Engineering 36
Chp 2: Force
DeComposition
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Engineering-36: Engineering Mechanics - Statics
1
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-02_Fa12_Forces_as_Vectorspptx
Force Defined
 Force: Action Of One Body On
Another; Characterized By Its
• Point Of Application
• Magnitude (intensity)
• Direction
Line of Action
Magnitude
Direction
 The DIRECTION of
a Force Defines its
Line of Action (LoA)
Engineering-36: Engineering Mechanics - Statics
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Point of
Application
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-02_Fa12_Forces_as_Vectorspptx
Newton’s Law of Gravitation
 Consider two massive bodies
Separated by a distance r
M
F
• Newton’s Gravitation Equation
GMm
F 2
r
-F
( F a SCALAR)
m
– Where
 F ≡ mutual force of attraction between 2 bodies
 G ≡ 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
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-02_Fa12_Forces_as_Vectorspptx
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)
GMm
 GM 
F
but R  h  F  m  2   mg 
2
R  h 
 R 
 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
Engineering-36: Engineering Mechanics - Statics
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W  mg
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-02_Fa12_Forces_as_Vectorspptx
Earth Facts
 D  7 926 miles (12 756 km)
 M  5.98 x 1024 kg
• About 2x1015 Empire
State Buildings
 Density,   5 520 kg/m3
• water  1 027 kg/m3
• steel  8 000 kg/m3
• glass  5 300 kg/m3
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-02_Fa12_Forces_as_Vectorspptx
Gravitation Example
 Jupiter Moon Europa
Europa Statistics
Simon Marius & Galileo Galilei
Discovered by
1610
Date of discovery
4.8e+22
Mass (kg)
8.0321e-03
Mass (Earth = 1)
1,569
Equatorial radius (km)
2.4600e-01
Equatorial radius (Earth = 1)
3010
Mean density (kg/m^3)
670,900
Mean distance from Jupiter (km)
3.551181
Rotational period (days)
3.551181
Orbital period (days)
13.74
Mean orbital velocity (km/sec)
0.009
Orbital eccentricity
0.470
Orbital inclination (degrees)
2.02
Escape velocity (km/sec)
0.64
Visual geometric albedo
5.29
Magnitude (Vo)
Engineering-36: Engineering Mechanics - Statics
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• Find Your Weight
on Europra
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-02_Fa12_Forces_as_Vectorspptx
Europa Weight
 Since your MASS is
SAME on both Earth
and Europa need to
Find only geu and
compare it to gea
 Recall
GM
g 2
R
 Europa Statistics
from table:
 Meu = 4.8x1022 kg
 Reu = 1 569 km
Engineering-36: Engineering Mechanics - Statics
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 Then geu
geu  6.673  10
11
m3 4.8  1022 kg
kg  s 2 1569  103 m 2


6.673  10 4.8  10 

11
geu

22
2.462  1012
m3  kg
kg  s 2  m2
g eu  1.301 m s 2
 With %Weu = geu/gea
1.301
%Weu 
 13.27%
9.807
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-02_Fa12_Forces_as_Vectorspptx
Contact Forces
 Normal Contact
Force
• When two Bodies
Come into Contact
the Line of Action is
Perpendicular to the
Contact Surface
Engineering-36: Engineering Mechanics - Statics
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 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
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-02_Fa12_Forces_as_Vectorspptx
Contact Forces
 Fluid Force
 Tension Force
• In Fluid Statics the
Pressure exerted by
the fluid acts
NORMAL to the
contact Surface
• A PULLING force
which tends to
STRETCH an object
upon application of
the force
pa
HMS Bounty
d
p = pa + gd
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-02_Fa12_Forces_as_Vectorspptx
Contact Forces
 Compression Force
• A PUSHING force
which tends to
SMASH an object
upon application of
the force
Engineering-36: Engineering Mechanics - Statics
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 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
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-02_Fa12_Forces_as_Vectorspptx
Recall Free-Body Diagrams
 SPACE DIAGRAM 
A Sketch Showing
The Physical
Conditions Of The
Problem
Engineering-36: Engineering Mechanics - Statics
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 FREE-BODY
DIAGRAM  A
Sketch Showing
ONLY The Forces
On The Selected
Body
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-02_Fa12_Forces_as_Vectorspptx
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
FBD showing
forces P, Q, R, S
Force Polygon
if Static
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-02_Fa12_Forces_as_Vectorspptx
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 Oriented
 Unit Vectors may be
indicated with “Carets”
i  iˆ
j  ˆj k  kˆ u  uˆ λ  ̂
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-02_Fa12_Forces_as_Vectorspptx
Example: FBD & Force-Polygon
EYE, Not
Pulley
 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
 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?
Engineering-36: Engineering Mechanics - Statics
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• 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
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-02_Fa12_Forces_as_Vectorspptx
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.
TAC
TAB
3500 lb


sin 120 sin 2 sin 58
TAB  3570 lb
TAC  144 lb
Engineering-36: Engineering Mechanics - Statics
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A pretty Tough Pull
for the Guy at C
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-02_Fa12_Forces_as_Vectorspptx
Vector Notation – Vector ID
 In Print and Handwriting We Must
Distinguish Between
• VECTORS
• SCALARS
 These are Equivalent Vector Notations
PPPP
• Boldface Preferred for Math Processors
• Over Arrow/Bar Used for Handwriting
• Underline Preferred for Word Processor
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-02_Fa12_Forces_as_Vectorspptx
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.:
P  P  P  P  P  Mag of P
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-02_Fa12_Forces_as_Vectorspptx
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 with
LoA of Pt1→Pt2
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-02_Fa12_Forces_as_Vectorspptx
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]
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-02_Fa12_Forces_as_Vectorspptx
Magnitude-Angle Form
 The Magnitude of the Force is
Proportional to the Geometric
Length of its vector representation:
F  L where L is the Pythagore an Length :
L
x2  x1    y2  y1   z2  z1 
2
2
 Note that if Pt1 is at the
ORIGIN and Pt2 has L 
CoOrds (x, y, z) then
Engineering-36: Engineering Mechanics - Statics
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2
x y z
2
2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-02_Fa12_Forces_as_Vectorspptx
2
Magnitude-Angle Form
 Then calculate SPACE
ANGLES as
 x2  x1 
 y2  y1 
 z2  z1 
 x  arccos
  y  arccos
  z  arccos

 L 
 L 
 L 
 By the 3D cos  2  cos  2  cos  2  1
x
y
z
Trig ID
• Find Δx, Δ y, Δ z using Direction Cosines
 x2  x1 
 y2  y1 
 z2  z1 
cos  x  
 cos  y  
 cos  z  

 L 
 L 
 L 
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-02_Fa12_Forces_as_Vectorspptx
Magnitude-Angle Form
 Thus the Vector
Representation of a Force is
Fully Specified by the
LENGTH and SPACE ANGLES
L
x2  x1    y2  y1   z2  z1 
2
2
2
 x2  x1 
 y2  y1 
 z2  z1 
cos  x  
 cos  y  
 cos  z  

 L 
 L 
 L 
• Note: Can use the Trig ID to find
the third θ if the other two are known
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-02_Fa12_Forces_as_Vectorspptx
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
cos  x  sin  cos   cos   cos  z
cos  y
cos  y  sin  sin   tan  
cos  x
cos  z  cos 
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-02_Fa12_Forces_as_Vectorspptx
Rectangular Force Components
 Using Rt-Angle Parallelogram
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
F  Fx  Fy  iFx  jFy
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-02_Fa12_Forces_as_Vectorspptx
Rectangular Vectors in 3D
 Extend the 2D
F  Fx  Fy  Fz
Cartesian concept to 3D
 Introducing the 3D Unit
Vector Triad (i, j, k)
 Then F  Fx i  Fy j  Fz k
 Where
Fx  F cos x
Fy  F cos y
Fz  F cos z
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-02_Fa12_Forces_as_Vectorspptx
Rectangular Vectors in 3D
 Thus Fxi, Fyj, and Fzk are
the PROJECTION of F
onto the CoOrd Axes
 Can Rewrite F  Fx i  Fy j  Fz k
F  F cos  x i  F cos  y j  F cos  z k
Note : Fm  F cos  m  And
F  F cos  x i  cos  y j  cos  z k 
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-02_Fa12_Forces_as_Vectorspptx
Rectangular Vectors in 3D
 Next DEFINE a UNIT
Vector, u, that is Aligned
with the LoA of the Force
vector, F. Mathematically
u  cos  x i  cos  y j  cos  z k
 Recall F from Last Slide to Rewrite in
terms of u (note unit Vector Notation û)
F  F cos  xiˆ  cos  y ˆj  cos  z kˆ  F uˆ

 F xiˆ  F yˆj  F xkˆ
Engineering-36: Engineering Mechanics - Statics
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
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-02_Fa12_Forces_as_Vectorspptx
Rectangular Vectors in 3D
 Find ||F|| by the
Pythagorean Theorem
F1  Fx2  Fy2
2
F  Fx2  Fy2  Fz2
 Can use ||F|| to
determine the
Direction Cosines
Engineering-36: Engineering Mechanics - Statics
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Fx
cos x
F
Fy
cos y
F
Fz
cos z
F
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-02_Fa12_Forces_as_Vectorspptx
2D Case
 In 2D: θz = 90° → cos θz = 0 → Fz = 0
 In this Case
F  Fx i  Fy j
F  F cos xi  F cos yj
Engineering-36: Engineering Mechanics - Statics
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tan x Fy Fx
F F  cos xi  sin  xj
u  cos xi  cos yj
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-02_Fa12_Forces_as_Vectorspptx
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
θ
Engineering-36: Engineering Mechanics - Statics
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• Use 2D Relations
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-02_Fa12_Forces_as_Vectorspptx
Example – 2D REcomposition
 The force
Description in
Component form
F  700lbi  1500lbj
 Find θ by atan
 Now use Fy =
||F||sinθ to find ||F||
Fy
1500lb
F 

sin  sin65
F  1655lb
1500lb 15  Or by Pythagorus
tan  


Fx
700lb
7 F  700lb 2  1500lb 2
15
F  1655lb
   arctan
 64.98
7
Fy
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-02_Fa12_Forces_as_Vectorspptx
Example – 3D DeComposition
 A guy-wire is connected
by a bolt to the
anchorage at Pt-A
 The Tension in the
wire is 2500 N
 Find
û
• 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
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-02_Fa12_Forces_as_Vectorspptx
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
 AB = Lxi + Lyj +Lzk
 With the CoOrd
origin as shown the
components of AB
AB  AB  L  L2x  L2y  L2z
• In this case
– Lx = –40 m
– Ly = +80 m
– Lz = +30 m
 Then the Distance
L = AB = ||AB||
L
 40m 2  80m 2  30m 2
AB  94.3m
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-02_Fa12_Forces_as_Vectorspptx
Example – 3D DeComposition
 Then the Vector AB
in Component form
 Note that ||F|| was
given at 2500 N
ˆ  80m  ˆj  30m  ˆj



40
m
i
AB   40mi  80mj  30mk F  2500 N
 Then the UNIT
Vector in the
direction of AB & F
u  AB AB  AB AB
 Recall
AB
AB
F  F uˆ  F
 2500 N
AB
AB
Engineering-36: Engineering Mechanics - Statics
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94.3m
F   1060N i  2120N j  795N k
 Thus the components
• Fx = −1060 N
• Fy = 2120 N
• Fz = 795 N
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-02_Fa12_Forces_as_Vectorspptx
Example – 3D DeComposition
 Now Find the
Force-Direction
Space-Angles
 Using Direction
Cosines
Fm
cos m 
F
 Note that ||F|| was
given at 2500 N
Engineering-36: Engineering Mechanics - Statics
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 Using Component
Values from Before
 1060 N
cos x
2500 N
2120 N
cos y
2500 N
795 N
cos z
2500 N
 Using arccos find
• θx = 115.1°
• θ y = 32.0°
• θ z = 71.5°
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-02_Fa12_Forces_as_Vectorspptx
WhiteBoard Work
Lets Work
This nice
Problem
TA  500lb
TB  400lb
100
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
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-02_Fa12_Forces_as_Vectorspptx
  wy
wy
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-02_Fa12_Forces_as_Vectorspptx
Engineering-36: Engineering Mechanics - Statics
38
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-02_Fa12_Forces_as_Vectorspptx
References
 Good “Forces” WebPages
• http://www.engin.brown.edu/courses/en3/N
otes/Statics/forces/forces.htm
• http://www.pt.ntu.edu.tw/hmchai/Biomecha
nics/BMmeasure/StressMeasure.htm
 Vectors
• http://www.netcomuk.co.uk/~jenolive/home
vec.html
Engineering-36: Engineering Mechanics - Statics
39
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-02_Fa12_Forces_as_Vectorspptx
Some Unit Vectors
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-02_Fa12_Forces_as_Vectorspptx