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Engineering Mechanics: StaticsEngineering Mechanics: Statics
Chapter 2: Force Vectors
Chapter 2: Force Vectors
ObjectivesObjectives
To show how to add forces and resolve them into components using the Parallelogram Law.
To express force and position in Cartesian vector form and explain how to determine the vector’s magnitude and direction.
To introduce the dot product in order to determine the angle between two vectors or the projection of one vector onto another.
Chapter OutlineChapter Outline
1. Scalars and Vectors2. Vector Operations3. Vector Addition of Forces4. Addition of a System of
Coplanar Forces5. Cartesian Vectors
Chapter OutlineChapter Outline
6. Addition and Subtraction of Cartesian Vectors7. Position Vectors8. Force Vector Directed along
a Line9. Dot Product
2.1 Scalars and Vectors2.1 Scalars and Vectors
Scalar – A quantity characterized by a positive or negative number – Indicated by letters in italic such as AEg: Mass, volume and length
2.1 Scalars and Vectors2.1 Scalars and Vectors
Vector – A quantity that has both magnitude and directionEg: Position, force and moment – Represent by a letter with an arrow over it such as or A– Magnitude is designated as or simply A– In this subject, vector is presented as A and its magnitude (positive quantity) as A
A
A
2.1 Scalars and Vectors2.1 Scalars and Vectors
Vector – Represented graphically as an arrow – Length of arrow = Magnitude of Vector – Angle between the reference axis and arrow’s line of action = Direction of Vector – Arrowhead = Sense of Vector
2.1 Scalars and Vectors2.1 Scalars and Vectors
ExampleMagnitude of Vector = 4 unitsDirection of Vector = 20° measured counterclockwise from the horizontal axisSense of Vector = Upward and to the right
The point O is called tail of the vector and the point P is called the tip or head
2.2 Vector Operations2.2 Vector Operations
Multiplication and Division of a Vector by a Scalar- Product of vector A and scalar a = aA- Magnitude = - If a is positive, sense of aA is the same as sense of A- If a is negative sense of aA, it is opposite to the sense of A
aA
2.2 Vector Operations2.2 Vector Operations
Multiplication and Division of a Vector by a Scalar- Negative of a vector is found by multiplying the vector by ( -1 )- Law of multiplication appliesEg: A/a = ( 1/a ) A, a≠0
2.2 Vector Operations2.2 Vector OperationsVector Addition
- Addition of two vectors A and B gives a resultant vector R by the parallelogram law- Result R can be found by triangle construction- CommunicativeEg: R = A + B = B + A
2.2 Vector Operations2.2 Vector Operations
Vector Addition
2.2 Vector Operations2.2 Vector OperationsVector Addition
- Special case: Vectors A and B are collinear (both have the same line of action)
2.2 Vector Operations2.2 Vector Operations
Vector Subtraction- Special case of additionEg: R’ = A – B = A + ( - B )- Rules of Vector Addition Applies
2.2 Vector Operations2.2 Vector Operations
Resolution of Vector- Any vector can be resolved into two components by the parallelogram law- The two components A and B are drawn such that they extend from the tail or R to points of intersection
2.3 Vector Addition of Forces
2.3 Vector Addition of Forces
When two or more forces are added, successive applications of the parallelogram law is carried out to find the resultantEg: Forces F1, F2 and F3 acts at a point O
- First, find resultant of F1 + F2
- Resultant,
FR = ( F1 + F2 ) + F3
2.3 Vector Addition of Forces
2.3 Vector Addition of Forces
Example Fa and Fb are forces exerting on the
hook.
Resultant, Fc can be found using the parallelogram law
Lines parallel to a and bfrom the heads of Fa and Fb are drawn to form a parallelogram Similarly, given Fc, Fa and Fb
can be found
2.3 Vector Addition of Forces
2.3 Vector Addition of Forces
Procedure for AnalysisParallelogram Law
- Make a sketch using the parallelogram law- Two components forces add to form the resultant force - Resultant force is shown by the diagonal of the parallelogram - The components is shown by the sides of the parallelogram
2.3 Vector Addition of Forces
2.3 Vector Addition of Forces
Procedure for AnalysisParallelogram Law
To resolve a force into components along two axes directed from the tail of the force- Start at the head, constructing lines parallel to the axes
- Label all the known and unknown force magnitudes and angles- Identify the two unknown components
2.3 Vector Addition of Forces
2.3 Vector Addition of Forces
Procedure for AnalysisTrigonometry
- Redraw half portion of the parallelogram- Magnitude of the resultant force can be determined by the law of cosines- Direction if the resultant force can be determined by the law of sines
2.3 Vector Addition of Forces
2.3 Vector Addition of Forces
Procedure for AnalysisTrigonometry
- Magnitude of the two components can be determined by the law of sines
2.3 Vector Addition of Forces
2.3 Vector Addition of Forces
Example 2.1The screw eye is subjected to two forces F1
and F2. Determine the magnitude and direction of the resultant force.
2.3 Vector Addition of Forces
2.3 Vector Addition of Forces
Solution Parallelogram LawUnknown: magnitude of FR and angle θ
2.3 Vector Addition of Forces
2.3 Vector Addition of Forces
Solution TrigonometryLaw of Cosines
N
N
NNNNFR
213
6.212
4226.0300002250010000
115cos1501002150100 22
2.3 Vector Addition of Forces
2.3 Vector Addition of Forces
Solution TrigonometryLaw of Sines
8.39sin
9063.06.212
150sin
115sin
6.212
sin
150
N
N
NN
2.3 Vector Addition of Forces
2.3 Vector Addition of Forces
Solution TrigonometryDirection Φ of FR measured from the horizontal
8.54
158.39
2.3 Vector Addition of Forces
2.3 Vector Addition of Forces
Example 2.2Resolve the 1000 N ( ≈ 100kg) force acting on the pipe into the components in the (a) x and y directions, (b) and (b) x’ and ydirections.
2.3 Vector Addition of Forces
2.3 Vector Addition of Forces
Solution(a) Parallelogram
Law
From the vector diagram,
yx FFF
NF
NF
y
x
64340sin1000
76640cos1000
2.3 Vector Addition of Forces
2.3 Vector Addition of Forces
Solution(b) Parallelogram
Law
'yx FFF
2.3 Vector Addition of Forces
2.3 Vector Addition of Forces
Solution(b) Law of Sines
NNF
NF
NNF
NF
y
y
x
x
108560sin
70sin1000
60sin
1000
70sin
6.88460sin
50sin1000
60sin
1000
50sin
'
'
NOTE: A rough sketch drawn to scale will give some idea of the relative magnitude of the components, as calculated here.
2.3 Vector Addition of Forces
2.3 Vector Addition of Forces
Example 2.3The force F acting on the frame has a magnitude of 500N and is to be resolved into two components acting along the members AB and AC. Determine the angle θ, measured below the horizontal, so that components FAC is directed from A towards C and has a magnitude of 400N.
2.3 Vector Addition of Forces
2.3 Vector Addition of Forces
SolutionParallelogram
LawACAB FFN 500
2.3 Vector Addition of Forces
2.3 Vector Addition of Forces
SolutionLaw of Sines
9.43
6928.0sin
60sin500
400sin
60sin
500
sin
400
N
N
NN
2.3 Vector Addition of Forces
2.3 Vector Addition of Forces
Solution
By Law of Cosines or Law of SinesHence, show that FAB
has a magnitude of 561N
1.769.4360180
,Hence
2.3 Vector Addition of Forces
2.3 Vector Addition of Forces
Solution F can be directed at an angle θ above the
horizontal to produce the component FAC. Hence, show that θ = 16.1° and FAB = 161N
2.3 Vector Addition of Forces
2.3 Vector Addition of Forces
Example 2.4The ring is subjected to two
forces F1 and F2. If it is required that the resultant force have a magnitude
of 1kN and be directed vertically downward, determine (a) magnitude of F1 and F2
provided θ = 30°, and (b) the magnitudes of F1 and F2 if
F2 is to be a minimum.
2.3 Vector Addition of Forces
2.3 Vector Addition of Forces
Solution(a) Parallelogram LawUnknown: Forces F1 and F2
View Free Body Diagram
2.3 Vector Addition of Forces
2.3 Vector Addition of Forces
SolutionLaw of Sines
NF
NF
NF
NF
446130sin
1000
20sin
643130sin
1000
30sin
2
2
1
1
2.3 Vector Addition of Forces
2.3 Vector Addition of Forces
Solution(b) Minimum length of F2
occur when its line of action is perpendicular to F1. Hence when
F2 is a minimum
702090
2.3 Vector Addition of Forces
2.3 Vector Addition of Forces
Solution(b) From the vector
diagram
NNF
NNF
34270cos1000
94070sin1000
2
1
2.4 Addition of a System of Coplanar
Forces
2.4 Addition of a System of Coplanar
ForcesFor resultant of two or more
forces: Find the components of the forces in the
specified axes Add them algebraically Form the resultant
In this subject, we resolve each force into rectangular forces along the x and y axes.
yx FFF
2.4 Addition of a System of Coplanar
Forces
2.4 Addition of a System of Coplanar
ForcesScalar Notation
- x and y axes are designated positive and negative- Components of forces expressed as algebraic scalarsEg: Sense of direction along positive x and y axes
yx FFF
2.4 Addition of a System of Coplanar
Forces
2.4 Addition of a System of Coplanar
ForcesScalar Notation
Eg: Sense of direction along positive x and negative y axes
yx FFF '''
2.4 Addition of a System of Coplanar
Forces
2.4 Addition of a System of Coplanar
ForcesScalar Notation
- Head of a vector arrow = sense of the vector graphically (algebraic signs not used)- Vectors are designated using boldface notations- Magnitudes (always a positive quantity) are designated using italic symbols
2.4 Addition of a System of Coplanar
Forces
2.4 Addition of a System of Coplanar
ForcesCartesian Vector Notation
- Cartesian unit vectors i and j are used to designate the x and y directions- Unit vectors i and j have dimensionless magnitude of unity ( = 1 )
- Their sense are indicated by a positive or negative sign (pointing in the positive or negative x or y axis) - Magnitude is always a positive quantity, represented by scalars Fx and Fy
2.4 Addition of a System of Coplanar
Forces
2.4 Addition of a System of Coplanar
ForcesCartesian Vector Notation
F = Fxi + Fyj F’ = F’xi + F’y(-j)
F’ = F’xi – F’yj
2.4 Addition of a System of Coplanar
Forces
2.4 Addition of a System of Coplanar
ForcesCoplanar Force Resultants
To determine resultant of several coplanar forces:- Resolve force into x and y components- Addition of the respective components using scalar algebra - Resultant force is found using the parallelogram law
2.4 Addition of a System of Coplanar
Forces
2.4 Addition of a System of Coplanar
ForcesCoplanar Force Resultants
Example: Consider three coplanar forces
Cartesian vector notationF1 = F1xi + F1yj
F2 = - F2xi + F2yj
F3 = F3xi – F3yj
Coplanar Force ResultantsVector resultant is thereforeFR = F1 + F2 + F3
= F1xi + F1yj - F2xi + F2yj + F3xi – F3yj
= (F1x - F2x + F3x)i + (F1y + F2y – F3y)j
= (FRx)i + (FRy)j
2.4 Addition of a System of Coplanar
Forces
2.4 Addition of a System of Coplanar
Forces
2.4 Addition of a System of Coplanar
Forces
2.4 Addition of a System of Coplanar
ForcesCoplanar Force ResultantsIf scalar notation are usedFRx = (F1x - F2x + F3x)
FRy = (F1y + F2y – F3y)
In all cases,FRx = ∑Fx
FRy = ∑Fy
* Take note of sign conventions
2.4 Addition of a System of Coplanar
Forces
2.4 Addition of a System of Coplanar
ForcesCoplanar Force Resultants- Positive scalars = sense of direction along the positive coordinate axes- Negative scalars = sense of direction along the negative coordinate axes- Magnitude of FR can be found by Pythagorean Theorem
RyRxR FFF 22
2.4 Addition of a System of Coplanar
Forces
2.4 Addition of a System of Coplanar
ForcesCoplanar Force Resultants- Direction angle θ (orientation of the force) can be found by trigonometry
Rx
Ry
F
F1tan
2.4 Addition of a System of Coplanar
Forces
2.4 Addition of a System of Coplanar
ForcesExample 2.5Determine x and y components of F1 and F2
acting on the boom. Express each force as a Cartesian vector
2.4 Addition of a System of Coplanar
Forces
2.4 Addition of a System of Coplanar
ForcesSolutionScalar Notation
Hence, from the slope triangle
NNNF
NNNF
y
x
17317330cos200
10010030sin200
1
1
12
5tan 1
2.4 Addition of a System of Coplanar
Forces
2.4 Addition of a System of Coplanar
ForcesSolutionAlt, by similar
triangles
Similarly,
NNF
N
F
x
x
24013
12260
13
12
260
2
2
NNF y 10013
52602
2.4 Addition of a System of Coplanar
Forces
2.4 Addition of a System of Coplanar
ForcesSolution
Scalar Notation
Cartesian Vector Notation
F1 = {-100i +173j }N
F2 = {240i -100j }N
NNF
NNF
y
x
100100
240240
2
2
2.4 Addition of a System of Coplanar
Forces
2.4 Addition of a System of Coplanar
ForcesExample 2.6The link is subjected to two forces F1
and F2. Determine the magnitude and
orientation of the resultant force.
2.4 Addition of a System of Coplanar
Forces
2.4 Addition of a System of Coplanar
ForcesSolutionScalar Notation
N
NNF
FF
N
NNF
FF
Ry
yRy
Rx
xRx
8.582
45cos40030sin600
:
8.236
45sin40030cos600
:
2.4 Addition of a System of Coplanar
Forces
2.4 Addition of a System of Coplanar
ForcesSolutionResultant Force
From vector addition,Direction angle θ is
N
NNFR629
8.5828.236 22
9.67
8.236
8.582tan 1
N
N
2.4 Addition of a System of Coplanar
Forces
2.4 Addition of a System of Coplanar
ForcesSolutionCartesian Vector NotationF1 = { 600cos30°i + 600sin30°j } N
F2 = { -400sin45°i + 400cos45°j } N
Thus, FR = F1 + F2
= (600cos30°N - 400sin45°N)i + (600sin30°N + 400cos45°N)j
= {236.8i + 582.8j}N
2.4 Addition of a System of Coplanar
Forces
2.4 Addition of a System of Coplanar
ForcesExample 2.7The end of the boom O is subjected to three concurrent and coplanar forces. Determine the magnitude and orientation of the resultant force.
2.4 Addition of a System of Coplanar
Forces
2.4 Addition of a System of Coplanar
ForcesSolutionScalar Notation
N
NNF
FF
NN
NNNF
FF
Ry
yRy
Rx
xRx
8.296
5
320045cos250
:
2.3832.383
5
420045sin250400
:
View Free Body Diagram
SolutionResultant Force
From vector addition,Direction angle θ is
2.4 Addition of a System of Coplanar
Forces
2.4 Addition of a System of Coplanar
Forces
N
NNFR485
8.2962.383 22
8.37
2.383
8.296tan 1
N
N
2.5 Cartesian Vectors2.5 Cartesian Vectors
Right-Handed Coordinate SystemA rectangular or Cartesian coordinate system is said to be right-handed provided:- Thumb of 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 towards the positive y axis
2.5 Cartesian Vectors2.5 Cartesian Vectors
Right-Handed Coordinate System- z-axis for the 2D problem would be perpendicular, directed out of the page.
2.5 Cartesian Vectors2.5 Cartesian VectorsRectangular Components of a Vector
- A vector A may have one, two or three rectangular components along the x, y and z axes, depending on orientation- By two successive application of the parallelogram law
A = A’ + Az
A’ = Ax + Ay
- Combing the equations, A can be expressed as
A = Ax + Ay + Az
2.5 Cartesian Vectors2.5 Cartesian Vectors
Unit Vector- Direction of A can be specified using a unit vector- Unit vector has a magnitude of 1- If A is a vector having a magnitude of A ≠ 0, unit vector having the same direction as A is expressed by
uA = A / ASo that
A = A uA
2.5 Cartesian Vectors2.5 Cartesian Vectors
Unit Vector- Since A is of a certain type, like force vector, a proper set of units are used for the description- Magnitude A has the same sets of units, hence unit vector is dimensionless- A ( a positive scalar) defines magnitude of A - uA defines the directionand sense of A
2.5 Cartesian Vectors2.5 Cartesian Vectors
Cartesian Unit Vectors- Cartesian unit vectors, i, j and k are used to designate the directions of z, y and z axes- Sense (or arrowhead) of these vectors are described by a plus or minus sign (depending on pointing towards the positive or negative axes)
2.5 Cartesian Vectors2.5 Cartesian Vectors
Cartesian Vector Representations- Three components of A act in the positive i, j and k directions
A = Axi + Ayj + AZk
*Note the magnitude and direction of each components are separated, easing vector algebraic operations.
2.5 Cartesian Vectors2.5 Cartesian Vectors
Magnitude of a Cartesian Vector - From the colored triangle,
- From the shaded triangle,
- Combining the equations givesmagnitude of A
222
22
22
'
'
zyx
yx
z
AAAA
AAA
AAA
2.5 Cartesian Vectors2.5 Cartesian Vectors
Direction of a Cartesian Vector- Orientation of A is defined as the coordinate direction angles α, β and γ measured between the tail of A and the positive x, y and z axes- 0° ≤ α, β and γ ≤ 180 °
2.5 Cartesian Vectors2.5 Cartesian Vectors
Direction of a Cartesian Vector- For angles α, β and γ (blue colored triangles), we calculate the direction cosines of A
A
Axcos
2.5 Cartesian Vectors2.5 Cartesian Vectors
Direction of a Cartesian Vector- For angles α, β and γ (blue colored triangles), we calculate the direction cosines of A
A
Aycos
2.5 Cartesian Vectors2.5 Cartesian Vectors
Direction of a Cartesian Vector- For angles α, β and γ (blue colored triangles), we calculate the direction cosines of A
A
Azcos
2.5 Cartesian Vectors2.5 Cartesian Vectors
Direction of a Cartesian Vector- Angles α, β and γ can be determined by the inverse cosines- Given
A = Axi + Ayj + AZk- then,
uA = A /A
= (Ax/A)i + (Ay/A)j + (AZ/A)k
where 222zyx AAAA
2.5 Cartesian Vectors2.5 Cartesian VectorsDirection of a Cartesian Vector
- uA can also be expressed as
uA = cosαi + cosβj + cosγk
- Since and magnitude of uA = 1,
- A as expressed in Cartesian vector form A = AuA = Acosαi + Acosβj + Acosγk
= Axi + Ayj + AZk
222zyx AAAA
1coscoscos 222
ExampleGiven: A = Axi + Ayj + AZk
and B = Bxi + Byj + BZk
Vector AdditionResultant R = A + B
= (Ax + Bx)i + (Ay + By )j + (AZ + BZ) kVector SubstractionResultant R = A - B
= (Ax - Bx)i + (Ay - By )j + (AZ - BZ) k
2.6 Addition and Subtraction of Cartesian
Vectors
2.6 Addition and Subtraction of Cartesian
Vectors
2.6 Addition and Subtraction of Cartesian
Vectors
2.6 Addition and Subtraction of Cartesian
VectorsConcurrent Force Systems
- Force resultant is the vector sum of all the forces in the system
FR = ∑F = ∑Fxi + ∑Fyj + ∑Fzk
where ∑Fx , ∑Fy and ∑Fz represent the algebraic sums of the x, y and z or i, j or k components of each force in the system
2.6 Addition and Subtraction of Cartesian
Vectors
2.6 Addition and Subtraction of Cartesian
Vectors
Force, F that the tie down rope exerts on the ground support at O is directed along the rope
Angles α, β and γ can be solved with axes x, y and z
2.6 Addition and Subtraction of Cartesian
Vectors
2.6 Addition and Subtraction of Cartesian
Vectors
Cosines of their values forms a unit vector u that acts in the direction of the rope
Force F has a magnitude of FF = Fu = Fcosαi + Fcosβj + Fcosγk
2.6 Addition and Subtraction of Cartesian
Vectors
2.6 Addition and Subtraction of Cartesian
VectorsExample 2.8Express the force F as Cartesian
vector
2.6 Addition and Subtraction of Cartesian
Vectors
2.6 Addition and Subtraction of Cartesian
VectorsSolutionSince two angles are specified, the third angle is found by
Two possibilities exit, namelyor
605.0cos
5.0707.05.01cos
145cos60coscos
1coscoscos
1
22
222
222
1205.0cos 1
2.6 Addition and Subtraction of Cartesian
Vectors
2.6 Addition and Subtraction of Cartesian
VectorsSolutionBy inspection, α = 60° since Fx is in the +x
directionGiven F = 200N
F = Fcosαi + Fcosβj + Fcosγk = (200cos60°N)i + (200cos60°N)j
+ (200cos45°N)k = {100.0i + 100.0j + 141.4k}N
Checking:
N
FFFF zyx
2004.1410.1000.100 222
222
2.6 Addition and Subtraction of Cartesian
Vectors
2.6 Addition and Subtraction of Cartesian
Vectors Example 2.9 Determine the magnitude and
coordinate direction angles of resultant force
acting on the ring
2.6 Addition and Subtraction of Cartesian
Vectors
2.6 Addition and Subtraction of Cartesian
VectorsSolutionResultant force
FR = ∑F
= F1 + F2 = {60j + 80k}kN
+ {50i - 100j + 100k}kN = {50j -40k + 180k}kN
Magnitude of FR is found by
kN
FR
1910.191
1804050 222
2.6 Addition and Subtraction of Cartesian
Vectors
2.6 Addition and Subtraction of Cartesian
VectorsSolutionUnit vector acting in the direction of FR
uFR = FR /FR
= (50/191.0)i + (40/191.0)j + (180/191.0)k
= 0.1617i - 0.2094j + 0.9422kSo that
cosα = 0.2617 α = 74.8° cos β = -0.2094 β = 102° cosγ = 0.9422 γ = 19.6°
*Note β > 90° since j component of uFR is negative
2.6 Addition and Subtraction of Cartesian
Vectors
2.6 Addition and Subtraction of Cartesian
VectorsExample 2.10Express the force F1 as a Cartesian
vector.
2.6 Addition and Subtraction of Cartesian
Vectors
2.6 Addition and Subtraction of Cartesian
VectorsSolutionThe angles of 60° and 45° are not
coordinate direction angles.
By two successive applications ofparallelogram law,
2.6 Addition and Subtraction of Cartesian
Vectors
2.6 Addition and Subtraction of Cartesian
VectorsSolutionBy trigonometry,
F1z = 100sin60 °kN = 86.6kNF’ = 100cos60 °kN = 50kNF1x = 50cos45 °kN = 35.4kNF1y = 50sin45 °kN = 35.4kN
F1y has a direction defined by –j, Therefore
F1 = {35.4i – 35.4j + 86.6k}kN
SolutionChecking:
Unit vector acting in the direction of F1
u1 = F1 /F1
= (35.4/100)i - (35.4/100)j + (86.6/100)k
= 0.354i - 0.354j + 0.866k
2.6 Addition and Subtraction of Cartesian
Vectors
2.6 Addition and Subtraction of Cartesian
Vectors
N
FFFF zyx
1006.864.354.35 222
21
21
211
Solutionα1 = cos-1(0.354) = 69.3°
β1 = cos-1(-0.354) = 111°
γ1 = cos-1(0.866) = 30.0°
Using the same method, F2 = {106i + 184j - 212k}kN
2.6 Addition and Subtraction of Cartesian
Vectors
2.6 Addition and Subtraction of Cartesian
Vectors
2.6 Addition and Subtraction of Cartesian
Vectors
2.6 Addition and Subtraction of Cartesian
VectorsExample 2.11Two forces act on the hook. Specify the coordinate direction angles of F2, so that the
resultant force FR acts along the positive y axis and has a magnitude of 800N.
2.6 Addition and Subtraction of Cartesian
Vectors
2.6 Addition and Subtraction of Cartesian
Vectors
SolutionCartesian vector formFR = F1 + F2
F1 = F1cosα1i + F1cosβ1j + F1cosγ1k
= (300cos45°N)i + (300cos60°N)j + (300cos120°N)k
= {212.1i + 150j - 150k}NF2 = F2xi + F2yj + F2zk
View Free Body Diagram
2.6 Addition and Subtraction of Cartesian
Vectors
2.6 Addition and Subtraction of Cartesian
VectorsSolutionSince FR has a magnitude of 800N and actsin the +j directionFR = F1 + F2
800j = 212.1i + 150j - 150k + F2xi + F2yj + F2zk
800j = (212.1 + F2x)i + (150 + F2y)j + (- 50 + F2z)k
To satisfy the equation, the corresponding components on left and right sides must be equal
2.6 Addition and Subtraction of Cartesian
Vectors
2.6 Addition and Subtraction of Cartesian
VectorsSolutionHence,
0 = 212.1 + F2x F2x = -212.1N 800 = 150 + F2y F2y = 650N 0 = -150 + F2z F2z = 150N
Since magnitude of F2 and its components are known, α1 = cos-1(-212.1/700) = 108° β1 = cos-1(650/700) = 21.8° γ1 = cos-1(150/700) = 77.6°
x,y,z Coordinates- Right-handed coordinate system- Positive z axis points upwards, measuring the height of an object or the altitude of a point- Points are measured relative to the origin, O.
2.7 Position Vectors2.7 Position Vectors
x,y,z CoordinatesEg: For Point A, xA = +4m along the x axis, yA = -6m along the y axis and zA = -6m
along the z axis. Thus, A (4, 2, -6) Similarly, B (0, 2, 0) and C (6, -1, 4)
2.7 Position Vectors2.7 Position Vectors
Position Vector- Position vector r is defined as a fixed vector which locates a point in space relative to another point. Eg: If r extends from the origin, O to point P (x, y, z) then, in Cartesian vector form
r = xi + yj + zk
2.7 Position Vectors2.7 Position Vectors
Position VectorNote the head to tail vector addition of the three components
Start at origin O, one travels x in the +i direction,
y in the +j direction and z in the +k direction, arriving at point P (x, y, z)
2.7 Position Vectors2.7 Position Vectors
2.7 Position Vectors2.7 Position Vectors Position Vector
- Position vector maybe directed from point A to point B - Designated by r or rAB
Vector addition givesrA + r = rB
Solving r = rB – rA = (xB – xA)i + (yB – yA)j + (zB –zA)k
or r = (xB – xA)i + (yB – yA)j + (zB –zA)k
Position Vector- The i, j, k components of the positive vector r may be formed by taking the coordinates of the tail, A (xA, yA, zA) and subtract them from the head B (xB, yB, zB)
Note the head to tail vector addition of the three components
2.7 Position Vectors2.7 Position Vectors
2.7 Position Vectors2.7 Position Vectors
Length and direction of cable AB can be found by measuring A and B using the x, y, z axes
Position vector r can be established
Magnitude r represent the length of cable
2.7 Position Vectors2.7 Position Vectors
Angles, α, β and γ represent the direction of the cable
Unit vector, u = r/r
2.7 Position Vectors2.7 Position Vectors
Example 2.12An elastic rubber band is attached to points A and
B. Determine its length and
its direction measured from
A towards B.
2.7 Position Vectors2.7 Position Vectors
SolutionPosition vector
r = [-2m – 1m]i + [2m – 0]j + [3m – (-3m)]k = {-3i + 2j + 6k}m
Magnitude = length of the rubber band
Unit vector in the director of ru = r /r = -3/7i + 2/7j + 6/7k
mr 7623 222
View Free Body Diagram
2.7 Position Vectors2.7 Position Vectors
Solutionα = cos-1(-3/7) = 115°
β = cos-1(2/7) = 73.4°
γ = cos-1(6/7) = 31.0°
2.8 Force Vector Directed along a Line
2.8 Force Vector Directed along a Line
In 3D problems, direction of F is specified by 2 points, through which its line of action lies
F can be formulated as a Cartesian vector
F = F u = F (r/r)
Note that F has units of forces (N) unlike r, with units of length (m)
2.8 Force Vector Directed along a Line
2.8 Force Vector Directed along a Line
Force F acting along the chain can be presented as a Cartesian vector by- Establish x, y, z axes- Form a position vector r along length of chain
2.8 Force Vector Directed along a Line
2.8 Force Vector Directed along a Line
Unit vector, u = r/r that defines the direction of both the chain and the force
We get F = Fu
2.8 Force Vector Directed along a Line
2.8 Force Vector Directed along a Line
Example 2.13The man pulls on the cord with a force of 350N. Represent this force
acting on the support A, as a Cartesian vector and determine its direction.
2.8 Force Vector Directed along a Line
2.8 Force Vector Directed along a Line
SolutionEnd points of the cord are A (0m, 0m,
7.5m) and B (3m, -2m, 1.5m)r = (3m – 0m)i + (-2m – 0m)j + (1.5m –
7.5m)k= {3i – 2j – 6k}m
Magnitude = length of cord AB
Unit vector, u = r /r = 3/7i - 2/7j - 6/7k
mmmmr 7623 222
2.8 Force Vector Directed along a Line
2.8 Force Vector Directed along a Line
SolutionForce F has a magnitude of 350N, direction specified by u
F = Fu = 350N(3/7i - 2/7j - 6/7k) = {150i - 100j - 300k} N
α = cos-1(3/7) = 64.6° β = cos-1(-2/7) = 107° γ = cos-1(-6/7) = 149°
2.8 Force Vector Directed along a Line
2.8 Force Vector Directed along a Line
Example 2.14The circular plate is partially supported by the cable AB. If the force of the cable on
the hook at A is F = 500N, express F as a Cartesian vector.
2.8 Force Vector Directed along a Line
2.8 Force Vector Directed along a Line
SolutionEnd points of the cable are (0m, 0m, 2m)
and B (1.707m, 0.707m, 0m)
r = (1.707m – 0m)i + (0.707m – 0m)j + (0m – 2m)k
= {1.707i + 0.707j - 2k}mMagnitude = length of cable AB
mmmmr 723.22707.0707.1 222
SolutionUnit vector,
u = r /r = (1.707/2.723)i + (0.707/2.723)j – (2/2.723)k
= 0.6269i + 0.2597j – 0.7345kFor force F,
F = Fu = 500N(0.6269i + 0.2597j – 0.7345k) = {313i - 130j - 367k} N
2.8 Force Vector Directed along a Line
2.8 Force Vector Directed along a Line
SolutionChecking
Show that γ = 137° and indicate this angle on the diagram
2.8 Force Vector Directed along a Line
2.8 Force Vector Directed along a Line
N
F
500
367130313 222
2.8 Force Vector Directed along a Line
2.8 Force Vector Directed along a Line
Example 2.15The roof is supported by cables. If the cables exert FAB = 100N and FAC = 120N
on the wall hook at A, determine the magnitude of the resultant force acting at A.
2.8 Force Vector Directed along a Line
2.8 Force Vector Directed along a Line
SolutionrAB = (4m – 0m)i + (0m – 0m)j + (0m –
4m)k = {4i – 4k}m
FAB = 100N (rAB/r AB)
= 100N {(4/5.66)i - (4/5.66)k} = {70.7i - 70.7k} N
mmmrAB 66.544 22
View Free Body Diagram
2.8 Force Vector Directed along a Line
2.8 Force Vector Directed along a Line
SolutionrAC = (4m – 0m)i + (2m – 0m)j + (0m – 4m)k
= {4i + 2j – 4k}m
FAC = 120N (rAB/r AB)
= 120N {(4/6)i + (2/6)j - (4/6)k} = {80i + 40j – 80k} N
mmmmrAC 6424 222
2.8 Force Vector Directed along a Line
2.8 Force Vector Directed along a Line
SolutionFR = FAB + FAC
= {70.7i - 70.7k} N + {80i + 40j – 80k} N = {150.7i + 40j – 150.7k} N
Magnitude of FR
N
FR
217
7.150407.150 222
2.9 Dot Product2.9 Dot ProductDot product of vectors A and B is
written as A·B (Read A dot B)Define the magnitudes of A and B and
the angle between their tails A·B = AB cosθ where 0°≤ θ
≤180°Referred to as scalar
product of vectors as result is a scalar
2.9 Dot Product2.9 Dot Product
Laws of Operation1. Commutative law
A·B = B·A2. Multiplication by a scalar
a(A·B) = (aA)·B = A·(aB) = (A·B)a
3. Distribution lawA·(B + D) = (A·B) + (A·D)
2.9 Dot Product2.9 Dot Product
Cartesian Vector Formulation- Dot product of Cartesian unit vectorsEg: i·i = (1)(1)cos0° = 1 andi·j = (1)(1)cos90° = 0- Similarly
i·i = 1 j·j = 1 k·k = 1 i·j = 0 i·k = 1 j·k = 1
2.9 Dot Product2.9 Dot Product
Cartesian Vector Formulation- Dot product of 2 vectors A and BA·B = (Axi + Ayj + Azk)· (Bxi + Byj + Bzk)
= AxBx(i·i) + AxBy(i·j) + AxBz(i·k)
+ AyBx(j·i) + AyBy(j·j) + AyBz(j·k)
+ AzBx(k·i) + AzBy(k·j) + AzBz(k·k)
= AxBx + AyBy + AzBz
Note: since result is a scalar, be careful of including any unit vectors in the result
2.9 Dot Product2.9 Dot Product
Applications- The angle formed between two vectors or intersecting lines
θ = cos-1 [(A·B)/(AB)] 0°≤ θ ≤180°
Note: if A·B = 0, cos-10= 90°, A is perpendicular to B
2.9 Dot Product2.9 Dot ProductApplications
- The components of a vector parallel and perpendicular to a line- Component of A parallel or collinear with line aa’ is defined by A║ (projection of A onto the line)
A║ = A cos θ
- If direction of line is specified by unit vector u (u = 1),
A║ = A cos θ = A·u
2.9 Dot Product2.9 Dot Product
Applications- If A║ is positive, A║ has a directional sense same as u- If A║ is negative, A║ has a directional sense opposite to u- A║ expressed as a vector
A║ = A cos θ u = (A·u)u
ApplicationsFor component of A perpendicular to line aa’ 1. Since A = A║ + A┴,
then A┴ = A - A║
2. θ = cos-1 [(A·u)/(A)]then A┴ = Asinθ
3. If A║ is known, by Pythagorean Theorem
2.9 Dot Product2.9 Dot Product
2||
2 AAA
2.9 Dot Product2.9 Dot Product For angle θ between the
rope and the beam A, - Unit vectors along the beams, uA = rA/rA
- Unit vectors along the ropes, ur=rr/rr
- Angle θ = cos-1 (rA.rr/rArr)
= cos-1 (uA· ur)
2.9 Dot Product2.9 Dot Product
For projection of the force along the beam A - Define direction of the beam
uA = rA/rA
- Force as a Cartesian vector
F = F(rr/rr) = Fur
- Dot product F║ = F║·uA
2.9 Dot Product2.9 Dot Product
Example 2.16The frame is subjected to a horizontal force F = {300j} N. Determine the components of this force parallel and perpendicular to the member AB.
2.9 Dot Product2.9 Dot Product
SolutionSince
Then
N
kjijuF
FF
kji
kjirr
u
B
AB
B
BB
1.257
)429.0)(0()857.0)(300()286.0)(0(
429.0857.0286.0300.
cos
429.0857.0286.0
362
362222
2.9 Dot Product2.9 Dot ProductSolutionSince result is a positive scalar,FAB has the same sense of
direction as uB. Express in Cartesian form
Perpendicular component
NkjikjijFFF
Nkji
kjiN
uFF
AB
ABABAB
}110805.73{)1102205.73(300
}1102205.73{
429.0857.0286.01.257
2.9 Dot Product2.9 Dot ProductSolutionMagnitude can be determined From F┴ or from Pythagorean
Theorem
N
NN
FFF AB
155
1.257300 22
22
2.9 Dot Product2.9 Dot Product
Example 2.17The pipe is subjected to F = 800N. Determine the angle θ between F and pipe segment BA, and the magnitudes of the components of F, which are parallel and perpendicular to BA.
2.9 Dot Product2.9 Dot Product
SolutionFor angle θrBA = {-2i - 2j + 1k}m
rBC = {- 3j + 1k}m
Thus,
5.42
7379.0
103
113202cos
BCBA
BCBA
rr
rr
View Free Body Diagram
2.9 Dot Product2.9 Dot Product
SolutionComponents of F
N
kjikj
uFF
kji
kjirr
u
BAB
AB
ABAB
590
3.840.5060
31
32
32
0.2539.758
.
31
32
32
3)122(
2.9 Dot Product2.9 Dot Product
SolutionChecking from trigonometry,
Magnitude can be determined From F┴
N
N
FFAB
590
5.42cos800
cos
NFF 5405.42sin800sin
2.9 Dot Product2.9 Dot Product
SolutionMagnitude can be determined from F┴ or
from Pythagorean Theorem
N
FFF AB
540
590800 22
22
Chapter SummaryChapter Summary
Parallelogram LawAddition of two vectorsComponents form the side and
resultant form the diagonal of the parallelogram
To obtain resultant, use tip to tail addition by triangle rule
To obtain magnitudes and directions, use Law of Cosines and Law of Sines
Chapter SummaryChapter Summary
Cartesian Vectors Vector F resolved into Cartesian vector
formF = Fxi + Fyj + Fzk
Magnitude of F
Coordinate direction angles α, β and γ are determined by the formulation of the unit vector in the direction of F
u = (Fx/F)i + (Fy/F)j + (Fz/F)k
222zyx FFFF
Chapter SummaryChapter Summary
Cartesian Vectors Components of u represent cosα, cosβ and
cosγ These angles are related by
cos2α + cos2β + cos2γ = 1
Force and Position Vectors Position Vector is directed between 2 points Formulated by distance and direction moved
along the x, y and z axes from tail to tip
Chapter SummaryChapter Summary
Force and Position Vectors For line of action through the two
points, it acts in the same direction of u as the position vector
Force expressed as a Cartesian vectorF = Fu = F(r/r)
Dot Product Dot product between two vectors A and
B A·B = AB cosθ
Chapter SummaryChapter Summary
Dot Product Dot product between two vectors A and B
(vectors expressed as Cartesian form)A·B = AxBx + AyBy + AzBz
For angle between the tails of two vectors θ = cos-1 [(A·B)/(AB)]
For projected component of A onto an axis defined by its unit vector u
A = A cos θ = A·u
Chapter ReviewChapter Review
Chapter ReviewChapter Review
Chapter ReviewChapter Review
Chapter ReviewChapter Review
Chapter ReviewChapter Review
Chapter ReviewChapter Review