JJ205
ENGINEERING MECHANICS
COURSE LEARNING OUTCOMES :
Upon completion of this course, students should be able to:
CLO 1. apply the principles of statics and dynamics to solve
engineering problems (C3)
CLO 2. sketch related diagram to be used in problem solving (C3)
CLO 3. study the theory of engineering mechanics to solve related
engineering problems in group (A3)
JJ205 ENGINEERING MECHANICS
CHAPTER 2:
FORCE VECTORS
CLO 1. apply the principles of statics and dynamics to
solve engineering problems (C3)
Prepared by:
THEEBENRAJ
2
Objectives:
• At the end of this chapter, student should be able to:
1. Understand scalars and vectors
a.
b.
Differentiate between scalars and vectors.
Distinguish free vectors, sliding vectors, fixed vectors.
2. Understand rectangular components
a.
Explain two forces acting on a particle.
3. Understand vectors and vector operations.
a.
b.
c.
Calculate addition of vectors.
Calculate subtraction of vectors.
Determine resolution of vectors.
4. Understand the resultant force of coplanar forces by
addition.
a.
b.
c.
Explain scalar notation.
Explain cartesian vector notation.
Determine coplanar forces and resultant force.
Objectives:
5. Understand Cartesian vectors.
a.
b.
c.
Explain right handed coordinate system.
Explain cartesian unit vector.
Apply cartesian vector representation.
6. Understand the magnitude of cartesian vector.
a.
Determine the direction of cartesian vector.
7. Understand resultant cartesian vector by addition and
substraction.
a.
Solve problems regarding concurrent force system.
8. Understand position vectors and x, y, z coordinates.
a.
Explain position vectors and x, y, z coordinate.
Objectives:
9. Understand the force vector directed along the line.
a.
b.
Explain the force vector directed along the line.
Determine the force vector directed along the line.
10. Understand the dot product.
a.
Apply laws of operation.
i.
ii.
iii.
b.
Commutative law
Multiplication by scalar
Distributive law
Formulate cartesian vector formulation.
Force On A Particle. Resultant of Two Forces
• A force represents the action of one body on
another and generally characterized by its
point of application, its magnitude, and its
direction.
• The magnitude of the force is characterized by
a certain number of units.
– SI units to measure the magnitude of a force are
the Newton (N).
– Multiple: kilonewton (kN) = 1000N
6
• The direction of a force is defined by the line
of action and the sense of the force.
a)
A
30˚
Magnitude of a force on particle A is 100N at 30˚.
b)
A
30˚
The different sense of the force but have same magnitude and same
direction.
7
• Two forces, P and Q acting on a particle A:
P
A
Q
• Can be replaced by a single force R which has
the same effect on the particle.
R
P
A
Q
R
=
A
• R is the resultant of the forces P and Q.
• The method is known as the parallelogram law
for the addition of two forces.
8
Example 2.1:
The two forces P and Q act on bolt A. Determine their resultant.
Solution:
a) Using graphical solution:
i.
A parallelogram with sides equal to P and Q is drawn
to scale.
ii. The magnitude and direction of the resultant are
measured and found to be
iii.
The triangle rule may also be used. Forces P and Q are drawn in tip-to-tail
fashion.
9
Continue…
b)
Trigonometric Solution.
i.
ii.
The triangle rule is again used; two sides and the included angle are known.
We apply the law of cosines:
iii.
Now, applying the law of sines, we write;
(1)
iv.
Solving equation (1) for sin A, we have;
v.
Using a calculator, we first compute the quotient,
then its are sines, and obtain;
10
Continue…
c) Alternative Trigonometric Solution.
i.
We construct the right triangle BCD and compute;
ii.
Then, using triangle ACD, we obtain;
iii.
Again,
11
Example 2.2
• The screw eye in Figure (a) is subjected to
two forces, F1 and F2. Determine the
magnitude and direction of the resultant
force.
– Solutions:
a) Parallelogram Law:
i. The parallelogram law of addition is shown
in Figure (b). The two unknowns are the
magnitude of FR and the angle θ (theta).
12
continue
b) Trigonometry:
-The vector triangle as shown in Fig. (c), is constructed from Fig. (b).
- FR is determined by using the law of cosines;
-The angle θ is determined by applying the law of sines;
-Thus, the direction ø (phi) of FR, measured from the horizontal, is;
13
Vectors
• Vectors is defined as mathematical
expressions possessing magnitude and
direction, which add according to the
parallelogram law.
• It is represented by arrows.
• The magnitude of a vector defines the length
of the arrow used to represent the vector.
14
• Types of vectors:
1) Fixed vectors:
cannot be moved without modifying the
conditions of the problem.
2) Free vectors:
couples, which represented by vectors which
may be freely moved in space.
3) Sliding vectors:
forces acting on a rigid body, which represented
by vectors which can be moved, or slid, along
their lines of action.
15
continue
4) Equal vectors:
two vectors have the same magnitude and the same
direction, whether or not they also have the same point of
application. They are denoted by same letter.
P
P
5) Negative vectors:
vector having the same magnitude but opposite direction.
P
-P
16
Addition of Vectors
• Vectors add according to the parallelogram law.
• The sum of two vectors P and Q is obtained by attaching the
two vectors to the same point A and constructing a
parallelogram, using P and Q as two sides of the
parallelogram.
•The diagonal that passes through A represents
the sum of the vectors P and Q which denoted as
P + Q.
P
A
Q
•However, the magnitude of the vector P+Q is NOT
In general, equal to the sum (P+Q) of the
magnitudes of the vectors P and Q.
•Since that, we conclude that the addition of two
Vectors is commutative, write as:
P+Q=Q+P
17
continue
• Triangle Rule is an alternative method for determining the
sum of two vectors from the parallelogram law.
• From the only half of the parallelogram;
Q
P
A
P
OR
A
Q
• From the figures shown above, it confirms the fact that
vectors addition is commutative.
18
Subtraction of Vectors
• Defined as the addition of the corresponding
negative vector.
• P-Q representing the difference between the vectors
P and Q is obtained by adding to P the negative
vector –Q. we write;
P – Q = P + (-Q)
-Q
P
A
19
Coplanar Vectors
• The sum of three vectors P, Q, and S was obtained graphically.
• The triangle rule was first applied to obtain the sum P+Q of the
vectors P and Q.
• It was applied again to obtain the sum of vectors P+Q and S.
Q
S
P
A
• For addition of vectors, Polygon Rule is applied by arranging the given vectors
in tip-to-tail fashion and connecting the tail of the first vector with the tip of
the last one.
Q
S
P
A
20
continue
• The result obtained would have been unchanged if the vectors
Q and S had been replaced by their sum Q + S. We may thus
write;
P + Q + S = (P + Q) + S = P + (Q + S)
Q
S
P
A
which expresses the fact that vector addition is associative.
• Recalling that vector addition has been shown, in the case of
two vectors, to be commutative, we write:
P + Q + S = (P + Q) + S = S + (P + Q)
= S + (Q + P) = S + Q + P
21
continue
• This expression, as well as others which may be obtained in
the same way, shows that the order in which several vectors
are added together is immaterial.
Q
S
P
A
P
S
Q
22
Resolution of A Force Into Components
• Components of the original force F, is a single force F acting on
a particle may be replaced by two or more forces which,
together, have the same effect on the particle.
• The process of substituting them for F is called resolving the
force F into components.
• There are two cases of particular interest:
1) One of the two components.
a)
b)
P is known.
Second component, Q, is obtained by applying the triangle rule and
joining the tip of P to the tip of F
Q
P
c)
F
A
Magnitude and direction of Q are determined graphically or by
trigonometry.
23
continue
2) The line of action of each component is known.
a) The magnitude and sense of the components are obtained by
applying the parallelogram law and drawing lines, through the tip
of F, parallel to the given lines of action.
Q
F
A
P
b) This process leads to two well-defined components, P and Q,
which can be determined graphically or computed
trigonometrycally by applying the law if sines.
24
Addition of a System of Coplanar Forces
• It will be found desirable to resolve a force into two
components which are perpendicular to each other.
• In figure below, the force F has been resolved into component
Fx along the x axis and a component Fy along the y axis.
y
y
Fy
OR
F
Fy
x
θ
Fx
θ
O
F
Fx
x
O
• The parallelogram drawn to obtain the two components is a
rectangle, and Fx and Fy are called rectangular components.
25
continue
• Cartesian Unit vectors:
– Two vectors of unit magnitude, directed respectively along the
positive x and y axes, will be introduced at this point. They are
denoted by i and j.
• Scalar components:
– The scalars Fx and Fy of forces F.
• Vector components:
– The actual component forces Fx and Fy of F.
y
j
i
x
26
continue
• Cartesian Vector Notation
– Note that the rectangular components Fx and Fy of a force F may be
obtained by multiplying respectively the unit vectors i and j by
appropriate scalars. We write;
Fx = Fx i
Fy = Fy j
and express F as the Cartesian vector,
F = Fx i + Fy j
y
Fy = Fy j
j
F
Fx = Fx i
θ
O
i
x
27
continue
• Scalar Notation.
– Indicates positive and negative Fx :
– Positive Fx when the vector component Fx has the same sense as the unit
vector i. (same sense as the positive x axis).
– Negative Fx when Fx has the opposite sense.
– The positive and negative Fy is same as Fx
y
Fy
F
θ
O
Fx
x
• Denoting by F the magnitude of the force F and by θ the angle
between F and the x axis, measured counterclockwise from the
positive x axis.
• This may express the scalar components of F as follows:
Fx = F cosθ
Fy = F sinθ
28
Example 2.3
A force of 800 N is exerted on a bolt A as shown in Figure (a). Determine the
horizontal and vertical components of the force.
(a)
The vector components of F are:
Fx = -(655 N) i
Fy = +(459 N) j
May write in Cartesian vector form:
F = -(655 N) i + (459 N) j
Example 2.4
• Resolve the 1000 N (≈ 100 kg) force acting on the pipe Fig. a, into
components in the (a) x and y directions, and (b) x’ and y directions.
• Solution:
– In each case the parallelogram law is used to resolve F into its two
components, and then the vector triangle is constructed to determine
the numerical results by trigonometry.
30
continue
– Part (a)
• The vector addition F = Fx + Fy is shown in Fig. b.
• In particular, note that the length of the components is scaled along
the x and y axes by first constructing lines from the tip of F parallel
to the axes in accordance with the parallelogram law.
• From the vector triangle, Fig. c,
31
continue
• Part (b)
– The vector addition F = Fx + Fy is shown in Fig. d.
– Note carefully how the parallelogram is constructed.
– Applying the law of sines and using the data listed on the
vector triangle, Fig. e, yields:
32
Example 2.5
The force F acting on the frame shown in Fig.2-12a has a
magnitude of 500 N and is to be resolved into two
components acting along members AB and AC. Determine
the angle θ, measured below the horizontal, so that the
component FAC is directed from A toward C and has a
magnitude of 400 N.
Solution:
i. by using the parallelogram law, the vector addition of the
two components yielding the resultant is shown in Fig.b.
ii. Note carefully how the resultant force is resolved into two
components FAB and FAC, which have specified lines of
action.
iii. The corresponding vector triangle is shown in Fig.c.
33
continue
iv. The angle ø can be determined by using the law of sines:
Hence;
34
continue
• Using this value for θ, apply the law of cosines or the law of sines and
show that FAB has a magnitude of 561 N.
• Notice that F can also be directed at an angle θ above the horizontal, as
shown in Fig. d, and still produce the required component FAC.
• Show that in this case θ = 16.1˚ and FAB = 161 N.
35
Example 2.6
The ring shown in Figure a is subjected to two forces, F1 and
F2. if it is required that the resultant force have a magnitude of
1 kN and be directed vertically downward, determine (a) the
magnitudes of F1 and F2 provided θ = 30˚, and (b) the
magnitudes of F1 and F2 if F2 is to be a minimum.
36
continue
• Solution:
Part (a):
i. A sketch of the vector addition according to the
parallelogram law is shown in Fig. b.
ii. From the vector triangle constructed in Fig. c, the
unknown magnitudes F1 and F2 are determined by
using the law of sines:
37
continue
Part (b):
i. If θ is not specified, then by the vector triangle, Fig. d,
F2 may be added to F1 in various ways to yield the
resultant 1000 N force.
ii. The minimum length or magnitude of F2 will occur
when its line of action is perpendicular to F1.
iii. Any other direction, such as OA or OB, yields a larger
value for F2.
iv. Hence, when θ = 90˚ - 20˚ = 70˚, F2 is minimum.
v. From the triangle shown in Fig. e, it is seen that;
F1 = 1000 sin 70˚ N = 940 N
F2 = 1000 cos 70˚ N = 342 N
38
• Coplanar Force Resultants.
– To determine the resultant of several coplanar forces
1. Resolved each force into its x and y components.
2. The respective components are added using scalar algebra since
they are colinear.
3. The resultant force is then formed by adding the resultants of the
x and y components using parallelogram law.
Example:
Given three concurrent forces below:
y
F2
F1
x
F3
To solve this problem using Cartesian
vector notation, each force is first
represented as a Cartesian vector, i.e;
F1 = F1x i + F1y j
F2 = - F2x i + F2y j
F3 = F3x i - F3y j
The vector resultant is therefore;
FR = F1 + F2+ F3
= F1x i + F1y j - F2x i + F2y j + F3x i - F3y j
= (F1x - F2x i + F3x )i + (F1y + F2y - F3y ) j
= (FRx )i + (FRy )j
39
continue
• To solve this problem using Scalar notation,
from figure shown, since x is positive to the
right and y is positive upward, we have;
y
F2y
F1y
F1x
F2x
F3x
x
F3y
The vector resultant is therefore;
FR = = (FRx )i + (FRy )j
In the general case, the x and y components of the resultant of any number
of coplanar forces can be represented symbolically by the algebraic sum of
the x and y components of all the forces;
40
continue
• When applying these equations, it is important to use the sign convention
establish for the components.
– Components having a directional sense along the positive coordinate
axes are considered positive scalars.
– Components having a directional sense along the negative coordinate
axes are considered negative scalars.
• If this convention is followed, then the signs of the resultant components
will specify the sense of these components.
– For example; positive result indicates that the component has a
directional sense which is in positive coordinate direction.
y
FRy
FR
θ
FRx
x
41
continue
•
Once the resultant components are determined, they may be sketched along the x
and y axes in their proper direction, and the resultant force can be determined
from vector addition.
y
FRy
FR
θ
FRx
x
•
Then, the magnitude of FR can be found from the Pythagorean Theorem; which is:
•
Also, the direction angle θ, which specifies the orientation of the force is
determined from trigonometry:
42
Example 2.7
Four forces act on bolt A as shown. Determine
the resultant of the forces on the bolt.
Force Magnitude, N
x Component, N
F1
150
+ 129.9
F2
80
- 27.4
F3
110
0
F4
100
+ 96.6
Rx = +199.1
y Component, N
+ 75.0
+ 75.2
- 110.0
- 25.9
Ry= + 14.3
Thus, the resultant R of the four forces is;
43
continue
The magnitude and direction of the resultant may now be determined.
From the triangle shown, we have:
44
Example 2.8
Determine the x and y components of F1 and F2 acting on the boom
shown in Fig. (a). Express each force as a Cartesian vector.
45
continue
• Solution:
• Scalar Notation:
F1 is resolved into x and y
components using
parallelogram law as shown in
Fig. (b).
The magnitude of each
component is determined by
trigonometry.
Then, we have;
46
continue
• Solution:
F2 is resolved into x and y
components as shown in Fig.
(c).
The slope of the line of action
is indicated and could obtain
the angle θ:
Then, determine the
magnitudes of the components;
47
continue
• Solution:
The magnitude of the horizontal component, F2x, was
obtained by multiplying the force magnitude by the ratio of
the horizontal leg of the slope triangle divided by the
hypotenuse.
The magnitude of the vertical component, F2y, was obtained
by multiplying the force magnitude by the ratio of the
vertical leg divided by the hypotenuse.
Hence, using scalar notation;
48
continue
• Solution:
• Cartesian Vector Notation:
The magnitudes and directions of the components of each
force is determined.
Thus, express each force as a Cartesian vector;
49
Example 2.9
The link in Figure (a) below is subjected to two forces F1 and F2.
determine the magnitude and orientation of the resultant force.
50
continue
• Solution 1:
• Scalar Notation: (*Can be solved using
parallelogram law.)
Each force is resolved into its x
and y components, Figure (b).
These components are summed
algebraically.
The “positive” sense of the x and
y force components alongside
each equation is indicated;
51
continue
• Solution 1:
• Scalar Notation: (*Can be solved using
parallelogram law.)
Magnitude of the resultant force,
shown in Figure (c);
From the vector addition, Figure
(c), the direction angle θ is;
52
continue
• Solution 2:
• Cartesian Vector Notation:
From Figure (b), each force is
expressed as a Cartesian
vector;
Thus,
The magnitude and direction of FR are determined in the same
53
manner in solution 1.
Example 2.10
The end of the boom O in Figure (a) below is subjected to three
concurrent and coplanar forces. Determine the magnitude and
orientation of the resultant force.
54
continue
• Solution :
Each force is resolved into its x and y components, Figure (b).
Summing the x components;
The “negative” sign indicates that FRx acts to the left, as noted
by the small arrow.
Summing the y components yields;
55
continue
• Solution :
Magnitude of the resultant force,
shown in Figure (c);
From the vector addition, Figure
(c), the direction angle θ is;
56
Exercises
2.31. Determine the x and y
components of the 800 N force
2.33. Determine the magnitude of force
F so that the resultant FR of the three
forces is as small as possible.
2.34. Determine the magnitude of the
resultant force and its direction,
measured counterclockwise from the
positive x axis.
57
Cartesian Vectors
• Right handed coordinate system.
– A right handed coordinate system will
be used for developing the theory of
vector algebra that follows.
– A rectangular or Cartesian coordinate
system is said to be right-handed
provided the thumb of the right hand
points in the direction of the positive z
axis when the right-hand fingers are
curled about this axis and directed
from the positive x toward the positive
y axis.
58
continue
• Rectangular Components of a Vector
– A vector A may have one, two, or three rectangular
components along the x, y, and z coordinate axes,
depending on how the vector is oriented relative to the
axes.
When A is directed within an octant of
the
x, y, z frame, then, by two successive
applications of the parallelogram law,
we may resolve the vector into
components as;
A = A’ + Az and A = Ax + Ay
z
Az
A
Ay
Ax
x
A’
Combining these equations, A
y represented by the vector sum of its
three rectangular components;
A = Ax + Ay + Az
59
continue
• Unit vector
– Specified as the direction of A since it has a
magnitude of 1.
– If A is a vector having a magnitude A ≠ 0, then the unit
vector having the same direction as A is represented
by;
2.1
• From equation 2.1, the unit vector will be dimensionless
since the unit will cancel out.
2.2
• Equation 2.2 therefore indicates that vector A may be
expressed in terms of both its magnitude and direction
separately
– Eg: A positive scalar defines the magnitude of A.
– uA (a dimensionless vector) defines the direction and sence of A.
60
continue
• Cartesian unit vectors.
– In 3D, the set of Cartesian unit vectors i, j, k, is used to
designate the directions of the x, y, z axes
respectively.
z
k
y
i
j
The sense (or arrowhead)
of these
vectors will be described
analytically by a plus or
minus sign, depending on
whether they are pointing
along the positive or
negative x, y, or z axes.
x
Figure shows the positive Cartesian unit vectors.
61
continue
• Cartesian Vector Representation.
z
Az k
A
k↑
j→
i↙
Ay j
y
Ax i
x
• Magnitude of Cartesian Vector
62
continue
• Direction of a Cartesian Vector
– The orientation of A is defined by the coordinate
direction angle α (alpha), β (beta), and γ (gamma),
measured between the tail of A and the positive x, y, z
axes located at the tail of A.
– Each of the angles will be between 0˚ and 180˚.
z
A
Az
y
Ax
x
Ay
63
continue
Direction cosine of A
z
z
z
A
A
A
Az
y
y
y
Ax
x
x
Ay
x
64
continue
Direction angles:
Since the magnitude of a vector is equal to the positive square
root of the sum of the squares of the magnitudes of its
components, and uA has a magnitude of 1.
• A expressed in Cartesian vector form as:
65
Example 2.11
• Determine the magnitude and the coordinate direction angles
of the resultant force acting on the ring in Fig. a.
66
continue
• Solution:
– Since each force is represented in Cartesian vector form, the resultant force,
shown in Fig. b, is:
– The magnitude of FR is found from equation above;
67
continue
– The coordinate direction angles α, β, ɣ are determined from
the components of unit vector acting in the direction of FR.
– So that,
– These angles are shown in Figure b. in particular, note that
β > 90˚ since the j component of uFR is negative.
68
Position Vectors
• x, y, z coordinates.
– Right-handed coordinate system is used to reference
the location of points in space.
– In many technical books, to require the positive z axis
to be directed upward (the zenith direction) so that is
measures the height of an object or the altitude of the
point.
– The x and y axes then lie in the horizontal plane.
– Points in space are located relative to the origin of
coordinates, O, by successive measurements along the
x, y, z axes.
69
z
2m
C
B
y
4m
4m
2m
1m
6m
x
A
•From the figure above, coordinate at point A:
xA = +4 m along the x-axis
yA = +2 m along the y-axis
zA = -6 m along the z-axis
• thus,
•A (4,2,-6)
•B (0,2,0)
•C (6,-1,4)
70
continue
• Position vector.
– The position vector, r, is
defined as a fixed vector
which locates a point in
space relative to another
point.
– For example, if r extends
from the origin of
coordinates, O, to point
P(x,y,z), then, r can be
expressed in Cartesian
vector form as:
z
zk
A
r
O
yj
y
xi
x
r = xi + yj + zk
71
continue
z
B (xB, yB, zB)
r
A (xA, yA, zA)
rB
rA
y
x
•By the head-to-tail vector addition, we require:
rA + r = rB
•Solving for r and expressing rA and rB in cartesian vector
form as:
r = rB - rA
= (xB i +yB j + zB k) – (xA i + yA j + zA k)
72
continue
z
r
(xB – xA) A
rB
B
(zB – zA)
rA
y
(yB – yA)
x
•In other way in solving for r and expressing rA and rB in
Cartesian vector:
[+ i direction]
[+ j direction]
[+ k direction]
73
Example 2.12
An elastic rubber band is attached to points A and B as shown in Fig.
(a). Determine its length and its direction measured from A toward B.
74
continue
• Solution:
– First, establish a position vector
A to B, Figure (b).
– The coordinates of the tail
A (1 m, 0, -3 m) are substracted
from the coordinates of the head
B (-2 m, 2 m, 3 m), which yields;
r=B–A
= [-2 m – 1 m] i + [2 m – 0] j + [3 m – (-3 m)] k
= { -3i + 2j + 6k} m
75
continue
• Solution:
– The components of r can be determined directly by
realizing from fig. (a) that they represent the direction and
distance one must go along each axis in order to move from
A to B.
– The magnitude of r represents the length of the rubber
band.
– Formulating a unit vector in the direction of r, we have;
76
continue
• Solution:
– The components of this unit vector yield the coordinate
direction angles:
– These angles are measured from the positive axes of a
localized coordinate system placed at the tail of r, point A
as shown in Fig. (c).
77
Force vector directed along a line
– The direction of a force is specified by two points
through which its line of action passes.
z
F
B
Force F is directed along the cord AB.
A
y
x
– We can formulate F as a Cartesian vector by realizing
that it has the same direction and sense as the
position vector r directed from point A to point B on
the cord.
– This common direction is specified by the unit vector
u=r/r. Hence, F = Fu = F (r/r)
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Example 2.13
The man shown in Figure (a) pulls on chord with a force of 350 N.
Represent this force, acting on the support A, as a Cartesian vector and
determine its direction.
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continue
• Solution:
– Force F is shown in Figure (b). The direction of this vector,
u, is determined from the position vector r, which extends
from A to B.
– The coordinates of the end points
of the cord are:
A (0, 0, 7.5 m)
B (3 m, -2 m, 1.5 m)
− Forming the position vector, we
have;
r=B–A
= (3 – 0) i + (-2 – 0) j + (1.5 – 7.5) k
= { 3i – 2j – 6k} m
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continue
– Magnitude of r, (represent the length of cord AB):
– Forming the unit vector that defines the direction and sense
of both r and F yields:
– Since F has a magnitude of 350 N and a direction specified
by u, then,
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continue
– Coordinate direction angles:
 Measured between r (or F) and the positive axes of a
localized coordinate system with origin placed at A.
 From the components of the unit vector:
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Dot Product
• Dot product defines a particular method for
“multiplying” two vectors and is used to solve the 3
Dimensional problems.
A
B
• Dot product of vectors A and B is A . B
– Defined as the product of the magnitudes of A and B and
the cosine of the angle θ between their tails.
– Equation form:
A . B = AB cos θ
(Eqn. 2.3)
Where 0˚ ≤ θ ≤ 180˚
– Dot product is often referred to as the scalar product of
vectors; since the result is a scalar, NOT a vector.
83
continue
• Laws of Operation
– Commutative Law
– Multiplication by a scalar
– Distributive Law
It is easy to prove the first and second laws by using Eqn. 2.3
84
continue
• Cartesian Vector Formulation
– Equation 2.1 may be used to find the dot product for
each of the Cartesian unit vectors.
– Example:
– In similar manner:
• Should not be memorized, but understood.
85
continue
– Consider now the dot product of two general vectors A and
B which are expressed in Cartesian vector form. We have:
– Carrying out the dot-product operations, the final result
becomes:
– Thus, to determine the dot product of two Cartesian vectors,
multiply their corresponding x,y,z components and sum their
products algebraically.
– Since the result is a scalar, NOT to include any unit vector in
final result.
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References:
• R.C. Hibbeler. (2004). Engineering Mechanics
Statics – Third Edition.
• F.P. Beer, E.R. Johnston, Jr, E.R. Eisenberg.
(2004). Vector Mechanics for Engineers.
Statics. Seventh Edition in SI Units.
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