Mech chapter 03

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  1. 1. Engineering Mechanics: Statics in SI Units, 12e3Equilibrium of a ParticleCopyright 2010 Pearson Education South Asia Pte Ltd
  2. 2. Chapter Outline3.1 Condition for the Equilibrium of a Particle3.2 The Free-Body Diagram3.3 Coplanar Systems3.4 Three-Dimensional Force Systems Copyright 2010 Pearson Education South Asia Pte Ltd
  3. 3. 3.1 Condition for the Equilibrium of a Particle Particle at equilibrium if - At rest - Moving at a constant velocity Newtons first law of motion F = 0 where F is the vector sum of all the forces acting on the particle Copyright 2010 Pearson Education South Asia Pte Ltd
  4. 4. 3.2 The Free-Body Diagram A sketch showing the particle free from thesurroundings with all the forces acting on it -- Consider two common connections in this subject Spring Cables and Pulleys Copyright 2010 Pearson Education South Asia Pte Ltd
  5. 5. 3.2 The Free-Body Diagram Spring Linear elastic spring: change in length is directly proportional to the force acting on it spring constant or stiffness k / : defines the elasticity of the spring Magnitude of force when spring is elongated or compressed F = ksCopyright 2010 Pearson Education South Asia Pte Ltd
  6. 6. 3.2 The Free-Body Diagram Cables and Pulley Cables (or cords) are assumed negligible weight andcannot stretch Tension always acts in the direction of the cable Tension force must have a constant magnitude forequilibrium For any angle , the cableis subjected to a constant tension TCopyright 2010 Pearson Education South Asia Pte Ltd
  7. 7. 3.2 The Free-Body Diagram Procedure for Drawing a FBD 1. Draw outlined shape 2. Show all the forces - Active forces: particle in motion - Reactive forces: constraints that prevent motion 3. Identify each forces - Known forces () with proper magnitude and direction - Letters () used to represent magnitude and directions of unknown forces ()Copyright 2010 Pearson Education South Asia Pte Ltd
  8. 8. Example (3-2)The sphere has a mass of 6kg and is supported.Draw a free-body diagram of the sphere, the cord CE andthe knot at C. Copyright 2010 Pearson Education South Asia Pte Ltd
  9. 9. SolutionFBD at SphereTwo forces acting, weight and theforce on cord CE.Weight of 6kg (9.81m/s2) = 58.9N Copyright 2010 Pearson Education South Asia Pte Ltd
  10. 10. SolutionCord CETwo forces acting: sphere and knotNewtons 3rd Law:FCE is equal but oppositeFCE and FEC pull the cord in tensionFor equilibrium, FCE = FECCopyright 2010 Pearson Education South Asia Pte Ltd
  11. 11. SolutionFBD at Knot C3 forces acting: cord CBA, cord CE and spring CDImportant to know that the weight of the sphere does notact directly on the knot but subjected to by the cord CE Copyright 2010 Pearson Education South Asia Pte Ltd
  12. 12. 3.3 Coplanar Systems A particle is subjected to coplanar forces in the x-yplane Resolve into i and j components for equilibriumFx = 0Fy = 0 Scalar equations of equilibriumrequire that the algebraic sumof the x and y components toequal to zeroCopyright 2010 Pearson Education South Asia Pte Ltd
  13. 13. 3.3 Coplanar Systems Procedure for Analysis 1. Free-Body Diagram - Establish the x, y axes - Label all the unknown and known forces2. Equations of Equilibrium - Apply F = ks to find spring force - Negative result force is the reserve - Apply the equations of equilibriumFx = 0Fy = 0Copyright 2010 Pearson Education South Asia Pte Ltd
  14. 14. Example (3-3)Determine the required length of the cord AC so that the8kg lamp is suspended.The undeformed length of the spring AB is lAB = 0.4m,and the spring has a stiffness of kAB = 300N/m. Copyright 2010 Pearson Education South Asia Pte Ltd
  15. 15. SolutionFBD at Point AThree forces acting, force by cable AC,force in spring AB and weight of the lamp.If force on cable AB is known, stretch ofthe spring is found by F = ks.+ Fx = 0; TAB TAC cos30 = 0+Fy = 0; TABsin30 78.5N = 0Solving,TAB = 136.0kNTAC = 157.0kNCopyright 2010 Pearson Education South Asia Pte Ltd
  16. 16. SolutionTAB = kABsAB 136.0 (N) = 300 (N/m) sAB (m)sAB = 0.453 mFor stretched length,lAB = lAB+ sABlAB = 0.4m + 0.453m= 0.853 mFor horizontal distance BC,2m = lACcos30 + 0.853mlAC = 1.32 m (Ans.)Copyright 2010 Pearson Education South Asia Pte Ltd
  17. 17. Test (3-1, 3-2, 3-3) Determine the tension in cables AB, BC, and CD,necessary to support the 10-kg and 15-kg traffic lights atB and C, respectively. Also, find the angle .Copyright 2010 Pearson Education South Asia Pte Ltd
  18. 18. 3.4 Three-Dimensional Force Systems For particle equilibriumF = 0 Resolving into i, j, k componentsFxi + Fyj + Fzk = 0 Three scalar equations representing algebraic sums ofthe x, y, z forces Fxi = 0 Fyj = 0 Fzk = 0 Copyright 2010 Pearson Education South Asia Pte Ltd
  19. 19. 3.4 Three-Dimensional Force Systems Procedure for AnalysisFree-body Diagram - Establish the z, y, z axes - Label all known and unknown forceEquations of Equilibrium- Apply Fx = 0, Fy = 0 and Fz = 0- Substitute vectors into F = 0 and set i, j, k components = 0- Negative results indicate that the sense of the force isopposite to that shown in the FBD.Copyright 2010 Pearson Education South Asia Pte Ltd
  20. 20. Example (3-4)Determine the force developed in each cable used tosupport the 40kN crate. Copyright 2010 Pearson Education South Asia Pte Ltd
  21. 21. SolutionFBD at Point ATo expose all three unknown forces in the cables.Equations of EquilibriumExpressing each forces in Cartesian vectors,FB = FB uB = FB (rB / rB) = -0.318FBi 0.424FBj + 0.848FBkFC = FC (rC / rC) = -0.318FCi + 0.424FCj + 0.848FCkFD = FDiW = -40k Copyright 2010 Pearson Education South Asia Pte Ltd
  22. 22. SolutionFor equilibrium,F = 0;FB + FC + FD + W = 0-0.318FB i 0.424FB j + 0.848FB k-0.318FC i + 0.424FC j + 0.848FC k +Fd i-40 k = 0Fx = 0; -0.318FB - 0.318FC + FD = 0Fy = 0; -0.424FB +0.424FC = 0Fz = 0;0.848FB +0.848FC - 40 = 0Solving,FB = FC = 23.6kN FD = 15.0kN (Ans.)Copyright 2010 Pearson Education South Asia Pte Ltd
  23. 23. Test (3-4) Determine the tension developed in cables AB, AC, andAD. Copyright 2010 Pearson Education South Asia Pte Ltd
  24. 24. END OF CHAPTER 3 Copyright 2010 Pearson Education South Asia Pte Ltd
  25. 25. 3.1 Condition for the Equilibrium of a Particle Newtons second law of motion F = ma When the force fulfill Newtons first law of motion,ma = 0a=0 therefore, the particle is moving in constant velocity or at rest Copyright 2010 Pearson Education South Asia Pte Ltd