ANALYTICAL METHOD OF VECTOR ADDITIONVectorsCP-02Page 28Rev. 0ForceMagnitudeAnglexcomponentycomponentF90 lbf39F = 90 cos 39F = 90 sin 3911x 1yo o oF = (90) (.777)F = (90) (.629)1x 1yF = 69.9 lbfF = 56.6 lbf1x 1yF50 lbf120F = 50 cos 120F = 50 sin 12022x 2yo o oF = (50) (-.5)F = (50) (.866)2x 2yF = -25 lbfF = 43.3 lbf2x 2yF125 lbf250F = 125 cos 250F = 125 sin 25033x 3yo o oF = (125) (-.342)F = (125) (-.94)3x 3yF = -42.8 lbfF = -117.5 lbf3x 3yHere is an example using this model. Follow it through step by step.Example:Given three forces acting on an object, determine the magnitude and direction ofthe resultant force F ._{R}F = 90 lbf at 391oF = 50 lbf at 1202oF = 125 lbf at 2503oStep 1:First draw x and y coordinate axes on a sheet of paper. Then, draw F , F ,1 2and F from the point of origin. It is not necessary to be totally accurate3in placing the vectors in the drawing. The approximate location in theright quadrant is all that is necessary. Label the drawing as in the model(Figure 26).Step 2:Resolve each force into its rectangular coordinates.