1. 2.6 Addition and Subtraction
of Cartesian Vectors
Example
Given: A = Axi + Ayj + AZk
and B = Bxi + Byj + BZk
Vector Addition
Resultant R = A + B
= (Ax + Bx)i + (Ay + By )j + (AZ + BZ) k
Vector Substraction
Resultant R = A - B
= (Ax - Bx)i + (Ay - By )j + (AZ - BZ) k

2. 2.6 Addition and Subtraction
of Cartesian Vectors
Concurrent Force Systems
- Force resultant is the vector sum of all
the forces in the system
FR = ∑F = ∑Fxi + ∑Fyj + ∑Fzk
where ∑Fx , ∑Fy and ∑Fz represent the
algebraic sums of the x, y and z or i, j or k
components of each force in the system

3. 2.6 Addition and Subtraction
of Cartesian Vectors
Force, F that the tie down rope exerts on the
ground support at O is directed along the rope
Angles α, β and γ can be solved with axes x, y
and z

4. 2.6 Addition and Subtraction
of Cartesian Vectors
Cosines of their values forms a unit vector u that
acts in the direction of the rope
Force F has a magnitude of F
F = Fu = Fcosαi + Fcosβj + Fcosγk

5. 2.6 Addition and Subtraction
of Cartesian Vectors
Example 2.8
Express the force F as Cartesian vector

6. 2.6 Addition and Subtraction
of Cartesian Vectors
Solution
Since two angles are specified, the third
angle is found by
cos 2 α + cos 2 β + cos 2 γ = 1
cos 2 α + cos 2 60o + cos 2 45o = 1
cos α = 1 − (0.5) − (0.707 ) = ±0.5
2 2
Two possibilities exit, namely
α = cos −1 (0.5) = 60o or α = cos −1 (− 0.5) = 120o

7. 2.6 Addition and Subtraction
of Cartesian Vectors
Solution
By inspection, α = 60° since Fx is in the +x
direction
Given F = 200N
F = Fcosαi + Fcosβj + Fcosγk
= (200cos60°N)i + (200cos60°N)j
+ (200cos45°N)k
= {100.0i + 100.0j + 141.4k}N
Checking: F = Fx2 + Fy2 + Fz2
= (100.0) + (100.0) + (141.4)
2 2 2
= 200 N

8. 2.6 Addition and Subtraction of
Cartesian Vectors
Example 2.9
Determine the magnitude and coordinate
direction angles of resultant force acting on
the ring

9. 2.6 Addition and Subtraction
of Cartesian Vectors
Solution
Resultant force
FR = ∑F
= F1 + F2
= {60j + 80k}kN
+ {50i - 100j + 100k}kN
= {50j -40k + 180k}kN
Magnitude of FR is found by
FR = (50)2 + (− 40)2 + (180)2
= 191.0 = 191kN

10. 2.6 Addition and Subtraction
of Cartesian Vectors
Solution
Unit vector acting in the direction of FR
uFR = FR /FR
= (50/191.0)i + (40/191.0)j +
(180/191.0)k
= 0.1617i - 0.2094j + 0.9422k
So that
cosα = 0.2617 α = 74.8°
cos β = -0.2094 β = 102°
cosγ = 0.9422 γ = 19.6°
*Note β > 90° since j component of uFR is negative

11. 2.6 Addition and Subtraction
of Cartesian Vectors
Example 2.10
Express the force F1 as a Cartesian vector.

12. 2.6 Addition and Subtraction
of Cartesian Vectors
Solution
The angles of 60° and 45° are not coordinate
direction angles.
By two successive applications of
parallelogram law,

13. 2.6 Addition and Subtraction
of Cartesian Vectors
Solution
By trigonometry,
F1z = 100sin60 °kN = 86.6kN
F’ = 100cos60 °kN = 50kN
F1x = 50cos45 °kN = 35.4kN
kN
F1y = 50sin45 °kN = 35.4kN
F1y has a direction defined by –j,
Therefore
F1 = {35.4i – 35.4j + 86.6k}kN

14. 2.6 Addition and Subtraction
of Cartesian Vectors
Solution
Checking:
F1 = F12 + F12 + F12
x y z
= (35.4)2 + (− 35.4)2 + (86.6)2 = 100 N
Unit vector acting in the direction of F1
u1 = F1 /F1
= (35.4/100)i - (35.4/100)j + (86.6/100)k
= 0.354i - 0.354j + 0.866k

15. 2.6 Addition and Subtraction
of Cartesian Vectors
Solution
α1 = cos-1(0.354) = 69.3°
β1 = cos-1(-0.354) = 111°
γ1 = cos-1(0.866) = 30.0°
Using the same method,
F2 = {106i + 184j - 212k}kN

16. 2.6 Addition and Subtraction
of Cartesian Vectors
Example 2.11
Two forces act on the hook. Specify the
coordinate direction angles of F2, so that the
resultant force FR acts along the positive y axis
and has a magnitude of 800N.

17. 2.6 Addition and Subtraction
of Cartesian Vectors
Solution
Cartesian vector form
FR = F1 + F2
F1 = F1cosα1i + F1cosβ1j + F1cosγ1k
= (300cos45°N)i + (300cos60°N)j
+ (300cos120°N)k
= {212.1i + 150j - 150k}N
F2 = F2xi + F2yj + F2zk

18. 2.6 Addition and Subtraction
of Cartesian Vectors
Solution
Since FR has a magnitude of 800N and acts
in the +j direction
FR = F1 + F2
800j = 212.1i + 150j - 150k + F2xi + F2yj + F2zk
800j = (212.1 + F2x)i + (150 + F2y)j + (- 50 + F2z)k
To satisfy the equation, the corresponding
components on left and right sides must be equal

19. 2.6 Addition and Subtraction
of Cartesian Vectors
Solution
Hence,
0 = 212.1 + F2x F2x = -212.1N
800 = 150 + F2y F2y = 650N
0 = -150 + F2z F2z = 150N
Since magnitude of F2 and its components
are known,
α1 = cos-1(-212.1/700) = 108°
β1 = cos-1(650/700) = 21.8°
γ1 = cos-1(150/700) = 77.6°