Download Chapter 2

Announcements
2-D Vector Addition
Today’s Objectives
• Understand the difference between scalars and vectors
• Resolve a 2-D vector into components
• Perform vector operations
Class Activities
• Applications
• Scalar and vector definitions
• Triangle and Parallelogram laws
• Vector resolution
• Examples
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Applications
There are four
concurrent cable forces
acting on the bracket.
How do you determine
the resultant force acting
on the bracket?
Scalars and Vectors
• Scalars
o Magnitude - positive or negative
o Addition rule - simple arithmetic
o Special Notation - none
o Examples – speed, mass, volume
• Vectors
o Magnitude and direction – positive or negative
o Addition rule – parallelogram law
o Special notation - Bold font, line, arrow, or “carrot”
o Examples – force, velocity
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Vector Operations
• Multiplication and division
Vector Addition
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Vector Subtraction
Resolution of a Vector
“Resolution” of a vector is breaking it up into components. It
is kind of like using the parallelogram law in reverse.
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Triangle Laws
Textbook Problem 2-4
Determine the magnitude of the resultant
force FR = F1 + F2 and its direction, measured
counterclockwise from the positive u axis.
Answer: Fr = 605N
θ = 85.4°
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Textbook Problem 2-16
Resolve the force F2 into components acting along
the u and v axes and determine the magnitudes of
the components.
Answer: F2v = 77.6N
F2u = 150N
Summary
• Understand the difference between scalars and vectors
• Resolve a 2-D vector into components
• Perform vector operations
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Announcements
Cartesian Vector Notation
Today’s Objectives
• Resolve 2-dimensional vectors into x and y components
• Find the resultant of a 2-dimensional force system
• Express vectors in Cartesian form
Class Activities
• Cartesian Vector Notation
• Addition of vectors
• Examples
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Cartesian Vector Notation
• We ‘resolve’ vectors into
components using the x and y
axes system
• Each component of the vector is
shown as a magnitude and a
direction
• The directions are based on the x and y axes. We use the
“unit vectors” i and j to designate the x and y axes,
respectively.
Cartesian Vector Notation - continued
For example,
F = Fx i + Fy j
F' = F'x i + F'y j
The x and y axes are always perpendicular to each
other. Together,they can be directed at any inclination.
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Addition of Several Vectors
• Step 1 is to resolve each force
into its components.
• Step 2 is to add all the x
components together and add
all the y components together.
These two totals become the
resultant vector.
Addition of Several Vectors - continued
• Step 3 is to find the magnitude
and angle of the resultant vector.
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Magnitude and Angle
You can also represent a 2-D vector with a
magnitude and an angle.
Example A
Given: Three concurrent forces
acting on a bracket.
Find:
The magnitude and angle
of the resultant force.
Plan:
a) Resolve the forces into their x and y components
b) Add the respective components to get the resultant vector
c) Find magnitude and angle from the resultant components
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Example A - continued
F1 = { 15 sin 40° i + 15 cos 40° j } kN
= { 9.642 i + 11.49 j } kN
F2 = { -(12/13)26 i + (5/13)26 j } kN
= { -24 i + 10 j } kN
F3 = { 36 cos 30° i – 36 sin 30° j } kN
= { 31.18 i – 18 j } kN
Example A - continued
Summing up all the i and j components respectively, we get,
FR = { (9.642 – 24 + 31.18) i + (11.49 + 10 – 18) j } kN
= { 16.82 i + 3.49 j } kN
FR = ((16.82)2 + (3.49)2)1/2 = 17.2 kN
θ = tan-1(3.49/16.82) = 11.7°
Engr221 Chapter 2
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FR
θ
x
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Example B
Given: Three concurrent
forces acting on a
bracket.
Find:
The magnitude and
angle of the
resultant force.
Plan:
a) Resolve the forces into their x and y components
b) Add the respective components to get the resultant vector
c) Find magnitude and angle from the resultant components
Example B - continued
F1 = { (4/5) 850 i - (3/5) 850 j } N
= { 680 i - 510 j } N
F2 = { -625 sin(30°) i - 625 cos(30°) j } N
= { -312.5 i - 541.3 j } N
F3 = { -750 sin(45°) i + 750 cos(45°) j } N
{ -530.3 i + 530.3 j } N
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Example B - continued
Summing up all the i and j components respectively, we get,
FR = { (680 - 312.5 - 530.3) i + (-510 - 541.3 + 530.3) j }N
= { - 162.8 i - 521 j } N
y
FR = ((162.8)2 + (521)2) ½ = 546 N
θ
θ = tan–1(521/162.8) = 72.64° or
From the positive x axis: θ = 180 + 72.64 = 253°
x
FR
Textbook Problem 2-55
If F2 = 150 lb and θ = 55º, determine the magnitude and
orientation, measured clockwise from the positive x axis, of
the resultant force of the three forces acting on the bracket.
Answer: FR = 161 lb
θ = 38.3°
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Summary
• Resolve 2-dimensional vectors into x and y components
• Find the resultant of a 2-dimensional force system
• Express vectors in Cartesian form
Announcements
• Homework feedback
– Graph paper
– Problem statements
• Week of prayer schedule
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3 Dimensional Vectors
Today’s Objectives
• Represent a 3-D vector in a Cartesian coordinate system
• Find the magnitude and coordinate angles of a 3-D vector
• Add vectors is 3-D space
Class Activities
• Applications
• Unit vector definition
• 3-D vector terms
• Adding vectors
• Examples
Applications
Many problems in real-life
involve 3-Dimensional space.
How do you represent each of
the cable forces in Cartesian
vector form?
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Applications - continued
Given the forces in the cables, how do you determine the
resultant force acting at D, the top of the tower?
Applications - continued
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Right–Handed Coordinate System
A 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.
A Unit Vector
For a vector A with a magnitude of
A, a unit vector UA is defined as
UA = A / A
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 axis
system are i, j, and k. They are unit
vectors along the positive x, y, and z
axes respectively.
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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 + AY2)1/2 .
The magnitude of the vector A can now be obtained as:
A = ((A´)2 + AZ2) ½ = (AX2 + AY2 + AZ2) ½
Terminology - continued
The direction or orientation of vector A is defined by the angles α, β, and γ.
These angles are measured between the vector and the positive x, y and z axes,
respectively. Their values range from 0° to 180°.
Using trigonometry, “direction cosines”, and coordinate direction angles are
found using the formulas:
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Terminology - continued
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 coordinate direction angles
and the unit vector. Recall, the formula
for finding the unit vector is:
or written another way: uA = cos α i + cos β j + cos γ k
Cartesian Vector Math
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
and
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 coordinate direction
angles, or
b) Magnitude and projection angles
You should be able to use both types of
information to change the representation
of the vector into the Cartesian form, i.e.,
F = {10 i – 20 j + 30 k} N
Example A
G
Given: Two forces F and G are applied
to a hook. Force F is shown in
the figure and it makes a 60°
angle with the x-y plane. Force
G is pointing up and has a
magnitude of 80 lb with α =
111° and β = 69.3°.
Find: The resultant force in Cartesian
vector form.
Plan:
1) Using geometry and trigonometry, write F and G in
Cartesian vector form
2) Add the two forces together
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Example A - continued
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
Example A - continued
Next, resolve force G.
We are given only α and β. 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°
G
Now using the coordinate direction angles, we can get UG, and determine G
from the formula: G = 80UG 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
Finally, find the resultant vector R = F + G or
R = {6.69 i – 7.08 j + 156 k} lb
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Example B
Given: The screw eye is subjected
to two forces.
Find:
The magnitude and the
coordinate direction angles
of the resultant force.
Plan:
1) Using geometry and trigonometry, write F1 and F2 in
Cartesian vector form
2) Add F1 and F2 to get FR
3) Determine the magnitude and α, β, and γ of FR
Example B - continued
First resolve force F1.
F1z
F1z = 300 sin 60° = 259.8 N
F´
F´ = 300 cos 60° = 150.0 N
F´ can be further resolved as,
F1x = -150 sin 45° = -106.1 N
F1y = 150 cos 45° = 106.1 N
Now we can write :
F1 = {-106.1 i + 106.1 j + 259.8 k } N
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Example B - continued
Force F2 can be represented in
Cartesian vector form as:
F2 = 500{ cos 60° i + cos 45° j +
cos 120° k } N
= { 250 i + 353.6 j – 250 k } N
FR = F1 + F2
= { 143.9 i + 459.6 j + 9.81 k } N
FR = (143.9 2 + 459.6 2 + 9.81 2) ½ = 481.7 = 482 N
α = cos-1 (FRx / FR) = cos-1 (143.9/481.7) = 72.6°
β = cos-1 (FRy / FR) = cos-1 (459.6/481.7) = 17.4°
γ = cos-1 (FRz / FR) = cos-1 (9.81/481.7) = 88.8°
Questions
1. What is not true about the unit vector, UA ?
A) It is dimensionless
B) Its magnitude is one
C) It always points in the direction of positive x axis
D) It always points in the direction of vector A
2. If F = {10 i + 10 j + 10 k} N and
G = {20 i + 20 j + 20 k } N, then F + G = { ____ } N
A) 10 i + 10 j + 10 k
B) 30 i + 20 j + 30 k
C) -10 i - 10 j - 10 k
D) 30 i + 30 j + 30 k
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Textbook Problem 2-68
Determine the magnitude and coordinate
direction angles of the resultant force.
Answers:
FR = 369 N
α = 19.5°
β = 20.6°
γ = 69.5°
Textbook Problem 2-78
Two forces F1 and F2 act on the bolt. If the resultant force
Fr has a magnitude of 50 lb and coordinate direction
angles α = 110º and β = 80º as shown, determine the
magnitude of F2 and its coordinate direction angles.
Answers:
F2 = 32.4 lb
α2 = 122º
β2 = 74.5º
γ2 = 144º
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Summary
• Represent a 3-D vector in a Cartesian coordinate system
• Find the magnitude and coordinate angles of a 3-D vector
• Add vectors (forces) in 3-D space
Announcements
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Review
Vector: A = (AX i + AY j + AZ k)
Direction cosines:
Trig. Law: cos ² α + cos ² β + cos ² γ = 1
Unit vector:
or written another way: u A = cos α i + cos β j + cos γ k
Position and Force Vectors
Today’s Objectives
• Represent a position vector in Cartesian vector form using given geometry
• Represent a force vector directed along a line
Class Activities
• Applications
• Position vectors
• Force vectors
• Examples
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Applications
How can we
represent the force
along the wing
strut in a 3-D
Cartesian vector
form?
Wing strut
Position Vector
A position vector is defined as a
fixed vector that locates a point
in space relative to another
point.
Consider two points, A & B, in
3D space. Let their coordinates be
(XA, YA, ZA) and (XB, YB, ZB ),
respectively.
The position vector directed from A to B, rAB , is defined as
rAB = {( XB – XA ) i + ( YB – YA ) j + ( ZB – ZA ) k } m
Note that B is the ending point and A is the starting point.
Always subtract the “tail” coordinates from the “tip” coordinates.
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Force Vector Along a Line
If a force is directed along a line,
then the force can be represented
in Cartesian coordinates by using a
unit vector and the force
magnitude. So we need to:
a) Find the position vector, rAB , along two points on
that line
b) Find the unit vector describing the line’s direction,
uAB = (rAB/rAB)
c) Multiply the unit vector by the magnitude of the
force, F = F uAB
Example A
Given: 400 lb force along the
cable DA
Find:
The force FDA in
Cartesian vector form
Plan:
1. Find the position vector rDA and the unit vector uDA
2. Obtain the force vector as FDA = 400 lb uDA
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Example A - continued
We can find rDA by subtracting the
coordinates of D from the coordinates of A.
D = (2, 6, 0) ft and A = (0, 0, 14) ft
rDA = {(0 – 2) i + (0 – 6) j + (14 – 0) k} ft
= {-2 i – 6 j + 14 k} ft
We can also calculate rDA by realizing 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.
rDA = (22 + 62 + 142)0.5 = 15.36 ft
uDA = rDA/rDA and FDA = 400 uDA lb
FDA = 400{(-2 i – 6 j + 14 k)/15.36} lb
= {-52.1 i – 156 j + 365 k} lb
Example B
Given: Two forces are acting on
a pipe as shown in the
figure
Find:
The magnitude and the
coordinate direction
angles of the resultant
force
Plan:
1) Find the forces along CA and CB in Cartesian vector form
2) Add the two forces to get the resultant force, FR
3) Determine the magnitude and the coordinate angles of FR
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Example B - continued
FCA = 100 lb{rCA/rCA}
FCA = 100 lb(-3 sin 40° i + 3 cos 40° j – 4 k)/5
FCA = {-38.57 i + 45.96 j – 80 k} lb
FCB = 81 lb{rCB/rCB}
FCB = 81 lb(4 i – 7 j – 4 k)/9
FCB = {36 i – 63 j – 36 k} lb
FR = FCA + FCB = {-2.57 i – 17.04 j – 116 k} lb
FR = (2.572 + 17.042 + 1162)1/2 = 117.3 lb = 117 lb
α = cos-1(-2.57/117.3) = 91.3°, β = cos-1(-17.04/117.3) = 98.4°
γ = cos-1(-116/117.3) = 172°
Questions
1. Two points in 3-D space have coordinates of P(1, 2, 3) and Q
(4, 5, 6) meters. The position vector rQP is given by
A) {3 i + 3 j + 3 k} m
B) {- 3 i – 3 j – 3 k} m
C) {5 i + 7 j + 9 k} m
D) {- 3 i + 3 j + 3 k} m
E) {4 i + 5 j + 6 k} m
2. P and Q are two points in 3-D space. How are the position
vectors rPQ and rQP related?
A) rPQ = rQP
B) rPQ = - rQP
C) rPQ = 1/rQP
D) rPQ = 2 rQP
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Textbook Problem 2.90
Determine the magnitude and coordinate direction
angles of the resultant force.
Answers:
FR = 52.2 lb
α = 87.8º
β = 63.7º
γ = 154º
Summary
• Represent a position vector in Cartesian vector form using
given geometry
• Represent a force vector directed along a line
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Announcements
• Homework
– Excuses. Not!
– Due date and time
– Graph paper
– Do solutions on one side only
– Problem statements
Dot Product
Today’s Objectives
• Determine the angle between two vectors
• Determine the projection of a vector along a specified line
Class Activities
• Applications
• Dot product
• Angle determination
• Projection determination
• Examples
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Applications
For this geometry, can you
determine angles between the
pole and the cables?
For force F at Point A, what
component of it (F1) acts along the
pipe OA? What component (F2) acts
perpendicular to the pipe?
Dot Product Definition
The dot product of vectors A and B is defined as A•B = AB cos θ
Angle θ is the smallest angle between the two vectors and is
always in a range of 0º to 180º
Dot Product Characteristics:
1. The result of the dot product is a scalar (a positive or
negative number)
2. The units of the dot product will be the product of the units
of the A and B vectors
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Dot Product Definition - continued
Examples:
i • j = 0 i.e. (1)(1)cos(90°)
i • i = 1 i.e. (1)(1)cos(0°)
A • B = (Ax i + Ay j + Az k) • (Bx I + By j + Bz k)
= (Ax Bx (i • i) + Ax By (i • j) + Ax Bz (i • k) + ….)
= AxBx + AyBy + AzBz
Angle Between Two Vectors
For two vectors in Cartesian form, one can find the angle by:
a) Finding the dot product, A • B = (AxBx + AyBy + AzBz )
b) Finding the magnitudes (A & B) of the vectors A & B
c) Using the definition of dot product and solving for θ, i.e.
θ = cos-1 [(A • B)/(A B)], where 0º ≤ θ ≤ 180º
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Projection of a Vector
You can determine the components of a vector parallel and
perpendicular to a line using the dot product.
Steps:
1. Find the unit vector, Uaa´ along the line aa´
2. Find the scalar projection of A along line aa´ by
A|| = A • U = AxUx + AyUy + AzUz
Projection of a Vector - continued
3. If needed, the projection can be written as a vector, A|| , by
using the unit vector Uaa´ and the magnitude found in step 2.
A|| = A|| Uaa´
4. The scalar and vector forms of the perpendicular component
can easily be obtained by
A⊥ = (A 2 - A|| 2) ½ and
A⊥ = A – A||
(rearranging the vector sum of A = A⊥ + A|| )
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Example A
Given: The force acting on the pole
Find:
A
The angle between the force
vector and the pole, and the
magnitude of the projection
of the force along the pole
OA.
Plan:
1. Find rOA
2. θ = cos-1{(F • rOA)/(F rOA)}
3. FOA = F • uOA or F cos θ
Example A - continued
rOA = {2 i + 2 j – 1 k} m
rOA = (22 + 22 + 12)1/2 = 3 m
F = {2 i + 4 j + 10 k} kN
F = (22 + 42 + 102)1/2 = 10.95 kN
A
F • rOA = (2)(2) + (4)(2) + (10)(-1) = 2 kN·m
θ = cos-1{(F • rOA)/(F rOA)}
θ = cos-1 {2/(10.95 × 3)} = 86.5°
uOA = rOA/rOA = {(2/3) i + (2/3) j – (1/3) k}
FOA = F • uOA = (2)(2/3) + (4)(2/3) + (10)(-1/3) = 0.667 kN
or FOA = F cos θ = 10.95 cos(86.51°) = 0.667 kN
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Example B
Given: The force acting on the pole.
Find: The angle between the force
vector and the pole, and the
magnitude of the projection of
the force along the pole AO.
Plan:
1. Find rAO
2. θ = cos-1{(F • rAO)/(F rAO)}
3. FOA = F • uAO or F cos θ
Example B - continued
rAO = {-3 i + 2 j – 6 k} ft
rAO = (32 + 22 + 62)1/2 = 7 ft
F = {-20 i + 50 j – 10 k} lb
F = (202 + 502 + 102)1/2 = 54.77 lb
F • rAO = (-20)(-3) + (50)(2) + (-10)(-6) = 220 lb·ft
θ = cos-1{(F • rAO)/(F rAO)}
θ = cos-1 {220/(54.77×7)} = 55.0°
uAO = rAO/rAO = {(-3/7) i + (2/7) j – (6/7) k}
FAO = F • uAO = (-20)(-3/7) + (50)(2/7) + (-10)(-6/7) = 31.4 lb
or FAO = F cos θ = 54.77 cos(55.0°) = 31.4 lb
Engr221 Chapter 2
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Questions
1. The dot product of two vectors P and Q is
defined as
A) P Q cos θ
B) P Q sin θ
C) P Q tan θ
D) P Q sec θ
P
θ
Q
2. The dot product of two vectors results in a _________
quantity.
A) scalar
B) vector
C) complex
D) zero
Questions
1. If a dot product of two non-zero vectors is 0, then the
two vectors must be _____________ to each other.
A) parallel (pointing in the same direction)
B) parallel (pointing in the opposite direction)
C) perpendicular
D) cannot be determined
2. Find the dot product of the two vectors P and Q.
P = {5 i + 2 j + 3 k} m
Q = {-2 i + 5 j + 4 k} m
A) -12 m
B) 12 m
C) 12 m2
D) -12 m2
E) 10 m2
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Textbook Problem 2-127
Determine the angle θ between the edges of
the sheet-metal bracket.
Answer: 82.0°
Summary
• Determine an angle between two vectors
• Determine the projection of a vector along a specified line
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