Statics - SMJP1033 Engineering Mechanics PDF

STATICS – SMJP1033 ENGINEERING MECHANICS CHAPTER 2 STATICS OF PART ICLES Study of the effect of forces acting on a particle. - A mean to produce motion OR destroy/tend to destroy a motion: - Bodies where the s...

STATICS – SMJP1033 ENGINEERING MECHANICS CHAPTER 2 STATICS OF PART ICLES Study of the effect of forces acting on a particle. - A mean to produce motion OR destroy/tend to destroy a motion: - Bodies where the size & shape - Body at rest will not affect the solution. - All forces are assumed to be Body in motion applied at the same point. - Body in motion Body accelerate or retarded STATICS OF PART ICLES 2.1 Resultant 2D R≠0 2.2 Equilibrium 2D R=0 2.3 Resultant 3D R≠0 2.4 Equilibrium 3D R=0  RESULTANT 2D CHAPTER 2.1 - FORCE AS A VECTOR AND RESULTANT - Characteristics of Forces Acting on A Particle Direction (Line of action) Point of applications Replacing multiple forces acting on a particle with a single equivalent force or resultant force  RESULTANT 2D - FORCE AS A VECTOR AND RESULTANT - Problem Solving: 1. Free Body Diagram (FBD) Trigonometry Method 2. Analysis of problem; (Sin & Cosine Rules) Rectangular Component Method (Theorem Pythagoras)  Example 2.1 : Determine the resultant of the two forces acting at point A. Problem Solving: 1. Free Body Diagram (FBD) Ans: 165.2 N 2. Analysis of problem; Ꝋ : 83.4°  Example 2.2 : The two forces, P and Q act on a bolt A. Determine their resultant force. Problem Solving: 1. Free Body Diagram (FBD) Ans: 97.7 N 2. Analysis of problem; Ꝋ : 35°  Example 2.3: Four forces act on a bolt A as shown. Determine the resultant of the forces on the bolt. Problem Solving: 1. Free Body Diagram (FBD) Ans: 199.6 N 2. Analysis of problem; Ꝋ: 4.1°  Exercise 1 If θ = 60˚ and F = 450 N , determine the magnitude of the resultant force and its direction, measured counterclockwise from the positive x axis. Ans: FR = 497 N ø = 155˚  Exercise 2 Determine the magnitude of the resultant force FR = F1 + F2 and its direction, measured counterclockwise from the positive x axis. Ans: FR = 393 N ø = 353˚  Exercise 3 The vertical force, F acts downward at on the two-membered frame. Determine the magnitudes of the two components of F directed along the axes of AB and AC. Set F = 500 N. Ans: FAB = 448 N FAC = 366 N  Exercise 4 Determine the magnitude of the resultant force and its direction, measured counterclockwise from the positive x axis. Ans: FR = 217 N ø = 87.0˚  Exercise 5 Determine the magnitude of the resultant force and its direction measured counterclockwise from the positive x axis. Ans: FR =413N ø = 24.2˚  Exercise 6 Express each of the three forces acting on the support in Cartesian vector form and determine the magnitude of the resultant force and its direction, measured clockwise from positive x axis. Ans: FR = 54.2 N ø = 43.5˚  Exercise 7 Determine the magnitude of the resultant force and its direction, measured counterclockwise from the positive x axis. Ans: FR = 12.5 kN ø = 64.1˚  EQUILIBRIUM 2D CHAPTER 2.2 RESULTANT FORCE = 0 Newton’s First Law: If the resultant force on a particle is zero, the particle will remain at rest or will continue at constant speed in a straight line. F1 F2 F1 F2 R F3 R≠0 R=0 Trigonometry Method (Sin & Cosine Rules) Problem Solving: 1. Free Body Diagram (FBD) Rectangular Component Method 2. Analysis of problem (Theorem Pythagoras) Example 2.4 : Determine the tension in cables BA and BC for the system to be in equilibrium. 20º *Must Remember! ∑F = 0 1. Trigonometry Method (Sine & Cosine Rules) 2. Rectangular Component Method (Theorem Pythagoras) Ans: TBC = 1735.5 N TBA = 2128.9 N  Example 2.5 : Determine the tension in cables AB and AC for the system to be in equilibrium. *Must Remember! ∑F = 0 Problem Solving: Ans: 1. Free Body Diagram (FBD) TAB = 3.57 kN 2. Analysis of problem TAC = 0.144 kN Example 2.6 : Prove that the system is in equilibrium. *Must Remember! ∑F = 0 Rectangular Component Method (Theorem Pythagoras)  Example 2.7: The cable system is used to maintain the position of mass m. If a 500 N force is acting at B as shown, determine the tension in cables BC, CD and the mass m m Ans: Problem Solving: TBC = 2835.64 N 1. Free Body Diagram (FBD) TCD = 8290.86 N 2. Analysis of problem m = 794.18 kg Example 2.8: The system shown in the figure is in equilibrium. Determine the tension in cable AB and the angle ϴ Ans: TAB = 297.45 N ϴ = 58.5º CONCEPT PULLEY Example 2.9: Determine the FBD of the given pulley systems. Example 2.10: The force P is acting at point D to maintain the system in equilibrium. Determine the magnitude of force P. *Keyword: TBC = TCD P Ans: P = 69.1 N Example 2.11: Determine the mass m1 to maintain mass m2 = 10 kg at the position shown Ans: m1 = 19.39 kg  Exercise 1 If the tension in cable AC is 20N, determine the tension in cable BC. Determine also the magnitude and direction of force P so that the equilibrium is maintained. Ans: TBC = 30.7 N P = 24.7 𝛼 = 37.2º  Exercise 2 The load at E with a mass of 6 kg is supported as shown. Determine the tension in the spring and the tension in cable AB if the system is in equilibrium. Ans: TSpring = 34 N TAB = 48.1 N  Exercise 3 Determine the force F and angle ϴ if the system shown is in equilibrium Ans: F = 16.92 N TCD = 8291 N m = 794.2 kg  Exercise 4 Two cables are tied together at C and are loaded as shown. Determine the tension in cable AC, and tension in cable BC. Ans: TAC = 352 N TBC = 261 N  Exercise 5 Two cables are tied together at C and are loaded as shown. Knowing that 𝛼 = 30º, determine the tension in cable AC and in cable BC. Ans: TAC = 5.22 kN TBC = 3.45 kN  Exercise 6 Knowing that 𝛼 = 20º, determine the tension in cable AC and in rope BC. Ans: TAC = 1.244 kN RBC = 115.4 N  Exercise 7 A sailor is being rescued using a boatswain’s chair that is suspended from a pulley that can roll freely on the support cable ACB and is pulled at a constant speed by cable CD. Knowing that 𝛼 = 30º and β = 10º and that the combined wight of the boatswain’s chair and the sailor is 900 N, determine the tension in the support cable ACB and the traction cable CD. Ans: TACB = 1.213 kN TBC = 166.3 N  Exercise 7 A load Q is applied to pulley C which can roll on the cable ACB. The pulley is held in the position shown by a second cable CAD, which passes over the pulley A and supports a load P. Knowing that P = 750 N, determine the tension in cable ACB and the magnitude of load Q. Ans: TACB = 1293 N FQ = 2220 N THANK YOU