bruce mayer, pe licensed electrical & mechanical engineer [email protected]

[email protected] • ENGR-36_Lec-03_Vector_Math.ppt1

Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Engineering 36

Chp 2: Vector

Mathematics



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

kjiu zyxˆcosˆcosˆcosFˆFF

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

F

B

A

BBB zyx ,,

AAA zyx ,, Define DIRECTION

Vector AB kzz

jyy

ixxAB

AB

AB

AB

ˆ

ˆ Or using Δ Notation

zzjyixAB ABABAB ˆˆˆ



Unit Vector, u, from Geometry Next Calculate

the Geometric Length, AB, of The Direction Vector AB using Pythagorus

F

B

A

BBB zyx ,,

AAA zyx ,,

“Normalizing” AB to its Length will produce a vector of unit Length; i.e., u

222 ABABAB

AB

zyx

LABAB

Now AB has the same Direction as u, but a different length



Unit Vector, u, from Geometry Normalize AB to

Produce û F

B

A

BBB zyx ,,

AAA zyx ,,222

ABABAB

AB

zyxkzjyix

ABAB

LABu

ˆˆˆ

ˆ

Note That this Calc also yields the Direction CoSines ABz

ABy

ABx

LzLyLx

coscoscos



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

ACF



Example: û from AC Locate CoOrd

Triad at Pt-A Calculate the Δ’s

z

y

x ftzftyftx

AC

AC

AC

11278

16

Then AC

ftkjiAC ˆ16ˆ8ˆ16

Calc Distance AC

ftft

ftAC

24576

168162

2222



Example: û from AC Calc û from

AC & AC

z

y

x Finally û

ft

ftkji

ACACu

24

ˆ16ˆ8ˆ16

ˆ

Also the Space ’s

kjiu ˆˆˆˆ32

31

32

8.13132arccos53.7031arccos

8.13132arccos

z

y

x



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



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

zzjyixAB ABABAB ˆˆˆ

ABABLAB ABAB u

ABFuF

CoOrds B-Pt &A -Pt Find

z

y

x



Vector Mathematics VECTOR Parameter Possessing

Magnitude And Direction, Which Add AccordingTo 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

B

C

C

QPQP

PQQP • Vector Addition is

Commutative• Vector Subtraction →

Reverse Direction ofThe Subtrahend



Vector Addition cont. Addition Of Three Or More

Vectors Through RepeatedApplication Of The Triangle Rule

The Polygon Rule ForThe Addition Of ThreeOr More Vectors• Vector Addition Is Associative SQP

Multiplication by a Scalar• Scales the Vector LENGTH

SQPSQP



Recall Vector/Force Components

Using Rt-Angle Parallelogram to Resolve Force Into Perpendicular Componentsyx FFF

yxyx FF jiFFF

Define Perpendicular UNIT Vectors Which Are Parallel To The Axes

Vectors May then Be Expressed as Products Of The Unit VectorsWith The SCALAR MAGNITUDESOf The Vector Components



Adding by Components Find: Sum, R, of 3+ Vectors

SQPR

yyyxxx

yxyxyxyx

SQPSQP

SSQQPPRR

ji

jijijiji

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

yyyyy

xxxxx

VSQPR

VQPPR

x

yyx R

RRRR 122 tan

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 & Direction1004 V

1501 V802 V

1103 V Do U

sing

MAT

LAB

Com

man

d W

indo

w



MATLA

B 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 + V4VR = 199.1348 14.2935 >> [alpha,Rm] = cart2pol(VR(1),VR(2))alpha = 0.0717Rm = 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



Example: Component Add Resolve V’s Into Rectangular

ComponentsVector No. Mag x-Comp y-Comp)1 150 129.9 75.02 80 -27.4 75.23 110 0.0 -110.04 100 96.6 -25.9

= 199.1 14.3

j.i.jiR 3141199 yx RR

The Resulting Vector Sum in Component form



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

Magnitude & Direction Form

NRNR 6.199cos

1.1991.199cos

1.41.199

3.14tan

4.16.199 NR



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 WhoseLoA’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

SQPR

FQP

VECTOR FORCE COMPONENTS Two, or More, Force VectorsWhich, Together, HaveThe Same Effect As An Original, Single Force Vector



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 Vector3D 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 Constant Speed In A Straight Line

Particle acted on by 2 Forces• Equal Magnitude• Same Line of Action• Opposite Direction 00 yx FFFR

Particle acted upon 3+ Forces• Graphical Soln → Closed Polygon• Algebraic Solution



Example: Resultant

a) the tension in each of the ropes for = 45o,

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



Example: Cable Tension Graphical Solution

• Parallelogram Rule With Known Resultant Direction & Magnitude, Known DIRECTIONS For Sides– By Scaling With a Ruler (cf. Engr-22)

lbf2600lbf3700 21 TT

105sinlbf5000

30sin45sin21 TT

lbf2590lbf3660 21 TT

• Analytical Solution– Triangle Rule with

Law of Sines



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 OccursWhen T1 & T2 Are Perpendicular 30sinlbf50002T lbf25002 T

30coslbf50001T lbf43301 T

3090 60

30º

LoA for T 1

• Analysis Simplified by Rt-Triangle



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.



Hull Drag Example Choosing the hull-eye as the

free body, draw a free-body diagram. Analyze Geometry.

25.60

75.1ft 4ft 7tan

56.20

375.0ft 4ft 1.5tan

0 DAEACAB FTTTR

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

iF

jTji

jiTji

jiT

DD

AE

ACAC

ACACAC

AB

F

TTTT

lb 069363.03512.0

56.20cos56.20sinlb 84.19lb 73.34

26.60coslb 4026.60sinlb 40

j

ijiR

609363.084.193512.073.34

000

AC

DAC

TFT



Hull Drag Example

j

iR

609363.084.193512.073.34

0

AC

DAC

TFT

609363.084.1900

3512.073.3400

ACy

DACx

TF

FTF

lb 66.19lb 9.42

D

AC

FT

This equation is satisfied only if EACH COMPONENT of the resultant is equal to ZERO



WhiteBoard Work

Let’s WorkThis NiceProblem

= 200 NW=400 N

Find θ so that the Cart will NOT Roll BackWards (down) or be Pushed-Up the

15%-Grade Ramp



Bruce Mayer, PERegistered Electrical & Mechanical Engineer

[email protected]

Engineering 36

Appendix



WhiteBoard Work

Let’s WorkThis NiceProblem

1F

2F

kN 3 BAAB TT

Show that F2 = −F1



MATLA

B C

ode% Bruce Mayer, PE% ENGR36 * 15Jul12% file = H13e_P2_105.m%clear; clc; % Clear Out Memory & Screen%% CoOrds for Pts A & B by Geometry & TrigA = [5, 3*cosd(20), -3*sind(20)]; % ftB = [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 VectorMagV = @(V) sqrt(dot(V,V))disp(' ')% Find Magnitude of vector AVbABm = MagV(ABv) % in ft% find uAB as ABv/ABmuAB = ABv/ABm;Fm = 12; % the Magnitude of F in lbs% find the Answer by Fm*uABFv = Fm*uAB;% Extra Credit: Find the Space Angles for Fv%* AKA the CoOrd Direction AnglesQ = acosd(uAB);%% Display ResultsFvDisp = 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 Fdisp(' ')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)

bruce mayer, pe licensed electrical & mechanical engineer [email protected]

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