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First ClassWednesday, January 20, 201007:55Statics: a subject in mechanicsMechanics: the study of forces on bodies and their resulting motion (Isaac Newton a great contributor)Emphasis on bodies which are not moving (or moving at constant speed)TAM 212: dynamics - moving bodiesBiomechanics: muskoskeletal mechanicsTAM 251: Introductory Solid Mechanics - strength of materials - study of how things deform - twist,compress, stretch ME 340: dynamics of buildingsME 310: fluid dynamicsME 330: behavior of materialsME 360: controls - roboticsME 320: heat transferME 370: designME 350: design for manufacturabilityNotes Page 1

SyllabusWednesday, January 20, 201008:19Forces and bending momentsMultiple bodiesBeams, structural analysis, trussesFrictionCan switch to TAM 211 at this pointForces on submerged bodiesNotes Page 2

No course website yet - on mechse website will beGrades on CompassHMWKS due Fridays at 08:00Ms Thomas's know what language we speakNotes Page 3

Precision: 3 significant figures ALWAYS (like IB math)Exams are during lectures (not in this room)Notes Page 4

Vectors, Newton's Laws of Motion, Gravity, Calculations, andAnalysisFriday, January 22, 201007:56 A1-A5 (vectors) in bookReadNewton's Three Laws of Motion1. A particle at rest or moving with constant velocity will remain in this state unless acted upon by anunbalanced force.2. A [article acted upon by an unbalanced force F experiences an acceleration a that is proportionalto the particle mass, m.3. The mutlual forces of action and reaction between two particles are equal opposite and collinear.Newton's law of gravitational attractionThe mutual force F of gravitation between two particles of mass m1 and m2 is given byWeight is the force exerted by the earth on a particle at the earth's surface:Units of MeasurementSystemLength TimeMassForceSImskgNUS Customary (FPS) ftssluglbNewton and slug are derived unitsUnit conversion- 1 ft 0.3048m (exact)- 1 slug 14.59 kg- 1 lb 4.448 NNumerical Calculations- Dimensional homogeneity Equations must be dimensionally homogenousNotes Page 5

Equations must be dimensionally homogenousSignificant Figures- Use 3 significant figures for final answers, more (usually 4) for intermediate calculations This is engineering convention (assume all dimensions are to 3 sf if not specified, 1 ft 1.00ft)General Procedure for analysis1. Read the problem carefully, write it down carefully.2. Draw given diagrams neatly and construct additional figures as necessary.3. Apply principles needed.4. Solve problem symbolically, then substitute numbers. Proved proper units throughout. Checksignificant figures. Box the final answer(s).5. See if answer is reasonable.Starting Monday: class meets in 228 Natural History BuildingNotes Page 6

ForcesMonday, January 25, 201007:57Office hours are now online - all are in 429 GraingerForce: the action of one body on another - can be treated as a vector, since forces obey all the rules thatvectors do. Forces have magnitude, direction, and sense.Problem: Finda unit vector in the xy plane that is perpendicular to the force 3i-4j 12k N.Answers in back of book: don't necessarily have all solutions - if there are two explainthatProblem: Find two 80 lb forces whose sum is the force 40i lb.Notes Page 7

Find a vector that is simultaneously in the plane of F1 and F2 and in the plane of F3 and F4.Discussions this weekHMWK due FridayNotes Page 8

Forces, components, and division thereofWednesday, January 27, 201007:58Forces: the action of one body on another - magnitude, sense, and direction - vectorsForce components: Law of sines and cosines can be used in 2-d problems, but using force components isusually more straightforward.- Can also solve problems graphicallyAdding vectors - simple- Pick a good system - but keep it once you have itPROBLEML Determine the magnitude of force F so that the resultant force of the three is as small aspossible. What is the minimum magnitude?Question: how do we know this is the minimum rather than the maximum?- Cannot be a maximum, since the force could be large as possible- Note: could solve it geometrically, but much more difficultForce along a line: often a force is described as acting along a line - often said to be a line connecting Aand B.Example: the window is held open by cable AB. Determine the length of the cable andexpress the 30N force at A along cable.Notes Page 9Cantilever bending - force vertical on lug nut

Components of forces parallel and perpendicular to a line:- The scalar component A // of a vector A along a line with unit vector u is given by A // Acos(theta) A.u. Thus the vector component is A// A//u (A.u)u, Aperp A-A//.Problem:Determine the angle theta betweeen the two cables. Determine the projected component of the ForceF 12lb in direction of ACNotes Page 10

Equilibrium, Free body diagrams, and IdealizationsFriday, January 29, 201007:59Equilibrium:- According to Newton's first law, a particle will be in equilibrium if and only if, ΣF 0, that is the sumof all forces on a particle is zero.- Generally, in 3D: Equilibrium requires: ΣF ΣFxi ΣFy j ΣFzk 0. - can be separated into vectorequations.- Note: this is valid in inertial reference frames - basically one in which we can regard as fixed- Many problems are also in only 1 or 2 dimensions - the remaining can be ignoredFree Body diagram- A free body diagram is a drawing of a body, or part of a body which all the forces acting on thebody are shown. Relevant dimensions may also be required- No judgment regarding equilibrium or lack of it is made when constructing the diagram. Forcesnot directly acting on the body are not shown.Idealization:- Pulleys are usually regarded as frictionless, then the tension in a rope or cord around the pulley isthe same on either side- Springs are usually regarded as linearly elastic then the tension is proportional to the change inlength, s.Notes Page 11

Example ProblemsFriday, January 29, 201008:03In reality either rope could break, as there is a distribution of ropesbreak at slightly different values - they follow a normal curveNotes Page 12

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Systems of ParticlesMonday, February 01, 201007:59Some practical engineering problems involve the use of statics of interacting or interconnected particles.To solve them, we use Newton's first law, ΣF 0, on multiple free body diagrams of particles or groups ofparticles.Notes Page 14If you want to find the value of a force you must drawa free body diagram that includes that force.

Example ProblemsMonday, February 01, 201007:59The five ropes can each take 1500 N without reaking. How heavy can W be without breaking any.A man having a weight of 175 lb attempts to lift himself using one of the two methods shown,determine the total force he must exert on bar AB in each case and the normal reaction he exerts on theplatform at C. The platform has a weight of 30 lb.Notes Page 15

Never forget to draw the overall free body diagram - it simplifiesthe problems greatlyThe collar can slide along the rod without friction. The spring, which is attached to the collar and to theceiling exerts a force kΔ proportional to its stretch Δ, where k is the modlus of the spring. If k 50lb/in,W 50 lb and the collar weiths 20 lb find how much the string is stretched in the given position.Notes Page 16

Moment of a forceWednesday, February 03, 201007:57Hour exam next Friday - cumulative so far- Vectors, Application of vectors - It will be in 2 different rooms - here and 100 MSEB - there will be assigned seatingThe moment, MP, of a force, F about a point P is equal to the product, MP Fd, where d is theperpendicular distance from P to the line of action of the force. The force F can be acting at any pointalong this line of action.Principle of transmittivity: F can be acting anywhere along the line of actionNotes Page 17Line of action of F

Example ProblemsWednesday, February 03, 201008:01Three forces act on the plate. Determine the sum of the moments of the thre forces about point P.Note: a torque is a special kind of momentNotes Page 18

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Moments of Forces and CouplesFriday, February 05, 201008:00This is used in post hole augers - things to dig holes, you put it in and then twist it - apply equal andopposite forces to turn itMoment of a Force about a Line:- For a point: MP rxF- Ml MP.ul- Ml (MP.ul)ul- Note the perpendicular one is the total minus the parallelCouples:- A couple is a pair of equal, opposite and noncollinear forces- The moment of a couple is given by M Fd (scalar)- M rxF, it is independent of the origin you choose- If several couples act, MR Σ(rxF).- This is a special case: note that the total force is equal to 0Notes Page 20

Example problemsFriday, February 05, 201008:00Example: determine moment about Oa axisDetermine the couple moment - express as a cartesian vectorNotes Page 21

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Equipollent force systemsMonday, February 08, 201007:58A force system is a collection of forces and couples applied to a body. Two force systems are said to beequipollent if they have the same resultant force, ΣF and the same resultant moment, ΣMP with respectto any point P. Here, for each system.ΣMP denotes the moment of all applied couples plus the sum of moments of applied forces withrespect to point PNote: equivalent is often used instead of equipollent.Pronunciation: ee-quee-pah-lentNotes Page 23Hour exam:- Assigned seating- On Friday- Homework-type problems, plus understanding questions- No calculators or electronics- Bring self, pencil, eraser- Through Friday's lecture- Do not need to use given-find-solution, only solution- 3 problems- Either pen or pencil is OK - pencil preferable- 50 minutes long, starts at precisely 8:00

Example problemsPost tensioning ram on a forkliftMonday, February 08, 201007:58Can rephrase as: Replace the force and couple system by its resultant at point Q.Replace the force and couple system by an equipollent force and couple moment at point Q.Note: it does not matter where this couple is applied for 8kNotes Page 24

Resultants of Force SystemsCheck your seat before examCourse notes posted outside 429 GraingerWednesday, February 10, 201007:57Resultants:- A resultant is an equipollent force system consisting of only one force Fr and one couple MrP: Must specify PFr (ΣF) 1MrQ (ΣM)1F can be anywhere along its line of action, couple can beapplied anywhereNotes Page 25

Example problemsWednesday, February 10, 201007:58The belt passing over the f pulley is subjected to two forces each having a magnitude of 40 N, force F1acts in the -k direction. Teplace thse by a resultant force system at point A. theta is 45 degrees.Notes Page 26

Hour ExamMonday, February 15, 201008:06Make sure to explain work - use wordsAlso simplifying importantExams by Friday hopefullyNotes Page 27

ReductionsAutonomous Materials GroupMonday, February 15, 201007:59Reduction classes:- Single resultant force (often possible) Concurrent forces with no couples Coplanar forces with couples perpendicular to force plane Parallel forces with couples perpendicular to forces- Screwdriver (always possible) - consists of a single force and a couple in the same directionA resultant is an equipollent force system consisting of only one force, Fr and one couple MrP - mustspecify pointSingle resultant forces- If FR is perpendicular to MrP with the force not zero, then an equipollent system consisting of onlya single force can always be found.1. Concurrent forces (no couples)i. Act at the same point, just sum the forces2. Coplanar, with couples perpendicular to planei. Add the forces, that will be the forceii. Find MrP and then divide by the force and move the force over that much - that will be thesingle force resultant3. Parallel forces with couples perpendiculari. Add the forces for the resultant forceii. Find moment about P, then move the force over by MrP divided by the forceiii. Common with a plate with wires and weightsScrewdriver- Most general case - always possible- Consists of a force and a couple in the same direction - push and twistMove the force over in order to cancel the perpendicularforce - then it will just be the parallel moment and the forceNotes Page 28

Example problemMonday, February 15, 201008:42Replace the 3 forces by a screwdriverNotes Page 29

Distributed forcesScrewdriver is referred to as a wrench by some texts.Wednesday, February 17, 201008:16Distributed forces- In structural analysis we are often presented with a distributed load (force/unit length) w(x) andwe need to find the equipollent loading F.Notes Page 30

Example ProblemsWednesday, February 17, 201008:16Replace the distributed load by an equipollent resultant force and specify its location on the beammeasured from the pin at CAn engineer measures the forces exerted by the soil on a 10m section of a building foundation and findsthey are described byDeterminea. The magnitude of the total force exerted on the foundationb. The magnitude of the moment about A due to the distributed load.Notes Page 31

Simple Distributed LoadingFriday, February 19, 201007:59Total force given by area under loading diagramIt is located at the centroid of the areaNotes Page 32

Equilibrium of a rigid bodyFriday, February 19, 201008:10We regard a rigid body as a collection of particles:Force equilibrium:Let Fi the resultant external force on particle iLet fij the internal force on particle i by particle jLet fji the internal force on particle j by particle iIn general: 2D problems have 3 unknowns, 3 equations (2 forces, 1 moment)3D problems have 6 unknowns, 6 equations (3 forces, 3 moments)Notes Page 33

Example problemsFriday, February 19, 201008:28Draw the free-body diagram of the automobile, which is being towed at constant velocity up the inclineusing the cable at C. The automobile has a mass of 5 Mg and a center of mass at G. The wheels are freeto roll. Explain