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# 1 Class Note for ME 27000 with Professor Jones at Purdue

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ME 270 Basic Mechanics I Prof Jones M39 222B Ph 4945 691 Email jonesj dpurdueedu 1023 INTRODUCTION Learning Objectives 1 To introduce and de ne the subject of mechanics 2 To introduce Newton39s Laws and to understand the signi cance of these laws 3 The review modeling dimensional consistency unit conversions and numerical accuracy issues 4 To review basic vector algebra ie vector addition and subtraction and scalar multiplication Definitions Mechanics Study of forces acting on a rigid body a Statics body remains at rest b Dynamics body moves Newton39s Laws First Law Given no net force a body at rest will remain at rest and a body moving at a constant velocity will continue to do so along a straight path 0 2M 0 Second Law Given a net force is applied a body will experience an acceleration in the direction of the force which is proportional to the net applied force 2F ma Third Law For each action there is an equal and opposite reaction FAB FBA Models of Physical Systems Develop a model that is representative of a physical system Particle a body of in nitely small dimensions conceptually a point Rigid Body a body occupying more than one point in space in which all the points remain a xed distance apart Deformable Body a body occupying more that one point in space in which the points do not remain a xed distance apart Dimensional Consistency If you add together two quantities x C1 V Cza these quantities need to have the same dimensions units eg if x V and a have units of L UT and LTZ then C1 and C2 must have units of T and T2 to maintain dimensional consistency Many times algebraic errors in analysis lead to dimensional inconsistency Use dimensional consistency as a check on your algebra Unit Conversions Use a logical process in your unit conversions For example to convert 60 mph to ft sec 60 mlles X hour X 520ft 882 hour 3600 s mlle s Accuracy of Numerical Answers Say that you do the following calculations X z t Y Say that you have used four signi cant digits for X and y but only two signi cant digits for t The numerical value of z will not have more than two signi cant digits and likely less than two Examples of Signi cant F igures 3851 four signi cant gures 3 851 four signi cant gures 003851 four signi cant gures 3851 X 107 four signi cant gures 704 X 10394 three signi cant gures 255 three signi cant gures 051 two signi cant gures 000005 one signi cant gure 27855 ve signi cant gures 891 X 104 three signi cant gures 2200 May have two three or four signi cant gures depending on the accuracy of the measurement that obtained the number Where such doubt may eXist it is better to write the number as 22 X 103 to show two signi cant gures or as 220 X 103 to show three signi cant gures 55 two signi cant gures 550 three signi cant gures The zero is signi cant in this case since it is not otherwise needed to show proper location of the decimal point Retention of Signi cant F igures 1 In recording measured data only one doubtful digit is retained and it is considered to be a signi cant gure 2 In dropping gures that are not signi cant the last gure retained should be increased by 1 if the rst gure dropped is 5 or greater 3 In addition and subtraction do not carry the result beyond the rst column that contains a doubtful gure For example 113 3 l 43 4273 427 the 3 in the hundredth s place is insigni cant since the 113 gure can only be measured to the closest tenth 4 In multiplication and diVision carry the result to the same number of signi cant gures that there are in the quantity entering into the calculation that has the least number of signi cant gures For example 155 X 54 8370 83 because there are only 2 signi cant gures in the lower number m Unless stated otherwise we shall retain three signi cant gures unless one or more digits are lost through additions for example 902 901 01 Thus given a dimension of2 ft assume it is actually 200 ft Vectors Vectors are characterized by both a magnitude and direction Examples of vectors position velocity acceleration force moment Vectors add by the parallelogram law For example KEC Says that vector A has the same magnitude and direction as the sum of vectors E andC More the tail of 3 to the head of E A is then found by connecting tail of B to the head of C See the text for the mechanics of adding and subtracting vectors and for multiplying a scalar with a vector Law of Cosines C VAZ B2 ZABCOS c Law of Sines A B C Sinai sinb sinc 2428 VECTOR DEFINITIONS DIRECTION COSINES DIRECTION ANGLES Learning Objectives 1 To determine and understand the differences between position vectors unit vectors and force vectors 2 To determine the direction cosines and direction angles 4 To do an engineering estimate of these quantities 10 Definitions Position Vector Q a vector used to identify the of a point in space relative to a reference point AB Unit Vector a a dimensionless vector of unit magnitude that is often used to describe the of a vector of interest uAB Force Vector FAB a vector used to represent the and of a force FAB Let unit vector pointing in the Xdirection unit vector pointing in the ydirection WI u unit vector pointing in the zdirection 11 Conseguences i The vector A can be written in terms of its quotcomponentsquot and the unit vectors and E as see gure K AXIijAZE ii The sum of two vectors is accomplished by adding together the respective components X E AXi Ay AZE BXi By BZE AX BXiAy ByjAZ BZE iii The magnitude length of X is given by using Pythagorean Theorem 2 2 2 A AXAYAZ iv A unit vector u pointing in the same direction as A is given by dividing A its magnitude A A A AZ u 1 k y A A W W 12 Direction Angles a cos 1Ax IKI angle between l and K vectors measured in the l K plane 3 COS 1Ay X angle between and K vectors measured in the JT K plane y cos 1AZ IX angle between E andX vectors measured in the k A plane Direction Cosines cos 05 Ax MEI cos Ay K cos y AZ X Note cos 002 cos B2 cos y2 1 l3 Conseguences i Rearranging we can see AX lelcosa Ay Xcos AZ Xcos7 ii Therefore lve can vvrite AAXiAy iAZ k 2K cos aiK cos jK cos yE Xcos aicos jcos yE iii Also fronl before cosaicos jcosyf IAI iv If AX gt 0 then 0 lt a lt 90 If AX lt 0 then 90 lt a lt 180 Final Remark Practice making the threedimensional vectors Show direction angles in gures Visualization is the rst step in understanding 14 29 DOT PRODUCT Scalar Product Learning Objectives 1 To use the dot product to determine i the projection ofa vector in another direction ii the angle 6 between two vectors and 2 To do an engineering estimate of these quantities 15 Definitions KB 2 E K scalar quantity In words A E the magnitude K times the magnitude of the projection of B on A ie IEI 009 Projection of a Vector A gtI u u Angle 9 Between Two Vectors e 16 Mechanics K E AXi Ay AZE BXi By BZE AXBX Ay By AZ BZ pABp Kp 1 1 1ar to multlpllcatlon ABCABAC 2 If gt 0 then the projection of E is in the direction of A ie 0 g e 3 90 If Kim then the project of E is in the opposite direction of A ie 900 g e 3 180 f KE O gt 6 90 and K is perpendicular to E 1fKEzlf mzand ispara eltoB 4K XX2 17 31 34 EQUILIBRIUM OF A PARTICLE Learning Objectives 1 To draw a free body diagram FBD of an object that is modeled as a particle 2 To evaluate the forces required for static equilibrium of an object that is modeled as a particle 3 To do an engineering estimate of these quantities 18 Force Classifications External Forces applied forces which are typically known or prescribed eg forces due to cables springs gravity etc Reaction Forces constraining forces at supports intended to prevent motion usually nonexistent unless system is externally loaded Free Bodv Diagram FBD Free Body Diagram FBD a graphical sketch of the system showing a coordinate system all externalreaction forces and moments and key geometric dimensions Bene ZZZS 1 Provides a coordinate system to establish a solution methodology 2 Provides a graphical display of all forcesmoments acting on the rigid body 3 Provides a record of geometric dimensions needed for establishing moments of the forces 19 Newton s First Law Given no net force a body at rest will remain at rest and a body moving at a constant velocity will continue to do so along a straight path i Xi 6 Static Eguilibrium Vector Equation FR 2 ii ZFXTZFJZFZE6 11 Component Equations Z 0 Z 0 2 0 20 Problem Solving 1 Draw complete FBD 2 Choose an xyz reference frame 3 Evaluate the geometrical parameters 4 Write equations of static equilibrium 5 Count number of scalar equations and number of unknowns 6 Solve equations of static equilibrium 21 4144 CROSS PRODUCT Vector Product Learning Objectives 1 To use cross products to i calculate the angle 6 between two vectors and ii determine a vector normal to a plane 2 To do an engineering estimate of these quantities 22 Purpose Four primary uses of the cross product are to I calculate the angle 6 between two vectors 2 determine a vector normal to a plane 3 calculate the moment of a force about a point and 4 calculate the moment of a force about a line Definition Kgtltl3l3gtltK In words IX X El the component of E perpendicular to X multiplied by the magnitude of A or Vice versa Angle 9 Between Two Vectors Unit Normal Vector Unit Normal 2 n 23 Mechanics assuming a righthanded coordinate system X x E Axi Ayi AZE gtlt BXi By BZE K xE AyBZ ByAZi AXBZ BXAZ AXBy BXAyE Recall ixijxjExEo XL 3 Exi3 Ixjk jxiz k M2 324 Note Given 6 EX E E is perpendicular to the plane containing vectors A and B Basic Properties 1 X X E E X X Use right hand rule to determine direction of resultant vector 2 PiKXE pXgtltEXx BE Ax CAxEAxc X XEXgtltE XE 3If AgtltE0 he 2 If leEAB9thao 4 XXXO 24 4144 MOMENT ABOUT A POINT Learning Objectives 1 To use cross products to calculate the moment of a force about a point 3 To do an engineering estimate of this quantity 25 Definition MomenlAboul a Point a measure of the tendency of a force to turn a body to which the force is applied VoriF9D sm z Of Mean lm mm 1 Of 0p 2 la ryj gtlt in ij rXFy ryFXE 26 Comments 1 Direction of the moment can be determined by the right hand rule 2 Point P can be any point along the line of action of the force without altering the resultant moment M o 3 Use trigonometric relationships for calculating the moment about a point for 2D problems 4 Use vectors and cross products when calculating the moment about a point for 3D problems 27 45 MOMENT ABOUT A LINE Learning Objectives 1 To use cross products and dot products to evaluate the moment about a line 2 To do an engineering estimate of this quantity 28 Definition MomentAbout a Line The vector component of the moment of the force about any point on the line which is parallel to the line MAB M0 moment of the force F about any point along the line AB ie MO rep x F uAB unit vectors along line AB Procedure 1 Calculate the moment about any point 0 along line AB ie M r x 1 0 0p 39 2 De ne a unit vector along line AB ie UAB 3 Find the magnitude of the projection of MO onto line AB by nding the dot product of Mo and UAB ie M0 39UAB 4 Find the moment projection along line AB by multiplying the magnitude of the projection part 3 by the same unit vector as used in part 3 ie M0 39uAB uAB 29 Comments 1 The moment of the force M O may be taken about any point along the line AB Choose the point that will provide the simplest cross product 2 The moment about line AB will be zero if the force F passes through any point along line AB 3 The moment about line AB will be zero if the force F if parallel to line AB 30 4648 FORCE COUPLES AND EQUIVALENT SYSTEMS Learning Objectives 1 To determine a resultant torque of a system of force couples 2 To determine an equivalent forcecouple of a system of forces and moments 3 To do an engineering estimate of this quantity 31 Definition Force Couple a pair of forces which are i equal in magnitude ii parallel and iii opposite in direction IMI IFI 01 where d perpendicular distance between the lines of action of the forces forming the couple M TABXF where TAB any position vector between the lines of action of the forces forming the couple Comments 1 Force couples cause no nelforce ie 2F 2 0 2 The moment due to a force couple is the same regardless of the point the moment is summed about This not true of a noncouple 32 Definition Equivalent Systems two forcecouple systems which exert i the same net force on a body and ii the same net moment or torque Force Condition ZFM Z Moment Condition ZMp1 2MP2 Note Selection of point P is arbitrary Thus choose a point that will simplify the moment equation ForceCouple Equivalent Force E 21 Moment MrP sz Equivalent ForceCouple System FR R0 about point P 33 59 FLUID STATICS LINE AND SURFACE LOADING Learning Objectives 2 To determine the resultant of a given line load and to evaluate the support reactions acting on the body that carries such a load 3 To do an engineering estimate of the equivalent load and its location Distributed Parallel Line Loads R fwd dx Resultant Force L MrP J1 X0 WXdX Resultant Moment about pt P L I Xc WX dX M P i IL WX dX R Line of action for equivalent force Note 0 FR area under the curve qX o i centroid of the area under the curve qX 0 Determine direction of FR and MR by inspection 0 Use your intuition to check your answer 34 5155 FREE BODY DIAGRAMS FBDs Learning Objectives 1 To inspect the supports of a rigid body in order to determine the nature of the reactions and to use that information to draw a ee body diagram FBD ForceMoment Classi cations External Forces lIoments applied forcesmoments which are typically known or prescribed eg forcesmoments due to cables springs gravity etc Reaction F orces lomentsz constraining forcesmoments at supports intended to prevent motion usually nonexistent unless system is externally loaded Free Bodv Diagram FBD Free Body Diagram FBD a graphical sketch of the system showing a coordinate system all extemalreaction forces and moments and key geometric dimensions Bene ZZZS 1 Provides a coordinate system to establish a solution methodology 2 Provides a graphical display of all forcesmoments acting on the rigid body 3 Provides a record of geometric dimensions needed for establishing moments of the forces 35 5155 STATIC EQUILIBRIUM OF RIGID BODIES 2D Lea ming im 1 To evaluate the unknown reactions holding a rigid body in equilibrium by solving the equations of static equilibrium 2 To recognize situations of partial and improper constraint as well as static indeterminacy on the basis of the solvability of the equations of static equilibrium Newton s First Law Given no net force a body at rest will remain at and a body moving at a constant velocity will continue to do so along a straight path Definitions ZeroForce Members structural members that support no loading but aid in the stability of the truss T woF orce Members structural members that are a subject to no applied or reaction moments and b are loaded only at two pin joints along the member MultiForce Members structural members that have a applied or reaction moments or b are loaded at more than two points along the member 36 Vector Eguations R MR ZM O 6 where O is any arbitrary point Component Eguations There are three alternate forms of equilibrium equations for 2D problems i Two component force equations X and y are moment equation z ii One component force equation X or y and two moment equations both about different points in the z direction 20 20 20 iii Three moment equations points A B and C cannot be collinear 37 Static DeterminacvPartial and Improper Constraints Static Indeterminacy occurs when a system has more constraints than is necessary to hold the system in equilibrium ie the system is overconstrained and thus has redundant reactions Static Determinancy occurs when a system has a su cient number of constraints to prevent motion without any redundancy Partial Constraint occurs when there is an insu cient number of reaction forces to prevent motion of the system ie the system is partially constrained Improper Constraint occurs when a system has a su cient number of reaction forces but one or more are improperly applied so as not to prevent motion of the system ie the system is improperly constrained Comments 1 Equations i are the equilibrium eqns most commonly used 2 NEVER attempt to use MORE THAN THREE equilibrium equations from a single planar FBD Only three independent equations can exist for a single planar FBD 3 If you have more than three unknown forces in your three equations then consider breaking the system or structure into smaller systems and write down equilibrium equations for each substructure If this is not possible you may have an indeterminate structure ie the evaluation of member forces requires consideration of deformation of the members resulting from the loading 38 4 If all forces act through a single point then the moment equation for any point will not provide any more new information 39 5155 STATIC EQUILIBRIUM OF RIGID BODIES 3D Lea ming im 1 To evaluate the unknown reactions holding a rigid body in equilibrium by solving the equations of static equilibrium 2 To recognize situations of partial and improper constraint as well as static indeterminacy on the basis of the solvability of the equations of static equilibrium Newton s First Law Given no net force a body at rest will remain at rest and a body moving at a constant velocity will continue to do so along a straight path FR Z 21 Z 0gt MR Z 2M0 Z O O Definitions ZeroForce Members structural members that support no loading but aid in the stability of the truss T woF orce Members structural members that are a subject to no applied or reaction moments and b are loaded only at two pin joints along the member MultiForce Members structural members that have a applied or reaction moments or b are loaded at more than two points along the member Vector Eguations 6 ml 2 ml Ki ll4 gl 2 Component Eguations where O is any arbitrary point For 3D problems there are SIX component equations 41 Static DeterminacvPartial and Improper Constraints Static Indeterminacy occurs when a system has more constraints than is necessary to hold the system in equilibrium ie the system is overconstrained and thus has redundant reactions Static Determinancy occurs when a system has a su cient number of constraints to prevent motion without any redundancy Partial Constraint occurs when there is an insu cient number of reaction forces to prevent motion of the system ie the system is partially constrained Improper Constraint occurs when a system has a su icient number of reaction forces but one or more are improperly applied so as not to prevent motion of the system ie the system is improperly constrained Problem Solving 1 Select body or bodies to be isolated in a FBD 2 Choose an xyz coordinate system 3 Complete FBD slowing all external reaction forcesmoments 4 2M0 O select point 0 to eliminate some unknown constraint forces and simplify the cross products 5 2 F 0 6 Solve simultaneous equations 42 61 TRUSSES Learning Objectives 1 To identify zeroforce members in a structure 2 To recognize planar and space ie threedimensional truss structures 3 To understand the assumptions made in modeling trusses 4 To understand why structures are often designed as trusses Definitions ZeroForce Members structural members that support loading but aid in the stability of the truss T woF orce Members structural members that are a subject to no applied or reaction and b are loaded only at pin joints along the member MultiForce Members structural members that have a applied or reaction moments or b are loaded at more than two points along the member Truss a rigid framework of straight lightweight members that are joined together at their ends Frame a rigid framework of straight andor curved members intended to be a stationary structure for supporting a load 43 Machine an assembly of rigid members designed to do mechanical work by transmitting a given set of input loading forces into another set of output forces Dynamics Simple and Compound Trusses Simple Truss a truss whose number of members is given by m 2j 3 where m no of members andj no ofjoints For simple space trusses the relationship is given by m 3j 6 Compound Truss a truss formed from two or more simple trusses Newton s Third Law For each action there is an and reaction l e D FABodyl FABody2 Assumptions for Modeling 1 All members are 2 All connections are modeled as 3 The centerlines of all members must be at the joint 4 External loads act only at the 5 Weight of members is compared with external loads Advantage of Truss Structures Truss structures can span long distances without intermediate supports eg bridge and roof trusses and can carry heavy loads with lightweight members 44 Applications See text book Two Methods of Solutions 0 Method of Joints 0 Method of Sections Static IndeterminacvPartial Constraint A truss is internally indeterminate if m gt 2j 3 for planar trusses where m no of members m gt 3j 6 for space trusses where j no of joints A truss is improperly constrained if m lt 2j 3 for planar trusses where m no of members m lt 3j 6 for space trusses where j no of joints 45 6163 METHOD OF JOINTS Learning Objectives 1 To employ the methods of joints to evaluate the axial force carried by each member in a truss 2 To identify zeroforce members in a truss 3 To do an engineering estimate of the load distribution in a truss 46 Procedure 1 Draw a FBD of the entire truss showing the reaction forces at the supports and the external loads Write the equilibrium equations and solve for as many unknowns as possible 2 Identify any zeroforce members and any members that carry the same load as other members or external loads 3 Draw a FBD of each joint in the truss Be sure to abide by Newton s Third Law reactions between interacting members are equal and opposite 4 Make a plan for solving the member loads Start with the joint with the least number of unknowns this frequently occurs at the supports In solving the equilibrium equations avoid joints that have more than two unknowns acting on it Remember that since the forces at each joint are concurrent ie they intersect at the joint only two equilibrium equations can be utilized 2 FX 2 0 and ZFy 0 no moment equation exists 5 When through solving go back and state whether each member is in tension or compression That is if a negative value is found for a member Then you assumed the wrong direction ILAT When drawing the FBDs of the joints assume all members are initially in ie show all member forces acting away om the joint Then 0 ifload is positive gt member is in o ifload is negative gt member is in 47 64 METHOD OF SECTIONS Learning Objectives 1 To employ the method of sections to evaluate the axial force carried by selected members in a truss 2 To do an engineering estimate of the load in select members of a truss Procedure 1 Draw a FBD of the entire truss showing the reaction forces at the supports and the external loads Write the equilibrium equations and solve for as many unknowns as possible 2 Locate the force members to be evaluated Identify whether any of these forces can be determined by observation eg zeroforce members 3 Identify section to be used and draw a FBD of the section including any support reactions external loads and internal forces of sectioned members Remember the cutting plane must cut through the members of interest Also the cutting plane need not be straight it may be curved 4 Write the equilibrium equations for one of the two sections The equations for either half of the section will yield the same member forces 5 Three equilibrium equations are available so up to three unknowns can be solved with a single section 6 At times more than one section may be necessary 48 66 FRAMES AND MACHINES Learning Objectives 1 To evaluate the unknown reactions at the supports and the interaction forces at the connection points of a rigid frame in equilibrium by solving the equations of static equilibrium of the overall structure and each individual member 2 To do an engineering estimate of these quantities Definitions T woForce Member a structural member that is loaded only at two pin joints along the member MultiForce Member a structural member that is loaded at more than two points along the member Truss a rigid framework of straight lightweight twoforce members that are joined together at their ends Frame an assembly of rigid members of which at least one is a member intended to be a stationary structure for supporting a load Machine an assembly of rigid members designed to do mechanical work by transmitting a given set of input loading forces into another set of output forces Newton s Third Law Newton s Third Law For each action there is an and 1 reactlon eABOdyl FABody ZJ 49 Frames In frames we are often interested not only in the reaction forces at the supports but also in the interaction forces between members and the loads carried by any twoforce members Procedure 1 Inspect structure for twoforce members 2 Draw FBDs of the entire structure and of each member Be sure the interaction forces between members are equal in magnitude opposite in direction and collinear ie satisfy Newton s Third Law 3 Count the number of unknowns and equations available for each FBD Successiyely write and solve the equilibrium equating corresponding to the FBDs of interest M 1 For a structure composed of N members will be N 1 sets of equilibrium equations and FBDs Only N sets of equations are independent 2 If all external reactions on a frame can be determined then the internal forces between members may be determined from either member 3 If there are more unknowns than available equations gt Statistically Indeterminate This is not always true Sometimes by disassembling the frame the forces can be determined using the equilibrium equations 50 8182 FRICTION Learning Objectives 1 To understand the principles of Coulomb dry friction 2 To evaluate the friction forces required to hold a system in static equilibrium 3 To determine the properties of a system or its loads for which a system will be in a condition of impending motion Introduction 1 Contact surfaces are rough and dry 2 Friction opposes motion or impending motion 51 Dry Friction l Coulomb 5 Law a Friction forces act in a direction opposite to that in which the surfaces move or would tend to move relative to each other b Consider a block of weight W on a rough horizontal surface subjected to a horizontal force P 52 i For no relative motion ie for static equilibrium FPSyN ii For impending motion ie at the instant before slipping FPuNF The maximum value of F for static equilibrium is F zyN iii After motion begins Fiz 53 2 Systems with Friction a Given external loads evaluate the friction force and determine if the body is in static equilibrium Note to be in static equilibrium FSyN i Procedure 1 Assume static equilibrium 2 Solve for F from equilibrium equations 3 Check assumption 0 If F gt SN then assumption incorrect and F kN o If F 3 SN then assumption correct and calculated F is correct b Given condition of impending motion FFyN Note F must be indicated in correct direction 54 8182 FRICTION SLIPPING VS TIPPING Learning Objectives 1 To evaluate two types of impending motion slipping vs tipping to determine which will occur rst 2 To do an engineering estimate of whether a system will slip or tip Two Tvpes of Impending Motion 1 Slipping 2 Tipping 55 Free Body Diagrams Imgending T igging Imgending Sligging Assumptions Imgending T igging Imgending Sligging 0NampFactat edge 0FFS o F S 0 Location ofN is Results Imgending T igging Imgending Sligging 0 Find PT and F 0 Find PS and q 0 HF 3 F5 then tipping o If q lt b then slipping assumption is correct P PT assumption is correct P PS 0 HF gt FS then slipping o If q gt b then tipping occurs rst then tipping occurs rst Comments 1 Generally easiest to assume tipping rst unless you re reasonably sure the system will slip 56 83 WEDGES Learning Objectives 1 To determine the force required to insert andor remove a wedge 2 To determine whether a wedge is selflocking 3 To determine the minimum coe cient of friction necessary for a wedge to be selflocking 4 To determine the minimum force necessary to hold a non selflockmg wedge in place Definitions Wedge a simple machine designed to affect a small change in the position of a system Wedges often experience large normal and friction forces SelfLocking Wedge a wedge in which the friction forces large enough to prevent it from being out Remarks 1 Friction forces always the direction of impending motion 2 Evaluating the condition of impending motion is the only way to determine if a wedge is selflocking 3 F FS SN for impending motion 4 The weight of the wedge is often neglected because of the large normal and frictional forces acting on the wedge 5 Wedges typically have half angles 3 of about 6 so that u between the wood and wedge need be only about 01 to be selflocking 57 85 FLEXIBLE FLAT BELTS Learning Objectives 1 To analyze the forces needed to lift and lower an object with a cable system 2 To determine the maximum torque which can be applied to a pulley system without causing the belt to slip De nition Bells devices designed to transfer forces between bodies eg pulleys sheaves drums etc F lat Belts where T2 larger tension T1 smaller tension 3 angle of wrap in radians r l1 5 static coefficient of friction for a nonslipping bell k kinetic coefficient of friction for a slipping bell 58 2 3 4 5 6 Remarks 1 The angle of wrap 3 must be measured in not degrees The larger tension T2 is always in the direction of The angle of wrap B is easier to determine when measured from a angle close to B The equations above are valid for impending slipping or slipping only As H andor 3 increase the difference between T2 and T1 increases T2 The tension ratio is independent of the pulley radius 59 91 CENTROID AND CENTER OF MASS BY INTEGRATION Learning Objectives 1 To determine the volume mass centroid and center of mass using integral calculus 2 To do an engineering estimate of the volume mass centroid and center of mass of a body Definitions Centroid center of a line area or volume Center of Mass center of a line area or volume The centroid and center of mass coincide when the is throughout the part 60 Centroid bV Integration a M LZL L chdL L chdL b m AZA A Ixch Ayjych c Volume VZV V jxcdv vyjycdv Vizzch where X y Z represent the centroid of the line area or volume X Y 10 represent the centroid of the differential C17 017 element under consideration 61 Center of Mass bV Integration mZmZ dV 1 1 1XG Ixc dmIXcpdV Note 0 For a homogeneous body p constant thus mZdVpZVpV 0 Tabulated values of the centroid and center of mass of several standard shapes can be found on the back inside cover of the textbook 62 92 CENTROID AND CENTER OF MASS BY COMPOSITE PARTS Learning Objectives 1 To evaluate the volume mass centroid and center of mass of a composite body 2 To do an engineering estimate of the volume mass centroid and center mass of a composite body Definitons Centroid geometric center of a line area or volume Center of Mass gravitational center of a line area or volume The centroid and center of mass coincide when the density is uniform throughout the part 63 Centroid bV Composite Parts a Line 11 n n LLi LXZLiXci LvZLiyci 1 b Area 14212141 A Aixci AYZZIIAi ye c Volume 11 Il VZV1 ViZViltxcgti i1 i1 VVZViyci VEZVJZCL i1 i1 where X Y Z centroid of line area or volume Xcia yo i Z91 centroid of individual parts 64 Center of Mass bv Composite Parts m i Xci mi Xci pi Vi n ioamimmmvgt Il n zmxmimxmv i1 X y Z center of mass of the composite body Xe i ye i Z91 center of mass of individual parts Note 0 Tabulated values of centroid and center of mass of several standard shapes can be found on the back inside cover of the textbook 65 9495 FLUID STATIC S Learning Objectives 1 To evaluate the hydrostatic pressure loading acting on a body that is immersed in a liquid 2 To determine the resultant of a given line pressure or body load on a submerged body and to evaluate the reaction acting on the body that carries such a load 3 To do an engineering estimate of the equivalent loading Assumptions 1 The liquid is ie p constant 2 The hydrostatic pressure loading always acts to any submerged surface regardless of orientation 3 The hydrostatic gage pressure at a point in a liquid is proportional to the h below the free surface Hvdrostatic Pressure Distribution p p hydrostatic gage pressure pgage patS patm p density of the liquid g acceleration of gravity h height below the free surface of the liquid 66 121 DYNAMICS Learning Objectives 1 To introduce and de ne the subject of dynamics 2 To introduce Newton s Second Law and understand its signi cance to dynamics 3 To describe the motion of a particle in rectangular cylindrical and path coordinates 4 To convert between coordinate systems Definitions Dynamics the study of bodies in motion and the forces that create the motion Kinematics the study of the of a body ie position velocity and acceleration Kinetics a study of the relationship between the and acting on a body Newton s Second Law Particle a body of in nitely small size such that the spatial distribution of the body is immaterial A body of nite size can be treated as a particle if it purely translates without rotating ie any rotational energy is negligibly small Newton s Second Law Given a net force is applied a body will experience an in the direction of the force which is to the net applied force F 2 1115 67 121123 KINEMATICS OF PARTICLES Learning Objectives 1 To understand the relationship between position X velocity V and acceleration a 2 To describe the rectilinear motion of a particle for constant acceleration 3 To understand how to use the separation of variables techniques to solve rstorder ordinary differential equations General Relationships dr V P P a velocity of point P dZE d aP dtz dt acceleration of point P Alternately 0 W0 Z t dt 0 lttgtltogtZlttgtdt 0 68 Rectilinear Straight Line Motion Constant Acceleration Consider particle P moving along a straight line in the X direction rsi P dr ds V P 1X1V1 F dt t d2r dZST a 2P2 212312a1 P dt dt If the acceleration a constant 69 Separation of Variables Consider seven cases a at V Vt a as V Vs a aV V constant a constant a a at acceleration is a known function of time aat Va v0 21mm b a as acceleration is a known function of position dV dV ds a as EE a chain rule dV a as V E Zltsgtds Zvdv V2 V5 70 c a aV acceleration is a known function of speed dV a aV E or a aV acceleration is a known function of speed dVdV ds aaVEE a cha1nru1e dV aaV V ds Zis2s s0 Ldv 0 0 V a constant VV0at V2 2V3 2as so at2 SZSO 71 e V Vt velocity is a known function of time VVt st 5 s0 ZOVtdt f V Vs velocity is a known function of position VVS Zitzt toz i 0 0 s g V constant g dt Ziszs sO ZitVt t0 0 0 V 72 24 RECTANGULAR CARTESIAN COORDINATES Learning Objectives 1 To describe the motion of a particle in rectangular cartesian coordinates Position Vector TXiyjzE Velocity Vector d D D D X1yJzk v a Acceleration Vector dV d2 a 2 a a azgijE Definitions aX i X component of acceleration ay y y component of acceleration aZ Z z component of acceleration 73 Projectile Motion Neglecting Wind Resistance l ZFX 0max m i ZFy mgmay my gt gt2 2y 2 Treat each orthogonal direction independently and use constant acceleration equations Horizontal Motion aE 0 VX VOX aXt 2 VX VOXZ 23XX XO 2 X XO V0Xt aXt Vertical Motion a 2 Vy VOy ayt VX VOX 2 VX 2 VOL XX V t vy my gt 2 2 2 2 VyVoy 2ayy yo 3 Vy V0y 2gy yo 1 2 y yO V0yt ayt 1 FY V tgt2 0 0 y 2 74 127 PATH VARIABLES Normal and Tangential Components Learning Objectives 1 To describe the motion of a particle in path variables Geometrv and Unit Vectors s position along path rop position vector from the original to poing P along the path ut tangential unit vector un normal unit vector Position Vector E as mso Velocity Vector V ri Ti drep E quot p dt ds dt dQZU Note ds t vpzvtutz Vt tangential velocity 75 Acceleration Vector i D n11 D apzvp sut sutsdt dEs v Nole dt un un 0 0 where R 2 3 d p dly radius of curvature of the path 12 12 V2 2 lap 392 002 magnitude of the acceleration of particle p Definitions D DD at tangential acceleration V S rate of change of speed D2 2 V s an normal acceleration F F centripetal acceleration 76 ab binormal acceleration O 77 128 POLAR DESCRIPTION Learning Objectives 1 To describe the motion of a particle in cylindrical coordinates Geometrv and Unit Vectors r radial distance 6 transverse angle z axial distance xrcos6 yrsin6 r 1 X2 y2 e tan 1yX Position Vector r0pru zk 1 Note Unit vectors ureand u change orientation as the angle 6 changes but the unit vector E is always in the vertical direction 78 Velocity Vector r irFZE f3rfzEzE P dt 0P dt r r r E Note Egg 7 Where 9 dt angular velocity radianssecond 39 vp ev9u 42v u v k 1 Definitions Vr radial velocity Ve tranverse velocity Vz vertical velocity 79 Acceleration Vector d d D a V ru r6u Zk p dtp dt T e i D D1 Di 39r39urrurr6uer6uer6ueik2k 9 P D D Note 119 9 Ur Definitions ar radian acceleration 39 transverse acceleration a2 vertical acceleration D 2 T 9 Coriolis acceleration 80 121128 JOINT KINEMATICAL DESCRIPTIONS Learning Objectives To use joint kinematical descriptions to solve dynamics problems Position r0p Xi yjzk rur zk Velocity Vp X1 y Zk Cartesian Coordinates D rur r e 119 Zk Cylindrical Coordinates ut Path Variables Acceleration ap Cartesian Coord Cylindrical Coord Path Variables Note 1 If two vectors are equal their components are also m before components can be equated both vectors must be written in terms of the same unit vectors 82 Solution Procedure for Joint Kinematical Descriptions 1 2 3 4 5 6 7 Read problem carefully and express all quantitative information in equation form Draw a free body diagram and make a sketch of the path if needed Choose appropriate coordinate systems based on given and requested information Write velocity in terms of each description and substitute in for any parameters that are given or can be found by geometry Express one set of unit vectors in terms of the other and substitute into velocity formulas Match like components of velocity and solve resulting equations Repeat procedure for the acceleration equation 83 1291210 RELATIVE MOTION Learning Objectives 1 To evaluate the relative motion between two moving objects 2 To evaluate the absolute motion of a moving object from the perspective of a second moving object Absolute Quantities TB VB 33 84 Relative Quantities rBAv VBAa aBA 1 13A VBA aBA where TB VB 33 absolute position velocity and acceleration of point B TA VA aA absolute position velocity and acceleration of point A rBA VBA aBA relative position velocity and acceleration of B with respect A 85 1291210 CABLEPULLEY SYSTEMS Learning Objectives 1 To learn a simple solution procedure for solving cable pulley systems Assumptions 1 Cables are assumed 2 Cables are assumed Solution Procedure 1 Write an equation for the constrained length of the cable 2 Locate each of the connected bodies by its distance from a xed point of reference 3 Sum the length of each free portion of the cable ignoring any constant values and equate the sum to a constant Note Positive unit vectors are in direction of increasing s 86 131134 KINETICS OF A PARTICLE Learning Objectives 1 To introduce Newton s Second Law and understand its signi cance to dynamics 2 To describe the rectilinear motion of a particle for non constant acceleration 3 To understand how to separate variables Definitions Dynamics the study of bodies in motion and the forces that create the motion Kinematics the study of the motion of a body ie position velocity and acceleration Kinetics a study of the relationship between the and acting on a body Newton s Second Law Particle a body of in nitely small size such that the spatial distribution of the body is immaterial A body of nite size can be treated as a particle if it purely translates without rotating ie any rotational energy is negligibly small Newton s Second Law Given a net force is applied a body will experience an acceleration in the direction of the force which is proportional to the net applied force GE 2 m5 87 Equations of Rectilinear Motion Vector Equation 2 F ma Scalar Equations ZFt mat mlt1m 2F 2 man 2 m EKG since p 00 Z Fb mab 0 by de nition Separation of Variables Consider particle P moving along a straight line in the X direction r si P dr dST 39 39 V P 1X1V1 P t dt 2 2 drPdST39 39 a 1 51 a1 88 Separation of Variables Consider seven cases a at V Vt a as V Vs a aV V constant a constant a a at acceleration is a known function of time aat Va v0 21mm b a as acceleration is a known function of position dV dV ds a as EE a chain rule dV a as V E Zltsgtds ZVdV V2 V5 89 c a aV acceleration is a known function of speed dV a aV E or a aV acceleration is a known function of speed dVdV ds aaVEE a cha1nru1e dV aaV V ds Zis2s s0 Ldv 0 0 V a constant VV0at V2 2V3 2as so at2 SZSO 90 e V Vt velocity is a known function of time VVt st 5 s0 ZOVtdt f V Vs velocity is a known function of position VVS Zitzt toz i 0 0 s g V constant g dt Ziszs sO ZitVt t0 0 0 V 91 131136 EQUATIONS OF CURVILINEAR MOTION Learning Objectives 1 To review the descriptions of the acceleration of a particle in cartesian cylindrical and path coordinates 2 To recall Newton s Second Law and understand how it de nes a relationship between the net force and acceleration of a particle 3 To de ne the kinetic relationships of a particle in rectangular cylindrical and path coordinates 4 To examine the degenerate case of trajectory problems Acceleration azaXiayjaZEiiyjik D2 D1 D aruraeueazkr r9 urr62r9 ueik 92 Newton s Second Law Given a net force is applied a body will experience an acceleration in the direction of the force which is proportional to the net applied 2 F 2 m5 Equate 3 components of the vector equation Kinetics of a Particle 1 Path Coordinates ZFt mat ZFn 2 man 2 ZFb mab Where ab 2 binorrnal acceleration that is the acceleration normal to at and an By de nition ab O 93 2 Rectangular Cartesian Coordinates ZFX maX ZFy may ZFZ maZ 3 Polar Coordinates 2Fr 2 mar ZF0 ma0 ZFZ maZ 94 Projectile Motion Neglecting Wind Resistance 1ZFXOmaXmX 3 i0 ZFyzmgzmayzmji gt jizg 2 Treat each orthogonal direction independently and use constant acceleration equations Horizontal Motion aE 0 VX VOX aXt VX VOX 2 VX VOX2 23XX XO 3 VX V0X 1 2 XXOVO t aXt XX Vt X 2 o 0 X Vertical Motion a g y VOy ayt Vy V0y gt V 2 Vy Voy2 2ayy yo 3 Vyz Voy2 2gy y0 1 2 1 yy0voyt ayt yy0V0yt Egt2

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