Chapter 1
Principles of Statics
Basic Principles of
Statics of Rigid Bodies
Chapter 1
Principles of Statics
Statics is a branch of mechanics which studies the effects and distribution of forces of
rigid bodies which are and remain at rest. In this area of mechanics, the body in which
forces are acting is assumed to be rigid. The deformation of non-rigid bodies is treated
in Strength of Materials.
1.1.1
DEFINITION
Mechanics – Mechanics is the physical science which deals with the effects of
forces on objects. The subject of mechanics is logically divided into two parts:
1.
2.
STATICS –primarily concerned to system of forces applied to body at rest.
DYNAMICS – is the study of body in motion.
Force – is the action of one body on another. A force tends to move a body in the
direction of its action. The action of a force is characterized by its magnitude, by the
direction of its action, and by its point of application.
External Forces – are those which are applied to the boundary of a structure. This
includes explicit externally applied forces as well as the forces that are applied by the
supports to restrain the structure.
Internal forces – are those that are exerted on a portion of a structure by the rest of
the structure. You would only see these forces if you were to make a cut in the
structure and separate it into two free body diagrams.
Particle – is a body of negligible dimensions. In the mathematical sense, a particle is a
body whose dimensions are considered to be near zero so that we may analyze it as a
mass concentrated at a point.
Rigid body – A body is considered rigid when the change in distance between any two
of its points is negligible for the purpose at hand
1
Basic Principles of
Statics of Rigid Bodies
1.1.2
Chapter 1
Principles of Statics
SCALAR AND VECTOR QUANTITIES
Scalar quantities are those with which only a magnitude is associated. Examples of
scalar quantities are time, volume, density, speed, energy, and mass.
Vector quantities are those possess direction and magnitude, and must obey the
parallelogram law of addition as described later in this article. Examples of vector
quantities are displacement, velocity, acceleration, force, moment, and momentum.
1.1.3
UNITS OF MEASUREMENT / UNITS OF CONVERSION
Table 1.1 – System Units
Name
Length
International
meter
System of Units
m
(SI)
Foot
U.S. Customary
(English Units)
ft
Time
Second
Mass
Kilogram
Force
Newton
s
Kg
N
second
slug
lb ∙ s 2
(
)
ft
pound
s
BASIC CONVERSION:
SI/Metric Units:
kg  m
1 N =1
s2
1 kN = 1000 N
g = 9.81 m/s2
English Units:
lb  s2
ft
1 kip = 1000 lb (Note: kip means kilo pounds)
1 slug = 1
g = 32.2 ft/s2
2
lb
Basic Principles of
Statics of Rigid Bodies
Chapter 1
Principles of Statics
1.2
BASES PRINCIPLE OF STATICS
1.2.1
PRINCIPLE OF TRANSMISSIBILITY
A force may be moved anywhere along its line of action without changing its external
effects on a rigid body.
1.2.2
PARALLELOGRAM LAW
The parallelogram is formed by drawing a line from the head of F1 that is parallel to
F2, and another line from the head of F2 that is parallel to F1. The resultant FR extends
to where these lines intersect at point A, Refer Figure 1.1.
A
𝑣
FR
F2
𝑢
θ
F1
Figure 1.1 – Force polygon
1.3
COMPONENT OF A FORCE
Forces acting at some angle from the coordinate axes can be resolved into mutually
perpendicular forces called components. The two components along the x and y axes,
the components are then called rectangular components.
y
Horizontal Component of Force F, Fx :
Fx = F cos θ
F
Fy
θ
Fx
Eq. 1.1
Vertical Component of Force F, Fy :
x
Fy = F sin θ
Eq. 1.2
Figure 1.2 – Component of a Force
3
Basic Principles of
Statics of Rigid Bodies
Chapter 1
Principles of Statics
SAMPLE PROBLEM 1.1
The force P = 450 N is shown in Figure SP-1.1.
a) Determine the X-component of P in Newton.
b) Determine the Y-component of P in Newton.
c) Determine the component of P in the X’-axis in Newton.
d) Determine the component of P in the Y’-axis in Newton.
Y
Y′
P
30°
30°
30°
𝐅𝐢𝐠𝐮𝐫𝐞 SP-1.1
X
X′
Answer: FX = 389.71 N, FY = 225 N, FX’ = 225 N, FY’ = 389.71 N
SOLUTION:
1.4
4
MOMENT OF A FORCE
Basic Principles of
Statics of Rigid Bodies
Chapter 1
Principles of Statics
Moment is the measure of the capacity or ability of the force to produce twisting or
turning effect about an axis. The magnitude of moment is equal to the product of the
force and the perpendicular distance from the axis to the line of action of the force.
𝑀O = F × d
Eq. 1.3
𝑧
x & y plane
MO
x & y plane
F
F
d
d
O
MO
O
Figure 1.3 – Moment of a Force about a Point
Where:
MO = Magnitude of moment about point O.
F = Force which is perpendicular to the moment arm.
d = Moment arm which is the perpendicular distance from the moment
center (point O) to the line of action of the force.
5
Basic Principles of
Statics of Rigid Bodies
Chapter 1
Principles of Statics
SAMPLE PROBLEM 1.2
Calculate the magnitude of the moment about the base point O of the 600-N force
in four different ways.
1.5 m
A
40°
600 N
4m
O
Figure SP-1.2
Answer: MO = 2417.02 N-m, clockwise
SOLUTION:
6
Basic Principles of
Statics of Rigid Bodies
Chapter 1
Principles of Statics
SAMPLE PROBLEM 1.3
Determine the resultant moment produced by the forces about point O.
500 N
300 N
45°
O
1m
2m
600 N
Figure SP-1.2
Answer: MO = 1253.6 N-m counter clockwise
SOLUTION:
7
Basic Principles of
Statics of Rigid Bodies
Chapter 1
Principles of Statics
SAMPLE PROBLEM 1.4
In Figure SP-1.3, assuming clockwise moments as positive, compute the moment of
force F = 20 kN and force P = 16 kN about points A, B, C, and D.
A
F
0.3 m
P
B
0.3 m
Figure SP-1.4
Answer: MA = –18.01 kN-m, MB = 24.81 kN-m
SOLUTION:
8
Basic Principles of
Statics of Rigid Bodies
Chapter 1
Principles of Statics
1.5
COUPLE
The moment produced by two equal, opposite, and non collinear forces is called
a couple. Couple is a system of forces whose magnitude of the resultant is zero and yet
has a moment sum. Geometrically, couple is composed of two equal forces that are
parallel to each other and acting in opposite direction. The magnitude of the couple is
given by
C = Fd
Eq. 1.4
Where F are the two forces and d is the moment arm, or the perpendicular distance
between the forces.
F
d
F
Couple is independent of the moment center, thus, the effect is unchanged in the
following conditions.
• The couple is rotated through any angle in its plane.
• The couple is shifted to any other position in its plane.
• The couple is shifted to a parallel plane.
9
Basic Principles of
Statics of Rigid Bodies
Chapter 1
Principles of Statics
SAMPLE PROBLEM 1.5
Compute the combined moment of the two 90-lb forces about
1. Point O
2. Point A
𝑦
90 lb
A
8 ft
𝑥
O
4 ft
4 ft
90 lb
Answer: 1440 lb-ft, clockwise
SOLUTION:
10
Basic Principles of
Statics of Rigid Bodies
Chapter 1
Principles of Statics
Instruction: Do as required. Present in writing the full details of your answer or
solutions. The grades you earn for each item will be based on correctness,
completeness and clarity of presentation.
𝑦
Situation 1 – Resolve the 200 lb force into
component. Refer Figure AP-1.1.
1. Along the x and y-axis
Answer: 173.21 lb, 100 lb
2. Along the x’ and y’-axis
Answer: 128.56 lb, 153.21 lb
200 lb
30°
𝑥
20°
𝑥′
Figure AP-1.1
Situation 2 – A block is resting on an incline of
slope 5:12 as shown in Figure AP-1.5. It is
subjected to a force F = 500 N on a slope of
3:4.
3. Determine the component of F parallel to
the incline.
Answer: 253.85 N
4. Determine the component of F
perpendicular to the incline.
Answer: –430.77 N
Situation 3 – Refer Figure AP-1.11.
5. Determine the X & Y component of force F in
lbs.
Answer: –150 lb & 360 lb
6. Determine the X & Y component of force T.
Answer: –300.42 lb, 257.12 lb
7. Determine the X & Y component of force P.
Answer: 259.81 lb, 150 lb
F = 500 N
3
4
5
12
Figure AP-1.5
𝑦
F = 390 lb
12
5
𝑥
40°
T = 400 lb
30°
P = 300 lb
Figure AP-1.11
11
Basic Principles of
Statics of Rigid Bodies
Chapter 1
Principles of Statics
Situation 4 – The cable AB prevents bar OA from
rotating clockwise about the pivot O. If the cable
tension is 750 N,
8. Determine the n-component of this force
acting on point A of the bar.
Answer: 332.65 N
9. Determine the t-component of this force
acting on point A of the bar
Answer: –672.19 N
𝑡
𝑛
A
P
O
60°
1.2 m
Figure SP-1.35
500 N
10. Determine the moment of the
force about point O.
3m
Answer: 1060.66 N-m
45°
O
Figure AP-1.16
250 N
11. Calculate the moment of the
250-N force on the handle of
the monkey wrench about the
center of the bolt. Assume
clockwise as positive.
Answer: 46.36 N-m
12
15°
200 mm
30 mm
Figure AP-1.17
Basic Principles of
Statics of Rigid Bodies
Chapter 1
Principles of Statics
Situation 5 – In Figure SP-1.4, assuming clockwise moments as positive, compute the
moment of force F = 20 kN and force P = 16 kN about the following points.
12. At point C.
Answer: MC = 2.03 kN-m
13. At point D
Answer: MD = 26.78 kN-m
F
C
0.3 m
P
D
0.3 m
Figure SP-1.4
Situation 6 – Two parallel 60-N forces are applied as shown to the corners A and C of a
200-mm square plate. Determine the moment of the couple formed by the two forces
14. by multiplying their magnitude by their perpendicular distance,
Answer: 16.39 N-m
15. by resolving each force into horizontal and vertical components and adding the
moments of the two resulting couples.
Answer: 16.39 N-m
D
C
60°
200 mm
60 N
60 N
60°
B
A
200 mm
13