ENGI 1313 Mechanics I Lecture 04: Force Vectors and System of Coplanar Forces Shawn Kenny, Ph.D., P.Eng. Assistant Professor Faculty of Engineering and Applied Science Memorial University.
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ENGI 1313 Mechanics I Lecture 04: Force Vectors and System of Coplanar Forces Shawn Kenny, Ph.D., P.Eng. Assistant Professor Faculty of Engineering and Applied Science Memorial University of Newfoundland [email protected] Tutorial Questions SI Units and Use Section 1.4 Page 9 Use a single prefix Magnitude between 0.1 and 1000 Pa N m 2 100 kN 5 mm 2 10 MPa MN m 2 2 10 5 10 2 3 6 2 10 N 3 2 10 N 10 3 © 2007 S. Kenny, Ph.D., P.Eng. 2 3 6 N 2 5 10 m m 3 2 mm mm N N GN 11 0 . 2 10 20 20 GPa 2 2 2 5m m m 6 10 mm m 10 N 10 mm 6 2 6 2 10 6 6 mm 2 ENGI 1313 Statics I – Lecture 04 N N mm 2 Tutorial Questions SI Units and Use Section 1.4 Page 9 Use a single prefix Magnitude between 0.1 and 1000 50 kN 60 nm 50 10 3 N 60 10 9 m 50 60 10 3 10 3 3 3 9 N m 3 © 2007 S. Kenny, Ph.D., P.Eng. N m 3 10 10 N m 3 6 mN m ENGI 1313 Statics I – Lecture 04 Tutorial Questions SI Units and Use Section 1.4 Page 9 Do not use compound prefixes kg 10 4 6 10 g 10 3 © 2007 S. Kenny, Ph.D., P.Eng. 6 3 g 10 g mg 3 ENGI 1313 Statics I – Lecture 04 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 5 © 2007 S. Kenny, Ph.D., P.Eng. ENGI 1313 Statics I – Lecture 04 Lecture 04 Objectives to sum force vectors, determine force resultants, and resolve force components for 2D vectors using Scalar or Cartesian Vector Notation to demonstrate by example 6 © 2007 S. Kenny, Ph.D., P.Eng. ENGI 1313 Statics I – Lecture 04 Why an Alternate Approach? Application of Parallelogram Law Cumbersome with a large number of coplanar forces due to successive application • Recall Lecture 02 (Slide 13) 7 © 2007 S. Kenny, Ph.D., P.Eng. ENGI 1313 Statics I – Lecture 04 Parallelogram Law (Lecture 02) Multiple Force Vectors F2 F1 8 F F1 F2 F2 © 2007 S. Kenny, Ph.D., P.Eng. FR F1 F2 F3 F1 F2 F3 F1 F2 F3 ENGI 1313 Statics I – Lecture 04 What is the Alternate Approach? Resolve Force Components Algebraic Summation Force Vectors 9 © 2007 S. Kenny, Ph.D., P.Eng. Component Vectors F Rx F Rx F x F Ry F Ry F y Recall Lecture 02 (Slide 9) ENGI 1313 Statics I – Lecture 04 What is the Alternate Approach? Resolve Force Components Algebraic Summation Form Resultant Force Force Vectors 10 © 2007 S. Kenny, Ph.D., P.Eng. Component Vectors Resultant Force Vector ENGI 1313 Statics I – Lecture 04 Coplanar Force Vector Summation How to resolve a system of forces into rectangular components and determine the resultant force? Two Notations Used (1) Scalar Notation • More familiar approach (2) Cartesian Vector Notation • Useful in applications of linear algebra • Advantageous over scalar notation for 3D 11 © 2007 S. Kenny, Ph.D., P.Eng. ENGI 1313 Statics I – Lecture 04 Characteristics Rectangular coordinate system Unique spatial position Vector algebra Analytical geometry 12 © 2007 S. Kenny, Ph.D., P.Eng. Ordinate Cartesian Coordinate System Abscissa ENGI 1313 Statics I – Lecture 04 Rectangular Force Components Axes Must be Orthogonal Axes Orientation Does not Matter F = Fx + Fy 13 © 2007 S. Kenny, Ph.D., P.Eng. ENGI 1313 Statics I – Lecture 04 Resolve Force Components Known: Force Vector and Orientation Angle F Fx Fy F x F x F cos 14 © 2007 S. Kenny, Ph.D., P.Eng. F y F y F sin ENGI 1313 Statics I – Lecture 04 Resolve Force Components Known: Force Vector and Slope F Fx Fy L F x F x F x Lh Lh Ly Lx 15 © 2007 S. Kenny, Ph.D., P.Eng. Ly F y F y F Lh ENGI 1313 Statics I – Lecture 04 Determine Resultant Force Known: Force Components Resultant Force Magnitude Pythagorean theorem F 2 2 Fx Fy F Resultant Force Direction Trigonometry tan 16 y 1 Fy Fy Fx Fx © 2007 S. Kenny, Ph.D., P.Eng. x ENGI 1313 Statics I – Lecture 04 Notation – Summation Coplanar Forces Scalar Notation FR FX + FY Common 1. Magnitude: FX & FY 2. Sense: + & - 3. Direction: Orthogonal X & Y axes FR Cartesian Vector Notation ^ +Y +j ^ FY F = FX i + FY j -X -i 3. Direction: Unit vectors 17 ^ Unit Vector; i = FX ^ Unit Vector; j = FY FX FY © 2007 S. Kenny, Ph.D., P.Eng. +X +i FX ENGI 1313 Statics I – Lecture 04 -Y -j Unit Vector Lecture 3 Scalar • Magnitude and sense (+,-) Vector • Magnitude, sense (+,-) and direction Unit Vector A 4 units Vector • Magnitude 4 units Positive X-axis +x • Sense • Direction 18 A uˆ A © 2007 S. Kenny, Ph.D., P.Eng. A A ENGI 1313 Statics I – Lecture 04 Coplanar Force Vector Summation Step 1: Define System of Forces Rectangular coordinate system Force vectors F1, F2 and F3 Force Vectors 19 © 2007 S. Kenny, Ph.D., P.Eng. ENGI 1313 Statics I – Lecture 04 Coplanar Force Vector Summation Step 2: Resolve Component Forces Fnx Fnx Fn cos Fn Fnx Fny Fny Fny Fn sin Force Vectors 20 Component Vectors © 2007 S. Kenny, Ph.D., P.Eng. ENGI 1313 Statics I – Lecture 04 Coplanar Force Vector Summation Step 3: Sum System Force Components Obtain resultant force vector components FRx FRx N F x FRy FRy n 1 Force Vectors 21 F y n 1 Component Vectors © 2007 S. Kenny, Ph.D., P.Eng. N Resultant Force Vector ENGI 1313 Statics I – Lecture 04 Coplanar Force Vector Summation Step 3: Sum System Force Components Scalar notation N F x FRx F1 x F2 x F3 x F y F Ry F1 y F 2 y F 3 y n 1 Force Vectors 22 Component Vectors © 2007 S. Kenny, Ph.D., P.Eng. Resultant Force Vector ENGI 1313 Statics I – Lecture 04 Coplanar Force Vector Summation Step 4: Determine Resultant Force Vector Magnitude, sense and direction F 2 2 Fx Fy Force Vectors 23 tan Component Vectors © 2007 S. Kenny, Ph.D., P.Eng. 1 Fy Fx Resultant Force Vector ENGI 1313 Statics I – Lecture 04 Coplanar Force Vector Summation Step 3: Sum System Force Components Cartesian vector notation F R F1 F 2 F 3 FR F1 x i F1 y j F2 x i F2 y j F3 x i F3 y j FR F1 x F2 x F3 x i F1 y F2 y F3 y j Unit Vector; ^i = FX FX Force Vectors 24 Component Vectors © 2007 S. Kenny, Ph.D., P.Eng. Resultant Force Vector ENGI 1313 Statics I – Lecture 04 Comprehension Quiz 4-01 y Resolve F along x and y axes in Cartesian F ĵ = -80 cos 30 vector notation. F = { ___________ } N y x 30° F = 80 N Fx = 80 sin30 A) 80 cos 30° i - 80 sin 30° j B) 80 sin 30° i + 80 cos 30° j C) 80 sin 30° i - 80 cos 30° j D) 80 cos 30° i + 80 sin 30° j C) 25 F 80 sin 30 i 80 cos 30 j N © 2007 S. Kenny, Ph.D., P.Eng. ENGI 1313 Statics I – Lecture 04 Comprehension Quiz 4-02 Determine the magnitude of the resultant force when F j R F1 F1 F1 = { 10 î + 20 ĵ } N F2 = { 20 î + 20 ĵ } N F2 i A) 30 N B) 40 N C) 50 N D) 60 N E) 70 N F F 26 10 20 i 20 20 j N 30 i 40 j N 30 N 40 N 50 N 2 2 © 2007 S. Kenny, Ph.D., P.Eng. C) 50 N ENGI 1313 Statics I – Lecture 04 Example Problem 4-01 Find the magnitude and angle of the resultant force acting on the bracket. Solution Plan 27 Step 1: Define system of forces Step 2: Resolve component forces Step 3: Sum system force components Step 4: Determine resultant force vector, magnitude and direction © 2007 S. Kenny, Ph.D., P.Eng. ENGI 1313 Statics I – Lecture 04 Example Problem 4-01 (cont.) Step 2: Resolve Components F1 Cartesian vector form, F1 F1x = 15kN sin 40 15 sin 40 i 15 cos 40 j kN F1y = 15kN cos 40 F1x F1y 28 © 2007 S. Kenny, Ph.D., P.Eng. ENGI 1313 Statics I – Lecture 04 Example Problem 4-01 (cont.) Step 2: Resolve Components F2 Cartesian vector form, F2 26 12 5 i 26 13 13 F2x = -26kN (12/13) F2y = 26kN (5/13) j kN F2y F2x 29 © 2007 S. Kenny, Ph.D., P.Eng. ENGI 1313 Statics I – Lecture 04 Example Problem 4-01 (cont.) Step 2: Resolve Components F3 Cartesian vector form, F3 F3x = 36kN cos 30 36 cos 30 i 36 sin 30 j kN F3y = 36kN sin 30 F3x F3y 30 © 2007 S. Kenny, Ph.D., P.Eng. ENGI 1313 Statics I – Lecture 04 Example Problem 4-01 (cont.) Step 2: Resolve Components F1 F2 F3 F1 F2 F3 31 Cartesian vector form 15 sin 40 i 15 cos 40 j kN 12 5 i 26 j kN 26 13 13 36 cos 30 i 36 sin 30 j kN Therefore 9 . 642 i 11 . 491 j kN 24 i 10 j kN 31 . 18 i 18 j kN © 2007 S. Kenny, Ph.D., P.Eng. F1x F1y F2y F3x F2x ENGI 1313 Statics I – Lecture 04 F3y Example Problem 4-01 (cont.) Step 3: Sum Collinear Forces Collinear Cartesian vector form F1 F2 F3 FR 32 F1y 9 . 642 i 11 . 491 j kN 24 i 10 j kN 31 . 18 i 18 j kN F2y F1x F3x F2x F3y 9 . 642 24 31 . 18 i 11 . 49 10 18 j kN © 2007 S. Kenny, Ph.D., P.Eng. ENGI 1313 Statics I – Lecture 04 Example Problem 4-01 (cont.) Step 3: Sum Collinear Forces Resultant components Cartesian vector form F1 F2 F3 9 . 642 i 11 . 491 j kN 24 i 10 j kN 31 . 18 i 18 j kN FR FRy 9 . 642 24 31 . 18 i 11 . 49 10 18 j kN FR 16 . 82 i 3 . 49 j kN 33 FRx © 2007 S. Kenny, Ph.D., P.Eng. ENGI 1313 Statics I – Lecture 04 Example Problem 4-01 (cont.) Step 4: Determine Resultant Force Vector F 16 . 82 tan 1 kN 2 3 . 49 N 17 . 2 kN 3 . 49 kN 2 11 . 7 16 . 82 kN ccw x axis FR = 11.7 FRy FRx 34 © 2007 S. Kenny, Ph.D., P.Eng. ENGI 1313 Statics I – Lecture 04 Group Problem 4-01 Find the magnitude and angle of the resultant force acting on the bracket. Solution Plan 35 Step 1: Define system of forces Step 2: Resolve component forces Step 3: Sum system force components Step 4: Determine resultant force vector, magnitude and direction © 2007 S. Kenny, Ph.D., P.Eng. ENGI 1313 Statics I – Lecture 04 Group Problem 4-01 (cont.) Step 2: Resolve Force Components F1x F1 36 F1y 4 3 i 850 j N 850 5 5 680 i 510 j N © 2007 S. Kenny, Ph.D., P.Eng. ENGI 1313 Statics I – Lecture 04 Group Problem 4-01 (cont.) Step 2: Resolve Force Components F1 F2 37 4 3 i 850 j N 850 5 5 680 i 510 j N F2y F2x 625 sin 30 i 625 cos 30 312 . 5 i 541 . 3 j N © 2007 S. Kenny, Ph.D., P.Eng. j N ENGI 1313 Statics I – Lecture 04 Group Problem 4-01 (cont.) Step 2: Resolve Force Components F1 F2 38 F3y 4 3 i 850 j N 850 5 5 680 i 510 j N 625 sin 30 i 625 cos 30 312 . 5 i 541 . 3 j N 750 cos 45 i 750 sin 45 530 . 3 i 530 . 3 j N F3 F3x j N j N © 2007 S. Kenny, Ph.D., P.Eng. ENGI 1313 Statics I – Lecture 04 Group Problem 4-01 (cont.) Step 3: Sum Collinear Forces F1 F2 F3 FRi 4 3 i 850 j N 850 5 5 680 i 510 j N 625 sin 30 i 625 cos 30 j N 312 . 5 i 541 . 3 j N 750 cos 45 i 750 sin 45 j N 530 . 3 i 530 . 3 j N FRj FR FR 680 312 . 5 530 . 3 i 510 541 . 3 530 . 3 j N FR 39 162 . 8 i 521 j N © 2007 S. Kenny, Ph.D., P.Eng. ENGI 1313 Statics I – Lecture 04 Group Problem 4-01 (cont.) Step 4: Determine Resultant Force Vector FR 162 . 8 i 521 j N F 162 . 8 N 2 tan 40 1 521 N 546 N 521 N FR 2 72 . 6 ( local ) 253 ( ccw x axis ) 162 . 8 N © 2007 S. Kenny, Ph.D., P.Eng. ENGI 1313 Statics I – Lecture 04 Classification of Textbook Problems 41 Hibbeler (2007) Problem Set Concept Degree of Difficulty Estimated Time 2-31 to 2-32 Vector Addition Parallelogram Law Medium 10-15min 2-33 to 2-38 Vector Addition Parallelogram Law Easy 5-10min 2-39 to 2-41 Resultant Force Easy 5-10min 2-42 to 2-55 Resultant & Components Medium 10-15min 2-56 Resultant & Components Hard 20min © 2007 S. Kenny, Ph.D., P.Eng. ENGI 1313 Statics I – Lecture 04 References Hibbeler (2007) http://wps.prenhall.com/esm_hibbeler_eng mech_1 en.wikipedia.org 42 © 2007 S. Kenny, Ph.D., P.Eng. ENGI 1313 Statics I – Lecture 04