Bruce Mayer, PE Licensed Electrical & Mechanical Engineer [email protected]

[email protected] • ENGR-36_Lec-02_Fa12_Forces_as_Vectorspptx1

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

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Engineering 36

Chp 2: ForceDeCompositi

on



Force Defined Force: Action Of One Body On

Another; Characterized By Its • Point Of Application• Magnitude (intensity)• Direction

The DIRECTION of a Force Defines its Line of Action (LoA)

Magnitude

Line of Action

Direction

Point of Application



Newton’s Law of Gravitation Consider two massive bodies

Separated by a distance rM

m

-F

F• Newton’s Gravitation Equation

SCALAR) a (2 Fr

GMmF

– WhereF ≡ mutual force of attraction between 2 bodiesG ≡ universal constant known as the

constant of gravitation (6.673x10−11 m3/kg-s2)M, m ≡ masses of the 2 bodies r ≡ distance between the 2 bodies



Weight Consider An Object of mass, m, at a

modest Height, h, Above the Surface of the Earth, Which has Radius R• Then the Force on the Object (e.g., Yourself)

gm

RGMmF

hRGMmF

22 hRbut

mgW

This Force Exerted by the Earth is called Weight• While g Varies Somewhat With the Elevation &

Location, to a Very Good Approximation– g 9.81 m/s2 32.2 ft/s2



Earth Facts D 7 926 miles (12 756 km) M 5.98 x 1024 kg

• About 2x1015 EmpireState Buildings

Density, 5 520 kg/m3 • water 1 027 kg/m3

• steel 8 000 kg/m3

• glass 5 300 kg/m3



Gravitation Example Jupiter Moon Europa

Europa Statistics Discovered by Simon Marius & Galileo Galilei Date of discovery 1610 Mass (kg) 4.8e+22 Mass (Earth = 1) 8.0321e-03 Equatorial radius (km) 1,569 Equatorial radius (Earth = 1) 2.4600e-01 Mean density (kg/m^3) 3010 Mean distance from Jupiter (km) 670,900 Rotational period (days) 3.551181 Orbital period (days) 3.551181 Mean orbital velocity (km/sec) 13.74 Orbital eccentricity 0.009 Orbital inclination (degrees) 0.470 Escape velocity (km/sec) 2.02 Visual geometric albedo 0.64 Magnitude (Vo) 5.29

• Find Your Weight on Europra



Europa Weight Since your MASS

is SAME on both Earth and Europa need to Find only geu and compare it to gea

Recall2R

GMg Europa Statistics

from table: Meu = 4.8x1022 kg Reu = 1 569 km

Then geu

23

22

2

311

1015691084106736

mkg

skgmgeu

..

22

3

12

2211

1046221084106736

mskgkgmgeu

...

23011 smgeu .

With %Weu = geu/gea

%...% 271380793011Weu



Contact Forces Normal Contact Force

• When two Bodies Come into Contact the Line of Action is Perpendicular to the Contact Surface

Friction Force• a force that resists the

relative motion of objects that are in surface contact– Generation of a Friction

Force REQUIRES the Presence of a Normal force



Contact Forces Fluid Force

• In Fluid Statics the Pressure exerted by the fluid acts NORMAL to the contact Surface

Tension Force• A PULLING force

which tends to STRETCH an object upon application of the force

pa

p = pa + gd

HMS Bounty

d



Contact Forces Compression Force

• A PUSHING force which tends to SMASH an object upon application of the force

Shear Force• a force which acts

across a object in a way that causes one part of the structure to slide over an other when it is applied



Recall Free-Body Diagrams

SPACE DIAGRAM A Sketch Showing The Physical Conditions Of The Problem

FREE-BODY DIAGRAM A Sketch Showing ONLY The Forces On The Selected Body



Concurrent Forces CONCURRENT FORCES ≡ Set Of Forces

Which All Pass Through The Same Point When Forces intersect at ONE point then

NO TWISTING Action is Generated In Equil the Vector Force POLYGON must

CLOSE

Force Polygonif Static

FBD showing forces P, Q, R, S



Vector Notation – Unit Vectors Unit Vectors have,

by definition a Magnitude of 1 (unit Magnitude)

Unit vectors may be • Aligned with the

CoOrd Axes to form a Triad

• Arbitrarily Oriented̂λˆˆˆˆi uukkjji

Unit Vectors may be indicated with “Carets”



Example: FBD & Force-PolygonEYE, Not Pulley

A 3500-lb automobile is supported by a cable. A rope is tied to the cable and pulled to center the automobile over its intended position. What is the tension in the rope?

SOLUTION PLAN:• Construct a free-body

diagram for the rope eye at the junction of the rope and cable.– i.e., Make a FBD for the

connection Ring-EYE• Apply the conditions for

equilibrium by creating a closed polygon from the forces applied to the connecting eye.

• Apply trigonometric relations to determine the unknown force magnitudes



Example Solution Construct A Free-body

Diagram For The Eye At A. Apply The Conditions For

Equilibrium. Solve For The Unknown Force

Magnitudes Using the Law of the Sines.

58sinlb3500

2sin120sinACAB TT

lb3570ABT

lb144ACT A pretty Tough Pull for the Guy at C



Vector Notation – Vector ID In Print and Handwriting We Must

Distinguish Between• VECTORS• SCALARS

These are Equivalent Vector NotationsPPP P

• Boldface Preferred for Math Processors• Over Arrow/Bar Used for Handwriting• Underline Preferred for Word Processor



Vector Notation - Magnitude The Magnitude of a vector is its

Intensity or Strength• Vector Mag is analogous to Scalar

Absolute Value → Mag is always positive– Abs of Scalar x → |x|– Mag of Vector P → ||P|| =

We can indicate a Magnitude of a vector by removing all vector indicators; i.e.:

PP of Mag PPPP



Force Magnitude & Direction Forces can be represented as Vectors

and so Forces can be Defined by the Vector MAGNITUDE & DIRECTION

Given a force F with magnitude, or intensity, ||F|| and direction as defined in 3D Cartesian Space withLoA of Pt1→Pt2



Angle Notation: Space ≡ Direction The Text uses [α,β,γ] to denote the

Space/Direction Angles Another popular Notation set is [θx,θy,θz] We will consider these Triads as

Equivalent Notation: [α,β,γ] ≡ [θx,θy,θz]



Magnitude-Angle Form The Magnitude of the Force is

Proportional to the Geometric Length of its vector representation:

: Lengthan Pythagore theis where LLF

212

212

212 zzyyxxL

Note that if Pt1 is at the ORIGIN and Pt2 has CoOrds (x, y, z) then

222 zyxL



Magnitude-Angle Form Then calculate SPACE

ANGLES as

L

zzL

yyL

xxzyx

121212 arccosarccosarccos

By the 3D Trig ID

1222 zyx coscoscos

• Find Δx, Δ y, Δ z using Direction Cosines

L

zzL

yyL

xxzyx

121212 coscoscos



Magnitude-Angle Form Thus the Vector

Representation of a Force isFully Specified by the LENGTH and SPACE ANGLES

L

zzL

yyL

xxzyx

121212 coscoscos

212

212

212 zzyyxxL

• Note: Can use the Trig ID to find the third θ if the other two are known



Spherical CoOrdinates A point in Space Can Be Specified by

• Cartesian CoOrds → (x, y, z)• Spherical CoOrds → (r, θ, φ)

Relations between θx, θy, θz, θ, φ

coscoscoscos

tansinsincos

coscoscossincos

z

x

yy

zx



Rectangular Force Components

Using Rt-Angle Parallelogram Resolve Force Into Perpendicular Components

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



Rectangular Vectors in 3D Extend the 2D

Cartesian concept to 3D zyx FFFF

Introducing the 3D Unit Vector Triad (i, j, k)

Then kjiF zyx FFF

Where

zz

yy

xx

FFF

cosFcosFcosF



Rectangular Vectors in 3D Thus Fxi, Fyj, and Fzk are

the PROJECTION of F onto the CoOrd Axes

Can Rewrite

kjiFF

F

kFjFiFF

zyx

mm

zyx

F

coscoscos

And cos :Note

coscoscos

kjiF zyx FFF



Rectangular Vectors in 3D Next DEFINE a UNIT

Vector, u, that is Aligned with the LoA of the Force vector, F. Mathematically

Recall F from Last Slide to Rewrite in terms of u (note unit Vector Notation û)

kcosjcosicosu zyx

kFjFiF

ukji

xyx

zyx

ˆˆˆ

ˆˆcosˆcosˆcos

FFF



Rectangular Vectors in 3D Find ||F|| by the

Pythagorean Theorem

Can use ||F|| to determine the Direction Cosines

2222zyx FFF F

221 yx FF F

Fcos

Fcos

Fcos

zz

yy

xx

F

F

F



2D Case In 2D: θz = 90° → cos θz = 0 → Fz = 0

In this Case

jcosFicosFFjiF

yx

yx FF

jiujiFF

yx

xx

xyx FF

coscossincos

tan



Example – 2D REcomposition Given Bolt with Rectilinear Appiled

Forces For this Loading

Determine• Magnitude of the

Force, ||F||• The angle, θ, with

respect to the x-axis Game Plan

• State F in Component form

• Use 2D Relationsθ



Example – 2D REcomposition The force

Description in Component form

Now use Fy = ||F||sinθ to find ||F||

jiF lblb 1500700 Find θ by atan

715

7001500

lblb

FF

x

ytan

98647

15 .arctan

lb

lbFy

165565

1500

Fsinsin

F

Or by Pythagorus

lb

lblb1655

1500700 22

FF



Example – 3D DeComposition

û

A guy-wire is connected by a bolt to the anchorage at Pt-A

The Tension in the wire is 2500 N

Find • The Components Fx, Fy,

Fz of the force acting on the bolt at Pt-A

• The Space Angles θx, θy, θz for the Force LoA



Example – 3D DeComposition The LoA of the force runs from A to B.

Thus Direction Vector AB has the same Direction Cosines and Unit Vector as F

With the CoOrd origin as shown the components of AB

AB = Lxi + Lyj +Lzk• In this case

– Lx = –40 m– Ly = +80 m– Lz = +30 m

Then the Distance L = AB = ||AB||

mAB

mmmL

LLLLABAB zyx

394308040 222

222

.



Example – 3D DeComposition Then the Vector

AB in Component form kji mmmAB 308040

Then the UNIT Vector in the direction of AB & F

ABABABAB u

Recall

ABABN

ABABu 2500 FˆFF

Note that ||F|| was given at 2500 N

m

jmjmimN394

3080402500.

ˆˆˆF

kjiF NNN 79521201060

Thus the components• Fx = −1060 N• Fy = 2120 N• Fz = 795 N



Example – 3D DeComposition Now Find the

Force-Direction Space-Angles

Using Direction Cosines

Note that ||F|| was given at 2500 N

Using Component Values from Before

Fcos m

mF

Using arccos find

• θx = 115.1°• θ y = 32.0°• θ z = 71.5°

NNNNNN

z

y

x

2500795cos25002120cos25001060cos



WhiteBoard Work

Lets WorkThis niceProblem

100

lbTA 500

lbTB 400

a. Express in Vector Notation the force that Cable-A exerts on the hook at C1

b. Express in Vector Notation the force that Cable-B exerts on the U-Bracket at C2



wy wy





References Good “Forces” WebPages

• http://www.engin.brown.edu/courses/en3/Notes/Statics/forces/forces.htm

• http://www.pt.ntu.edu.tw/hmchai/Biomechanics/BMmeasure/StressMeasure.htm

Vectors• http://www.netcomuk.co.uk/~jenolive/

homevec.html

http://www.engin.brown.edu/courses/en3/Notes/Statics/forces/forces.htm

http://www.engin.brown.edu/courses/en3/Notes/Statics/forces/forces.htm

http://www.pt.ntu.edu.tw/hmchai/Biomechanics/BMmeasure/StressMeasure.htm

http://www.pt.ntu.edu.tw/hmchai/Biomechanics/BMmeasure/StressMeasure.htm



Some Unit Vectors

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Bruce Mayer, PE Licensed Electrical & Mechanical Engineer [email protected]

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