1) A force vector acting along a line between two points can be represented as a Cartesian vector using the position vector r from one point to the other and the unit vector u=r/r. 2) Example problems show calculating the force vectors for cables supporting structures by determining r between the cable endpoints, taking the unit vector, and multiplying by the force magnitude. 3) The resultant force on a wall hook supported by two cables was found by adding the Cartesian force vectors for each cable.

1. 2.8 Force Vector Directed
along a Line
In 3D problems, direction of F is specified by
2 points, through which its line of action lies
F can be formulated as a Cartesian vector
F = F u = F (r/r)
Note that F has units of
forces (N) unlike r, with
units of length (m)

2. 2.8 Force Vector Directed
along a Line
Force F acting along the chain can be
presented as a Cartesian vector by
- Establish x, y, z axes
- Form a position vector r along length of
chain

3. 2.8 Force Vector Directed
along a Line
Unit vector, u = r/r that defines the direction
of both the chain and the force
We get F = Fu

4. 2.8 Force Vector Directed
along a Line
Example 2.13
The man pulls on the cord
with a force of 350N.
Represent this force acting
on the support A, as a
Cartesian vector and
determine its direction.

5. 2.8 Force Vector Directed
along a Line
Solution
End points of the cord are A (0m, 0m, 7.5m)
and B (3m, -2m, 1.5m)
r = (3m – 0m)i + (-2m – 0m)j + (1.5m –
7.5m)k
= {3i – 2j – 6k}m
Magnitude = length of cord AB
r = (3m ) + (− 2m ) + (− 6m ) = 7m
2 2 2
Unit vector, u = r /r
= 3/7i - 2/7j - 6/7k

6. 2.8 Force Vector Directed
along a Line
Solution
Force F has a magnitude of 350N, direction
specified by u
F = Fu
= 350N(3/7i - 2/7j - 6/7k)
= {150i - 100j - 300k} N
α = cos-1(3/7) = 64.6°
β = cos-1(-2/7) = 107°
γ = cos-1(-6/7) = 149°

7. 2.8 Force Vector Directed
along a Line
Example 2.14
The circular plate is
partially supported by
the cable AB. If the
force of the cable on
the
hook at A is F = 500N,
express F as a
Cartesian vector.

8. 2.8 Force Vector Directed
along a Line
Solution
End points of the cable are (0m, 0m, 2m) and B
(1.707m, 0.707m, 0m)
r = (1.707m – 0m)i + (0.707m – 0m)j
+ (0m – 2m)k
= {1.707i + 0.707j - 2k}m
Magnitude = length of cable AB
r= (1.707m )2 + (0.707m )2 + (− 2m )2 = 2.723m

9. 2.8 Force Vector Directed
along a Line
Solution
Unit vector,
u = r /r
= (1.707/2.723)i + (0.707/2.723)j – (2/2.723)k
= 0.6269i + 0.2597j – 0.7345k
For force F,
F = Fu
= 500N(0.6269i + 0.2597j – 0.7345k)
= {313i - 130j - 367k} N

10. 2.8 Force Vector Directed
along a Line
Solution
Checking
F= (313) + (130) + (− 367 )
2 2 2
= 500 N
Show that γ = 137° and
indicate this angle on the
diagram

11. 2.8 Force Vector Directed
along a Line
Example 2.15
The roof is supported by
cables. If the cables exert
FAB = 100N and FAC = 120N
on the wall hook at A,
determine the magnitude of
the resultant force acting at
A.

12. 2.8 Force Vector Directed
along a Line
Solution
rAB = (4m – 0m)i + (0m – 0m)j + (0m – 4m)k
= {4i – 4k}m
rAB = (4m )2 + (− 4m )2 = 5.66m
FAB = 100N (rAB/r AB)
= 100N {(4/5.66)i - (4/5.66)k}
= {70.7i - 70.7k} N

13. 2.8 Force Vector Directed
along a Line
Solution
rAC = (4m – 0m)i + (2m – 0m)j + (0m – 4m)k
= {4i + 2j – 4k}m
rAC = (4m )2 + (2m )2 + (− 4m )2 = 6m
FAC = 120N (rAB/r AB)
= 120N {(4/6)i + (2/6)j - (4/6)k}
= {80i + 40j – 80k} N

14. 2.8 Force Vector Directed
along a Line
Solution
FR = FAB + FAC
= {70.7i - 70.7k} N + {80i + 40j – 80k} N
= {150.7i + 40j – 150.7k} N
Magnitude of FR
FR = (150.7 )2 + (40)2 + (− 150.7 )2
= 217 N