The Magnitde of Fc: Graphically

Sum of components of vectors in x and y direction must be zero. You can also find magnitude and angle using law of cosines

 

F_A + F_B + F_C = 0.

In components form,
F_Ax + F_Bx + F_Cx = 0. .......(1)
and
F_Ay + F_By + F_Cy = 0 ......(2)

Now subtitute the values in equation 1 and 2.

 

Graphically you draw a triangle of forces, with side lengths proportional to the magnitude of the forces. One side of length corresponding to Fb (80 mm for example), the other (Fa) 100 mm and the angle between them 30°. Then measure the length of the third side with a ruler and measure the angle alpha with a compass.

If you want to calculate the length of the third force, the simplest way is the law of cosines: Fc² =Fa²+Fb²-2FaFbcos(30°). And you get the angle alpha by the law of sines:sin(alpha)/sin(30)=Fb/Fc.

But you can calculate also with the horizontal and vertical components, as you started, using that the sum of both the horizontal (x) and vertical (y) components must be zero. (Fbx+Fax+Fcx=0, Fby+Fay+Fcy=0). For that calculation, you need to take care of the signs: Fbx is negative, Fbx=-69.3 N. When you get the components Fcx and Fcy, the magnitude of Fc is Fc=√(Fcx²+Fcy²)and the angle alpha = arctan(|Fcy/Fcx|)

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