ANALYTICAL METHOD OF VECTOR ADDITION
Vectors
CP-02
Page 28
Rev. 0
Force
Magnitude
Angle
x component
y component
F
90 lbf
39
F = 90 cos 39
F = 90 sin 39
1
1x
1y
o
o
o
F = (90) (.777)
F = (90) (.629)
1x
1y
F = 69.9 lbf
F = 56.6 lbf
1x
1y
F
50 lbf
120
F = 50 cos 120
F = 50 sin 120
2
2x
2y
o
o
o
F = (50) (-.5)
F = (50) (.866)
2x
2y
F = -25 lbf
F = 43.3 lbf
2x
2y
F
125 lbf
250
F = 125 cos 250
F = 125 sin 250
3
3x
3y
o
o
o
F = (125) (-.342)
F = (125) (-.94)
3x
3y
F = -42.8 lbf
F = -117.5 lbf
3x
3y
Here 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 of
the resultant force F .R
F = 90 lbf at 39
1
o
F = 50 lbf at 120
2
o
F = 125 lbf at 250
3
o
Step 1:
First draw x and y coordinate axes on a sheet of paper. Then, draw F , F ,
1
2
and F from the point of origin. It is not necessary to be totally accurate
3
in placing the vectors in the drawing. The approximate location in the
right quadrant is all that is necessary. Label the drawing as in the model
(Figure 26).
Step 2:
Resolve each force into its rectangular coordinates.
