ENGI 1313 Mechanics I - Faculty of Engineering and Applied ... · Shawn Kenny, Ph.D., P.Eng....
Transcript of ENGI 1313 Mechanics I - Faculty of Engineering and Applied ... · Shawn Kenny, Ph.D., P.Eng....
Shawn Kenny, Ph.D., P.Eng.Assistant ProfessorFaculty of Engineering and Applied ScienceMemorial University of [email protected]
ENGI 1313 Mechanics I
Lecture 06: Cartesian and Position Vectors
2 ENGI 1313 Statics I – Lecture 06© 2007 S. Kenny, Ph.D., P.Eng.
Chapter 2 Objectives
to review concepts from linear algebrato sum forces, determine force resultants and resolve force components for 2D vectors using Parallelogram Lawto express force and position in Cartesian vector formto introduce the concept of dot product
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Lecture 06 Objectives
to further examine 3D Cartesian vectorsto define a position vector in Cartesian coordinate systemto determine force vector directed along a line
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Example Problem 6-01
Problem 2-77 (Hibbeler, 2007). The bolt is subjected to the force F, which has components acting along the x, y, z axes as shown. If the magnitude of F is 80 N, and α = 60° and γ = 45°, determine the magnitudes of its components.
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Example Problem 6-01
Known
Find
N80F =r
o60=α
o45=γ
xF yF zF
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Example Problem 6-01 (cont.)
Find Angle β
Find component magnitudes
oo 45cos60cos1cos 221 −−= −β
oo 120or60=β
N40cosFFx == αr
N40cosFFy == βr
N6.56cosFFz == γr
Fz
Fy
Fx
α = 60°γ = 45°
N80F =r
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Position Vectors – General
3D CoordinatesUnique position in spaceRight-hand coordinate system• A(4,2,-6)• B(0,2,0)• C(6,-1,4)
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Position Vectors – Origin to a Point
Fixed vector locating a point P(x,y,z) in space relative to another point (origin) within a defined coordinate system.
Right-hand Cartesian coordinate systemTip-to-tail vector component technique
kzjyixrr OP ++==rr
zk
yj
xii j⇒ k⇒
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Position Vector – General Case
Two Points in SpaceRectangular Cartesian coordinate system• Origin O
Point A and Point B
A(xA, yA, zA)
B(xB, yB, zB)
x
y
z
O(0, 0, 0)
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Position Vector – General Case
Establish Position VectorsFrom Point O to Point A (rOA = rA) From Point O to Point B (rOB = rB)From Point A to Point B (rAB = r )
A(xA, yA, zA)
B(xB, yB, zB)
x
y
z
O(0, 0, 0)rOA
rOB
rAB
OBABOA rrrrrr
=+ Recall “tip-to-tail”vector addition laws
⇒
11 ENGI 1313 Statics I – Lecture 06© 2007 S. Kenny, Ph.D., P.Eng.
Position Vector – General Case
Define Position Vector (rAB = r )“tip – tail” or B(xB, yB, zB) – A(xA, yA, zA)
A(xA, yA, zA)
x
y
z
O(0, 0, 0)rOA
rOB
rAB
^(xB – xA) i
^(zB – zA) k
^(yB – yA) j
r = rABB(xB, yB, zB)
( ) ( ) ( )kzzjyyixxrrrr ABABABABAB −+−+−=−==rrrr
BABA rrrrrr
=+
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Comprehension Quiz 6-01Two points in 3D 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} mB) {-3 i - 3 j - 3 k} mC) { 5 i + 7 j + 9 k} mD) {-3 i + 3 j + 3 k} mE) { 4 i + 5 j + 6 k} m
Answer: B ⇒ {-3 i - 3 j - 3 k} m
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Comprehension Quiz 6-02
P and Q are two points in a 3-D space. How are the position vectors rPQ and rQPrelated?
A) rPQ = rQPB) rPQ = -rQPC) rPQ = 1/rQPD) rPQ = 2rQP
Answer: B
Q(xB, yQ, zQ)
x
y
z
P(xP, yP, zP) rPQ = -rQP
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Comprehension Quiz 6-03If F is a force vector (N) and r is a position vector (m), what are the units of the expression
A) NB) DimensionlessC) mD) N⋅mE) The expression is algebraically illegal
Answer: A
⎟⎟⎠
⎞⎜⎜⎝
⎛
rrF r
rr
⎟⎟⎠
⎞⎜⎜⎝
⎛==
rrFuFF r
rrrr
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Example 6-01
Express the force vector FDA in Cartesian form
Known:A(0,0,14) ftD(2,6,0) ftFDA = 400 lb
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Example 6-01 (cont.)
Find Position Vector rDAThrough point coordinates
DADA rrrrrrrr
−==
( ) ( ) ( )kzzjyyixxr DADADADA −+−+−=r
( ) ( ) ( ){ }ftk014j60i20rDA −+−+−=r
{ }ftk14j6i2rDA +−−=r
rDA
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Example 6-01 (cont.)
Find Position Vector |rDA|Magnitude
{ }ftk14j6i2rDA +−−=r
ft36.151462r 222DA =++=r
rDA
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Example 6-01 (cont.)
Find unit vector uDA
DA
DADA r
ru r
rv
=
{ } { }k9115.0j3906.0i1302.0ft36.15
ftk14j6i2uDA +−−=+−−
=v
uDA
{ }ftk14j6i2rDA +−−=r
ft36.151462r 222DA =++=r
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Example 6-01 (cont.)
Find Unit Vector uDAMagnitude
Confirm unity uDA
{ }k9115.0j3906.0i1302.0uDA +−−=v
000.19115.03906.01302.0u 222DA =++=v
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Example 6-01 (cont.)
Find Force Vector FDA
or
{ }lbk9115.0j3906.0i1302.0400FDA +−−=r
{ }lbk365j156i1.52FDA +−−=r
DADADA uFFrr
=
DADA
DADADADA r
r
FuFF
rr
rrr
==
{ } { }lbk365j156i1.52ftk14j6i2ft36.15
lb400FDA +−−=+−−=r
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Group Problem 6-01
Find the resultant force magnitude and coordinate direction
PlanCartesian vector form of FCA and FCB
Sum concurrent forcesObtain solution
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Group Problem 6-01 (cont.)
Position Vectors and MagnituderCA
rCB
{ }ftk4j40cos3i40sin3rCA −+−= oor
{ }ftk4j7i4rCB −−=r
ft9474r 222CB =++=r
ft54298.2928.1r 222CA =++=r
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Group Problem 6-01 (cont.)
Force Vectors and MagnitudeFCA
FCB
{ }ftk4j40cos3i40sin3ft5lb100FCA −+−= oo
r
rr
FuFF
rr
rrr
==
{ }lbk80j96.45i57.38FCA −+−=r
{ }ftk4j7i4ft9lb81FCB −−=
r
{ }lbk36j63i36FCA −−=r
lb100FCA =⇔r
lb81FCA =⇔r
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Group Problem 6-01 (cont.)
Force Resultant Vector Magnitude & Orientation
( ) ( ) ( ){ }lbk3680j6396.45i3657.38FR −−+−++−=r
{ }lbk116j0.17i57.2FR −−−=r
lb117FR =⇔r
CBCAR FFFrrr
+=
ooo 1723.117
116cos4.983.11704.17cos3.91
3.11757.2cos 111 =
−==
−==
−= −−− γβα
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-50
5-20 -10 0 10
-120
-100
-80
-60
-40
-20
0
20
X-AxisY-Axis
Z-A
xis
Group Problem 6-01 (cont.)
Force Resultant Vector Magnitude & Orientation
{ }lbk116j0.17i57.2FR −−−=r
ooo 1723.117
116cos4.983.11704.17cos3.91
3.11757.2cos 111 =
−==
−==
−= −−− γβα
F1F2
FR
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Classification of Textbook Problems
Hibbeler (2007)
15-20minMediumResultant force vectors2-85 to 2-90
15-20minMediumResultant force & position vectors2-91 to 2-96
5-10minEasyPosition vectors2-79 to 2-84
Resultant force & position vectors
Resultant force & position vectors
Position vectors
Concept
30minHard2-100
10-15minEasy2-97 to 2-99
15-20minMedium2-101 to 2-106
Estimated Time
Degree of DifficultyProblem Set
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References
Hibbeler (2007)http://wps.prenhall.com/esm_hibbeler_engmech_1