Force Vector (cont’)
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
 Unit vector, u = r/r that defines the
direction of both the chain and the force
 We get F = Fu
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.
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
r
3m 2   2m 2   6m 2
 7m
Magnitude = length of cord AB
Unit vector,
u = r /r
= 3/7i - 2/7j - 6/7k
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°
2.9 Dot Product
 Dot product of vectors A and B is
written as A·B (Read A dot B)
 Define the magnitudes of A and B and
the angle between their tails
A·B = AB cosθ where 0°≤ θ ≤180°
 Referred to as scalar product of vectors
as result is a scalar
2.9 Dot Product
 Laws of Operation
1. Commutative law
A·B = B·A
2. Multiplication by a scalar
a(A·B) = (aA)·B = A·(aB) = (A·B)a
3. Distribution law
A·(B + D) = (A·B) + (A·D)
2.9 Dot Product
 Cartesian Vector Formulation
- Dot product of Cartesian unit vectors
i·i = (1)(1)cos0° = 1
i·j = (1)(1)cos90° = 0
- Similarly
i·i = 1
j·j = 1
k·k = 1
i·j = 0
i·k = 1 j·k = 1
2.9 Dot Product
 Cartesian Vector Formulation
 Dot product of 2 vectors A and B
A·B = AxBx + AyBy + AzBz
 Applications
 The angle formed between two vectors or
intersecting lines.
θ = cos-1 [(A·B)/(AB)]
0°≤ θ ≤180°
 The components of a vector parallel and
perpendicular to a line.
Aa = A cos θ = A·u
Example 2.17
The frame is subjected to a horizontal
force F = {300j} N. Determine the
components of this force parallel and
perpendicular to the member AB.
Solution
Since




r
2i  6 j  3k

u B  B 
rB
22  62  32



 0.286i  0.857 j  0.429k
Thus


FAB  F cos 





 F .u B  300 j   0.286i  0.857 j  0.429k 
 (0)(0.286)  (300)(0.857)  (0)(0.429)
 257.1N
Solution
Since result is a positive scalar, FAB has the
same sense of direction as uB. Express in
Cartesian form

 
FAB  FAB u AB



 257.1N 0.286i  0.857 j  0.429k 



 {73.5i  220 j  110k }N
Perpendicular component



 





F  F  FAB  300 j  (73.5i  220 j  110k )  {73.5i  80 j  110k }N
Solution
Magnitude can be determined from F┴ or
from Pythagorean Theorem,

F 

2  2
F  FAB
300 N 2  257.1N 2
 155 N
Quiz
1. Which one of the following is a scalar quantity?
A) Force
B) Position C) Mass D) Velocity
2. For vector addition, you have to use ______ law.
A) Newton’s Second B) the arithmetic C) Pascal’s
D) the parallelogram
3. Can you resolve a 2-D vector along two directions, which are not at 90° to
each other?
A) Yes, but not uniquely.
B) No.
C) Yes, uniquely.
4. Can you resolve a 2-D vector along three directions (say at 0, 60, and 120°)?
A) Yes, but not uniquely.
B) No.
C) Yes, uniquely.
5. Resolve F along x and y axes and write it in vector form. y
A) 80 cos (30°) i – 80 sin (30°) j
B) 80 sin (30°) i + 80 cos (30°) j
C) 80 sin (30°) i – 80 cos (30°) j
30°
D) 80 cos (30°) i + 80 sin (30°) j
x
F = 80 N
6. Determine the magnitude of the resultant (F1 + F2) force in N when
F1={ 10i + 20j }N and F2={ 20i + 20j } N .
A) 30 N
B) 40 N
C) 50 N
D) 60 N
E) 70 N
7. Vector algebra, as we are going to use it, is based on
a ___________ coordinate system.
A) Euclidean B) Left-handed C) Greek D) Right-handed
E) Egyptian
8. The symbols , , and  designate the __ of a 3-D Cartesian vector.
A) Unit vectors
B) Coordinate direction angles
C) Greek societies
D) X, Y and Z components
9. What is not true about an unit vector, uA ?
A) It is dimensionless.
B) Its magnitude is one.
C) It always points in the direction of positive X- axis.
D) It always points in the direction of vector A.
10. If F = {10 i + 10 j + 10 k} N and G = {20 i + 20 j + 20 k } N, then
+ G = { ____ } N
A) 10 i + 10 j + 10 k
B) 30 i + 20 j + 30 k
C) – 10 i – 10 j – 10 k
D) 30 i + 30 j + 30 k
F
12. A force of magnitude F, directed along a unit vector U, is given by F =
A) F (U)
B) U / F
C) F / U
D) F + U
(E) F – U
13. P and Q are two points in a 3-D space. How are the position vectors rPQ and
rQP related?
A) rPQ = rQP
B) rPQ = - rQP
C) rPQ = 1/rQP
D) rPQ = 2 rQP
14. If F and r are force vector and position vectors, respectively, in SI units,
what are the units of the expression (r * (F / F)) ?
A) Newton
B) Dimensionless
C) Meter
D) Newton – Meter
E) The expression is algebraically illegal.
15. Two points in 3 – D space have coordinates of P (1, 2, 3) and Q (4, 5, 6)
meters. The position vector rQP is given by
A) {3 i + 3 j + 3 k} m
B) {– 3 i – 3 j – 3 k} m
C) {5 i + 7 j + 9 k} m
D) {– 3 i + 3 j + 3 k} m
E) {4 i + 5 j + 6 k} m
16. Force vector, F, directed along a line PQ is given by
A) (F/ F) rPQ
B) rPQ/rPQ
C) F(rPQ/rPQ)
D) F(rPQ/rPQ)
17. The dot product of two vectors P and Q is defined as
A) P Q cos 
B) P Q sin 
C) P Q tan  D) P Q sec 
18. The dot product of two vectors results in a _________ quantity.
A) Scalar
B) Vector
C) Complex
D) Zero
19. If a dot product of two non-zero vectors is 0, then the two vectors must be
_____________ to each other.
A) Parallel (pointing in the same direction)
B) Parallel (pointing in the opposite direction)
C) Perpendicular
D) Cannot be determined.
20. If a dot product of two non-zero vectors equals -1, then the vectors must
be ________ to each other.
A) Parallel (pointing in the same direction)
B) Parallel (pointing in the opposite direction)
C) Perpendicular
D) Cannot be determined.
21. The dot product can be used to find all of the following except ____ .
A) sum of two vectors
B) angle between two vectors
C) component of a vector parallel to another line
D) component of a vector perpendicular to another line
22. Find the dot product of the two vectors P and Q.
P = {5 i + 2 j + 3 k} m
Q = {-2 i + 5 j + 4 k} m
A) -12 m
D) -12 m2
B) 12 m
E) 10 m2
C) 12 m2
Next Chapter...
Equilibrium of a particle
Chapter Objectives
Concept of the free-body diagram for a
particle
Solve particle equilibrium problems using
the equations of equilibrium
Chapter Outline
1. Condition for the Equilibrium of a Particle
2. The Free-Body Diagram
3. Coplanar Systems
4. Three-Dimensional Force Systems
3.1 Condition for the Equilibrium of a Particle
Particle at equilibrium if
- At rest
- Moving at constant a constant velocity
Newton’s first law of motion
∑F = 0
where ∑F is the vector sum of all the
forces acting on the particle
3.1 Condition for the Equilibrium of a Particle
Newton’s second law of motion
∑F = ma
When the force fulfill Newton's first law
of motion,
ma = 0
a=0
therefore, the particle is moving in
constant velocity or at rest
3.2 The Free-Body Diagram
Best representation of all the unknown
forces (∑F) which acts on a body
A sketch showing the particle “free” from
the surroundings with all the forces
acting on it
Consider two common connections in
this subject –
 Spring
 Cables and Pulleys
3.2 The Free-Body Diagram
Spring
 Linear elastic spring: change in length is
directly proportional to the force acting on it
 spring constant or stiffness k: defines the
elasticity of the spring
 Magnitude of force when spring
is elongated or compressed
F = ks
3.2 The Free-Body Diagram
Cables and Pulley
 Cables (or cords) are assumed negligible
weight and cannot stretch
 Tension always acts in the direction of the
cable
 Tension force must have a constant
magnitude for equilibrium
 For any angle θ, the cable
is subjected to a constant tension T
Procedure for Drawing a FBD
1. Draw outlined shape
2. Show all the forces
- Active forces: particle in motion
- Reactive forces: constraints that prevent motion
3. Identify each forces
- Known forces with proper magnitude and
direction
- Letters used to represent magnitude and
directions
Example 3.1
The sphere has a mass of 6kg and is
supported. Draw a free-body diagram of
the sphere, the cord CE and the knot at C.
Solution
FBD at Sphere
Two forces acting, weight and the
force on cord CE.
Weight of 6kg (9.81m/s2) = 58.9N
Cord CE
Two forces acting: sphere and knot
Newton’s 3rd Law:
FCE is equal but opposite
FCE and FEC pull the cord in tension
For equilibrium, FCE = FEC
Solution
FBD at Knot
3 forces acting: cord CBA, cord CE and spring CD
Important to know that the weight of the sphere
does not act directly on the knot but subjected to
by the cord CE
3.3 Coplanar Systems
A particle is subjected to coplanar forces
in the x-y plane
Resolve into i and j components for
equilibrium
∑Fx = 0
∑Fy = 0
Scalar equations of equilibrium
require that the algebraic sum
of the x and y components to
equal to zero
Procedure for Analysis
1. Free-Body Diagram
- Establish the x, y axes
- Label all the unknown and known forces
2. Equations of Equilibrium
- Apply F = ks to find spring force
- When negative result force- indicates its sense is
reverse of that shown on the free body diagram
- Apply the equations of equilibrium
∑Fx = 0
∑Fy = 0
Example 3.4
Determine the required length of the cord
AC so that the 8kg lamp is suspended. The
undeformed length of the spring AB is
l’AB = 0.4m, and the spring has a stiffness
of kAB = 300N/m.
Solution
1. Draw FBD at Point A
Three forces acting, force by cable AC, force in
spring AB and weight of the lamp.
If force on cable AB is known, stretch of the
spring is found by F = ks.
+→ ∑Fx = 0; TAB – TAC cos30º = 0
+↑ ∑Fy = 0; TABsin30º – 78.5N = 0
Solving,
TAC = 157.0kN
TAB = 136.0kN
Solution
TAB = kABsAB;
136.0N = 300N/m(sAB)
sAB = 0.453N
For stretched length,
lAB = l’AB+ sAB
lAB = 0.4m + 0.453m
= 0.853m
For horizontal distance BC,
2m = lACcos30° + 0.853m
lAC = 1.32m
3.4 Three-Dimensional Force Systems
For particle equilibrium
∑F = 0
Resolving into i, j, k components
∑Fxi + ∑Fyj + ∑Fzk = 0
Three scalar equations representing
algebraic sums of the x, y, z forces
∑Fxi = 0
∑Fyj = 0
∑Fzk = 0
Procedure for Analysis
Free-body Diagram
- Establish the z, y, z axes
- Label all known and unknown force
Equations of Equilibrium
- Apply ∑Fx = 0, ∑Fy = 0 and ∑Fz = 0
- Substitute vectors into ∑F = 0 and set i, j, k
components = 0
- Negative results indicate that the sense of
the force is opposite to that shown in the
FBD.
Example 3.7
Determine the force developed in each cable
used to support the 40kN crate.
Solution
1. Draw FBD at Point A
To expose all three unknown forces in the cables.
2. Equations of Equilibrium
Expressing each forces in Cartesian vectors,
FB = FB(rB / rB)
 FB [
 3i  4 j  8k
]
(3) 2  (4) 2  (8) 2
= -0.318FBi – 0.424FBj + 0.848FBk
FC = FC (rC / rC)
= -0.318FCi – 0.424FCj + 0.848FCk
FD = FDi
W = -40k
Solution
For equilibrium,
∑F = 0;
FB + FC + FD + W = 0
-0.318FBi – 0.424FBj + 0.848FBk - 0.318FCi
– 0.424FCj + 0.848FCk + FDi - 40k = 0
∑Fx = 0;
∑Fy = 0;
∑Fz = 0;
-0.318FB - 0.318FC + FD = 0
– 0.424FB – 0.424FC = 0
0.848FB + 0.848FC - 40 = 0
Solving,
FB = FC = 23.6kN
FD = 15.0kN
QUIZ
1. When a particle is in equilibrium, the
sum of forces acting on it equals ___ .
(Choose the most appropriate answer)
A) A constant
B) A positive number
C) Zero
D) A negative number
E) An integer
2. For a frictionless pulley and cable,
tensions in the cables are related as
A) T1 > T2
B) T1 = T2
C) T1 < T2
D) T1 = T2 sin 
T1
T2
100 N
100 N
(A)
(B)
100 N
(C)
3. Assuming you know the geometry of the ropes,
you cannot determine the forces in the cables in
which system above?
4. Why?
A) The weight is too heavy.
B) The cables are too thin.
C) There are more unknowns than equations.
D) There are too few cables for a 100 kg weight.
5. Select the correct FBD of particle A.
30
A 40
100 kg
F1
A
A)
F2
B)
30
40°
100 kg
A
F
C)
30
°
D)
A
100 kg
F1
30
°
100 kg
F2
40°
A
6. Using this FBD of Point C, the sum of
forces in the x-direction ( FX) is ___ .
Use a sign convention of +  .
A) F2 sin 50°
B) F2 cos 50°
C) F2 sin 50°
D) F2 cos 50°
– 20
– 20
– F1
+ 20
=0
=0
=0
=0
50o
20kN
F2
C
7. In 3-D, when a particle is in equilibrium,
which of the following equations apply?
A) ( Fx) i + ( Fy) j + ( Fz) k = 0
B)  F = 0
C)  Fx =  Fy =  Fz = 0
D) All of the above.
E) None of the above.
8. In 3-D, when you know the direction of a
force but not its magnitude, how many
unknowns corresponding to that force
remain?
A) One
B) Two
C) Three
D) Four
9. In 3-D, when you don’t know the
direction or the magnitude of a force,
how many unknowns do you have
corresponding to that force?
A) One
B) Two
C) Three
D) Four
10. Four forces act at point A and point A
is in equilibrium. Select the correct
force vector P.
A) {-20 i + 10 j – 10 k}N
z
B) {-10 i – 20 j – 10 k} N
F = 10 N
P
F = 10 N
C) {+ 20 i – 10 j – 10 k}N
A
y
D) None of the above.
3
2
F1 = 20 N
x
Answer
1.C
2.B
3.C
4.C
5.D
6.B
7.D
8.A
9.C
10.D