Incorrect Answer Calculated For Force Calculation

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]

Impossible to say without a diagram of the problem.

[NTG]

Just updated with diagram

Answer 1

[David Hill]

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]

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]

This problem is defined (by the figure) in a Cartesion frame.

No sense in creating confusion with polar coordinates.

Answer 2

[Jim Riggs]

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