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Shawn Kenny, Ph.D., P.Eng. Assistant Professor Faculty of Engineering and Applied Science Memorial University of Newfoundland spkenny@engr.mun.ca ENGI 1313 Mechanics I Lecture 03:Force Vectors and Parallelogram Law

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ENGI 1313 Statics I – Lecture 03© 2007 S. Kenny, Ph.D., P.Eng. 2 Revised – Course Method of Evaluation 6 Tutorial Quizzes15% During week 38, 39, 40, 43, 44, & 45 Best 5 out of 6 toward final Mid-Term Exam30% Oct. 18 Final Exam55% Dec. 6

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ENGI 1313 Statics I – Lecture 03© 2007 S. Kenny, Ph.D., P.Eng. 3 Tutorial Sessions Teaching Assistants Kenton Pike (kenton@engr.mun.ca) Nasser Daiyan (daiyann@engr.mun.ca) YanZhen Ou (yanzhen@engr.mun.ca) Section 123456 Day MonThu Fri Time 3–3:502–2:504–4:5010–10:503–3:504–4:50 Room EN1040 EN2007EN1040

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ENGI 1313 Statics I – Lecture 03© 2007 S. Kenny, Ph.D., P.Eng. 4 Chapter 2 Objectives to review concepts from linear algebra to sum forces, determine force resultants and resolve force components for 2D vectors using Parallelogram Law to express force and position in Cartesian vector form to introduce the concept of dot product

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ENGI 1313 Statics I – Lecture 03© 2007 S. Kenny, Ph.D., P.Eng. 5 Lecture 03 Objectives to review concepts from linear algebra to sum force vectors, determine force resultants, and resolve force components for 2D vectors using Parallelogram Law

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ENGI 1313 Statics I – Lecture 03© 2007 S. Kenny, Ph.D., P.Eng. 6 Introductory Concepts Scalar Magnitude (value) and sense (positive, negative) No direction Examples Mass Volume Length Temperature Speed

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ENGI 1313 Statics I – Lecture 03© 2007 S. Kenny, Ph.D., P.Eng. 7 Direction Magnitude ►Sense Vector Magnitude Sense (+, -) Direction or orientation Convention Textbook is boldface, A PowerPoint notation typically A Examples Force Velocity Introductory Concepts

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ENGI 1313 Statics I – Lecture 03© 2007 S. Kenny, Ph.D., P.Eng. 8 Scalar Multiplication and Division Change in Magnitude Change in Sense

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ENGI 1313 Statics I – Lecture 03© 2007 S. Kenny, Ph.D., P.Eng. 9 Vector Operations Engineering Need Determine resultant force due to applied forces Resolve force into components Method Parallelogram law Triangle construction

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ENGI 1313 Statics I – Lecture 03© 2007 S. Kenny, Ph.D., P.Eng. 10 Vector Addition Parallelogram Law Graphical construction Vector Tip Vector Tail Resultant Vector (F R ) Component Vectors (F 1, F 2 ) Resultant Vector forms the Parallelogram Diagonal

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ENGI 1313 Statics I – Lecture 03© 2007 S. Kenny, Ph.D., P.Eng. 11 Vector Addition Parallelogram Law Special case Collinear vectors Algebraic addition

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ENGI 1313 Statics I – Lecture 03© 2007 S. Kenny, Ph.D., P.Eng. 12 Vector Addition Parallelogram Law Triangle construction “Tip-to-Tail” technique Parallelogram

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ENGI 1313 Statics I – Lecture 03© 2007 S. Kenny, Ph.D., P.Eng. 13 Vector Addition Parallelogram Law Triangle construction “Tip-to-Tail” technique Parallelogram

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ENGI 1313 Statics I – Lecture 03© 2007 S. Kenny, Ph.D., P.Eng. 14 Vector Subtraction Parallelogram Law Triangle Construction “Tip-to-Tail” technique

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ENGI 1313 Statics I – Lecture 03© 2007 S. Kenny, Ph.D., P.Eng. 15 Parallelogram Law Multiple Force Vectors

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ENGI 1313 Statics I – Lecture 03© 2007 S. Kenny, Ph.D., P.Eng. 16 Vector Summation Resultant Force Magnitude Cosine law

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ENGI 1313 Statics I – Lecture 03© 2007 S. Kenny, Ph.D., P.Eng. 17 Vector Summation Resultant Force Direction or Magnitude of Component Forces Sine law

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ENGI 1313 Statics I – Lecture 03© 2007 S. Kenny, Ph.D., P.Eng. 18 Applications Lifting Devices

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ENGI 1313 Statics I – Lecture 03© 2007 S. Kenny, Ph.D., P.Eng. 19 Applications Guyed Towers

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ENGI 1313 Statics I – Lecture 03© 2007 S. Kenny, Ph.D., P.Eng. 20 Applications Cable Stayed Bridge

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ENGI 1313 Statics I – Lecture 03© 2007 S. Kenny, Ph.D., P.Eng. 21 Applications Offshore Platform Foundation Connections

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ENGI 1313 Statics I – Lecture 03© 2007 S. Kenny, Ph.D., P.Eng. 22 Applications Towing

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ENGI 1313 Statics I – Lecture 03© 2007 S. Kenny, Ph.D., P.Eng. 23 Comprehension Quiz 2-01 Scalar or Vector? Force Time Mass Position Scalar or Vector? Force Vector Time Scalar Mass Scalar Position Vector

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ENGI 1313 Statics I – Lecture 03© 2007 S. Kenny, Ph.D., P.Eng. 24 Comprehension Quiz 2-02 Q: Is this the correct application of the parallelogram law to determine the resultant force vector (F R )? F 1 = 4 kN F 2 = 10 kN 30 4 kN 90 FRFR X Y

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ENGI 1313 Statics I – Lecture 03© 2007 S. Kenny, Ph.D., P.Eng. 25 Comprehension Quiz 2-02 (cont.) A: No “Tip-to-Tail” triangle construction technique F 1 = 4 kN R = 180 – (180 – 30 – 90 ) = 120 FRFR F 1 = 4 kN F 2 = 10 kN X Y 30 RR 11 22

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ENGI 1313 Statics I – Lecture 03© 2007 S. Kenny, Ph.D., P.Eng. 26 Comprehension Quiz 2-02 (cont.) Determine Resultant Force Magnitude Cosine Law Therefore F R = 12.5 kN F 1 = 4 kN FRFR F 2 = 10 kN X Y 30 RR 11 22 R = 120

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ENGI 1313 Statics I – Lecture 03© 2007 S. Kenny, Ph.D., P.Eng. 27 Comprehension Quiz 2-02 (cont.) Determine Resultant Force Direction Sine Law F 1 = 4 kN FRFR F 2 = 10 kN X Y 30 RR 11 22 R = 120 Therefore 43.9 from horizontal (clockwise) 43.9

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ENGI 1313 Statics I – Lecture 03© 2007 S. Kenny, Ph.D., P.Eng. 28 Example Problem 3-01 Determine the component magnitudes (F X and F Y ) of the 700-lb force resultant (F R ) FxFx FYFY XX YY RR F R = 700 lb X Y 60 30 F R = 700 lb X Y Vector Triangle 60 30

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ENGI 1313 Statics I – Lecture 03© 2007 S. Kenny, Ph.D., P.Eng. 29 Example Problem 3-01 (cont.) Determine Interior Angles of Vector Triangle XX YY RR X Y 60 30 Y = 60 - 30 = 30 30 = 90 - 30 = 60 R = 180 - 60 - 30 = 90 X = = 60

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ENGI 1313 Statics I – Lecture 03© 2007 S. Kenny, Ph.D., P.Eng. 30 Example Problem 3-01 (cont.) Determine the component magnitudes (F x and F y ) of the resultant 700-lb force FxFx FYFY 60 30 90 F R = 700 lb X Y 60 30

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ENGI 1313 Statics I – Lecture 03© 2007 S. Kenny, Ph.D., P.Eng. 31 Example Problem 3-02 Problem 2-12 from Hibbeler (2007) The component of force F acting along line aa is required to be 30 lb. Determine the magnitude of F and its component along line bb. Given:

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ENGI 1313 Statics I – Lecture 03© 2007 S. Kenny, Ph.D., P.Eng. 32 Example Problem 3-02 (cont.) Problem 2-12 from Hibbeler (2007) Draw force vectors a a b b F 80 60 F a = 30lb FbFb bb FF F = 180 - 1 - b = 180 - 80 - 60 = 40 2 = b = 60

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ENGI 1313 Statics I – Lecture 03© 2007 S. Kenny, Ph.D., P.Eng. 33 Example Problem 3-02 (cont.) Problem 2-12 from Hibbeler (2007) Magnitude of F & F b from sine law F 80 FbFb F a = 30lb 60 40 F = 40 2 = b = 60 1 = a = 80

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ENGI 1313 Statics I – Lecture 03© 2007 S. Kenny, Ph.D., P.Eng. 34 Vector Summation Methods Studied Parallelogram Law Vector triangle construction Sine law Cosine law Limitations Resultant of multiple vectors determined through successive summation of two vectors Cumbersome for large systems

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ENGI 1313 Statics I – Lecture 03© 2007 S. Kenny, Ph.D., P.Eng. 35 Representative Problems Hibbeler (2007) Textbook Problem SetConcept Degree of Difficulty Estimated Time 2-1 to 2-10Vector Addition Parallelogram LawEasy5-10min 2-11 to 2-19Vector Addition Parallelogram LawMedium10-15min 2-20 to 2-24Vector Addition Parallelogram LawEasy5-10min 2-25 to 2-30Vector Addition Parallelogram LawMedium10-15min

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ENGI 1313 Statics I – Lecture 03© 2007 S. Kenny, Ph.D., P.Eng. 36 References Hibbeler (2007) http://wps.prenhall.com/esm_hibbeler_engmech_1 www.hanessupply.com www.sabrecom.com en.wikipedia.org www.caldwellinc.com www.atlantia.com www.c-core.ca www.straylight.ca/greglocke/hibernia.htm www.hibernia.ca

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Students will be able to : a) Resolve a 2-D vector into components

Students will be able to : a) Resolve a 2-D vector into components

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