# 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? ##### 2 CommentsShowHide 1 older comment
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
##### 2 CommentsShowHide 1 older comment
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