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8/14/2019 Lecture 4- 1page
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© 2003 The McGraw-Hill Companies, Inc. All rights reserved.
Vector Mechanics for Engineers: Statics S ev en t h
E d i t i on
2 - 1
Rectangular Components in Space (3-D Vectors)
• The vector is
contained in the
plane OBAC .
F r
• Resolve into
horizontal and
vertical components.
F r
y y F F θ cos=
yh F F θ sin=
• Resolve into
rectangular componentshF
φ θ
φ
φ θ
φ
sinsin
sin
cossin
cos
y
h y
y
h x
F
F F
F
F F
=
=
==
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Vector Mechanics for Engineers: Statics S ev en t h
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Rectangular Components in Space
• With the angles between and the axes,F r
( )
k ji
F
k jiF
k F jF iF F
F F F F F F
z y x
z y x
z y x
z z y y x x
rrrr
r
rrr
rrrr
θ θ θ λ
λ
θ θ θ
θ θ θ
coscoscos
coscoscos
coscoscos
++==
++=
++=
===
• is a unit vector along the line of action of
and are the directioncosines for
F r
F r
λ r
z y x θ θ θ cosand ,cos,cos
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A UNIT VECTOR
Characteristics of a unit vector :
a) Its magnitude is 1.
b) It is dimensionless.
c) It points in the same direction as the
original vector ( A).
The unit vectors in the Cartesian axissystem are i, j, and k. They are unit
vectors along the positive x, y, and z
axes respectively.
For a vector A with a magnitude ofA, an unit vector is defined as U A =
A / A .
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3-D CARTESIAN VECTOR TERMINOLOGY
Consider a box with sides AX,
AY, and AZ meters long.
The vector A can be defined as
A = (AX i + AY j + AZ k) m
The projection of the vector A in the x-y plane is A´. The
magnitude of this projection, A´, is found by using the same
approach as a 2-D vector: A´ = (AX2 + AY
2)1/2 .
The magnitude of the position vector A can now be obtained as
A = ((A´)2
+ AZ2
) ½
= (AX2
+ AY2
+ AZ2
) ½
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The direction or orientation of vector A isdefined by the angles α, β, and γ.
These angles are measured between the vector
and the positive X, Y and Z axes,
respectively. Their range of values are from
0° to 180°Using trigonometry, “direction cosines” are found using the formulas
These angles are not independent. They must satisfy the following equation.
cos ² α + cos ² β + cos ² γ = 1
This result can be derived from the definition of a coordinate direction anglesand the unit vector. Recall, the formula for finding the unit vector of any
position vector:
or written another way, u A = cos α i + cos β j + cos γ k .
TERMS (continued)
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ADDITION/SUBTRACTION OF VECTORS (Section 2.6)
Once individual vectors are written in Cartesian form, it is easy
to add or subtract them. The process is essentially the same as
when 2-D vectors are added.
For example, if
A = AX i + AY j + AZ k and
B = BX i + BY j + BZ k , then
A + B = (AX + BX) i + (AY + BY) j + (AZ + BZ) k
or
A – B = (AX - BX) i + (AY - BY) j + (AZ - BZ) k .
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IMPORTANT NOTES
Sometimes 3-D vector information is given as:
a) Magnitude and the coordinate direction angles, or
b) Magnitude and projection angles.
You should be able to use both these types of
information to change the representation of
the vector into the Cartesian form, i.e.,
F = {10 i – 20 j + 30 k} N .
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EXAMPLE
Given:Two forces F and G are applied
to a hook. Force F is shown in
the figure and it makes 60°
angle with the X-Y plane. ForceG is pointing up and has a
magnitude of 80 lb with α =
111° and β = 69.3°.
Find: The resultant force in the
Cartesian vector form.
Plan:
1) Using geometry and trigonometry, write F and G in the
Cartesian vector form.
2) Then add the two forces.
G
S E
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Solution : First, resolve force F.Fz = 100 sin 60° = 86.60 lb
F' = 100 cos 60° = 50.00 lb
Fx = 50 cos 45° = 35.36 lb
Fy = 50 sin 45° = 35.36 lb
Now, you can write:
F = {35.36 i – 35.36 j + 86.60 k} lb
S E
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Now resolve force G.
We are given only α and β. Hence, first we need to find the value
of γ.
Recall the formula cos ² (α) + cos ² (β) + cos ² (γ) = 1.
Now substitute what we know. We have
cos ² (111°) + cos ² (69.3°) + cos ² (γ) = 1.
Solving, we get γ = 30.22° or 120.2°. Since the vector is pointing
up, γ = 30.22°
Now using the coordinate direction angles, we can get U G, and
determine G = 80 U G lb.
G = {80 ( cos (111°) i + cos (69.3°) j + cos (30.22°) k )} lb
G = {- 28.67 i + 28.28 j + 69.13 k } lb
Now, R = F + G or
R = {6.69 i – 7.08 j + 156 k} lb
V M h i f E i S i S E
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POSITION VECTOR
A position vector is defined asa fixed vector that locates a
point in space relative to
another point.
Consider two points, A & B, in
3-D space. Let their coordinates
be (XA, YA, ZA) and ( XB,
YB, ZB ), respectively.
The position vector directed from A to B, r AB , is defined as
r AB
= {( XB
– XA
) i + ( YB
– YA
) j + ( ZB
– ZA
) k }m
Please note that B is the ending point and A is the starting point.
So ALWAYS subtract the “tail” coordinates from the “tip”
coordinates!
V t M h i f E i St ti S E
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FORCE VECTOR DIRECTED
ALONG A LINE (Section 2.8)
If a force is directed along a line,
then we can represent the forcevector in Cartesian Coordinates by
using a unit vector and the force
magnitude. So we need to:
a) Find the position vector, r AB , along two points on
that line.
b) Find the unit vector describing the line’s direction,u AB = ( r AB/r AB).
c) Multiply the unit vector by the magnitude of the
force, F = F u AB .
V t M h i f E i St ti S E
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Rectangular Components in Space
Direction of the force is defined by
the location of two points,
( ) ( )222111 ,,and ,, z y x N z y x M
( )
d Fd F
d Fd F
d Fd F
k d jd id d
F F
z zd y yd x xd
k d jd id
N M d
z z
y y
x x
z y x
z y x
z y x
===
++=
=
−=−=−=
++==
rrrr
rr
rrr
r
1
and joiningvector
121212
λ
λ
V t M h i f E i St ti S
E
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APPLICATIONS
Many problems in real-lifeinvolve 3-Dimensional Space.
How will you represent
each of the cable forces in
Cartesian vector form?
V t M h i f E i St ti S e
E d
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APPLICATIONS (continued)
Given the forces in the cables, how will you determine
the resultant force acting at D, the top of the tower?
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Sample Problem 2.7
SOLUTION:• Determine the unit vector pointing from A
towards B.
( ) ( ) ( )
( ) ( ) ( )
m3.94
m30m80m40
m30m80m40
222
=
++−=
++−=
AB
k ji ABrrr
• Determine the components of the force.
( )( )
( ) ( ) ( )k ji
k ji
F F
rrr
rrr
rr
N795 N2120 N1060
318.0848.0424.0 N2500
++−=
++−=
= λ
k ji
k ji
rrr
rrrr
318.0848.0424.0
3.9430
3.9480
3.9440
++−=
⎟ ⎠ ⎞⎜⎝ ⎛ +⎟ ⎠ ⎞⎜⎝ ⎛ +⎟ ⎠ ⎞⎜⎝ ⎛ −=λ
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Sample Problem 2.7
The tension in the guy wire is 2500 N.
Determine:
a) components F x , F y , F z of the force
acting on the bolt at A,
b) the angles θ x , θ y , θ z defining the
direction of the force
SOLUTION:
• Based on the relative locations of the
points A and B, determine the unit
vector pointing from A towards B.
• Apply the unit vector to determine the
components of the force acting on A.
• Noting that the components of the unitvector are the direction cosines for the
vector, calculate the corresponding
angles.
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Sample Problem 2.7
• Noting that the components of the unit vectorare the direction cosines for the vector, calculate
the corresponding angles.
k ji
k ji z y xrrr
rrrr
318.0848.0424.0
coscoscos
++−=++= θ θ θ λ
o
o
o
5.71
0.321.115
=
==
z
y
x
θ
θ θ
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EXAMPLE
Given: 400 lb force along the
cable DA.
Find: The force F DA in the
Cartesian vector form.
Plan:
1. Find the position vector r DA and the unit vector u DA.
2. Obtain the force vector as F DA = 400 lb u DA .
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EXAMPLE (continued)
The figure shows that when relating D
to A, we will have to go -2 ft in the x-
direction, -6 ft in the y-direction, and
+14 ft in the z-direction. Hence, r DA = {-2 i – 6 j + 14 k} ft.
We can also find r DA by subtracting
the coordinates of D from the
coordinates of A.
r DA = (22
+ 62
+ 142
)0.5
= 15.36 ftu DA = r DA/r DA and F DA = 400 u DA lb
F DA = 400{(-2 i – 6 j + 14 k)/15.36} lb
= {-52.1 i – 156 j + 365 k} lb