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


Lecture 06 Objectives

to further examine 3D Cartesian vectorsto define a position vector in Cartesian coordinate systemto determine force vector directed along a line


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.


Example Problem 6-01

Known

Find

N80F =r

o60=α

o45=γ

xF yF zF


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


Position Vectors – General

3D CoordinatesUnique position in spaceRight-hand coordinate system• A(4,2,-6)• B(0,2,0)• C(6,-1,4)


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⇒


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)



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

⇒



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

=+


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


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


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


Example 6-01

Express the force vector FDA in Cartesian form

Known:A(0,0,14) ftD(2,6,0) ftFDA = 400 lb


Example 6-01 (cont.)

Find Position Vector rDAThrough point coordinates

DADA rrrrrrrr

−==

( ) ( ) ( )kzzjyyixxr DADADADA −+−+−=r

( ) ( ) ( ){ }ftk014j60i20rDA −+−+−=r

{ }ftk14j6i2rDA +−−=r

rDA



Find Position Vector |rDA|Magnitude


ft36.151462r 222DA =++=r

rDA



Find unit vector uDA

DA

DADA r

ru r

rv

=

{ } { }k9115.0j3906.0i1302.0ft36.15

ftk14j6i2uDA +−−=+−−

=v

uDA


ft36.151462r 222DA =++=r



Find Unit Vector uDAMagnitude

Confirm unity uDA

{ }k9115.0j3906.0i1302.0uDA +−−=v

000.19115.03906.01302.0u 222DA =++=v



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


Group Problem 6-01

Find the resultant force magnitude and coordinate direction

PlanCartesian vector form of FCA and FCB

Sum concurrent forcesObtain solution


Group Problem 6-01 (cont.)

Position Vectors and MagnituderCA

rCB

{ }ftk4j40cos3i40sin3rCA −+−= oor

{ }ftk4j7i4rCB −−=r

ft9474r 222CB =++=r

ft54298.2928.1r 222CA =++=r



Force Vectors and MagnitudeFCA

FCB

{ }ftk4j40cos3i40sin3ft5lb100FCA −+−= oo

r

rr

FuFF

rr

rrr

==

{ }lbk80j96.45i57.38FCA −+−=r

{ }ftk4j7i4ft9lb81FCB −−=

r

{ }lbk36j63i36FCA −−=r

lb100FCA =⇔r

lb81FCA =⇔r



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


-50

5-20 -10 0 10

-120

-100

-80

-60

-40

-20

0

20

X-AxisY-Axis

Z-A

xis


Force Resultant Vector Magnitude & Orientation

{ }lbk116j0.17i57.2FR −−−=r

ooo 1723.117

116cos4.983.11704.17cos3.91

3.11757.2cos 111 =

−==

−==

−= −−− γβα

F1F2

FR


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


References

Hibbeler (2007)http://wps.prenhall.com/esm_hibbeler_engmech_1

engi 1313 mechanics i - faculty of engineering and applied ... · shawn kenny, ph.d., p.eng....

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