# Incorrect Answer Calculated For Force Calculation

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NTG on 5 Feb 2020
Commented: Jim Riggs on 6 Feb 2020
I might be making a really stupid mistake but in order to find the magnitude and angle of a 3rd force vector when 2 other forces are given, I used the following bit of code
syms F3 y
F1 = 9;
F2 = 18;
a = 60;
b = 45;
[solF3,soly] = solve((F3*sind(y)== -(F1*sind(a)-cosd(b))), F3*cosd(y)== -(F1*cosd(a)-F2*sind(b)))
For some reason it is returning the incorrect answer. Is there another way I'm supposed to code a solution to this? Jim Riggs on 5 Feb 2020
Impossible to say without a diagram of the problem.
NTG on 5 Feb 2020
Just updated with diagram

David Hill on 5 Feb 2020
Vector addition, should reference all angles from same location.
[x,y]=pol2cart(60*pi/180,9);
[xx,yy]=pol2cart(-135*pi/180,18);
[c,F]=cart2pol(-(x+xx),-(y+yy));
c=c*180/pi;%convert c to degrees

David Hill on 6 Feb 2020
When using polar coordinates, the angle must be reference to the same location. If you don't use polar coordinates, then where the angles are referenced does not matter. Fundamentally, this is just a vector addition problem.
Jim Riggs on 6 Feb 2020
This problem is defined (by the figure) in a Cartesion frame.
No sense in creating confusion with polar coordinates.

Jim Riggs on 5 Feb 2020
Edited: Jim Riggs on 6 Feb 2020
It is assumed that you want to find F3 and y such that the system is in equilibrium (this is not stated)
(and I disagree with David Hill - you may define the angle any way that you like. You have made a good diagram to work from that clearly defines your terms)
You have two equations with two unkbowns that can be solved to get the solution.
( Note that in your first equation, you are missing a "F2" factor s/b: "F2*cosd(b)" )
The two equations are:
F3*cosd(y) + F1*cosd(a) - F2*sind(b) == 0 % X-direction
F3*sind(y) + F1*sind(a) - F2*cosd(b) == 0 % Y-direction
or
F3*cosd(y) == -F1*cosd(a) + F2*sind(b) % X-direction
F3*sind(y) == -F1*sind(a) + F2*cosd(b) % Y-direction