1. Chp 2: Force DeComposition
Engineering 36
Chp 2: Force DeComposition
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer

2. 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)
Point of Application

3. Newton’s Law of Gravitation
Consider two massive bodies Separated by a distance r M m -F F
Newton’s Gravitation Equation
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

4. 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)
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

5. Earth Facts D  7 926 miles (12 756 km) M  5.98 x 1024 kg
About 2x1015 Empire State Buildings
Density,   kg/m3
water  kg/m3
steel  kg/m3
glass  kg/m3

6. Gravitation Example
Jupiter Moon Europa
Find Your Weight on Europra

7. Europa Weight
Since your MASS is SAME on both Earth and Europa need to Find only geu and compare it to gea
Recall
Then geu
Europa Statistics from table:
Meu = 4.8x1022 kg
Reu = km
With %Weu = geu/gea

8. Contact Forces Normal Contact Force Friction 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

9. Contact Forces Fluid Force Tension 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

10. Contact Forces Compression Force Shear 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

11. 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

12. Force Polygon if Static
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

13. 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”

14. 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
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
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?

15. 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.
If angle =
15° => Tac = 128 lb
0° =>122 lb
A pretty Tough Pull for the Guy at C

16. Vector Notation – Vector ID
In Print and Handwriting We Must Distinguish Between
VECTORS
SCALARS
These are Equivalent Vector Notations
Boldface Preferred for Math Processors
Over Arrow/Bar Used for Handwriting
Underline Preferred for Word Processor

17. 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.:

18. 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

19. 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]

20. Magnitude-Angle Form
The Magnitude of the Force is Proportional to the Geometric Length of its vector representation:
Note that if Pt1 is at the ORIGIN and Pt2 has CoOrds (x, y, z) then

21. Magnitude-Angle Form Then calculate SPACE ANGLES as By the 3D Trig ID
Find Δx, Δ y, Δ z using Direction Cosines

22. Magnitude-Angle Form
Thus the Vector Representation of a Force is Fully Specified by the LENGTH and SPACE ANGLES
Note: Can use the Trig ID to find the third θ if the other two are known

23. 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

24. 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

25. Rectangular Vectors in 3D
Extend the 2D Cartesian concept to 3D
Introducing the 3D Unit Vector Triad (i, j, k)
Then
Where

26. Rectangular Vectors in 3D
Thus Fxi, Fyj, and Fzk are the PROJECTION of F onto the CoOrd Axes
Can Rewrite

27. 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)

28. Rectangular Vectors in 3D
Find ||F|| by the Pythagorean Theorem
Can use ||F|| to determine the Direction Cosines

29. 2D Case
In 2D: θz = 90° → cos θz = 0 → Fz = 0
In this Case

30. 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 Relations θ

31. Example – 2D REcomposition
The force Description in Component form
Now use Fy = ||F||sinθ to find ||F||
Find θ by atan
Or by Pythagorus

32. Example – 3D DeCompositionû
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

33. 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||

34. Example – 3D DeComposition
Then the Vector AB in Component form
Note that ||F|| was given at 2500 N
Then the UNIT Vector in the direction of AB & F
Thus the components
Fx = −1060 N
Fy = 2120 N
Fz = 795 N
Recall

35. Example – 3D DeComposition
Now Find the Force-Direction Space-Angles
Using Direction Cosines
Using Component Values from Before
Using arccos find
θx = 115.1°
θ y = 32.0°
θ z = 71.5°
Note that ||F|| was given at 2500 N

36. Lets Work This nice Problem
WhiteBoard Work
Lets Work
This nice Problem
S&T
Express in Vector Notation the force that Cable-A exerts on the hook at C1
Express in Vector Notation the force that Cable-B exerts on the U-Bracket at C2

37. wy wy

39. References Good “Forces” WebPages Vectors
Vectors

40. Some Unit Vectors