Shawn Kenny, Ph.D., P.Eng.Assistant ProfessorFaculty of Engineering and Applied ScienceMemorial University of [email protected]
ENGI 1313 Mechanics I
Lecture 03: Force Vectors and Parallelogram Law
2 ENGI 1313 Statics I – Lecture 03© 2007 S. Kenny, Ph.D., P.Eng.
Revised – Course Method of Evaluation
6 Tutorial Quizzes 15%During week 38, 39, 40, 43, 44, & 45 Best 5 out of 6 toward final
Mid-Term Exam 30%Oct. 18
Final Exam 55%Dec. 6
3 ENGI 1313 Statics I – Lecture 03© 2007 S. Kenny, Ph.D., P.Eng.
Tutorial Sessions
Teaching AssistantsKenton Pike ([email protected])Nasser Daiyan ([email protected])YanZhen Ou ([email protected])
FriFriThuThuThuMonDay
EN1040EN1040EN2007EN1040EN1040EN1040Room
4–4:503–3:5010–10:504–4:502–2:503–3:50Time
654321Section
4 ENGI 1313 Statics I – Lecture 03© 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
5 ENGI 1313 Statics I – Lecture 03© 2007 S. Kenny, Ph.D., P.Eng.
Lecture 03 Objectives
to review concepts from linear algebrato sum force vectors, determine force resultants, and resolve force components for 2D vectors using Parallelogram Law
6 ENGI 1313 Statics I – Lecture 03© 2007 S. Kenny, Ph.D., P.Eng.
Introductory Concepts
ScalarMagnitude (value) and sense (positive, negative)No direction
ExamplesMassVolumeLengthTemperatureSpeed
kg22kg12kg10CBAC
=+=+=
7 ENGI 1313 Statics I – Lecture 03© 2007 S. Kenny, Ph.D., P.Eng.
Direction
Magnitude
►SenseVectorMagnitudeSense (+, -)Direction ororientationConvention• Textbook is boldface, A• PowerPoint notation typically A
Examples• Force• Velocity
Introductory Concepts
8 ENGI 1313 Statics I – Lecture 03© 2007 S. Kenny, Ph.D., P.Eng.
Scalar Multiplication and Division
amFrr
=Change in MagnitudeChange in Sense
9 ENGI 1313 Statics I – Lecture 03© 2007 S. Kenny, Ph.D., P.Eng.
Vector Operations
Engineering NeedDetermine resultant force due to applied forcesResolve force into components
MethodParallelogram law• Triangle construction 4321R FFFFF
rrrrr+++=
10 ENGI 1313 Statics I – Lecture 03© 2007 S. Kenny, Ph.D., P.Eng.
Vector Addition
Parallelogram LawGraphical construction 21R FFF
rrr+=
2Fr Vector Tip
1Fr Vector Tip
Vector Tail
Resultant Vector (FR)
Component Vectors (F1, F2)
RFr Resultant Vector
forms the Parallelogram Diagonal
11 ENGI 1313 Statics I – Lecture 03© 2007 S. Kenny, Ph.D., P.Eng.
Vector Addition
Parallelogram LawSpecial case• Collinear vectors• Algebraic addition
12 ENGI 1313 Statics I – Lecture 03© 2007 S. Kenny, Ph.D., P.Eng.
Vector Addition
Parallelogram LawTriangle construction• “Tip-to-Tail” technique 21R FFF
rrr+=
1Fr 2F
r
RFr
Parallelogram
13 ENGI 1313 Statics I – Lecture 03© 2007 S. Kenny, Ph.D., P.Eng.
Vector Addition
Parallelogram LawTriangle construction• “Tip-to-Tail” technique 12R FFF
rrr+=
1Fr
2Fr
RFr
Parallelogram
14 ENGI 1313 Statics I – Lecture 03© 2007 S. Kenny, Ph.D., P.Eng.
Vector Subtraction
Parallelogram LawTriangle Construction• “Tip-to-Tail” technique
( )2121 FFFFFrrrrr
−+=−=
1Fr
2Fr
RFr
15 ENGI 1313 Statics I – Lecture 03© 2007 S. Kenny, Ph.D., P.Eng.
Parallelogram Law
Multiple Force Vectors ( ) 321
321
FFFFFFFrrr
rrrr
++=++=
3Fr2F
r1Fr
2Fr
21 FFrr
+ RFr
21 FFrr
+
16 ENGI 1313 Statics I – Lecture 03© 2007 S. Kenny, Ph.D., P.Eng.
Vector Summation
Resultant Force MagnitudeCosine law
17 ENGI 1313 Statics I – Lecture 03© 2007 S. Kenny, Ph.D., P.Eng.
Vector Summation
Resultant Force Direction or Magnitude of Component Forces
Sine law
o180cba =++
18 ENGI 1313 Statics I – Lecture 03© 2007 S. Kenny, Ph.D., P.Eng.
Applications
Lifting Devices
19 ENGI 1313 Statics I – Lecture 03© 2007 S. Kenny, Ph.D., P.Eng.
Applications
Guyed Towers
20 ENGI 1313 Statics I – Lecture 03© 2007 S. Kenny, Ph.D., P.Eng.
Applications
Cable Stayed Bridge
21 ENGI 1313 Statics I – Lecture 03© 2007 S. Kenny, Ph.D., P.Eng.
Applications
Offshore Platform Foundation Connections
22 ENGI 1313 Statics I – Lecture 03© 2007 S. Kenny, Ph.D., P.Eng.
Applications
Towing
23 ENGI 1313 Statics I – Lecture 03© 2007 S. Kenny, Ph.D., P.Eng.
Comprehension Quiz 2-01
Scalar or Vector?ForceTimeMassPosition
Scalar or Vector?Force ⇒ VectorTime ⇒ ScalarMass ⇒ ScalarPosition ⇒ Vector
24 ENGI 1313 Statics I – Lecture 03© 2007 S. Kenny, Ph.D., P.Eng.
Comprehension Quiz 2-02
Q: Is this the correct application of the parallelogram law to determine the resultant force vector (FR)?
F1 = 4 kN
F2 = 10 kN
30°
4 kN
90°
FR ( ) 30sinkN4
3090180sinFR =
−−
kN93.6FR =
X
Y
25 ENGI 1313 Statics I – Lecture 03© 2007 S. Kenny, Ph.D., P.Eng.
Comprehension Quiz 2-02 (cont.)
A: No“Tip-to-Tail” triangle construction technique
F1 = 4 kN
θR = 180° – (180° – 30° – 90°) = 120° FR
F1 = 4 kN
F2 = 10 kN
X
Y
30°
θR
θ1
θ2
26 ENGI 1313 Statics I – Lecture 03© 2007 S. Kenny, Ph.D., P.Eng.
Comprehension Quiz 2-02 (cont.)
Determine Resultant Force Magnitude
Cosine Law
Therefore FR = 12.5 kN
F1 = 4 kN
FR
F1 = 4 kN
F2 = 10 kN
X
Y
30°
θR
θ1
θ2
θR = 120°R21
22RR cosFF2FFFF
21θ−+==
r
( ) ( ) ( )( ) kN49.12120coskN10kN42kN10kN4F 22R =−+=
27 ENGI 1313 Statics I – Lecture 03© 2007 S. Kenny, Ph.D., P.Eng.
Comprehension Quiz 2-02 (cont.)
Determine Resultant Force Direction
Sine Law
F1 = 4 kN
FR
F1 = 4 kN
F2 = 10 kN
X
Y
30°
θR
θ1
θ2
θR = 120°R
R
1
1
sinF
sinF
θθ=
o11 09.16
kN49.12120sinkN4sin =⎟⎟
⎠
⎞⎜⎜⎝
⎛= −θ
Therefore 43.9° from horizontal (clockwise)
43.9°
28 ENGI 1313 Statics I – Lecture 03© 2007 S. Kenny, Ph.D., P.Eng.
Example Problem 3-01
Determine the component magnitudes (FX and FY) of the 700-lb force resultant (FR)
Fx
FY
θX
θY
θR
FR = 700 lb
X
Y
60°
30°
FR = 700 lb
X
Y
Vector Triangle
60°30°
29 ENGI 1313 Statics I – Lecture 03© 2007 S. Kenny, Ph.D., P.Eng.
Example Problem 3-01 (cont.)
Determine Interior Angles of Vector Triangle
θX
θY
θR
X
Y
60°30°
θY = 60° - 30° = 30°
30°
α = 90° - 30° = 60°
θR = 180° - 60° - 30° = 90° α
θX = α = 60°
30 ENGI 1313 Statics I – Lecture 03© 2007 S. Kenny, Ph.D., P.Eng.
Example Problem 3-01 (cont.)
Determine the component magnitudes (Fx and Fy) of the resultant 700-lb force
Fx
FY
60°
30°
90°
FR = 700 lb
X
Y
60°30°
lb60690sinlb70060sin
sinFsinF
R
RXX −===
θθ
lb35090sinlb70030sin
sinFsinF
R
RYY ===
θθ
31 ENGI 1313 Statics I – Lecture 03© 2007 S. Kenny, Ph.D., P.Eng.
Example Problem 3-02
Problem 2-12 from Hibbeler (2007)The component of force Facting along line aa is required to be 30 lb.
Determine the magnitude of Fand its component along line bb.
Given:
32 ENGI 1313 Statics I – Lecture 03© 2007 S. Kenny, Ph.D., P.Eng.
Example Problem 3-02 (cont.)
Problem 2-12 from Hibbeler (2007)Draw force vectors
a
a
b
b
F80°
60°
Fa= 30lb
Fb
Fa = 30lb
θb
θF
θF = 180° - θ1 - θb = 180° - 80° - 60° = 40°
θ2 = θb = 60°
33 ENGI 1313 Statics I – Lecture 03© 2007 S. Kenny, Ph.D., P.Eng.
Example Problem 3-02 (cont.)
Problem 2-12 from Hibbeler (2007)Magnitude of F & Fb from sine law
F80°
Fb
Fa = 30lb
60°
40°
θF = 40°
θ2 = θb = 60°
θ1 = θa = 80°
lb6.1980sinlb3040sin
sinFsinF
a
aF ===
θθ
lb4.2680sinlb3060sin
sinFsinF
a
abb ===
θθ
34 ENGI 1313 Statics I – Lecture 03© 2007 S. Kenny, Ph.D., P.Eng.
Vector Summation
Methods StudiedParallelogram Law• Vector triangle construction• Sine law• Cosine law
LimitationsResultant of multiple vectors determined through successive summation of two vectors• Cumbersome for large systems
35 ENGI 1313 Statics I – Lecture 03© 2007 S. Kenny, Ph.D., P.Eng.
Representative Problems
Hibbeler (2007) Textbook
10-15minMediumVector Addition Parallelogram Law2-11 to 2-19
5-10minEasyVector Addition Parallelogram Law2-1 to 2-10
Vector Addition Parallelogram Law
Vector Addition Parallelogram Law
Concept
10-15minMedium2-25 to 2-30
5-10minEasy2-20 to 2-24
Estimated Time
Degree of DifficultyProblem Set
36 ENGI 1313 Statics I – Lecture 03© 2007 S. Kenny, Ph.D., P.Eng.
ReferencesHibbeler (2007)http://wps.prenhall.com/esm_hibbeler_engmech_1www.hanessupply.comwww.sabrecom.comen.wikipedia.orgwww.caldwellinc.comwww.atlantia.comwww.c-core.cawww.straylight.ca/greglocke/hibernia.htmwww.hibernia.ca
