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JJ205ENGINEERING 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)
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JJ205 ENGINEERING MECHANICS
CHAPTER 2:
FORCE VECTORSCLO 1. apply the principles of statics and dynamics tosolve engineering problems (C3)
Prepared by:
THEEBENRAJ
2
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Objectives:
At the end of this chapter, student should be able to:
1. Understand scalars and vectorsa. Differentiate between scalars and vectors.
b. 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. Calculate addition of vectors.
b. Calculate subtraction of vectors.
c. Determine resolution of vectors.
4. Understand the resultant force of coplanar forces byaddition.
a. Explain scalar notation.
b. Explain cartesian vector notation.
c. Determine coplanar forces and resultant force.
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Objectives:
5. Understand Cartesian vectors.
a. Explain right handed coordinate system.
b. Explain cartesian unit vector.
c. 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 andx, y, z coordinates.a. Explain position vectors andx, y, z coordinate.
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Objectives:
9. Understand the force vector directed along the line.
a. Explain the force vector directed along the line.
b. Determine the force vector directed along the line.
10. Understand the dot product.
a. Apply laws of operation.i. Commutative law
ii. Multiplication by scalar
iii. Distributive law
b. Formulate cartesian vector formulation.
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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 arethe Newton (N).
Multiple: kilonewton (kN) = 1000N
6
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The direction of a force is defined by the line
of action and the sense of the force.
a)
b)
A 30
Magnitude of a force on particle A is 100N at 30.
A 30
The different sense of the force but have same magnitude and same
direction.
7
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Two forces, P and Qacting on a particle A:
Can be replaced by a single force R which has
the same effect on the particle.
R is the resultant of the forces P and Q.
The method is known as the parallelogram law
for the addition of two forces.
A
A
P
P
Q
Q
R=
A
R
8
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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
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Continue
b) Trigonometric Solution.
i. The triangle rule is again used; two sides and the included angle are known.
ii. We apply the law of cosines:
iii. Now, applying the law of sines, we write;
iv. Solving equation (1) for sin A, we have;
v. Using a calculator, we first compute the quotient,
then its are sines, and obtain;
(1)
10
i
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c) Alternative Trigonometric Solution.
i. We construct the right triangle BCD and compute;
ii. Then, using triangle ACD, we obtain;
iii. Again,
Continue
11
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Example 2.2 The screw eye in Figure (a) is subjected to
two forces, F1 and F2. Determine the
magnitude and direction of the resultantforce.
Solutions:
a) Parallelogram Law:
i. The parallelogram law of addition is shown
in Figure (b). The two unknowns are the
magnitude ofFR and the angle (theta).
12
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b) Trigonometry:
continue
13
-The vector triangle as shown in Fig. (c), is constructed from Fig. (b).
- FR is determined by using the law of cosines;
-The angle is determined by applying the law of sines;
-Thus, the direction (phi) ofFR, measured from the horizontal, is;
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Vectors
Vectors is defined as mathematicalexpressions possessing magnitude anddirection, which add according to theparallelogram law.
It is represented by arrows.
The magnitude of a vector defines the lengthof the arrow used to represent the vector.
14
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Types of vectors:
1) Fixed vectors:
cannot be moved without modifying the
conditions of the problem.
2) Free vectors:
couples, which represented by vectors whichmay 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
i
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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.
5) Negative vectors:
vector having the same magnitude but opposite direction.
continue
PP
P
-P16
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Addition of Vectors
Vectors add according to the parallelogram law.
The sum of two vectors P and Qis obtained by attaching the
two vectors to the same point A and constructing a
parallelogram, using P and Qas two sides of the
parallelogram.
A
P
Q
The diagonal that passes through A representsthe sum of the vectors P and Qwhich denoted as
P + Q.
However, the magnitude of the vector P+Qis NOT
In general, equal to the sum (P+Q) of themagnitudes of the vectors P and Q.
Since that, we conclude that the addition of two
Vectors is commutative, write as:
P + Q = Q + P17
ti
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Triangle Rule is an alternative method for determining the
sum of two vectors from the parallelogram law.
From the only half of the parallelogram;
From the figures shown above, it confirms the fact thatvectors addition is commutative.
continue
A
P
Q
OR
A
P
Q
18
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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
vectorQ. we write;
P Q = P + (-Q)
Subtraction of Vectors
A
-Q
P
19
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The sum of three vectors P, Q, and S was obtained graphically.
The triangle rule was first applied to obtain the sum P+Qof thevectors P and Q.
It was applied again to obtain the sum of vectors P+Qand S.
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 ofthe last one.
Coplanar Vectors
A
P
QS
A
P
QS
20
continue
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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)
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
continue
A
P
QS
21
continue
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This expression, as well as others which may be obtained in
the same way, shows that the order in which several vectorsare added together is immaterial.
A
P
QS
continue
SQ
P
22
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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 theforce F into components.
There are two cases of particular interest:
1) One of the two components.
a) P is known.
b) Second component, Q, is obtained by applying the triangle rule andjoining the tip of P to the tip ofF
c) Magnitude and direction ofQare determined graphically or bytrigonometry.
Resolution of A Force Into Components
A
P
Q
F
23
continue
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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
ofF, parallel to the given lines of action.
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.
continue
F
A
Q
P
24
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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.
The parallelogram drawn to obtain the two components is a
rectangle, and Fx and Fy are called rectangular components.
Addition of a System of Coplanar Forces
O
FFy
Fx
y
x
O
FFy
Fx
y
xOR
25
continue
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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 andj.
Scalar components:
The scalars Fx and Fy of forces F.
Vector components:
The actual component forces Fx and Fy ofF.
j
y
x
continue
i
26
continue
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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 andj by
appropriate scalars. We write;
Fx = Fxi Fy = Fyj
and express F as the Cartesian vector,
F = Fx
i+ Fy
j
O
Fj
y
x
continue
i
Fy = Fyj
Fx = Fxi
27
continue
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Scalar Notation. Indicates positive and negative Fx :
Positive Fx when the vector component Fx has the same sense as the unitvector 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
Denoting by F the magnitude of the force F and by the anglebetween F and the x axis, measured counterclockwise from thepositive x axis.
This may express the scalar components of F as follows:
Fx = F cos Fy = F sin
continue
O
F
y
x
Fy
Fx
28
Example 2 3
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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)
Example 2.3
The vector components ofF 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
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Resolve the 1000 N ( 100 kg) force acting on the pipe Fig. a, into
components in the (a)x andy directions, and (b)xandy 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
Example 2.4
continue
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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 ofF parallel
to the axes in accordance with the parallelogram law.
From the vector triangle, Fig. c,
31
continue
continue
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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
continue
Example 2 5
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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 , measured below the horizontal, so that thecomponent 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 ofaction.
iii. The corresponding vector triangle is shown in Fig.c.
33
Example 2.5
continue
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iv. The angle can be determined by using the law of sines:
Hence;
34
continue
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Using this value for , 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 above the horizontal, asshown in Fig. d, and still produce the required component FAC.
Show that in this case = 16.1 and FAB = 161 N.
35
Example 2 6
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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) themagnitudes ofF1 and F2 provided = 30, and (b) the
magnitudes ofF1 and F2 ifF2 is to be a minimum.
36
Example 2.6
continue
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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 F1and F2 are determined by
using the law of sines:
37
continue
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Part (b):
i. If 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 ofF2 will occur
when its line of action isperpendicularto F1.
iii. Any other direction, such as OA or OB, yields a larger
value for F2.
iv. Hence, when = 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.
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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:
F1
F3
F2
x
y
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 jF3 = 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
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To solve this problem using Scalar notation,
from figure shown, sincex is positive to the
right andy is positive upward, we have;
The vector resultant is therefore;
FR = = (FRx )i + (FRy )j
In the general case, thex andy components of the resultant of any number
of coplanar forces can be represented symbolically by the algebraic sum ofthex andy components of all the forces;
F1y
F3xx
y
F2x
F2y
F1x
F3y
40
continue
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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.
FR
x
y
FRy
FRx
41
continue
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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.
Then, the magnitude of FR can be found from the Pythagorean Theorem; which is:
Also, the direction angle , which specifies the orientation of the force isdetermined from trigonometry:
FR
x
y
FRy
FRx
42
Example 2 7
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43
Four forces act on bolt A as shown. Determine
the resultant of the forces on the bolt.
Example 2.7
Thus, the resultant R of the four forces is;
Force Magnitude, N x Component, N y Component, N
F1 150 + 129.9 + 75.0
F2 80 - 27.4 + 75.2F3 110 0 - 110.0
F4 100 + 96.6 - 25.9
Rx = +199.1 Ry= + 14.3
continue
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44
The magnitude and direction of the resultant may now be determined.
From the triangle shown, we have:
Example 2 8
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45
Example 2.8
Determine thex andy components ofF1 and F2 acting on the boom
shown in Fig. (a). Express each force as a Cartesian vector.
continue
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46
Solution:
Scalar Notation:
F1 is resolved into x and ycomponents using
parallelogram law as shown in
Fig. (b).
The magnitude of each
component is determined by
trigonometry.
Then, we have;
continue
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47
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 :
Then, determine the
magnitudes of the components;
continue
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48
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 thehypotenuse.
The magnitude of the vertical component, F2y, was obtained
by multiplying the force magnitude by the ratio of thevertical leg divided by the hypotenuse.
Hence, using scalar notation;
continue
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49
Solution:
Cartesian Vector Notation:
The magnitudes and directions of the components of eachforce is determined.
Thus, express each force as a Cartesian vector;
Example 2.9
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50
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.
continue
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51
Solution 1:
Scalar Notation: (*Can be solved usingparallelogram law.)
Each force is resolved into itsx
andy components, Figure (b).
These components are summed
algebraically.
The positive sense of thex and
y force components alongside
each equation is indicated;
continue
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52
Solution 1:
Scalar Notation: (*Can be solved usingparallelogram law.)
Magnitude of the resultant force,shown in Figure (c);
From the vector addition, Figure(c), the direction angle is;
continue
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53
Solution 2:
Cartesian Vector Notation:
From Figure (b), each force isexpressed as a Cartesian
vector;
Thus,
The magnitude and direction of FR are determined in the same
manner in solution 1.
Example 2.10
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54
p
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.
continue
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55
Solution :
Each force is resolved into itsx andy components, Figure (b).
Summing thex components;
The negative sign indicates that FRx acts to the left, as notedby the small arrow.
Summing the y components yields;
continue
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56
Solution :
Magnitude of the resultant force,
shown in Figure (c);
From the vector addition, Figure
(c), the direction angle is;
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Exercises
2.31. Determine thex andycomponents of the 800 N force
2.33. Determine the magnitude of force
F so that the resultant FR of the threeforces is as small as possible.
2.34. Determine the magnitude of the
resultant force and its direction,measured counterclockwise from the
positivex axis.
57
C t i V t
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Right handed coordinate system.
A right handed coordinate system willbe used for developing the theory ofvector algebra that follows.
A rectangular or Cartesian coordinatesystem is said to be right-handedprovided the thumb of the right handpoints in the direction of the positive zaxis when the right-hand fingers arecurled about this axis and directedfrom the positive x toward the positivey axis.
Cartesian Vectors
58
continue
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Rectangular Components of a Vector
A vector A may have one, two, or three rectangular
components along thex,y, andz coordinate axes,depending on how the vector is oriented relative to the
axes.
A
Ay
A
Ax
Az
z
y
x
When A is directed within an octant of
the
x, y, z frame, then, by two successiveapplications of the parallelogram law,
we may resolve the vector into
components as;
A = A + Az and A = Ax + Ay
Combining these equations, A
represented by the vector sum of its
three rectangular components;
A = Ax
+ Ay
+ Az
59
U i
continue
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Unit vector
Specified as the direction ofA since it has amagnitude of 1.
IfA is a vector having a magnitude A 0, then the unitvector having the same direction as A is representedby;
From equation 2.1, the unit vector will be dimensionlesssince the unit will cancel out.
Equation 2.2 therefore indicates that vector A may beexpressed in terms of both its magnitude and directionseparately
Eg: A positive scalar defines the magnitude ofA.
uA (a dimensionless vector) defines the direction and sence of A.
2.1
2.2
60
continue
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Cartesian unit vectors.
In 3D, the set of Cartesian unit vectors i,j,k, is used to
designate the directions of thex, y, z axesrespectively.
z
y
x
k
ji
The sense (or arrowhead)
of these
vectors will be describedanalytically by a plus or
minus sign, depending on
whether they are pointing
along the positive or
negativex, y, orz axes.
Figure shows the positive Cartesian unit vectors.
61
continue
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Cartesian Vector Representation.
Magnitude of Cartesian Vector
z
y
x
k
j
i
A
Az k
Ax i
Ay j
62
f
continue
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Direction of a Cartesian Vector
The orientation ofA is defined by the coordinate
direction angle (alpha), (beta), and (gamma),measured between the tail ofA and the positivex, y, z
axes located at the tail ofA.
Each of the angles will be between 0 and 180.z
y
x
A
Az
Ax
Ay 63
continue
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Direction cosine ofA
z
y
A
x
z
y
x
A
Ax
z
y
x
A
Ay
Az
64
continue
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A expressed in Cartesian vector form as:
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.
65
Example 2.11
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Determine the magnitude and the coordinate direction angles
of the resultant force acting on the ring in Fig. a.
66
continue
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Solution: Since each force is represented in Cartesian vector form, the resultant force,
shown in Fig. b, is:
The magnitude ofFR is found from equation above;
67
continue
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The coordinate direction angles ,, are determined from
the components of unit vector acting in the direction ofFR.
So that,
These angles are shown in Figure b. in particular, note that
> 90 since the j component ofuFR is negative.
68
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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 axisto be directed upward (the zenith direction) so that ismeasures the height of an object or the altitude of thepoint.
Thex andy axes then lie in the horizontal plane.
Points in space are located relative to the origin ofcoordinates, O, by successive measurements along the
x, y, z axes.
Position Vectors
69
z
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y
x
4 m
4 m
6 m1 m
2 m
2 m
From the figure above, coordinate at point A:
xA = +4 m along thex-axis
yA = +2 m along they-axis
zA = -6 m along thez-axis
thus,
A (4,2,-6)
B (0,2,0)
C (6,-1,4)
A
BC
70
continue
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Position vector.
The position vector, r, is
defined as a fixed vectorwhich locates a point inspace relative to anotherpoint.
For example, ifr extendsfrom the origin ofcoordinates, O, to pointP(x,y,z), then, r can beexpressed in Cartesian
vector form as:
r =xi +yj+zk
A
O
x i
z
y
x
yj
r
zk
71
continue
z
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y
x
r
rA
rB
B (xB, yB, zB)
A (xA, yA, zA)
By the head-to-tail vector addition, we require:
rA +r = rBSolving for r and expressingrA andrB in cartesian vector
form as:
r = rB -rA= (xB i +yB j + zB k)(xA i + yA j + zA k)
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z
continue
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y
x
r
rA
rB
B
A(xBxA)
(yByA)
(zBzA)
In other way in solving for r and expressingrA andrB in
Cartesian vector:
[+ i direction] [+j direction] [+ kdirection]
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Example 2.12
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An elastic rubber band is attached to pointsA andB as shown in Fig.
(a). Determine its length and its direction measured fromA towardB.
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Solution:
First, establish a position vector
A toB, 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 = BA
= [-2 m1 m]i
+ [2 m0]j
+ [3 m(-3 m)]k
= { -3i + 2j+ 6k} m
Solution:
continue
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Solution:
The components ofr 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 fromA toB.
The magnitude ofr represents the length of the rubber
band.
Formulating a unit vector in the direction ofr, we have;
Solution:
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Solution:
The components of this unit vector yield the coordinatedirection angles:
These angles are measured from thepositive axes of alocalized coordinate system placed at the tail ofr, point Aas shown in Fig. (c).
d d l l
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The direction of a force is specified by two pointsthrough which its line of action passes.
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)
Force vector directed along a line
z
y
x
A
B
F
Force F is directed along the cord AB.
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Example 2.13
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79
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.
Solution:
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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 = BA
= (30) i + (-20)j + (1.57.5) k
= { 3i2j6k} m
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Magnitude of r (represent the length of cord AB):
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Magnitude ofr, (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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Coordinate direction angles:
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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:
Dot Product
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Dot product defines a particular method formultiplying two vectors and is used to solve the 3Dimensional problems.
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 between their tails.
Equation form:
A . B = AB cos
Where 0 180
Dot product is often referred to as the scalar product ofvectors; since the result is a scalar, NOT a vector.
Dot Product
A
B
(Eqn. 2.3)
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Laws of Operation
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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
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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.
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Consider now the dot product of two general vectors A and
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Consider now the dot product of two general vectors A andB which are expressed in Cartesian vector form. We have:
Carrying out the dot-product operations, the final resultbecomes:
Thus, to determine the dot product of two Cartesian vectors,multiply their corresponding x,y,z components and sum theirproducts algebraically.
Since the result is a scalar, NOT to include any unit vector infinal result.
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R f
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References:
R.C. Hibbeler. (2004). Engineering MechanicsStatics Third Edition.
F.P. Beer, E.R. Johnston, Jr, E.R. Eisenberg.
(2004). Vector Mechanics for Engineers.Statics. Seventh Edition in SI Units.