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Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu

Engineering 36. Chp 2: Force DeComposition. Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu. Force: Action Of One Body On Another ; Characterized By Its Point Of Application Magnitude (intensity) Direction. Force Defined. Line of Action. Magnitude.

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Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu

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  1. Engineering 36 Chp2: ForceDeComposition Bruce Mayer, PE Licensed Electrical & Mechanical EngineerBMayer@ChabotCollege.edu

  2. Force: Action Of One Body OnAnother; Characterized By Its Point Of Application Magnitude (intensity) Direction Force Defined Line of Action Magnitude Direction • The DIRECTION of a Force Defines its Line of Action (LoA) Point of Application

  3. M F -F m Newton’s Law of Gravitation • Consider two massive bodies Separated by a distance r • 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. 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) Weight • 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 EmpireState Buildings • Density,   5 520 kg/m3 • water  1 027 kg/m3 • steel  8 000 kg/m3 • glass  5 300 kg/m3

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

  7. Since your MASS is SAME on both Earth and Europa need to Find only geu and compare it to gea Recall Europa Weight • Then geu • Europa Statisticsfrom table: • Meu = 4.8x1022 kg • Reu = 1 569 km • With %Weu = geu/gea

  8. Normal Contact Force When two Bodies Come into Contact the Line of Action is Perpendicular to the Contact Surface Contact Forces • 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. Fluid Force In Fluid Statics the Pressure exerted by the fluid acts NORMAL to the contact Surface Contact Forces • Tension Force • A PULLING force which tends to STRETCH an object upon application of the force

  10. Compression Force A PUSHING force which tends to SMASH an object upon application of the force Contact Forces • 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. 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 Polygonif Static

  13. 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 Vector Notation – Unit Vectors • Unit Vectors may be indicated with “Carets”

  14. Example: FBD & Force-Polygon • 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 EYE, 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?

  15. 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. Example Solution 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 withLoA 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 SPACEANGLES as • By the 3D Trig ID • Find Δx, Δ y, Δ zusing Direction Cosines

  22. Magnitude-Angle Form • Thus the Vector Representation of a Force isFully 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. Using Rt-Angle Parallelogram Resolve Force Into Perpendicular Components Rectangular Force Components • 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

  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 Fonto 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. Given Bolt with Rectilinear Appiled Forces θ Example – 2D REcomposition • 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

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

  32. û 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 • 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. The LoA of the force runs from A to B. Thus Direction VectorAB 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|| Example – 3D DeComposition

  34. Then the Vector AB in Component form Example – 3D DeComposition • 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. Now Find the Force-Direction Space-Angles Using Direction Cosines Example – 3D DeComposition • Using Component Values from Before • Using arccos find • θx = 115.1° • θy = 32.0° • θz = 71.5° • Note that ||F|| was given at 2500 N

  36. WhiteBoard Work Lets Work This niceProblem • 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

  38. 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

  39. Some Unit Vectors

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