Chp 2: Vector Mathematics Engineering 36 Chp 2: Vector Mathematics Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu

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 Cosθm is called the DIRECTION COSINE in m-Direction Use Geometry to find u and hence the Direction CoSines

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 mAB => From-A to-B Or using Δ Notation

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

Unit Vector, u, from Geometry Normalize AB to Produce û Note That this Calc also yields the Direction CoSines

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

Example: û from AC Locate CoOrd Triad at Pt-A Calculate the Δ’s Then AC Calc Distance AC

Example: û from AC Calc û from AC & AC Finally û Also the Space ’s

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 file = H13e_P2_105.m

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 use “norm”) ReCall Space Angle notation Equivalency See file: H13e_P2_105.m

MATLAB Code % Bruce Mayer, PE % ENGR36 * 15Jul12 * 24Aug16 % H13e_P2_105_norm_1608.m % clear; close; clc; % Clear Out Memory, Screen, and Fig Window % 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 % Find the Length (magnitude) of ABv using the norm command ABm = norm(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('check answer by finding the magnitude of Fv using norm command fcn') FmChk = norm(Fv); FmChkDisp = sprintf('FmChk = %f', FmChk); disp(FmChkDisp) MATLAB Code

Example: cartesian F by MATLAB ABv = -5.0000 -2.8191 7.0261 ABm = 9.0726 F components in lbs: Fx = -6.613293 Fy = -3.728678 Fz = 9.293079 SpaceAngles in °: Qx = 123.443042 Qy = 108.102815 Qz = 39.247227 check answer by finding the magnitude of Fv using norm command fcn FmChk = 12.000000

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

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

Vector Addition Parallelogram Rule For Vector Addition Examine Top & Bottom of The Parallelogram TriAngle Rule For Vector Addition B C Vector Addition is Commutative Vector Subtraction → Reverse Direction of The Subtrahend

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

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

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

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

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 Do Using MATLAB Command Window Use cart2pol to obtain polar (r, θ) form

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 = >> a = (180/pi)*atan2(VR(2),VR(1)) a = MATLAB Session

Example: Component Add Resolve V’s Into Rectangular Components Vector No. Mag x-Comp y-Comp) 1 150 129.9 75.0 2 80 -27.4 75.2 3 110 0.0 -110.0 4 100 96.6 -25.9  = 199.1 14.3 The Resulting Vector Sum in Component form

Example: R by Mag & Dir As Use Geometry & Trig to Find R in Magnitude & Direction Form

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

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 The Resultant would be applied at the point of concurrency

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 3D Vector 2D Vector

Particle Equilibrium When The Resultant Of All Forces Acting On A Particle Is Zero, The Particle Is In Equilibrium: Recall Newton’s 1st Law: If The Resultant Force On A Particle Is Zero, The Particle Will Remain At Rest Or Will Continue At a Constant Speed In a Straight Line If ALL forces have a point of concurrency then then the entire body can be “concentrated” into a “particle” located at the PoC 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

Fall 2016 Only 30Aug16 * Fa16 Start at Slide 29 in Lec-01 ENGR-36_Lec-03_Fa16_Vector_Math_H13e.pptx Fall 2016 Only

UC Berkeley Model Ship Tow-Tank Richmond Field Station http://fml.berkeley.edu/facilities/towtank.htm

UC Berkeley Model Ship Tow-tank Large Carriage http://fml.berkeley.edu/facilities/towtank.htm

Typical Model-Ship Tow Test http://www.shallowwater.ugent.be/EN/kc_fac_st_EN.htm

Example: Resultant 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. R = 5 kip 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 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. the tension in each of the ropes for a = 45o, the value of a for which the tension in rope 2 is a minimum. The angle for minimum tension in rope 2 is determined by applying the Triangle Rule and observing the effect of variations in a.

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 Scale factor could be 1000lb per Inch • Scale Parallelogram on white bd using meter stick • Draw TriAngle using Protractor

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 T1 & T2 Are Perpendicular 30º LoA for T1 Analysis Simplified by Rt-Triangle Shortest T2

Check Analytically Thre Value of the Minimum Force, 𝑇 2 Use Law of Sines 𝑇 2 sin 30° = 5 𝑘𝑖𝑝 sin 𝛼 And sin 30°= 1 2 so 𝑇 2 1 2 = 5 𝑘𝑖𝑝 sin 𝛼 ;or 𝑇 2 = 5 𝑘𝑖𝑝 2 ∙ 1 sin 𝛼 At T2min: 𝑑 𝑇 2 𝑑𝛼 =0; and 𝑑 𝑇 2 𝑑𝛼 = 5 𝑘𝑖𝑝 2 ∙ 𝑑 𝑑𝛼 1 sin 𝛼 30° T2 𝛼

Check Analytically Find 𝑑 𝑑𝛼 1 sin 𝛼 by MuPad 30° Thus if 𝑑 𝑇 2 𝑑𝛼 =0 need cos 𝛼 =0 cos 𝛼 =0 at ± 90° In this case 𝛼 must be Positive T2 𝛼

Check Analytically Thre Value of the Minimum Force, 𝑇 2 By Calculus: 𝛼=90° Recall: 𝑇 2 sin 30° = 5 𝑘𝑖𝑝 sin 𝛼 Using sin 30°= 1 2 and 𝛼= 90° → 𝑇 2 = 5 𝑘𝑖𝑝 2 ∙ 1 sin 90° = 5 𝑘𝑖𝑝 2 ∙ 1 1 Thus Confirmation of Geometric Answer 𝑇 2 = 5 𝑘𝑖𝑝 sin 30°=2.5 𝑘𝑖𝑝 𝑇 3 = 5 𝑘𝑖𝑝 cos 30° =4.33 𝑘𝑖𝑝 30° T2 90° T3

Analyitical Result Checks  Thus the Trig & Calculus based answer 60° 30° 2500 lb 4330 lb 90°

Hull Drag Example 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. 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. Note that I did a naval architecture lab that required TOWING of a ship hull at UCB Richmond Field station

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

Hull Drag Example Resolve the vector equilibrium equation into X-Y component equations. Solve for the two unknown cable tensions

Hull Drag Example This equation is satisfied only if EACH COMPONENT of the resultant is equal to ZERO Notice y-Eqn has only one ONE Variable, TAC. → Solve first

Let’s Work This Nice Problem WhiteBoard Work = 200 N W=400 N Let’s Work This Nice Problem Find θ so that the Cart will NOT Roll BackWards (down) or be Pushed-Up the 15%-Grade Ramp ST 4.5.54 • ENGR-36_Tutorial_Lab-03_Cart-Roll_Fa16.pptx

Registered Electrical & Mechanical Engineer BMayer@ChabotCollege.edu Engineering 36 Appendix Bruce Mayer, PE Registered Electrical & Mechanical Engineer BMayer@ChabotCollege.edu

Let’s Work This Nice Problem WhiteBoard Work Let’s Work This Nice Problem ST 4.5.60 • ENGR-36_Tutorial_Lab-03_Cart-Roll_Fa16.pptx Show that F2 = −F1

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) 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('check answer by finding the Mag of Fv using MagV fcn') FmChk = MagV(Fv); FmChkDisp = sprintf('FmChk = %f', FmChk); disp(FmChkDisp) disp('Find Vector Mag using built-in MATLAB command NORM') ABmByNORM = norm(Abv) MATLAB Code