Index in position 1 exceeds array bounds when it doesn't?

Hey guys,

This is my first time posting on Matlab forums, hopefully I don't break any rules, but please don't be harsh if I do.

I am a mechanical engineering student in Hungary, and I am struggling to get a code working. I am working on a 4 bar mechanism, and I am currently trying to get my kinetostatistics values for the joints. On bar1 I have a constant angular velocity torque providing motor, so I am trying to account for that with 'Mhajto' in my code, which stands for the torque of said motor.

I'll enclose my code, so you can have a look at it. If I remove the part "+Mhajto" the code works fine. As soon as I introduce it to my code, it falls apart. Please help, I have 2 weeks to fix this :D

%Dénes Péter ÓE BGK Mechanizmus elmélet 2. házifeladat

%-----------------------PISZKOZAT--------------------
%cosinus tétel BD-re
%BD^2=AB^2+AD^2-2*AB*AD*cos(fiABD)
%BD=AB^2+AD^2-2*AB*AD*cos(fiABD);
%rBD=[BD*cos(fiABD); BD*sin(fiABD); 0]
%-----------------------PISZKOZAT VÉGE-----------------


clc
clear


%Ismert adatok


w1=60;
e1=0;
m1=3;
m2=6;
m3=4;
F0=[3; 0; 0];
%fiABC= deg2rad(180-143.42);
fiBAD= deg2rad(10);%(input("Írj be egy számot, pl 10.0: "));
%fiBC= deg2rad(46.58);
%fiADC=deg2rad(180-59.22);
%fiABD= deg2rad(166.70); 
L1=0.25*1000;
L2=0.65*1000;
L3=0.6*1000;
DB=L3;
AD=1*1000;
aA=[ 0; 0; 0];
vAx=0;
vAy=0;
vDx=0;
vDy=0;
vA=[vAx;vAy;0];
aD=[0 ; 0; 0];
vD=[vDx ; vDy; 0];
AB=L1;
BA=AB;
BC=L2;
CB=BC;
CD=L3;
DC=CD;
om1=[0; 0; w1];

%KÖREGYENLETEK
"A csuklók pozíciói"
A= [0;0;0]
D= [1000;0;0];
%Innentől kéne a for ciklus
H01= [cos(fiBAD) -sin(fiBAD) 0; sin(fiBAD) cos(fiBAD) 0; 0 0 1];
H12=[1 0 L1; 0 1 0; 0 0 1];
%H23=[cos(fiABC) -sin(fiABC) 0; sin(fiABC) cos(fiABC) 0; 0 0 1];
re2=[0; 0; 1];
rB=H01*H12*re2; %Ez is jó
B=[rB(1,1); rB(2,1); 0]


syms x y 
eqn1=(x-B(1,1)).^2+(y-B(2,1)).^2 == L2^2;
eqn2=(x-D(1,1)).^2+(y-D(2,1)).^2 == L3^2;

megoldas = solve([eqn1, eqn2], [x,y]);
CX = double(megoldas.x);
CY = double(megoldas.y);
C=[CX(2,1); CY(2,1); 0]

fiABC=rad2deg(acos((AB^2+BC^2-vecnorm(C)^2)/(2*AB*BC)));
fiABC_kieg= deg2rad(180 - fiABC);
D

%BDhossz=sqrt((B(1,1)-D(1,1))^2+(B(2,1)-D(2,1))^2);



%CIKLUSOZÁS


%for loop ágyazással 
"Helyvektorok"
rAB=[B(1,1)-A(1,1); B(2,1)-A(2,1);0]
rBC=[C(1,1)-B(1,1); C(2,1)-B(2,1);0]
rDC=[C(1,1)-D(1,1); C(2,1)-D(2,1);0]
%rAB=[AB*cos(fiBAD); AB*sin(fiBAD); 0]
%rDC=[DC*cos(fiADC); DC*sin(fiADC); 0]
%rBC=[BC*cos(fiBC); BC*sin(fiBC); 0]
%idáig jó


%H01b= [cos(fiBAD) -sin(fiBAD) 0; sin(fiBAD) cos(fiBAD) 0; 0 0 1];
%H12b=[1 0 L1; 0 1 0; 0 0 1];
%H23b=[cos(fiABC_kieg) -sin(fiABC_kieg) 0; sin(fiABC_kieg) cos(fiABC_kieg) 0; 0 0 1];
%%re2b=[L2; 0; 1];
%rC=H01b*H12b*H23b*re2b; %Ez is jó
%%C=[rC(1,1); rC(2,1); 0];
%"Redeklaráció, javítandó!" 



"Sebességek"
format bank
om1_x_rAB=cross(om1,rAB);
vB=double([vAx+om1_x_rAB(1,1); vAy+om1_x_rAB(2,1);0])




%Egyenletrendszer felírása a C pontra
syms vCx vCy om2 om3 w2 w3 om2_x_rBC om3_x_rDC 
%javítás: syms változók között ismert értéket nem adunk meg
om2= [0; 0; w2];
om3= [0; 0; w3];
om2_x_rBC=cross(om2,rBC);
om3_x_rDC=cross(om3,rDC);

vCx= vB(1,1)+om2_x_rBC(1,1)==vD(1,1)+om3_x_rDC(1,1);
vCy= vB(2,1)+om2_x_rBC(2,1)==vD(2,1)+om3_x_rDC(2,1);

sol = solve([vCx,vCy],[w2, w3]);
  "w2:";
  w2_mo = double(sol.w2)
  "w3:";
  w3_mo = double(sol.w3)
  
om2= [0; 0; sol.w2];
om3= [0; 0; sol.w3];
om2_x_rBC=cross(om2,rBC);
om3_x_rDC=cross(om3,rDC);

vCx= vB(1,1)+om2_x_rBC(1,1);
vCy= vB(2,1)+om2_x_rBC(2,1);
vC=double([vCx; vCy; 0])

"Gyorsulás"
eps1=[0; 0; e1];
eps1_x_rAB=cross(eps1,rAB);
aBx=aA(1,1)+eps1_x_rAB(1,1)-w1^2*rAB(1,1);
aBy=aA(2,1)+eps1_x_rAB(2,1)-w1^2*rAB(2,1);
aB=double([aBx; aBy; 0])

syms e2 e3 eps2_x_rBC eps3_x_rDC eps2 eps3 aCx aCy
eps2= [0; 0; e2];
eps3= [0; 0; e3];
eps2_x_rAB= cross(eps2,rAB);
eps3_x_rDC= cross(eps3,rDC);


aCx=aD(1,1)+eps3_x_rDC(1,1)-w3_mo^2*rDC(1,1)==aBx+eps2_x_rAB(1,1)-w2_mo*rBC(1,1);
aCy=aD(2,1)+eps3_x_rDC(2,1)-w3_mo^2*rDC(2,1)==aBy+eps2_x_rAB(2,1)-w2_mo*rBC(2,1);

 sol = solve([aCx,aCy],[e2,e3]);
  "e2:";
  e2_mo = double(sol.e2)
  "e3:";
  e3_mo = double(sol.e3)
  
eps2= [0; 0; sol.e2];
eps3= [0; 0; sol.e3];
eps2_x_rAB= cross(eps2, rAB);
eps3_x_rDC= cross(eps3, rDC);
aCx=aD(1,1)+eps3_x_rDC(1,1)-w3_mo^2*rDC(1,1);
aCy=aD(2,1)+eps3_x_rDC(2,1)-w3_mo^2*rDC(2,1);
aC=double([aCx; aCy; 0])


%inerciaerők

mA=0.25*m1;
mP=0.75*m1;
FA=0; 
rAP=[2/3*rAB(1,1); 2/3*rAB(2,1); 0];
eps1_x_rAP=cross(eps1, rAP);
aPx= aA(1,1)+eps1_x_rAP(1,1)-w1^2*rAP(1,1);
aPy= aA(2,1)+eps1_x_rAP(2,1)-w1^2*rAP(2,1);
aP=[aPx; aPy; 0]
FP=-mP*aP;
Fi1=FA+FP 
Fi1hossz=vecnorm(Fi1);
lambda1=(rAP(1,1)*FP(2,1)-rAP(2,1)*FP(1,1))/(rAB(1,1)*Fi1(2,1)-rAB(2,1)*Fi1(1,1));
double(lambda1);
rAT=rAP

mD=0.25*m3;
mR=0.75*m3;
FD=0 ;
rDR=[2/3*rDC(1,1); 2/3*rDC(2,1); 0];
eps3_x_rDR=cross(eps3, rDR);
aRx= aD(1,1)+eps3_x_rDR(1,1)-w3_mo^2*rDR(1,1);
aRy= aD(2,1)+eps3_x_rDR(2,1)-w3_mo^2*rDR(2,1);
aR=double([aRx; aRy; 0]);
FR=-mR*aR;
Fi3=FD+FR
Fi3hossz=vecnorm(double(Fi3));
lambda3=(rDR(1,1)*FR(2,1)-rDR(2,1)*FR(1,1))/(rDC(1,1)*Fi3(2,1)-rDC(2,1)*Fi3(1,1));
double(lambda3);
rDT=lambda3* rDC


mB=0.25*m2;
mQ=0.75*m2;
FB=[mB*aB(1,1); mB*aB(2,1); 0];
rBQ=[2/3*rBC(1,1); 2/3*rBC(2,1); 0];
eps2_x_rBQ=double(cross(eps2, rBQ));
aQx= aB(1,1)+eps2_x_rBQ(1,1)-w2_mo^2*rBQ(1,1);
aQy= aB(2,1)+eps2_x_rBQ(2,1)-w2_mo^2*rBQ(2,1);
aQ=[aQx; aQy; 0];
FQ=[aQ(1,1)*-mQ; aQ(2,1)*-mQ;0]
Fi2=FB+FQ;
Fi2hossz=vecnorm(Fi2);
lambda2=(rBQ(1,1)*FQ(2,1)-rBQ(2,1)*FQ(1,1))/(rBC(1,1)*Fi2(2,1)-rBC(2,1)*Fi2(1,1));
rBT = rBC * lambda2

"Virtuális teljesítmény"
vT1= double(vA + cross(om1,rAP))
vT2= double(vB + cross(om2,rBT))
vT3= double(vD + cross(om3,rDT))
Fi1_vT=Fi1(1,1)*vT1(1,1)+Fi1(2,1)*vT1(2,1);
Fi2_vT=Fi2(1,1)*vT2(1,1)+Fi2(2,1)*vT2(2,1);
Fi3_vT=Fi3(1,1)*vT3(1,1)+Fi3(2,1)*vT3(2,1);
F0_vC= F0(1,1)*vC(1,1);
Mhajto=double([0; 0; (Fi1_vT+Fi2_vT+Fi3_vT+F0_vC)/w1])

"Kinetostatikai számítások"


rBA=rAB*-1;

rCB=-1*rBC;

rCD=-1*rDC;
rAB
rAT
rBT1=-(rAB-rAT)

rCB
rBT2=rBT


%rDC
%rDT
%rCT3=-(rDC-rDT)



syms FAx FAy FCx FCy 

FAx_mo = FAx == ((rBA(1,1)*FAy)+Mhajto)/rBA(2,1)==-Fi1(1,1)-Fi2(1,1)-FCx;
FAy_mo = FAy == ((rBA(2,1)*FAx)+Mhajto)/rBA(1,1)==-Fi1(2,1)-Fi2(2,1)-FCy;

FCx_mo = FCx == (rBT2(1,1)*Fi2(2,1)-rBT2(2,1)*Fi2(1,1)+rBC(1,1)*FAy-F0(1,1)*rBC(2,1))/rBC(2,1)==-Fi1(1,1)-Fi2(1,1)-FAx-F0(1,1);
FCy_mo = FCy == (rBT2(2,1)*Fi2(1,1)-rBT2(1,1)*Fi2(2,1)+rBC(2,1)*FAx+F0(2,1)*rBC(2,1))/rBC(1,1)==-Fi1(2,1)-Fi2(2,1)-FAy-F0(2,1);


megoldasegyenlet = solve([FAx_mo, FAy_mo, FCx_mo, FCy_mo],[FAx FAy FCx FCy]);

'Az egye cuklókban ébredő erők'
FA=double([megoldasegyenlet.FAx; megoldasegyenlet.FAy;0])
FC=double([megoldasegyenlet.FCx; megoldasegyenlet.FCy;0])


FBx=-FA(1,1)-Fi1(1,1);
FBy=-FA(2,1)-Fi1(2,1);
FB=double([FBx;FBy;0])

FDx=-FC(1,1)-Fi3(1,1);
FDy=-FC(2,1)-Fi3(2,1);
FD=double([FDx; FDy; 0])

[Ajay Kumar]

Where exactly is the error? try to debug line by line. You'll get to know where is it exactly.

[Peter Denes]

Well the error message is at line :

FBy=-FA(2,1)-Fi1(2,1);

but it's because FA becomes [0], instead of [FAx; FAy; 0]

[Steven Lord]

A couple stylistic suggestions that might help you determine why you're receiving the error and how to correct it:

Sections

Your code seems to consist of sections, delimited by a comment and/or a string that gets displayed when you run your script. Consider taking this one step further and define them as actual code sections. One benefit this will give you is the ability to run individual sections, independent of the other sections. Once you've isolated a particular section as containing the cause of your error you can run the previous sections, save the variables, and experiement with different fixes by running that one section. load the saved variables when or if a particular fix experiment fails and you'll be back in a known good state for your next experiment.

Functions

Taking things one step further than code sections, turning those sections into functions will let you isolate what steps are performed inside the section from what happens outside. The function can accept only those pieces of information it needs to perform its task as inputs and return only the finished product back to the main script. Because what happens inside the function doesn't affect the data stored outside, you can debug the function by calling it with the appropriate inputs. If the function fails and needs to be changed you change it then call it again. Since the inputs "live" outside the function, they weren't affected by what happened in the incorrectly fixed function.

Variable Names

I don't know if the variable names you're using are tied to the way you (or the textbook) have defined the problem you're trying to solve. It's possible they're the labels used in a diagram of the mechanism you're modeling. But I'd consider using names that are a bit longer and more descriptive of their purpose. Longer names would require more typing (or taking advantage of tab completion) but it may help you identify errors in the logic of your code.

As an example if you were trying to add two variables, one named barLength and the other named plateArea, the fact that your first variable's units are suggested (by the name) to be a length and your second variable's units are suggested (again by the name) to be a length squared should look suspicious.

Comments

When I'm writing a sufficiently complicated piece of code, my first step is not to start writing the code. I'll start by writing comments describing the steps I want to follow to achieve the task. I'll break the task into pieces that I think are manageable as a single function or maybe one screen full of code. Having those steps well defined before I start reduces the chances that I have to rework my code significantly as it's being developed. If I'm writing functions or defining variables, the descriptions of the steps in the comments can suggest appropriate and descriptive names for those functions or variables. With this approach, sometimes I can explain what the code does to my colleagues simply by having them read the "story" that I wrote using the comments and function/variable names.

I know these are much broader suggestions that I expect you were hoping to receive. But if you choose to adopt them the process of breaking your code into code sections or functions, using descriptive variable names, and adding comments may lead you to a "Wait, that doesn't look right ..." moment that can help you locate the cause of the error and determine how to correct it.

[John D'Errico]

I want to upvote Steve's comment at least a hundred times. And then a hundred more.

[Peter Denes]

Thank you Steve, although I already wrote this code, and I am not going to implement functions into it this time, I already started working on another project in the same subject, and your ideas came in very handy, so I used your suggestions. I even contemplated building a GUI on it, but unfortunately my upcoming exams are more important.

Answer 1

[Ameer Hamza]

Explicitly considering the error, you mentioned in your question. It happens because 'Mhajto' is a 3x1 vector. And when you add it in the expression of FAx_mo, then FAx_mo also has dimensions of 3x1. FCx_mo is not dependent on Mhajto on Mhajto, so its dimensions are 1x1. I don't understand the logic of your code so I am not sure what is the correct thing to do here but changing the value of Mhajto to the following line will correct this error

Mhajto=double([(Fi1_vT+Fi2_vT+Fi3_vT+F0_vC)/w1]) % Mhajto is 1x1

However, note that with the new of Mhajto, the solve() function returns an empty vector, i.e., no solution exist

megoldasegyenlet = solve([FAx_mo, FAy_mo, FCx_mo, FCy_mo],[FAx FAy FCx FCy]);

which will result in a different error on the line

FBx=-FA(1,1)-Fi1(1,1);
FBy=-FA(2,1)-Fi1(2,1);

because FA has dimensions of 1x1.

An alternate solution is to keep Mhajto as a 3x1 vector and use their single elements in the following equations

FAx_mo = FAx == ((rBA(1,1)*FAy)+Mhajto(1))/rBA(2,1)==-Fi1(1,1)-Fi2(1,1)-FCx;
FAy_mo = FAy == ((rBA(2,1)*FAx)+Mhajto(2))/rBA(1,1)==-Fi1(2,1)-Fi2(2,1)-FCy;

Which of the above solution is correct depends on the formula you are trying to implement.

[Peter Denes]

Thank you Ameer,

unfortunately I have already tried Mhajto(3,1), and it yielded the same result.

Also, I can't change Mhajto (English: M_Input), because it's torque, which is interpreted in Z, not X. Of course, I could sacrifice formality, but I have also tried a similar thing, which was, instead of adding Mhajto, I added a simple scalar number (1).

Since posting, I have found a forum entry, regarding the solve() function empty vector with & without warning phenomena, but I am pretty sure my example has a solution, as it is not an abstract mathematical one, but calculation for a real-life mechanism.

I can't really understand why it works fine without Mhajto, but crashes as soon as I try to add either Mhajto, or a simple scalar value.

Looking forward to your reply.

[Ameer Hamza]

If the equations are related to a mechanical system, then it mush have a solution. The other problem is there might be a mistake somewhere in the equations. It is difficult to pinpoint the exact reason why solve() return an empty vector without understanding the system of equations and closely going through the code to find out the mistake. I suggest you to write down the equations on a piece of paper and try to solve them for one specific value of inputs. Then add a breakpoint to the code and run it line by line to see where the deviation occurs from your paper calculations. That will give you clues about the problem

[Peter Denes]

Thank you Ameer, I qill get back to you once I did that!

[Peter Denes]

Alright, I did the proofing of the code, and now it doesn't give me the error message. However, it still gives me false numbers, regarding the solution to FA, FB, FC, FD.

The problem is probably still on my part, probably a missing negative sign or mixed up variable. However, could you please reassure me, that I'm using the correct way to define a system of equtations?

syms FAx FAy FCx FCy  

FAx_mo =  ((rBA(1,1)*FAy)-Mhajto(3,1))/rBA(2,1)==-Fi1(1,1)-Fi2(1,1)-FCx;
FAy_mo =  ((rBA(2,1)*FAx)+Mhajto(3,1))/rBA(1,1)==-Fi1(2,1)-Fi2(2,1)-FCy;

FCx_mo =  (rBT2(1,1)*Fi2(2,1)-rBT2(2,1)*Fi2(1,1)+rBC(1,1)*FCy)/rBC(2,1)==-Fi1(1,1)-Fi2(1,1)-FAx;
FCy_mo =  (-rBT2(1,1)*Fi2(2,1)+rBT2(2,1)*Fi2(1,1)+rBC(2,1)*FCx)/rBC(1,1)==-Fi1(2,1)-Fi2(2,1)-FAy;


megoldasegyenlet = solve([FAx_mo, FAy_mo, FCx_mo, FCy_mo],[FAx FAy FCx FCy]);

[Ameer Hamza]

Yes, the syntax is correct to define a symbolic equation. You can try to put the solution back into the equations to see if you get any indication of where might be a mistake in writing the equations.

[Peter Denes]

Dear Ameer,

what exactly do you mean by "put the solution back into the equations"?

Do you mean like in this part of my code?

%Egyenletrendszer felírása a C pontra  -> system of equtations at point C
syms vCx vCy om2 om3 w2 w3 om2_x_rBC om3_x_rDC 
%javítás: syms változók között ismert értéket nem adunk meg %note for myself, I initially made the mistake of giving sym value to already known values
om2= [0; 0; w2];
om3= [0; 0; w3];
om2_x_rBC=cross(om2,rBC);
om3_x_rDC=cross(om3,rDC);
vCx= vB(1,1)+om2_x_rBC(1,1)==vD(1,1)+om3_x_rDC(1,1);
vCy= vB(2,1)+om2_x_rBC(2,1)==vD(2,1)+om3_x_rDC(2,1);
sol = solve([vCx,vCy],[w2, w3]);
  "w2:";
  w2_mo = double(sol.w2)
  "w3:";
  w3_mo = double(sol.w3)
  
om2= [0; 0; sol.w2];
om3= [0; 0; sol.w3];
om2_x_rBC=cross(om2,rBC);
om3_x_rDC=cross(om3,rDC);

[Ameer Hamza]

Yes, like this. And then put the values of om2_x_rBC and om3_x_rDC in the equations. Both sides of the equations will be equal, but it may help you to figure out, why the solution is incorrect.

[Peter Denes]

I tried it like this, but it yields the same results as before? (Just to clarify: I do get numbers now, they just aren't correct.)

It could very well be, that my equtations are wrong, but a friend and I have looked them over several times, and they seem correct.

syms FAx FAy FCx FCy  

FAx_mo =  ((rBA(1,1)*FAy)-Mhajto(3,1))/rBA(2,1)==-Fi1(1,1)-Fi2(1,1)-FCx;
FAy_mo =  ((rBA(2,1)*FAx)+Mhajto(3,1))/rBA(1,1)==-Fi1(2,1)-Fi2(2,1)-FCy;

FCx_mo =  (rBT2(1,1)*Fi2(2,1)-rBT2(2,1)*Fi2(1,1)+rBC(1,1)*FCy)/rBC(2,1)==-Fi1(1,1)-Fi2(1,1)-FAx;
FCy_mo =  (-rBT2(1,1)*Fi2(2,1)+rBT2(2,1)*Fi2(1,1)+rBC(2,1)*FCx)/rBC(1,1)==-Fi1(2,1)-Fi2(2,1)-FAy;


megoldasegyenlet = solve([FAx_mo, FAy_mo, FCx_mo, FCy_mo],[FAx FAy FCx FCy]);

syms FAx FAy FCx FCy  

FAx_mo =  ((rBA(1,1)*megoldasegyenlet.FAy)-Mhajto(3,1))/rBA(2,1)==-Fi1(1,1)-Fi2(1,1)-megoldasegyenlet.FCx;
FAy_mo =  ((rBA(2,1)*megoldasegyenlet.FAx)+Mhajto(3,1))/rBA(1,1)==-Fi1(2,1)-Fi2(2,1)-megoldasegyenlet.FCy;

FCx_mo =  (rBT2(1,1)*Fi2(2,1)-rBT2(2,1)*Fi2(1,1)+rBC(1,1)*megoldasegyenlet.FCy)/rBC(2,1)==-Fi1(1,1)-Fi2(1,1)-megoldasegyenlet.FAx;
FCy_mo =  (-rBT2(1,1)*Fi2(2,1)+rBT2(2,1)*Fi2(1,1)+rBC(2,1)*megoldasegyenlet.FCx)/rBC(1,1)==-Fi1(2,1)-Fi2(2,1)-megoldasegyenlet.FAy;

He, on the other, tried a different approach:

syms FAx FAy FCx FCy

FA = [FAx; FAy; 0];
FC = [FCx; FCy; 0];

FA_mo = (cross(rBA,FA)+Mhajto)/rBA == FA+Fi1+Fi2+FC;
FC_mo = (cross(rBT2,Fi2)+cross(rBC,FC))/rBC == FA+Fi1+Fi2+FC;

megoldasegyenlet = solve([FA_mo, FC_mo],[FAx FAy FCx FCy]);

This actually gave the correct solution for FA, however, the other forces for point B C and D are still wrong. What's more concerning is that this produces a 3x3 sym, which makes it impossible to use the aforementioned "put it back into the equtation" method, since I cannot take crossproduct from a 1x3 and a 3x3 matrix.

[Ameer Hamza]

Can you share the equations you are trying to implement in mathematical form? It wight make it easy to spot the error.

[Peter Denes]

I am sorry Ameer, I failed to check my e-mails. Sorry for the delay.

[Peter Denes]

Also sorry for the handwriting, visuals are not really my strongsuit, especially not when they are just drafts. However, I did not know that I would be posting them online, so if you need, I can rewrite them tomorrow!

[Peter Denes]

I rewrote it anyway.

Answer 2

[Peter Denes]

[Ameer Hamza]

Looking at the equations and code, it is difficult to follow step by step, especially since I don't have any knowledge about this system of equations. As a last resort, I suggest you solve one set of the equation with hand and compare your results with the values from MATLAB. It will definitely help you to see which equation is giving an unexpected result.

[Peter Denes]

I just realized I messed up the equtations I inserted here as a picture yesterday...ouch.

Anyway, a friend of mine was able to help me, so I'll attach the code section in case someone needs it in the future.

syms FAx FAy FCx FCy FBx FBy FDx FDy i j k 


FA = [FAx; FAy; 0];
FC = [FCx; FCy; 0];

crossA = (cross(rBA,FA)+Mhajto);
crossC = (cross(rBT2,Fi2)+cross(rBC,FC));
FA_mox = crossA(3,1)/rBA(1,1) == 0;
FA_moy = crossA(3,1)/rBA(2,1) == 0;
FC_mox = crossC(3,1)/rBC(1,1) == 0;
FC_moy = crossC(3,1)/rBC(1,1) == 0;

sumFx = FA(1,1)+Fi1(1,1)+Fi2(1,1)+FC(1,1) == 0;
sumFy = FA(2,1)+Fi1(2,1)+Fi2(2,1)+FC(2,1) == 0;

megoldasegyenlet = solve([FA_mox, FA_moy, FC_mox,  FC_mox, sumFx, sumFy],[FAx FAy FCx FCy]);

fprintf('Az egyes csuklókban ébredő erők')
FA=double([megoldasegyenlet.FAx; megoldasegyenlet.FAy;0])
FC=double([megoldasegyenlet.FCx; megoldasegyenlet.FCy;0])


FB = [FBx; FBy; 0];
FD = [FDx; FDy; 0];

eqn2 = FA + Fi1 + FB  == 0;


eqn3 = +F0 - FC + Fi3 + FD == 0;

FB_FD_mo = solve([eqn2, eqn3], [FBx FBy FDx FDy]);
FB = double([FB_FD_mo.FBx; FB_FD_mo.FBy; 0])
FD = double([FB_FD_mo.FDx; FB_FD_mo.FDy; 0])

Also, thanks for all the help and feedback, it was nice talking to you!

[Ameer Hamza]

I am glad that the issue is resolved. Sometimes, when implementing mathematical equations, even a trivial mistake can make the code very difficult to debug.