6161103 ch02b
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Example
Given: A = Axi + Ayj + AZk
and B = Bxi + Byj + BZk
Vector Addition
2.6 Addition and Subtraction of Cartesian Vectors
2.6 Addition and Subtraction of Cartesian Vectors
Vector Addition
Resultant R = A + B
= (Ax + Bx)i + (Ay + By )j + (AZ + BZ) k
Vector Substraction
Resultant 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
� Concurrent 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 kcomponents 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, yand 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 F
F = Fu = Fcosαi + Fcosβj + Fcosγk
2.6 Addition and Subtraction of Cartesian Vectors
2.6 Addition and Subtraction of Cartesian Vectors
Example 2.8
Express the force F as Cartesian vector
2.6 Addition and Subtraction of Cartesian Vectors
2.6 Addition and Subtraction of Cartesian Vectors
Solution
Since two angles are specified, the third
angle is found by
Two possibilities exit, namely
or
( ) ( )
( ) o
oo
605.0cos
5.0707.05.01cos
145cos60coscos
1coscoscos
1
22
222
222
==
±=−−=
=++
=++
−α
α
α
γβα
( ) o1205.0cos 1 =−= −α
2.6 Addition and Subtraction of Cartesian Vectors
2.6 Addition and Subtraction of Cartesian Vectors
Solution
By inspection, α = 60° since Fx is in the +x
direction
Given 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 ringthe ring
2.6 Addition and Subtraction of Cartesian Vectors
2.6 Addition and Subtraction of Cartesian Vectors
Solution
Resultant force
FR = ∑F
= F1 + F2= 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 Vectors
Solution
Unit vector acting in the direction of FR
uFR = FR /FR
= (50/191.0)i + (40/191.0)j + (180/191.0)k(180/191.0)k
= 0.1617i - 0.2094j + 0.9422k
So 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 Vectors
Example 2.10
Express the force F1 as a Cartesian vector.
2.6 Addition and Subtraction of Cartesian Vectors
2.6 Addition and Subtraction of Cartesian Vectors
Solution
The angles of 60° and 45° are not coordinate
direction angles.
By two successive applications of
parallelogram law,
2.6 Addition and Subtraction of Cartesian Vectors
2.6 Addition and Subtraction of Cartesian Vectors
SolutionBy trigonometry,
F1z = 100sin60 °kN = 86.6kNF’ = 100cos60 °kN = 50kNF = 50cos45 kN = 35.4kNF1x = 50cos45 °kN = 35.4kNF1y = 50sin45 °kN = 35.4kN
F1y has a direction defined by –j, Therefore
F1 = {35.4i – 35.4j + 86.6k}kN
Solution
Checking:
2.6 Addition and Subtraction of Cartesian Vectors
2.6 Addition and Subtraction of Cartesian Vectors
( ) ( ) ( )
FFFF zyx
222
21
21
211
=+−+=
++=
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
( ) ( ) ( ) N1006.864.354.35 222 =+−+=
Solution
α1 = cos-1(0.354) = 69.3°
β1 = cos-1(-0.354) = 111°
γ1 = cos-1(0.866) = 30.0°
2.6 Addition and Subtraction of Cartesian Vectors
2.6 Addition and Subtraction of Cartesian Vectors
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
Example 2.11
Two forces act on the hook. Specify the
coordinate direction angles of F2, so that the
resultant force F acts along the positive y axis 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
Solution
Cartesian vector form
FR = F1 + F2
F = F cosα i + F cosβ j + F cosγ kF1 = F1cosα1i + F1cosβ1j + F1cosγ1k
= (300cos45°N)i + (300cos60°N)j
+ (300cos120°N)k
= {212.1i + 150j - 150k}N
F2 = F2xi + F2yj + F2zk
2.6 Addition and Subtraction of Cartesian Vectors
2.6 Addition and Subtraction of Cartesian Vectors
Solution
Since FR has a magnitude of 800N and acts
in the +j direction
FR = F1 + F2FR = 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 Vectors
SolutionHence,
0 = 212.1 + F2x F2x = -212.1N800 = 150 + F2y F2y = 650N0 = -150 + F2z F2z = 150N0 = -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
2.7 Position Vectors2.7 Position Vectors
the height of an object or the altitude of a point
- Points are measured relative to the origin, O.
� x,y,z Coordinates
Eg: 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)
2.7 Position Vectors2.7 Position Vectors
along the z axis. Thus, A (4, 2, -6)
Similarly, B (0, 2, 0) and C (6, -1, 4)
� Position Vector
- Position vector r is defined as a fixed vector which locates a point in space relative to another point.
2.7 Position Vectors2.7 Position Vectors
point.
Eg: If r extends from the
origin, O to point P (x, y, z)
then, in Cartesian vector
form
r = xi + yj + zk
� Position Vector
Note the head to tail vector addition of the
three components
2.7 Position Vectors2.7 Position Vectors
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
� Position Vector
- Position vector maybe directed from point A to point B
- Designated by r or rABAB
Vector addition gives
rA + 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 rmay be formed by taking the coordinates of the tail, A (xA, yA, zA) and subtract them from the head B (xB, yB, zB)
2.7 Position Vectors2.7 Position Vectors
head B (xB, yB, zB)
Note the head to tail vector addition of the
three components
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 axesthe 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 γrepresent the direction of the cableof the cable
� Unit vector, u = r/r
2.7 Position Vectors2.7 Position Vectors
Example 2.12
An elastic rubber band is
attached to points A and B. attached to points A and B.
Determine its length and its
direction measured from A
towards B.
2.7 Position Vectors2.7 Position Vectors
Solution
Position vector
r = [-2m – 1m]i + [2m – 0]j + [3m – (-3m)]k
= {-3i + 2j + 6k}m= {-3i + 2j + 6k}m
Magnitude = length of the rubber band
Unit vector in the director of r
u = r /r
= -3/7i + 2/7j + 6/7k
( ) ( ) ( ) mr 7623 222 =++−=
2.7 Position Vectors2.7 Position Vectors
Solution
α = cos-1(-3/7) = 115°
β = cos-1(2/7) = 73.4°
γ = cos-1(6/7) = 31.0°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 2.13
The man pulls on the cord
with a force of 350N. 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
Solution
End 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 –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
Solution
Force 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 2.14
The circular plate is
partially supported by
the cable AB. If the 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
Solution
End points of the cable are (0m, 0m, 2m) and B
(1.707m, 0.707m, 0m)
r = (1.707m – 0m)i + (0.707m – 0m)jr = (1.707m – 0m)i + (0.707m – 0m)j
+ (0m – 2m)k
= {1.707i + 0.707j - 2k}m
Magnitude = length of cable AB
( ) ( ) ( ) mmmmr 723.22707.0707.1 222 =−++=
Solution
Unit vector,
u = r /r
= (1.707/2.723)i + (0.707/2.723)j – (2/2.723)k
2.8 Force Vector Directed along a Line
2.8 Force Vector Directed along a Line
= (1.707/2.723)i + (0.707/2.723)j – (2/2.723)k
= 0.6269i + 0.2597j – 0.7345k
For force F,
F = Fu
= 500N(0.6269i + 0.2597j – 0.7345k)
= {313i - 130j - 367k} N
Solution
Checking
2.8 Force Vector Directed along a Line
2.8 Force Vector Directed along a Line
( ) ( ) ( )N
F
500
367130313 222
=
−++=
Show that γ = 137° and
indicate this angle on the
diagram
N500=
2.8 Force Vector Directed along a Line
2.8 Force Vector Directed along a Line
Example 2.15
The roof is supported by
cables. If the cables exert
F = 100N and F = 120NFAB = 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
Solution
rAB = (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 =−+=
2.8 Force Vector Directed along a Line
2.8 Force Vector Directed along a Line
Solution
rAC = (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
Solution
FR = FAB + FAC
= {70.7i - 70.7k} N + {80i + 40j – 80k} N
= {150.7i + 40j – 150.7k} N= {150.7i + 40j – 150.7k} N
Magnitude of FR
( ) ( ) ( )N
FR
217
7.150407.150 222
=
−++=
2.9 Dot Product2.9 Dot Product
� Dot 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 tailsangle 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 Operation
1. Commutative law
A·B = B·AA·B = B·A
2. Multiplication by a scalar
a(A·B) = (aA)·B = A·(aB) = (A·B)a
3. Distribution law
A·(B + D) = (A·B) + (A·D)
2.9 Dot Product2.9 Dot Product
� Cartesian Vector Formulation
- Dot product of Cartesian unit vectors
Eg: i·i = (1)(1)cos0° = 1 andEg: i·i = (1)(1)cos0° = 1 and
i·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 B
A·B = (Axi + Ayj + Azk)· (Bxi + Byj + Bzk)
= A B (i·i) + A B (i·j) + A B (i·k)= 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 linesintersecting 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 Product
� Applications- The components of a vector parallel and perpendicular to a line
- Component of A parallel or collinear with line aa’ is - 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 - 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
2.9 Dot Product2.9 Dot Product
then A┴ = A - A║2. θ = cos-1 [(A·u)/(A)]
then A┴ = Asinθ
3. If A║ is known, by Pythagorean Theorem
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/rAbeams, 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
u = r /ruA = 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 2.16
The frame is subjected to a horizontal force
F = {300j} N. Determine the components of
this force parallel and perpendicular to the this force parallel and perpendicular to the
member AB.
2.9 Dot Product2.9 Dot Product
Solution
Since
( ) ( ) ( )kji
r
ru
B
BB
362
362222 ++
++==
rrr
r
rr
Then
( ) ( ) ( )
( ) ( )
N
kjijuF
FF
kji
r
B
AB
B
1.257
)429.0)(0()857.0)(300()286.0)(0(
429.0857.0286.0300.
cos
429.0857.0286.0
362 222
=
++=
++⋅==
=
++=
++
rrrrrr
rr
rrr
θ
2.9 Dot Product2.9 Dot Product
Solution
Since result is a positive scalar,
FAB has the same sense of
direction as uB. Express in B
Cartesian form
Perpendicular component
( )( )
NkjikjijFFF
Nkji
kjiN
uFF
AB
ABABAB
}110805.73{)1102205.73(300
}1102205.73{
429.0857.0286.01.257
rrrrrrrrrr
rrr
rrr
rrr
−+−=++−=−=
++=
++=
=
⊥
2.9 Dot Product2.9 Dot Product
Solution
Magnitude can be determined
From F┴ or from Pythagorean TheoremTheorem
( ) ( )N
NN
FFF AB
155
1.257300 22
22
=
−=
−=⊥
rrr
2.9 Dot Product2.9 Dot Product
Example 2.17
The 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 magnitudes of the components of F, which are
parallel and perpendicular to BA.
2.9 Dot Product2.9 Dot Product
Solution
For angle θ
rBA = {-2i - 2j + 1k}m
rBC = {- 3j + 1k}mrBC = {- 3j + 1k}m
Thus,
( )( ) ( )( ) ( )( )
o
rr
rr
5.42
7379.0
103
113202cos
=
=
+−−+−=
⋅=
θ
θBCBA
BCBA
rr
rr
2.9 Dot Product2.9 Dot Product
Solution
Components of Fkji
r
ru
AB
ABAB 3
)122( +−−==
rrr
r
rr
( )
N
kjikj
uFF
kji
BAB
AB
590
3.840.5060
31
32
32
0.2539.758
.
31
32
32
=
++=
+
−+
−⋅+−=
=
+
−+
−=
rrrrr
rrr
rrr
2.9 Dot Product2.9 Dot Product
Solution
Checking from trigonometry,
FFAB cos=rr
θ
Magnitude can be determined
From F┴
N
N
540
5.42cos800
=
= o
NFF 5405.42sin800sin ===⊥o
rrθ
2.9 Dot Product2.9 Dot Product
Solution
Magnitude can be determined from F┴ or from Pythagorean Theorem
( ) ( )N
FFF AB
540
590800 22
22
=
−=
−=⊥rrr
Chapter SummaryChapter Summary
Parallelogram Law� Addition of two vectors
� Components form the side and resultant � Components 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 form
F = Fxi + Fyj + Fzk
� Magnitude of F� 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 vectorposition vector
� Force expressed as a Cartesian vector
F = 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 + AzBzA·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