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ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx 1 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Bruce Mayer, PE Licensed Electrical & Mechanical Engineer Engineering 36 Chp 5: 2D Equil Special Cases

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ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx 2 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics 2D Equil Special Cases PARTICLE: Size & Shape of the Object can be neglected as long as all applied Forces have a Point of Concurrency Covered in Detail in Chp03 TWO-FORCE MEMBER: A Structural Element of negligible Wt with only 2 Forces acting on it

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ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx 3 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics 2D Equil Special Cases THREE-FORCE MEMBER: A structural Element of negligible Wt with only 3 Forces acting on it The forces must be either concurrent or parallel. –In the PARALLEL Case the PoC is located at Infinity –The NONparallel Case can be Very Useful in Load Analysis

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ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx 4 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics 2D Equil Special Cases FRICTIONLESS PULLEY: For a frictionless pulley in static equilibrium, the tension in the cable is the same on both sides of the pulley Discussed Briefly in Chp03 –Will Prove the T 1 = T 2 = T Behavior Today

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ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx 5 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics 2D Planar System Equilibrium In 2D systems it is assumed that The System Geometry resides completely the XY Plane There is NO Tendency to –Translate in the Z-Direction –Rotate about the X or Y Axes These Conditions Simplify The Equilibrium Equations

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ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx 6 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics 2D Planar System: No Z-Translation NO Z-Directed Force: No X or Y Rotation NO X or Y Applied Moments

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ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx 7 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Special Case: 2-Force Member A 2-Force Member/Element is a Body with negligible Weight and Only two applied Forces. Some Special Properties of 2-Frc Eles the LoAs of the Two Forces MUST Cross and thus Produce a PoC –Treat as a PARTICLE The Crossed LoAs Define a PLANE –Treat as PLANAR System

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ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx 8 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics 2-Force Element Equilibrium Consider a L-Bracket plate subjected to two forces F 1 and F 2 For static equilibrium, the sum of moments about Pt-A must be zero. Thus the moment of F 2 About Pt-A must be zero. It follows that the line of action of F 2 must pass through Pt-A Similarly, the line of action of F 1 must pass through Pt-B for the sum of moments about Pt-B to be zero. Requiring that the sum of forces in any direction be zero leads to the conclusion that F 1 and F 2 must have equal magnitude but opposite sense.

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ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx 9 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Special Case: 2-Force Element Mathematically Since the Two Forces Must be Concurrent Since the System is in Equilibrium ΣFs =0. –Thus the two force are Equal and Opposite; that is, the forces CANCEL

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ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx 10 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Special Case: 3-Force Member A 3-Force Element is a PLANAR Body with negligible Weight with Exactly 3 applied Forces (No applied Moments). Claim: If a Planar 3-Force Element is in Equilibrium, Then the LoAs for the 3-Forces must be CONCURRENT If the Claim is TRUE, then the 3-Force Element can be treated as a PARTICLE

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ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx 11 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics 3-Force 2D Body Equilibrium Consider a Planar rigid body subjected to forces acting at only 3 points. The lines of action of intersect F 1 & F 2, at Pt-D. The moment of F 1 and F 2 about this point of intersection is zero. Since the rigid body is in equilibrium, the sum of the moments of F 1, F 2, and F 3 about ANY Pivot-Pt must be zero. It follows that the moment of F 3 about D must be zero as well and that the line of action of F 3 must pass through D. The lines of action of the three forces must be Concurrent OR Parallel.

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ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx 12 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics 3-Force 2D Body: Parallel Forces F1F1 F2F2 F3F3 d1d1 d2d2 d3d3 O If 3 Parallel Forces Maintain a Rigid Body in Static Equilibrium, The following Conditions MUST be Satisfied For Translation Equilibrium For Rotation Equilibrium x

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ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx 13 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Special Case: 3-Force Element Mathematically for ||-Forces Since a Body in Equil. Has NO Net Moment Since the System is in Equilibrium ΣFs =0. In Summary: The d m F m products and, 3 Forces, Sum to Zero

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ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx 14 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Special Case: 3-Force Element A Graphical Summary AB is 3F Member (BC is 2F Member)

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ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx 15 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Example Pole Raising Solution Plan Create a free-body diagram of the joist. –Note that the joist is a 3 force body acted upon by the ROPE, its WEIGHT, and the REACTION at A The three forces must be concurrent for static equilibrium. Therefore, the reaction R must pass through the intersection of the lines of action of the weight and rope forces. Determine the direction of the reaction force R. A man Raises a 10 kg Joist, of Length 4 m, by pulling on a rope. Find the TENSION in the rope and the REACTION at A.

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ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx 16 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Example Pole Raising Use LoAs & Trigonometry to Determine the direction of the reaction force R Create a free-body diagram of the joist A LARGE, SCALED Diagram is REALLY Useful in this Problem

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ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx 17 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics 70º Angle Analysis

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ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx 18 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Example Pole Raising Use the Law of the Sines to Find the Reaction Force R Draw the Force Triangle to Scale Solving find

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ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx 19 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Special Case: Frictionless Pulley A FrictionLess Pulley is Typically used to change the Direction of a Cable or Rope in Tension Pulley with PERFECT Axel (FrictionLess)

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ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx 20 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Special Case: FrictionLess Pulley A Perfect Axel Generates NO moment to Resist Turning. Consider the FBD for a Perfect Pulley Since the LoAs for F Ax & F Ay Pass Thru the Axel-Axis Pt-A they Generate No moment about this point. T 1 and T 2 have Exactly the SAME Lever arm, i.e., the Radius, R, of the Pulley

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ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx 21 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Special Case: FrictionLess Pulley Since the Pulley is in Equilibrium ΣM A = 0 Writing the Moment Eqn Thus for the NO-Friction Perfect Pulley

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ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx 22 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics FritionFilled Pulley Consider the case where we have a pulley that is NOT Free Wheeling; i.e., the pulley resists rotation Example: Automobile alternator changes thermal-mechanical energy into electrical energy

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ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx 23 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics FrictionFilled Pulley In Alternator Operation the generation of electricity produces a resisting moment that counters the direction of spin; The FBD in this case The ΣM A = 0 M Az

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ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx 24 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics FrictionFilled Pulley Thus a RESISTING Moment causes a DIFFERENCE between the two Tensions More on This when we Learn Chp08 M Az

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ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx 25 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics FrictionLess Pulley; 3F Mem In the System at Right Member ABC, which is a FOUR- Force System, can be reduced to a 3-Force System using and Equivalent Resultant-Couple System at the Pulley

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ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx 26 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics FrictionLess Pulley; 3F Mem Recall that Forces Can be MOVED to a new point on a Body as long as the Rotation Tendency caused by the move is accounted for by the Addition of a COUPLE-Moment at the new Point

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ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx 27 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics FrictionLess Pulley; 3F Mem Apply the Equivalent Loading Method to a FrictionLess Pulley From the Previous Discussion the MOMENT about the Axle (Pin) of a Frictionless pulley produced by the Tensions is ZERO Thus Can Move the Ts to the Pin with a Couple of ZERO

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ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx 28 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics FrictionLess Pulley; 3F Mem The Equivalent Systems by M A = 0

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ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx 29 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics FrictionLess Pulley; 3F Mem Moving the FrictionLess Pulley Force- Resultant to the Pin at Pt-A produces the FBD Shown At Right Now can Draw the Force Triangle

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ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx 30 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics FrictionLess Pulley - Important For a FrictionLess Pulley the Tension Forces and be to the Pulley Axel (Pin) WithOUT the Addition of a Couple =

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ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx 31 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Special Cases Summarized Particle: 3-Force Planar Element: 2-Force Element: FrictionLess Pulley:

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ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx 32 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics WhiteBoard Work Lets Work These Nice Problems

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ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx 33 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Bruce Mayer, PE Registered Electrical & Mechanical Engineer Engineering 36 Appendix

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ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx 34 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Jib Problem The upper portion of the crane boom consists of the jib AB, which is supported by the pin at A, the guy line BC, and the backstay CD, each cable being separately attached to the mast at C. If the 5-kN load is supported by the hoist line, which passes over the pulley at B, determine the magnitude of the resultant force the pin exerts on the jib at A for equilibrium, the tension in the guy line BC, and the tension T in the hoist line. Neglect the weight of the jib. The pulley at B has a radius of 0.1 m.

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ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx 35 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Disk Problem The smooth disks D and E have a weight of 200 lb and 100 lb, respectively. Determine the largest horizontal force P that can be applied to the center of disk E without causing the disk D to move up the incline.

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