Unit 1a: Forces & Equilibrium
(General Principles, Force Vectors, Force
System Resultants )
Introduction
•
Mechanics
–
–
•
Branch of the physical sciences concerned with the state of
rest or motion of bodies (rigid-body, deformable-body or
fluid) that are subjected to the action of forces.
We will look only rigid-body mechanics
Rigid-body Mechanics
–
–
a) Statics - body remains at rest and in equilibrium
b) Dynamics - body moves
2
Fundamental Concepts
Basic Quantities
•
Length
– needed to locate the position of a point in space and describe the
size of a physical system. (Units: meter, millimeter, centimeter)
•
Time
– Is conceived as a succession of events. Although the principles
of statics are time independent, this quantity plays an important
role in the study of dynamics. (Units: seconds, minutes)
4
Basic Quantities
•
Mass
–
•
Is a property of matter which we can compare the action of one
body with another. (Units: grams, kilograms, tonnes)
Force
–
–
–
–
Is considered as a ‘push’ or ‘pull’ exerted by one body on
another.
Can occur when there is direct contact between the bodies;
eg. a person pushing on a wall.
It can also occur through a distance when the bodies are
physically separated; eg. Gravitational, electrical and
magnetic forces.
In any case, a force is completely characterized by its
magnitude. (Units: Newton, kilo Newton)
5
Idealizations
•
•
Models or idealizations are used in mechanics in order
to simplify application of the theory
Types of models:
a. Particle – it has a mass, but a size that can be neglected
b. Rigid body – Can be considered as a combination of a large
number of particles in which all the particles remain at a fixed
distance from one another both before and after applying a
load.
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c. Concentrated load – applied to a point in the body.
Eg. Loads transferred by a column to a beam.
d. Linear distributed load – surface loading is applied
along a narrow area. Here the loading is measured as
having an intensity of force/length along the area. Eg.
Wall load applied on beams.
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Newton’s Three Law of Motions
•
•
First Law – A particle originally at rest, or moving in a straight
line with constant velocity, will remain in this state provided the
particle is not subjected to an unbalanced force.
Second Law – A particle acted upon by an unbalanced force, F
experiences an acceleration (a), that has the same direction as the
force and magnitude that is directly proportional to the force. If F
is applied to a particle of mass, m, this law may be expressed
mathematically as
F = m.a
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c. Third Law – The mutual forces of action and reaction
between two particles are equal, opposite and collinear
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Newton’s Law of Gravitational Attraction
• Gravitational attraction between any two particles
m1 m 2
FG
2
r
Where
F = force of gravitation between the two
particles
G = universal constant of gravitation;
= 66.73 (10-12) m3/(kg.s2)
m1, m2 = mass of each of the two particles
r = distance between two particles
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Weight (W)
W = m.g
Where
m
g
= mass of the object
= acceleration due to gravity
(g = 9.81 m/s2)
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Units of Measurements
•
SI Units
–
The International System of units, abbreviated SI is a modern
version of the metric system which has received worldwide
recognition.
Name
Length
Time
Mass
Force
Internation
al System of
Units (SI)
Meter (m)
Second (s)
Kilogram
(kg)
Newton (N)
𝒌𝒈∙𝒎
𝒔𝟐
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International System of Units
• Prefix
Name
Exponential
Form
Prefix
SI Symbol
1 000 000 000
109
Giga
G
1 000 000
106
Mega
M
1 000
103
kilo
k
0.001
10-3
milli
M
0.000 001
10-6
micro

0.000 000 001
10-9
nano
n
Multiple
Submultiple
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Example 1a-1(Conversion of Unit):
(a) Convert 2 km/h to m/s. How many ft/s is this?
TRY:
(b) Convert the followings units to the unit in bracket
(i) (50 mN)(6 GN) – (kN2)
(ii) (400 mm) (0.6 MN)2 - (mMN2 or Gm N2)
(iii) 45 MN3/900 Gg – (kN3/kg)
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Force Vectors
Scalar and Vectors
a. Scalar
•
•
•
A mathematical quantity possessing magnitude only. A
quantity characterized by a positive or negative number is called
a scalar.
Eg. Mass, volume & length.
Scalars are normally indicated by letters in italic, such as scalar
A
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b. Vector
•
•
•
•
•
A vector is a quantity that has both a magnitude and a direction.
Eg. Position, force & moment.
A vector is normally
indicated by a letter with an arrow written over

it such as A or in boldface A (book uses this).
Its magnitude, which is always a positive quantity, is written as A .
Book uses italics for all scalars.
A vector is represented graphically by an arrow, which is used to
define its magnitude, direction and sense.
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• Magnitude – length of the arrow
• Direction – angle between a reference axis and line of action
• Sense – Arrowhead
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Vector Operations
a. Scalar multiplications and divisions
b. Vector addition
•
The sum of 2 vectors can be obtained by attaching the 2
vectors to the same point and constructing a
parallelogram (parallelogram law)
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c. Vector subtraction
•
The addition of the corresponding negative vector
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d. Resolution of Vector
21
Trigonometry
A line intersecting parallel lines:
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Example 1a-2 (Resolution of vector):
What are the x and y components of R, if R = 500 N and θ = 30°?
(a)
y
4θ
R
x
θ
(b)
(c)
y
y
2.5θ
R
R
θ
x
θ
x
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Addition of a System of Coplanar Forces
• In two-dimensional Euclidean space, any force
vector can be resolved into its Cartesian
components.
• The Cartesian unit vectors i and j are used to
designate the directions of the x and y axes
• Thus,
F = Fxi + Fyj
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25
• If there are multiple force vectors acting on a
system, the resultant force vector is given by
where n is the number of force vectors
• The magnitude of the resultant is given by
Note: Only applicable
to Cartesian coordinate
system
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• The directional angle, θ, can be computed as
Note: Only applicable
to Cartesian coordinate
system
• We always adopt the Right Hand Rule convention,
i.e. → and ↑ are considered positive.
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Example 1a-3 (Magnitude and direction of resultant force in Cartesian Syetem) :
y
(a)
(b)
F2 = 40kN
F3 = 50kN
F1 = 30kN
40 50
30
x
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Example 1a-4 (Resolution of a force) :
If it is required that the resultant force have a magnitude of 1 kN and be
directed vertically downward, determine the magnitude of F1 and F2 if
(a)  = 30, and (b)
F 2 is to be a minimum.
Try out part (a)……..
F2
F1
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Example 1a-4 (Resolution of a force) (continued):
• When θ is not specified, the minimum length or
magnitude of F2 will occur when the line is
perpendicular to F1.
F2
θ = 70°
1 kN
F1
θ = 20°
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Q&A
31
3-Dimensions
Right Handed Coordinate System
32
3-D: Cartesian Force Representation
33
Important Relationships:
1. Cartesian vector representation: A = Axi + Ayj + Azk
2. Magnitude of a Cartesian vector:
 Ax2  Ay2  Az2
A
3. Direction cosines of a Cartesian vector A:
A
cos   x
A
cos  
Ay
A
Az
cos  
A
4.
α+
β+
γ=1
5. Direction of A can be represented by the unit vector (in term of cosine),
uA = cos αi + cos βj + cos γk
6. Magnitude and coordinate direction angles of A:
cos2
cos2
cos2
A = (A)(uA) = (A cos α)i + (A cos β)j + (A cos γ)k
= Axi + Ayj + Azk
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Example 1a-5 (Determine Force as the Cartesian Vector):
Express the force F as a Cartesian vector
35
Example 1a-5 (Determine Force as the Cartesian Vector) (continued):
Since only two coordinate direction angles are specified, angle α must be
determined.
cos 2   cos 2   cos 2   1
cos 2   cos 2 60   cos 2 45   1
cos   1  0.5  0.707   0.5
2
2
Hence, two possibilities exist, namely
α = cos-1(0.5) = 60° or
α = cos-1(-0.5) = 120°
By inspection of the figure, it is necessary that α = 60°, since Fx is in the +x
direction.
F = F cos αi + F cos βj + F cos γk
= (200 cos 60°)i + (200 cos 60°)j + (200 cos 45°)k
= 100.0i + 100.0j + 141.4k
36
Example 1a-6 (Resultant force of 3-D Cartesian Vectors) :
Determine the magnitude and the coordinate direction angles of the
resultant force acting on the ring (i.e. find the angle , , and ).
kN
kN
kN
37
Example 1a-6 (Resolution of a 3-D Cartesian Vectors) (continued) :
FR = ∑F = F1 + F2
= (60j + 80k) + (50i – 100j +100k) kN
= (50i -40j +180k) kN (Cartesian Vector form)
Magnitude of FR =
(50) 2  (40) 2  (180) 2
= 191 kN
50
40
180
The coordinate direction angles FR, FR

i
j
k
191
191
191
FR
= 0.2617i – 0.2094j + 0.9422k
The direction angles, α, β and γ are
cos α = 0.2617
cos β = –0.2094
cos γ = 0.9422
α = 74.8°
β = 102°
γ = 19.6°
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Moment of a Force
• The moment of a force about a point or axis provides a measure
of the tendency of the force to cause a body to rotate about the
point or axis. It is denoted M, and is a vector quantity.
M = ∑(F × d)
• d is perpendicular distance from the force to point or axis of
rotation
• Clockwise moment = negative (-ve)
• Counter-clockwise moment = positive (+ve)
• The direction of the moment can also be shown graphically cw or
ccw without the sign convention
39
40
Line of action F
r
F
d
θ
θ
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Example 1a-7 (Moment of one force about a point):
Determine the moments of the 800 N force acting on the
frame in the figure about points A, B, C and D.
MA = 800 × 2.5 = 2000 N.m
MB = 800 × 1.5 = 1200 N.m
MC = 800 × 0 = 0
MD = 800 × 0.5 = 400 N.m
42
Example 1a-8 (Moment of few forces about a point):
Determine the resultant moment of the four forces acting on
the rod about point O.
MO = (Fxd) = -50(2)-60(0)+20(3sin30)-40(4+3cos30)
= -334Nm
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Vector Formulation
A = Axi + Ayj + Azk

Dot Product: A · B = ABcos
= AxBx + AyBy + AzBz
B = Bxi + Byj + Bzk
The angle formed between vectors A and B, 
= cos-1 (A·B/AB)
b
Resolve vector to two lines (a & b axes)
perpendicular to each other
A

Aa = Acos ua
ua
Parallel component, Aa = Aa ua , where Aa = A cos
Vertical component, Ab = A – Aa, where Ab = (A2 – Aa2)½
44
a
Moment of a Force – Vector Formulation (use of vector
cross product & rotate about a point)
Cross Product: A x B = (ABsin )uc
MO = r × F =
i
rx
Fx
j
ry
Fy
k
rz
Fz
**Moment MO is perpendicular
to the plane containing O and F.
**Use right hand rule to
determine the direction of MO.
= (ryFz – rzFy)i - (rxFz – rzFx)j + (rxFy – ryFx)k
Where rx, ry, rz represent the x, y, z components of
the position vector drawn from point O to
any point on the line of action of the force
Fx, Fy, Fz represents the x, y, z components
of the force vector
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Moment of a Force – Vector Formulation (rotate about a
**Projection of MO onto
specified axis)
MO = ua.(r × F) =
Where
u ax
ua y
ua z
rx
ry
rz
Fx
Fy
Fz
rotation axis ua axis (use dot
product)
**MO = MOua
= u a (ryFz – rzFy) - u a (rxFz – rzFx) + u a (rxFy – ryFx)
ua , ua , ua represent the x, y, z components of the unit
vector defining the direction of the a axis
x
x
y
y
z
z
rx, ry, rz represent the x, y, z components of the
position vector drawn from point O to any point on
the line of action of the force
Fx, Fy, Fz represents the x, y, z components of the
force vector
46
• Position vector: A fixed vector which locates a
point in space relative to another point. z
B(xB, yB, zB)
A(xA, yA, zA)
rA
r
rB
y
Position Vector, r = xi + yj + zk
x
Position Vector, r = rAB
= (xB - xA)i+ (yB - yA)j+(zB - zA)k
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• Position vector: A fixed vector which locates a
point in space relative to another point. z
B(xB, yB, zB)
A(xA, yA, zA)
rA
r
rB
y
Position Vector, r = xi + yj + zk
x
Position Vector, r = rAB
= (xB - xA)i+ (yB - yA)j+(zB - zA)k
48
Example 1a-9 (Moment about an axis in 3-D Cartesian Vector):
The force, F = (-40i + 20j + 10k) acts at a point A shown. Determine the
moment about x and Oa axes.
B
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Example 1a-9 (Moment of a force about an axis) (continued):
For x-axis:
Position vector; rA = (–3i + 4j + 6k) and ux = i
1
0 0
Mx = ua.(r × F) =  3 4 6
 40 20 10
= 1[4(10) – 6(20)] – 0[(–3)(10) – 6(–40)] + 0[(–3)(20) – 4(–40)]
= – 80 N.m
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Example 1a-9 (Moment of a force about an axis) (continued):
For oa- axis:
3
5
4
5
uoa =  i  j
Moa = ua.(r × F) =
3

5
3
4
5
4
0
6
 40 20 10

3
410  620  4  310  6 40  0 320  4 40
5
5
= – 120 N.m
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Example 1a-10 (Moment about a point in 3-D Cartesian Vector):
The pole is subjected to a 60N force that is directed from C to B.
Determine the magnitude of the moment created by this force about the
support at A.
53
Example 1a-10 (Moment of a force about a point) (continued):
Position vectors: rB = (1i + 3j + 2k) m and rC = (3i + 4j) m
The force has a magnitude of 60N and a direction specified by the
unit vector uF directed from C to B. Therefore,
 1  3i  3  4j  2  0k 

F = (60).uF = 60

 22   12  22 
= (– 40i – 20j + 40k) N
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Example 1a-10 (Moment of a force about a point) (continued):
MA = r B × F =
i
1
j
3
k
2
 40  20 40
= [3(40) – 2(– 20)]i – [1(40) – 2(– 40)]j + [1(– 20) – 3(– 40)]k
or
MA = r C × F =
i
3
j
4
k
0
 40  20 40
= [4(40) – 0(– 20)]i – [3(40) – 0(– 40)]j + [3(– 20) – 4(– 40)]k
55
Example 1a-10 (Moment of a force about a point) (continued):
In both cases,
MA = (160i – 120j + 100k) N.m
MA 
1602   1202  1002
 224 N.m
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Q&A
57