Mech chapter 03

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Mech chapter 03

  1. 1. Engineering Mechanics: Statics in SI Units, 12e 3 Equilibrium of a Particle Copyright © 2010 Pearson Education South Asia Pte Ltd
  2. 2. Chapter Outline 3.1 Condition for the Equilibrium of a Particle 3.2 The Free-Body Diagram 3.3 Coplanar Systems 3.4 Three-Dimensional Force Systems Copyright © 2010 Pearson Education South Asia Pte Ltd
  3. 3. 3.1 Condition for the Equilibrium of a Particle 質點平衡的條件 • Particle at equilibrium if - At rest - Moving at a constant velocity • Newton’s 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 自由體圖 • A sketch showing the particle “free” from the surroundings with all the forces acting on it • 自由體圖 -- 將此質點從周圍的環境分離出來,並標示作 用在此質點之上的所有力 • 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 彈簧 – Linear elastic spring: change in length is directly proportional to the force acting on it – spring constant or stiffness k 彈簧常數 / 硬性 : defines the elasticity of the spring – Magnitude of force when spring is elongated or compressed  F = ks Copyright © 2010 Pearson Education South Asia Pte Ltd
  6. 6. 3.2 The Free-Body Diagram • Cables and Pulley 繩索與滑輪 – Cables (or cords) are assumed negligible weight and cannot stretch – Tension always acts in the direction of the cable – Tension force must have a constant magnitude for equilibrium – For any angle θ, the cable is subjected to a constant tension T Copyright © 2010 Pearson Education South Asia Pte Ltd
  7. 7. 3.2 The Free-Body Diagram Procedure for Drawing a FBD 畫自由體圖的步驟 1. Draw outlined shape 畫出外形 2. Show all the forces 標示所有力 - Active forces: particle in motion 使質點運動的作用力 - Reactive forces: constraints that prevent motion 阻止質點運動的反作用力 3. Identify each forces 辨識所有的力 - Known forces (已知力) with proper magnitude and direction - Letters (字母) used to represent magnitude and directions of unknown forces (未知力) 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 and the knot at C. Copyright © 2010 Pearson Education South Asia Pte Ltd
  9. 9. Solution FBD at Sphere Two forces acting, weight and the force on cord CE. Weight of 6kg (9.81m/s2) = 58.9N Copyright © 2010 Pearson Education South Asia Pte Ltd
  10. 10. Solution Cord CE Two forces acting: sphere and knot Newton’s 3rd Law: FCE is equal but opposite FCE and FEC pull the cord in tension For equilibrium, FCE = FEC Copyright © 2010 Pearson Education South Asia Pte Ltd
  11. 11. Solution FBD at Knot C 3 forces acting: cord CBA, cord CE and spring CD Important to know that the weight of the sphere does not act 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 共平面力系 • A particle is subjected to coplanar forces in the x-y plane • Resolve into i and j components for equilibrium ∑Fx = 0 ∑Fy = 0 • Scalar equations of equilibrium require that the algebraic sum of the x and y components to equal to zero Copyright © 2010 Pearson Education South Asia Pte Ltd
  13. 13. 3.3 Coplanar Systems • Procedure for Analysis 解題步驟 1. Free-Body Diagram 畫自由體圖 - Establish the x, y axes - Label all the unknown and known forces 2. Equations of Equilibrium 使用平衡方程式 - Apply F = ks to find spring force - Negative result force is the reserve 負值結果代表方向相反 - Apply the equations of equilibrium ∑Fx = 0 ∑Fy = 0 Copyright © 2010 Pearson Education South Asia Pte Ltd
  14. 14. Example (3-3) Determine the required length of the cord AC so that the 8kg lamp is suspended. The undeformed length of the spring AB is l’AB = 0.4m, and the spring has a stiffness of kAB = 300N/m. Copyright © 2010 Pearson Education South Asia Pte Ltd
  15. 15. Solution FBD at Point A Three forces acting, force by cable AC, force in spring AB and weight of the lamp. If force on cable AB is known, stretch of the spring is found by F = ks. +→ ∑Fx = 0; TAB – TAC cos30º = 0 +↑ ∑Fy = 0; TABsin30º – 78.5N = 0 Solving, TAB = 136.0kN TAC = 157.0kN Copyright © 2010 Pearson Education South Asia Pte Ltd
  16. 16. Solution TAB = kABsAB 136.0 (N) = 300 (N/m) sAB (m) sAB = 0.453 m For stretched length, lAB = l’AB+ sAB lAB = 0.4m + 0.453m = 0.853 m For horizontal distance BC, 2m = lACcos30° + 0.853m lAC = 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 at B and C, respectively. Also, find the angle Θ. Copyright © 2010 Pearson Education South Asia Pte Ltd
  18. 18. 3.4 Three-Dimensional Force Systems 三維力系 • For particle equilibrium ∑F = 0 • Resolving into i, j, k components ∑Fxi + ∑Fyj + ∑Fzk = 0 • Three scalar equations representing algebraic sums of the 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 Analysis Free-body Diagram - Establish the z, y, z axes - Label all known and unknown force Equations 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 is opposite 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 to support the 40kN crate. Copyright © 2010 Pearson Education South Asia Pte Ltd
  21. 21. Solution FBD at Point A To expose all three unknown forces in the cables. Equations of Equilibrium Expressing each forces in Cartesian vectors, FB = FB uB = FB (rB / rB) = -0.318FBi – 0.424FBj + 0.848FBk FC = FC (rC / rC) = -0.318FCi + 0.424FCj + 0.848FCk FD = FDi W = -40k Copyright © 2010 Pearson Education South Asia Pte Ltd
  22. 22. Solution For 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 = 0 ∑Fx = 0; -0.318FB - 0.318FC + FD = 0 ∑Fy = 0; -0.424FB +0.424FC = 0 ∑Fz = 0; 0.848FB +0.848FC - 40 = 0 Solving, 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, and AD. 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 • Newton’s second law of motion ∑F = ma • When the force fulfill Newtons first law of motion, ma = 0 a=0 therefore, the particle is moving in constant velocity or at rest Copyright © 2010 Pearson Education South Asia Pte Ltd

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