ME 2560 Statics
Position Vectors, Unit Vectors, and Forces Acting Along Lines
Position Vectors and Unit Vectors
o The diagram shows two points A and B and
their coordinates relative to a known origin.
o The position vectors of the two points relative
to the origin of the XYZ axes are
r A  xA i  y A j  z A k
r B  xB i  y B j  z B k
o Using the concept of vector addition, we can
write: r B  r A  r B / A
o Here, the vector r B/ A represents the position vector of B relative to A.
o So, given the coordinates of A and B, we can find the vector r B/ A as follows
r B / A  r B  r A   xB  x A  i   y B  y A  j   z B  z A  k
o The diagram also shows a unit vector u which points in the direction of r B/ A . We can
calculate u by dividing r B/ A by its magnitude
u
r B/ A
| r B/ A |

 xB  x A  i   y B  y A  j   z B  z A  k
| r B/ A |
| r B/ A |
| r B/ A |
o The magnitude of r B/ A is | r B / A |  ( xB  x A )2  ( y B  y A )2  ( z B  z A )2
o The angles that u makes with the X, Y, and Z axes are
  xB  x A  
 | r B / A | 


 x  cos 1 
  yB  y A  
 | r B / A | 


 y  cos 1 
  zB  z A  
 | r B / A | 


 z  cos 1 
Forces Acting Along Lines
o If a force F acts along the line from A to B, then we can write
F | F |u
Kamman – ME 2560: page 1/2
Example:
Given: Coordinates of A and B: A (8,2,4) (ft) and B(3,6,12) (ft)
Force F of magnitude 200 (lb) acts along the line from A to B
Find:
r B/ A the position vector of B relative to A; u the unit vector that points along the
line AB, and the force F , and the angles that F makes with the X, Y, and Z axes.
Solution:
o
r B/ A  (3  8) i  (6  2) j  (12  4) k  5 i  4 j  8 k
o | r B / A |  52  42  82  10.247 (ft)
5
8
4
o u   10.247
 i   10.247
 j   10.247
 k  0.4880 i  0.3904 j  0.7807 k
o
F  200 u  97.6 i  78.1 j  156 k (lb)
o  x  cos1  0.4880   119 (deg) ,  y  cos 1  0.3904   67 (deg) ,
 z  cos1  0.7807   38.7 (deg)
Kamman – ME 2560: page 2/2