1. BMayer@ChabotCollege.edu ENGR-36_Lec-03_Vector_Math.ppt 1 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu Engineering 36 Chp 2: Vector Mathematics

2. BMayer@ChabotCollege.edu ENGR-36_Lec-03_Vector_Math.ppt 2 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Unit Vector, u, from Geometry  For some structural elements such as cables and slender rods the LoA for the Force associated with the member is CoIncident with the element Geometry  Recall The Unit Vector Force DeComp Formulation  Use Geometry to find u and hence the Direction CoSines

3. BMayer@ChabotCollege.edu ENGR-36_Lec-03_Vector_Math.ppt 3 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Unit Vector, u, from Geometry  Consider a Force Vector F, with LoA running from Pt-A to Pt-B with CoOrds as shown  Define DIRECTION Vector AB  Or using Δ Notation

4. BMayer@ChabotCollege.edu ENGR-36_Lec-03_Vector_Math.ppt 4 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Unit Vector, u, from Geometry  Next Calculate the Geometric Length, AB, of The Direction Vector AB using Pythagorus  “Normalizing” AB to its Length will produce a vector of unit Length; i.e., u  Now AB has the same Direction as u, but a different length

5. BMayer@ChabotCollege.edu ENGR-36_Lec-03_Vector_Math.ppt 5 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Unit Vector, u, from Geometry  Normalize AB to Produce û  Note That this Calc also yields the Direction CoSines

6. BMayer@ChabotCollege.edu ENGR-36_Lec-03_Vector_Math.ppt 6 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Example: û from AC  During Construction a building wall is temporarily held in place by ropes staked to Ground as shown in the Space Diagram at right  Find the Unit Vector for the Force exerted on the stake by Rope AC  Solution Plan Define CoOrd System Calc û as AC/AC

7. BMayer@ChabotCollege.edu ENGR-36_Lec-03_Vector_Math.ppt 7 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Example: û from AC  Locate CoOrd Triad at Pt-A  Calculate the Δ’s  Then AC  Calc Distance AC

8. BMayer@ChabotCollege.edu ENGR-36_Lec-03_Vector_Math.ppt 8 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Example: û from AC  Calc û from AC & AC  Finally û  Also the Space  ’s

9. BMayer@ChabotCollege.edu ENGR-36_Lec-03_Vector_Math.ppt 9 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Example: cartesian F by MATLAB  The Cord AB exerts a force, F, of 12 lbs on the Pipe  Express the Force as a vector, F, in Cartesian form using MATLAB  By Physics: since this a cord/cable the Force LoA is CoIncident with the Cord/Cable Geometry

10. BMayer@ChabotCollege.edu ENGR-36_Lec-03_Vector_Math.ppt 10 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Example: cartesian F by MATLAB  Math used  Use MATLAB to perform these Operations file H13e_P2_105.m  Note the use of the DOT product in the construction of the Anon Fcn that Calcs the Magnitude of any Vector (or us “norm”)  Space Angle notation Equivalency

11. BMayer@ChabotCollege.edu ENGR-36_Lec-03_Vector_Math.ppt 11 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Vector Mathematics  VECTOR  Parameter Possessing Magnitude And Direction, Which Add According To The Parallelogram Law Examples: Displacements, Velocities, Accelerations, Forces  SCALAR  Parameter Possessing Magnitude But Not Direction Examples: Mass, Volume, Temperature  Vector Classifications FIXED Or BOUND Vectors Have Well Defined Points Of Application That Cannot Be Changed Without Affecting An Analysis

12. BMayer@ChabotCollege.edu ENGR-36_Lec-03_Vector_Math.ppt 12 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Vectors cont. FREE Vectors May Be Moved In Space Without Changing Their Effect On An Analysis SLIDING Vectors May Be Applied Anywhere Along Their Line Of Action Without Affecting the Analysis EQUAL Vectors Have The Same Magnitude And Direction NEGATIVE Vector Of a Given Vector Has The Same Magnitude but The Opposite Direction Equal Vectors Negative Vectors

13. BMayer@ChabotCollege.edu ENGR-36_Lec-03_Vector_Math.ppt 13 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Vector Addition  Parallelogram Rule For Vector Addition  Examine Top & Bottom of The Parallelogram Triangle Rule For Vector Addition B B C C Vector Addition is Commutative Vector Subtraction → Reverse Direction of The Subtrahend

14. BMayer@ChabotCollege.edu ENGR-36_Lec-03_Vector_Math.ppt 14 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Vector Addition cont.  Addition Of Three Or More Vectors Through Repeated Application Of The Triangle Rule  The Polygon Rule For The Addition Of Three Or More Vectors Vector Addition Is Associative  Multiplication by a Scalar Scales the Vector LENGTH

15. BMayer@ChabotCollege.edu ENGR-36_Lec-03_Vector_Math.ppt 15 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Recall Vector/Force Components  Using Rt-Angle Parallelogram to Resolve Force Into Perpendicular Components  Define Perpendicular UNIT Vectors Which Are Parallel To The Axes  Vectors May then Be Expressed as Products Of The Unit Vectors With The SCALAR MAGNITUDES Of The Vector Components

16. BMayer@ChabotCollege.edu ENGR-36_Lec-03_Vector_Math.ppt 16 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Adding by  Components  Find: Sum, R, of 3+ Vectors  Plan: Resolve Each Vector Into Components Add LIKE Components To Determine Resultant Use Trig to Fully Describe Resultant

17. BMayer@ChabotCollege.edu ENGR-36_Lec-03_Vector_Math.ppt 17 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Adding by Comp.  cont.  The Scalar Components Of The Resultant R Are Equal To The Sum Of The Corresponding Scalar Components Of The Given Forces  Use The Scalar Components Of The Result to Find By Trig the Magnitude & Direction for R

18. BMayer@ChabotCollege.edu ENGR-36_Lec-03_Vector_Math.ppt 18 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Example: Component Addition  Find the Sum for the Vectors with Magnitudes and Directions as Shown  Solution Plan Resolve The 4 Vectors Into Rectangular Components Determine the Sum Components By Adding the Corresponding Vector Components Calculate Sum Result, R, Magnitude & Direction

19. BMayer@ChabotCollege.edu ENGR-36_Lec-03_Vector_Math.ppt 19 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics MATLAB Session >> V1 = 150*[cosd(30) sind(30)] V1 = 129.9038 75.0000 >> V2 = 80*[cosd(110) sind(110)] V2 = -27.3616 75.1754 >> V3 = 110*[cosd(270) sind(270)] V3 = 0 -110 >> V4 = 100*[cosd(360-15) sind(360-15)] V4 = 96.5926 -25.8819 >> VR = V1 + V2 + V3 + V4 VR = 199.1348 14.2935 >> [alpha,Rm] = cart2pol(VR(1),VR(2)) alpha = 0.0717 Rm = 199.6471 >> alphaDeg = 180*alpha/pi alphaDeg = 4.1055 >> Rmag = norm(VR) Rmag = 199.6471 >> a = (180/pi)*atan2(VR(2),VR(1)) a = 4.1055

20. BMayer@ChabotCollege.edu ENGR-36_Lec-03_Vector_Math.ppt 20 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Example: Component Add  Resolve V’s Into Rectangular Components Vector No.Magx-Compy-Comp) 1150129.975.0 280-27.475.2 31100.0-110.0 410096.6-25.9  = 199.114.3  The Resulting Vector Sum in Component form

21. BMayer@ChabotCollege.edu ENGR-36_Lec-03_Vector_Math.ppt 21 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Example: R by Mag & Dir  As Use Geometry & Trig to Find R in Magnitude & Direction Form

22. BMayer@ChabotCollege.edu ENGR-36_Lec-03_Vector_Math.ppt 22 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Resultant of Two Forces  Force: Action Of One Body On Another; Characterized By Its Point Of Application Magnitude (Intensity) Direction  Experimental Evidence Shows That The Combined Effect Of Multiple Forces May Be Represented By A Single Resultant Force

23. BMayer@ChabotCollege.edu ENGR-36_Lec-03_Vector_Math.ppt 23 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Concurrent Force Resultant  CONCURRENT FORCES  Set Of Forces Whose LoA’s All Pass Through The Same Point A Set Of Concurrent Forces Applied To A body May Be Replaced By a Single Resultant Force Which Is The Vector Sum Of The Applied Forces  VECTOR FORCE COMPONENTS  Two, or More, Force Vectors Which, Together, Have The Same Effect As An Original, Single Force Vector

24. BMayer@ChabotCollege.edu ENGR-36_Lec-03_Vector_Math.ppt 24 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Resultant cont.  The Resultant Is Equivalent To The Diagonal Of A Parallelogram Which Contains The Two Forces In Adjacent Legs As Forces are VECTOR Quantities 2D Vector 3D Vector

25. BMayer@ChabotCollege.edu ENGR-36_Lec-03_Vector_Math.ppt 25 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Particle Equilibrium  When The Resultant Of All Forces Acting On A Particle Is Zero, The Particle Is In Equilibrium: Recall Newton’s 1 st Law: If The Resultant Force On A Particle Is Zero, The Particle Will Remain At Rest Or Will Continue At Constant Speed In A Straight Line  Particle acted on by 2 Forces Equal Magnitude Same Line of Action Opposite Direction  Particle acted upon 3+ Forces Graphical Soln → Closed Polygon Algebraic Solution

26. BMayer@ChabotCollege.edu ENGR-36_Lec-03_Vector_Math.ppt 26 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Example: Resultant a)the tension in each of the ropes for  = 45 o, b)the value of  for which the tension in rope 2 is a minimum. A barge is pulled by two tugboats. If the resultant of the forces exerted by the tugboats is 5000 lbf directed along the axis of the barge, then determine SOLUTION: Find a graphical solution by applying the Parallelogram Rule for vector addition. The parallelogram has sides in the directions of the two ropes and a diagonal in the direction of the barge axis and length proportional to 5000 lbf. The angle for minimum tension in rope 2 is determined by applying the Triangle Rule and observing the effect of variations in . Find a trigonometric solution by applying the Triangle Rule for vector addition. With the magnitude and direction of the resultant known and the directions of the other two sides parallel to the ropes, apply the Law of Sines to find the rope tensions. R = 5 kip

27. BMayer@ChabotCollege.edu ENGR-36_Lec-03_Vector_Math.ppt 27 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Example: Cable Tension  Graphical Solution Parallelogram Rule With Known Resultant Direction & Magnitude, Known DIRECTIONS For Sides –By Scaling With a Ruler (cf. Engr-22) Analytical Solution –Triangle Rule with Law of Sines

28. BMayer@ChabotCollege.edu ENGR-36_Lec-03_Vector_Math.ppt 28 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Example: Min Tension  Minimum Tension In Rope 2 Is Determined By Applying The Triangle Rule And Observing The Effect Of Variations in α. By Geometric Construction Minimum Rope-2 Tension 2 Occurs When T 1 & T 2 Are Perpendicular 30º LoA for T 1 Analysis Simplified by Rt-Triangle

29. BMayer@ChabotCollege.edu ENGR-36_Lec-03_Vector_Math.ppt 29 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Hull Drag Example  A model Sailboat Hull undergoes a drag test. Three cables align its bow on the channel centerline. For a given speed, the tension is 40 lbs in cable AB and 60 lbs in cable AE Find the drag force and the tension in cable AC.  SOLUTION PLAN Choosing the hull-eye (the point of concurrency) as the free body, draw a free-body diagram. Express the condition for equilibrium for the hull by writing that the sum of all forces must be zero Resolve the vector equilibrium equation into two component equations. Solve for the two unknown cable tensions.

30. BMayer@ChabotCollege.edu ENGR-36_Lec-03_Vector_Math.ppt 30 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Hull Drag Example  Choosing the hull-eye as the free body, draw a free-body diagram. Analyze Geometry.  Express the condition for equilibrium for the hull by noting that the sum of all forces must be zero

31. BMayer@ChabotCollege.edu ENGR-36_Lec-03_Vector_Math.ppt 31 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Hull Drag Example  Resolve the vector equilibrium equation into X-Y component equations. Solve for the two unknown cable tensions

32. BMayer@ChabotCollege.edu ENGR-36_Lec-03_Vector_Math.ppt 32 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Hull Drag Example  This equation is satisfied only if EACH COMPONENT of the resultant is equal to ZERO

33. BMayer@ChabotCollege.edu ENGR-36_Lec-03_Vector_Math.ppt 33 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics WhiteBoard Work Let’s Work This Nice Problem = 200 N W=400 N  Find θ so that the Cart will NOT Roll BackWards (down) or be Pushed-Up the 15%-Grade Ramp

34. BMayer@ChabotCollege.edu ENGR-36_Lec-03_Vector_Math.ppt 34 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics

35. BMayer@ChabotCollege.edu ENGR-36_Lec-03_Vector_Math.ppt 35 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Bruce Mayer, PE Registered Electrical & Mechanical Engineer BMayer@ChabotCollege.edu Engineering 36 Appendix

36. BMayer@ChabotCollege.edu ENGR-36_Lec-03_Vector_Math.ppt 36 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics WhiteBoard Work Let’s Work This Nice Problem  Show that F 2 = −F 1

37. BMayer@ChabotCollege.edu ENGR-36_Lec-03_Vector_Math.ppt 37 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics

38. BMayer@ChabotCollege.edu ENGR-36_Lec-03_Vector_Math.ppt 38 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics MATLAB Code % Bruce Mayer, PE % ENGR36 * 15Jul12 % file = H13e_P2_105.m % clear; clc; % Clear Out Memory & Screen % % CoOrds for Pts A & B by Geometry & Trig A = [5, 3*cosd(20), -3*sind(20)]; % ft B = [0 0 6]; % ft % find the AB vector = [delX*i, delY*j, delZ*k], ABv = B-A % % Make Anon Fcn to find the Magnitude of any Vector %* use the DOT product to find Mag^2 of the Vector MagV = @(V) sqrt(dot(V,V)) disp(' ')') % Find Magnitude of vector AVb ABm = MagV(ABv) % in ft % find uAB as ABv/ABm uAB = ABv/ABm; Fm = 12; % the Magnitude of F in lbs % find the Answer by Fm*uAB Fv = Fm*uAB; % Extra Credit: Find the Space Angles for Fv %* AKA the CoOrd Direction Angles Q = acosd(uAB); % % Display Results FvDisp = sprintf('F components in lbs: Fx = %f Fy = %f Fz = %f',Fv(1), Fv(2), Fv(3)); disp(FvDisp) disp(' ')') QDisp = sprintf('SpaceAngles in °: Qx = %f Qy = %f Qz = %f',Q(1), Q(2), Q(3)); disp(QDisp) % % do a quick check on the result for F disp(' ')') disp('check answer by finding the Mag of Fv using MagV fcn') FmChk = MagV(Fv); FmChkDisp = sprintf('FmChk = %f', FmChk); disp(FmChkDisp) % disp(' ')') disp('Find Vector Mag using built-in MATLAB command NORM') ABmByNORM = norm(Abv)