlab.

F1*cos(alpha1) + F2*cos(alpha2) = 0
F1*sin(alpha1) - F2*sin(alpha2) = F3
-F2*sin(alpha2)*a + F2*cos(alpha2)*b = F3*c

Variables: F1 | F2 | alpha2

a = 330, b=183, c=124, F3 = 1200, alpha1 = 50°

Is there a way to solve it?

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Answer 1

Walter Roberson 2013-3-28

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syms F1, F2, alpha2
a = 330; b=183; c=124; F3 = 1200; alpha1 = 50;
[F1sol, F2sol, alpha2sol] = solve( F1*cosd(alpha1) + F2*cosd(alpha2), F1*sind(alpha1) - F2*sind(alpha2) - F3, -F2*sind(alpha2)*a + F2*cosd(alpha2)*b - F3*c, {F1, F2, alpha2} );

You might want to simplify() the results.

The result might perhaps have a RootOf() in F2 and alpha2.

The result will be symbolic rather than numeric. If you want the numeric result, use double() around the answers.

Note: there are two answers. F1 is the same for both, F2 is positive of negative of the same value for both, and alpha2 differs between the two.

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Answer 2

Son Goku ssj4 2013-3-28

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I tried something similar the another time but I hit the wall with:

Undefined function 'cosd' for input arguments of type 'sym'.

I don't know what's the problem here. Can't have "sym" in cos / sin / tan etc.? So still not solved.

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Son Goku ssj4 2013-3-31

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Thanks, the numerics are working, but... I still don't know the solution. My equations are (a little rewritten to match my program):

 syms F_I F_C alpha_C_rad
 a = 482.23; b=155.09; c=198.61; d=a-b; Fg = 1200; alpha_I = 61.72;
 alpha_I_rad = alpha_I * pi/180; %degree to radians
display(' Uvolnění k bodu C ')
[F_C_sol, F_I_sol, alpha_C_sol] = solve( ...
    F_I*cos(alpha_I_rad) - F_C*cos(alpha_C_rad) ==0 , ...
    F_I*sin(alpha_I_rad) + F_C*sin(alpha_C_rad) - Fg == 0, ...
    -F_I*cos(alpha_I_rad)*(c)-F_I*sin(alpha_C_rad)*(a) + Fg*(d) == 0, ...
    F_I, F_C, alpha_C_rad)
 F_I = subs(F_I_sol);
    disp(['F_I = ' num2str(F_I) ' N']);
 F_C = subs(F_C_sol);
    disp(['F_C = ' num2str(F_C) ' N']);
 alpha_C_rad = subs(alpha_C_sol);
 alpha_C = alpha_C_rad / (pi/180);
    disp(['alpha_C = ' num2str(alpha_C) ' N']);

The answer was:

F_I = 1641.2206 N
F_C = -815.362 N
alpha_C = 162.4893 N
SR Suma(X) = -1.1369e-13 N ... (mělo by být = 0)
SR Suma(y) = 2.2737e-13 N ... (mělo by být = 0)
SR Suma(MoC) = -458.848 Nm ... (mělo by být = 0)
SR Suma(MoI) = -458.848 Nm ... (mělo by být = 0)

Where SR SUMA (X), (Y), (MoC) (MoI) should be ZERO. In this particular case it is zero for (X) and (Y) but not for the moment MoC / MoI.

I still don't have the answer. Starting to feel pretty desperate :-/ The problem is this:

http://songokussj4.abonas.com/screeny/2013-03-31_014622.jpg

Son Goku ssj4 2013-3-31

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SR Suma(X) is sum of forces affecting x orientation (horizontal)

SR Suma(Y( is sum of all forces affecting y orientation (vertical)

SR Suma(MoC) is sum of moments (force x didstance) to one point (C)

I've got the result.

 F_C must be cca 643.1N
 F_I must be 756.2N
 alpha_C must be 56.1° (degree)

My methods for this were two.

 (1)
 From the third equation express F_I. I've got 756,2N
 Then 1. and 2. equations do square and I've got:
 F_C^2 *cos^2(alpha_C) = F_I^2*cos^2(alpha_I)
 F_C^2 *sin^2(alpha_C) = (Fg - F_I * sin(alpha_I))
 1. -> add (sum?) -> 2. 
 F_C = sqrt(756.2^2*cos^2(61.72)+(1200-756.2*sin(61.72))^2) which is 643 and the alpha from 1. equation is 56.1°

The second method:

 (2)
 From the third equation express F_I -> 756,2N
 Then the 1st & 2nd equation divide (1)/(2) which get me:
 -cotan(alpha_C) = (F_I*cos(alpha_I))/(Fg - F_I*sin(alpha_I))
 alpha_C = arcus cotangens... -> 56.1°
 Then from 1st equation -> F_C = 643N

But I don't know how to do this only with the help of matlab. F_I = 1640 / -1050 | F_C = -2180 / -815 is incorrect. :-) But man I'm glad I found the right result.

Walter Roberson 2013-3-31

How can you use the third equation to expression F_I in one step? The third equation involves alpha_C_rad, so you would only be able to express it dependent on alpha_C_rad instead of getting a numeric value in one step ??

Son Goku ssj4 2013-3-31

在 MATLAB Online 中打开

Ah, sorry. I was referring to this system of equation:

http://songokussj4.abonas.com/screeny/2013-03-31_014622.jpg

One has

-F_I*cos(alpha_I)*c-F_I*sin(alpha_I)*a+Fg*d = 0

The other has

F_C*sin(alpha_C)*a-F_C*cos(alpha_C)*c-Fg*b = 0

The equations -must- be equal so I chose the first one.

If I chose the second, I don't seem too have the ability to solve the system of equations. And if I try it in matlab, it gives me different results (F_C, F_I, alpha_C) so I have chosen the first one where the only variable is F_I.

(Not the task I wrote in the first question, I know. Sorry.)

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