ENGINEERING STATICS
CHAPTER ONE:
SCALARS and VECTORS
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M.M
STATICS

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Statics deals with the equilibrium of bodies,
that are either at rest or move with a constant
velocity.
Introduction
 Mechanics is a physical science which deals with the
state of rest or motion of rigid bodies under the
action of forces.
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PARTS OF MECHANICS
Mechanics divided into three parts:
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PARTS OF MECHANICS. . .
 Statics: deals with the equilibrium of rigid bodies
under the action of forces.
 Dynamics: deals with the motion of rigid bodies
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caused by unbalanced force acting on them.
 Dynamics is further subdivided into two parts:
 Kinematics: dealing with geometry of motion of
bodies with out reference to the forces causing the
motion, and
 Kinetics: deals with motion of bodies in relation to
the forces causing the motion.
Basic Concepts:
(Statics)
 The bodies are assumed to be rigid,
 don’t deform or change its shape.
 But translation or rotation may exist.
 The loads are assumed to cause only external
movement, not internal. (In reality, the bodies may
deform.)
 (the changes in shapes are assumed to be minimal and
insignificant to affect the condition of equilibrium
(stability) or motion of the structure under load).
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Fundamental Principles
❖Newton laws
 First Law: A particle remains at rest or continues to
move in a straight line with uniform velocity if there is
no unbalanced force on it.
 Second Law: The acceleration of a particle is
proportional to the resultant force acting on it and is
in the direction of this force.
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F=mxa
Fundamental Principles. . . .
❖Newton laws. . .
 Third Law: The forces of action and reaction between
interacting bodies are equal in magnitude, opposite in
direction, and collinear.
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Fundamental Principles . . .
 Law of gravitation (by Newton), as it usual to
compute the weight of bodies. Accordingly:
thus the weight of a mass ‘m’
W = mg
- m1 & m2 are particle masses
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- G is the universal constant of gravitation, 6.673 x 10-11 m3/kg-s2
- r is the distance between the particles.
SCALARS and VECTORS
 Scalar quantities: -
✓ are physical quantities that can be completely
described (measured) by their magnitude alone.
✓ These quantities do not need a direction to point out
their application.
✓ They only need the magnitude and the unit of
measurement to fully describe them.
E.g. Time[s], Mass [Kg], Area [m2], Volume [m3],
Density [Kg/m3], Distance [m], etc.
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SCALARS AND VECTORS . . .
 Vector quantities: -
✓Have both a magnitude and direction
✓Vectors are represented by short arrows on top of
the letters designating them.
E.g. Force [N, Kg.m/s2],
Velocity [m/s],
Acceleration [m/s2], Momentum [N.s, kg.m/s],
etc.
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Types of Vectors
❖Generally there are three type vectors :
oFree Vectors
oFixed Vectors
oSliding Vectors:
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NB: a force can be applied any where
along its line of action on a rigid body
with out altering its external effect on
the body. This principle is known as
Principle of Transmissibility.
Representation of Vectors
A) Graphical representation
 Graphically,
➢directed line segment headed by an arrow.
➢The length of the line segment is equal to the magnitude
of the vector and
➢the arrow indicates the direction of the vector.
 The direction of the vector may be measured by an
angle θ from some known reference direction.
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Representation of Vectors . . .
B) Algebraic (arithmetic) representation
 Algebraically a vector is represented by the
components of the vector along the three dimensions.
E.g
Where ax, ay and az are components of the vector A along the
x, y and z axes respectively.
NB:- The vectors i, j and k are unit vectors along the respective
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axes.
Representation of Vectors . . .
B) Algebraic (arithmetic) representation . . .
 In our course:
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Representation of Vectors . . .
B) Algebraic (arithmetic) representation . . .
Using general definition,
iˆ  ˆj
Magnitude: (i )( j )(sin  )
Direction: R.H. Rule  kˆ
iˆ  ˆj = kˆ
i j = k ik = − j ii = 0
j  k = i j  i = −k j  j = 0
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k  i = j k  j = −i k  k = 0
Representation of Vectors. . .
B) Algebraic (arithmetic) representation. . .
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 ax = A cosθx = Al
, l = cosθx
 ay = A cosθy = Am
, m = cosθy
 az = A cosθz = An
, n = cosθz ,
where l, m, n are the directional cosines of the vector.
 Thus,
A2 = ax2 + ay2 + az2 ⇒ l 2 + m2 + n2 = 1
Properties of vectors
 Equality of vectors:
Two free vectors are said to be equal if and only if
they have the same magnitude and direction.
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Properties of vectors . . .
 The Negative of a vector: is a vector which has
equal magnitude to a given vector but opposite in
direction.
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Properties of vectors . . .
 Null vector: is a vector of zero magnitude. It has an
arbitrary direction.
 Unit vector: is any vector whose magnitude is unity.
 A unit vector along the direction of a certain vector,
say vector A (denoted by uA) can be determined by
dividing vector A by its magnitude.
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Properties of vectors . . .
 Generally, any two or more vectors can be aligned in
different manner.
they may be:
• Collinear-Having the same line of action.
• Coplanar- Lying in the same plane.
• Concurrent- Passing through a common point.
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Vector Addition or Composition of Vectors
 Composition of vectors is the process of adding two or
more vectors to get a single vector, (Resultant vector).
 There are different techniques of adding vectors
A) Graphical Method
oI. The parallelogram law
oII. The Triangle rule
oIII. Analytic method.
B. Component method of vector addition
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Vector Addition or Composition of Vectors…
A) Graphical Method
I. The parallelogram law
1st draw the parallelogram in to scale,
2nd fined the magnitude of the resultant by
measuring the diagonal,
3rd The direction of the resultant can be found by
measuring the angle the diagonal makes.
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Vector Addition or Composition of Vectors…
A) Graphical Method. . .
I. The parallelogram law . . .
Note: As we can see in the above figure.
A+B=R=B+A,
⇒ vector addition is commutative
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Vector Addition or Composition of Vectors…
A) Graphical Method. . .
The parallelogram law . . .
 The other diagonal of the parallelogram gives the
difference of the vectors, it represents either A− B or
B−A
I.
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Vector Addition or Composition of Vectors…
I. The parallelogram law . . .
 The diagonal vectors in the above figure are not
equal,
 one is the negative vector of the other,
 vector subtraction is not commutative.
i.e. A − B ≠ B − A
NB. Vector subtraction is addition of the negative of
one vector to the other.
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Vector Addition or Composition of Vectors…
II. The Triangle rule
 applied to more than two vectors at once.
 It states “If the two vectors, which are drawn on
scale, are placed tip (head) to tail, their resultant
will be the third side of the triangle which has tail at
the tail of the first vector and head at the head of the
last.”
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Vector Addition or Composition of Vectors…
II. The Triangle rule . . .
 This rule can be extended to more than two vectors
as, “If a system of vectors are joined head to tail,
their resultant will be the vector that completes the
polygon so formed, and it starts from the tail of the
first vector and ends at the head of the last vector.”
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NB. From the Triangle rule it can easily
seen that if a system of vectors when
joined head to tail and form a closed
polygon, their resultant will be a null
vector.
Vector Addition or Composition of Vectors…
III. Analytic method
 The resultant is found mathematically instead of
measuring it from the drawings as in the graphical
method.
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Vector Addition or Composition of Vectors…
 Consider triangle ABC
III. Analytic method. . .
 From cosine law,
A.Trigonometric rules:
 The resultant vectors can
be found analytically by
applying the cosine and
This is the magnitude of the
the sine rules.
RESULTANAT of the two
vectors
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➢resultant makes with
vector A
Vector Addition or Composition of Vectors…
III. Analytic method. . .
 Decomposition of vectors:
➢The process of getting the components of a given
vector along some other different axis.
➢The reverse of composition.
➢Consider the following vector A . And let our aim be
to find the components of the vector along the n and t
axes.
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Vector Addition or Composition of Vectors…
III. Analytic method. . .
 Decomposition of vectors:. . .
From Triangle ABC @ (B), α = 180 − (θ + φ )
From sine law then,
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Vector Addition or Composition of Vectors…
III. Analytic method. . .
 Decomposition of vectors. . .
 In most cases though, components are sought along
perpendicular axes, i.e. α=180-(θ+Φ) = 90
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Vector Addition or Composition of Vectors…
III. Analytic method. . .
B. Component method of vector addition
 most efficient method of vector addition, especially
when the number of vectors to be added are large.
 The sum of the components of each vector along each
axis will be equal to the components of their resultant
along the respective axes.
 Once the components of the resultant are found, the
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resultant can be found by parallelogram rule as
discussed above.
Vector Multiplication: Dot and Cross products
a)Multiplication of vectors by scalars
 Let n be a non-zero scalar and A be a vector, then
multiplying A by n gives as a vector whose magnitude
is n|A| and whose direction is in the direction of A if
n is positive or is in opposite direction to A if n is
negative.
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Vector Multiplication: Dot and Cross products . . .
a)Multiplication of vectors by scalars . . .
 Multiplication of vectors by scalars obeys the
following rules:
i. Scalars are distributive over vectors.
n(A + B) = nA + nB
ii. Vectors are distributive over scalars.
(n + m)A = nA + mA
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iii. Multiplication of vectors by scalars is associative.
(nm)A = n(mA) = m(nA)
Vector Multiplication: Dot and Cross products . . .
b )Multiplication of vector by a vector
 In mechanics there are a few physical quantities that
can be represented by a product of vectors.
 Eg. Work, Moment, etc
 There are two types of products of vector
multiplication
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Vector Multiplication: Dot and Cross products. . .
c) Cross Product: Vector Product
 The vector product of two vectors A and B that are θ
degrees apart denoted by AxB (A cross B)
i.e. AxB = || A || || B || sinθ , perpendicular to the
plane formed by A and B
 The sense of the resulting vector can be determined
by the right-hand rule.
 If the two vectors are represented analytically as,
 A= ax i +ay j+ az k and B =bx i +by j+ bz k
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Vector Multiplication: Dot and Cross products. . .
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Vector Multiplication: Dot and Cross products. . .
c) Cross Product: Vector Product . . .
 NB. Vector product is not commutative; in fact, AxB = −BxA
A B  B  A
In fact
A  B = −B  A
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Vector Multiplication: Dot and Cross products. . .
c) Cross Product: Vector Product . . .
 Moment of a Vector
 The moment of a vector V about any point O is given
by:
 Position vector r is defined as afixed vector that
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locates a point in space relative to another point in
space.
Example 1
 Determine graphically, the magnitude and
direction of the resultant of the two forces
using
(a) Parallelogram law ,and
(b) Triangle rule.
600 N
900 N
45o
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30o
Solution:
Solution
A parm. with sides equal to 900 N and 600 N is drawn to scale as shown.
The magnitude and direction of the resultant can be found by drawing to scale.
600N
600 N
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R
o
45o
30o
900 N
45o
30o
The triangle rule may also be used. Join the forces in a tip to tail fashion and
measure the magnitude and direction of the resultant.
600 N
45o
R
B
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900N
30o
135o C
900 N
Trignometric Solution
Using the cosine law:
R2 = 9002 + 6002 - 2 x 900 x 600 cos 1350
R
R = 1390.6 = 1391 N
Using the sine law:
B

R
600
600
sin
135
−1
=
i
.
e
.
B
=
sin
sin 135 sin B
1391
= 17.8
The angle of the resul tan t = 30 + 17.8 = 47.8
ie. R = 139N
47.8o
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30o
600N
135o
900 N
Example 2
 Two structural members P and Q are bolted to bracket
A.
Knowing that both members are in tension and
that P = 30 kN and Q
=
20 kN, determine the
magnitude and direction of the resultant force exerted
on the bracket.
P
25o
50o
Q
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Solution
Solution: Using Triangle rule:
75o
20 kN
30 kN
105o

25o
Q
R
R2 = 302 + 202 - 2 x 30 x 20 cos 105 0 - cosine law
R = 40.13 N
Using sine rule:
40.13 N
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=
Sin 
Sin 105o
and
Sin −1
Angle R = 28.8 o − 25o = 3.8 o
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i. e
R = 40.1 N ,
3.8 o
20 sin 105o
=
= 28.8 o
40.13
Example 3
 Determine the resultant of the three forces
below.
y
600 N
800 N
350 N
60o
45o
25o
x
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Solution
 F x = 350 cos 25o + 800 cos 70o - 600 cos 60o
= 317.2 + 273.6 - 300 = 290.8 N
 F y = 350 sin 25o + 800 sin 70o + 600 sin 60o
= 147.9 + 751 + 519.6 = 1419.3 N
i.e. F = 290.8 N i + 1419.3 N j
Resultant, F
y
800 N
600 N
350 N
F = 290.8 + 1419.3 = 1449 N
2
2
1419.3
 = tan
= 78.4 0
290.8
−1
60o
45o
25o
x
F = 1449 N
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78.4o
Example 4
 A hoist
trolley is subjected to the three
forces shown. Knowing that  = 40o ,
determine
(a) the magnitude of force, P for which the
resultant of the three forces is vertical
(b) the corresponding magnitude of the
resultant.

2000 N
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P

1000 N
Solution
(a) The resultant being vertical means that the
horizontal component is zero.
 F x = 1000 sin 40o + P - 2000 cos 40o = 0
P = 2000 cos 40o - 1000 sin 40o =
1532.1 - 642.8 = 889.3 = 889 kN
(b)
 Fy
= - 2000 sin 40o - 1000 cos 40o =
- 1285.6 - 766 = - 2052 N =
40o
50
2000 N
2052 N
P
40o
1000 N
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