ME 2560 Statics
Vector Components and Vector Addition (2D)
Cartesian Components of Vectors (2D)
o Given the magnitude of a vector and the direction of the
vector relative to a set of reference axes, the vector can be
expressed in terms of its components along those axes.
Y
o For our convenience, it is usually beneficial to have the
reference axes be mutually perpendicular.
o In the diagram, V x and V y represent the components of the
vector V along the mutually perpendicular X and Y axes.
X
o The parallelogram formed by V x and V y is now a rectangle, and the triangle formed
by V x and V y is now a right triangle.
o So, if the magnitude of the vector V is | V |  V , we now have
V  V x  V y  V cos( )  i  V sin( )  j
Example 1:
Given: A force F has magnitude F  100 (lbs) and angle   30 (deg) .
Find: Express the force F in terms of the unit vectors i and j .
Solution:
30o
F  100cos(30) i  100sin(30) j  86.6 i  50 j (lb)
Example 2:
Given: A force F has magnitude F  100 (lbs) and angle   120 (deg) .
Find: Express the force F in terms of the unit vectors i and j .
Solution:
F  100cos(120) i  100sin(120) j  50 i  86.6 j (lb)
120o
Kamman – ME 2560: page 1/3
Example 3:
Given: A force F  70 i  50 j (lb) .
?
Find: The magnitude and direction of F .
?
Solution:
 35.54 (deg)
F  702  502  86.0 (lbs) and   tan 1 (50 / 70)  
0.6202 (rad)
Example 4:
Given: A force F  50 i  70 j (lb) .
?
Find: The magnitude and direction of F .
?
Solution:
F  (50) 2  702  86.0 (lb)
54.46  180  125.5 (deg)
 0.9505    2.191 (rad)
  tan 1 (70 / 50)  
Kamman – ME 2560: page 2/3
Vector Addition using Cartesian Components (2D)
To add two or more vectors, simply express them in terms of the same unit vectors,
and then add like components.
Example 5:
Given: F1  150 (lbs) , 1  20 (deg)
F2  100 (lbs) , 2  60 (deg)
Find: a) The total force F acting on the support in terms of the unit vectors shown.
b) the magnitude and direction of F .
Solution:
a) The total force is the vector sum of the two forces.
F1  150cos(20) i  150sin(20) j  140.95 i  51.3 j (lb)
F2  100cos(60) i  100sin(60) j  50 i  86.6 j (lb)
F  F1  F2  (140.95  50) i  (51.3  86.6) j  90.95 i  137.9 j (lb)
b) F  90.952  137.92  165.2 (lb)
 56.59 (deg)
0.9877 (rad)
  tan 1 137.9 / 90.95   
Kamman – ME 2560: page 3/3