Chp 2: Force DeComposition Engineering 36 Chp 2: Force DeComposition Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu

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

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

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

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

Gravitation Example Jupiter Moon Europa 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 Recall Then geu Europa Statistics from table: Meu = 4.8x1022 kg Reu = 1 569 km With %Weu = geu/gea

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

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

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

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

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

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”

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?

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

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

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.:

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

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

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

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

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

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

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

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

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)

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

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

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

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

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||

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

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

Lets Work This nice Problem WhiteBoard Work Lets Work This nice Problem S&T 4.5.34 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

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References Good “Forces” WebPages Vectors 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

Some Unit Vectors