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Basics and statics of particles unit i - GE6253 PPT

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THANGA KASI RAJAN,S
ASSISTANT PROFESSOR,
KAMARAJ COLLEGE OF ENGINEERING AND TECHNOLOGY,
VIRUDHUNAGAR

Published in: Engineering
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Basics and statics of particles unit i - GE6253 PPT

  1. 1. ENGINEERING MECHANICS Unit – I Basics and Statics of Particles by S.Thanga Kasi Rajan Assistant Professor Department of Mechanical Engineering Kamaraj College of Engineering & Technology, Virudhunagar – 626001. Tamil Nadu, India Email : stkrajan@gmail.com
  2. 2. INTRODUCTION 02/01/2017 S.ThangaKasiRajan, stkrajan@gmail.com 2
  3. 3. INTRODUCTION To deal with the above situations, we need to know about engineering mechanics Mechanics is the foundation of most engineering sciences and is an indispensable prerequisite to their study. Mechanics is the science which describes and predicts the conditions of rest or motion of bodies under the action of forces. 02/01/2017 S.ThangaKasiRajan, stkrajan@gmail.com 3
  4. 4. Branches of Engineering Mechanics 02/01/2017 S.ThangaKasiRajan, stkrajan@gmail.com 4
  5. 5. Mechanics: The actions and effects of forces on bodies. Mechanics Statics: Bodies at rest, or in equilibrium Dynamics: Bodies in motion, or out of equilibrium IN EQUILIBRIUM Will be static, OR move with constant velocity OUT OF EQUILIBRIUM Will accelerate (velocity changing) Velocity=0 Velocity= constant Velocity changing with time Engineering Mechanics 02/01/2017 S.ThangaKasiRajan, stkrajan@gmail.com 5
  6. 6. Kinematics: how fast, how far, and how long it takes... Kinetics: What forces were involved to produce this motion? Dynamics: Kinematics: Study of motion without reference to forces producing motion: Relations between position, velocity, acceleration and time. Kinetics: Relations between unbalanced forces and the changes in motion they produce. E.g. Rollercoaster ride: - Weight - Friction - Aerodynamic drag What are the resulting accelerations?02/01/2017 S.ThangaKasiRajan, stkrajan@gmail.com 6
  7. 7. Six Fundamental Concept of Mechanics • Space - associated with the motion or the position of a point P given in terms of three coordinates measured from a reference point or origin. • Time - definition of an event requires specification of the time and position at which it occurred. • Mass - used to characterize and compare bodies, e.g., response to earth’s gravitational attraction and resistance to change in translational motion.02/01/2017 S.ThangaKasiRajan, stkrajan@gmail.com 7
  8. 8. Force A force is a push or pull. An object at rest needs a force to get it moving; a moving object needs a force to change its velocity. The magnitude of a force can be measured using a spring scale. Six Fundamental Concept of Mechanics 02/01/2017 S.ThangaKasiRajan, stkrajan@gmail.com 8
  9. 9. Six Fundamental Concept of Mechanics Particle A particle has a mass but size is neglected. When a body is idealised as a particle, the principles of mechanics reduces to a simplified form, since the geometry of the body will not be concerned in the analysis of the problem. 02/01/2017 S.ThangaKasiRajan, stkrajan@gmail.com 9
  10. 10. Six Fundamental Concept of Mechanics Rigid Body A combination of large number of Particles in which all the particles remain at a fixed distance from one another before and after application of load. Here mass & size of the bodies are considered when analysing the forces. 02/01/2017 S.ThangaKasiRajan, stkrajan@gmail.com 10
  11. 11. Basic Laws of Mechanics 1. Newton’s Law 2. Newton’s Law of Gravitation 3. Triangle Law 4. Parallellogram Law 5. Lami’s Theorem 6. Principle of Transmissibility 02/01/2017 S.ThangaKasiRajan, stkrajan@gmail.com 11
  12. 12. Basic Laws of Mechanics • Newton’s First Law: If the resultant force on a particle is zero, the particle will remain at rest or continue to move in a straight line. • Newton’s Third Law: The forces of action and reaction between two particles have the same magnitude and line of action with opposite sense. • Newton’s Second Law: A particle will have an acceleration proportional to a nonzero resultant applied force. amF   • Newton’s Law of Gravitation: Two particles are attracted with equal and opposite forces, 22 , R GM gmgW r Mm GF  02/01/2017 S.ThangaKasiRajan, stkrajan@gmail.com 12
  13. 13. Parallelogram Law Parallelogram Law If two vectors acting at a point be represented in magnitude and direction by the adjacent sides of a parallelogram, then their resultant is represented in magnitude and direction by the diagonal of the parallelogram passing through that point. 𝑅 = 𝑃2 + 𝑄2 + 2𝑃𝑄𝑐𝑜𝑠Ө Ө = tan−1 𝑄𝑠𝑖𝑛 𝛼 𝑃 + 𝑄 cos 𝛼 02/01/2017 S.ThangaKasiRajan, stkrajan@gmail.com 13
  14. 14. Triangle Law If two vectors acting at a point are represented by the two sides of a triangle taken in order, then their resultant is represented by the third side taken in opposite order 𝑅 = 𝑃 + 𝑄 𝑃 𝑄 𝑅 02/01/2017 S.ThangaKasiRajan, stkrajan@gmail.com 14
  15. 15. sine and cosine rules When adding forces it is often useful to consider solving the problem using geometric rules, rather than considering components       The sine rule states that  sinsinsin XBA  The cosine rule states that cos2222 BXXBA  cos2222 AXXAB  cos2222 ABBAX  02/01/2017 S.ThangaKasiRajan, stkrajan@gmail.com 15
  16. 16. Principle of Transmissibility According to this law the state of rest or motion of the rigid body is unaltered if a force acting on the body is replaced by another force of the same magnitude and direction but acting anywhere on the body along the line of action of the replaced force. • Principle of Transmissibility - Conditions of equilibrium or motion are not affected by transmitting a force along its line of action. NOTE: F and F’ are equivalent forces. 02/01/2017 S.ThangaKasiRajan, stkrajan@gmail.com 16
  17. 17. Coordinate System 1. An origin as the reference point 2. A set of coordinate axes with scales and labels 3. Choice of positive direction for each axis 4. Choice of unit vectors at each point in space Coordinate system: used to describe the position of a point in space and consists of Cartesian Coordinate System 17
  18. 18. Vector Representation of Forces A vector is a quantity that has both direction and magnitude. 18
  19. 19. Application of Vectors (1) Vectors can exist at any point P in space. (2) Vectors have direction and magnitude. (3) Vector Equality: Any two vectors that have the same direction and magnitude, are equal no matter where in space they are located. 19
  20. 20. Resolution and Composition of Forces Resolution of force is a reverse process in which a single force is expressed in terms of its components. These components are sometimes referred to as the resolved parts of the force. F x y  O X Y OX = F cos  OY = F sin  Splitting of forces into their components 30° 20 N
  21. 21. Examples X = -15  Cos 42 = -11.1 N Y = 15  Sin 42 = 10.0 N X = 35  Cos 62 = 16.4 N Y = -35  Sin 62 = - 30.9 N x y 42° 15 N x y 62° 35 N x y 32° 10 N X = -10  Sin 32 = - 5.30 N Y = -10  Cos 32 = -8.48 N 02/01/2017 S.ThangaKasiRajan, stkrajan@gmail.com 21
  22. 22. Resolution and Composition of Forces F F cosӨ F sinӨ F cosӨF cosӨ F sinӨ F sinӨ Ө Ө Ө F sinӨ F cosӨ Ө F F F 1. It is convenient to have Fx = F cos Ө Fy = F sin Ө and Always measure angle from horizontal reference(acute angle). 2. Assume force pointing Right and Top as positive otherwise negative 02/01/2017 S.ThangaKasiRajan, stkrajan@gmail.com 22
  23. 23. Procedure to find Resultant Force Procedure to find the magnitude and direction of the resultant force 1. Find ƩFx 2. Find ƩFy 3. Magnitude of the resultant force is given by 𝑅 = (Ʃ𝐹𝑥)2+(Ʃ𝐹𝑦)2 4. Plot ƩFx and ƩFy with its appropriate sign 5. Direction of the resultant force is given by tan Ө = Ʃ𝐹 𝑦 Ʃ𝐹𝑥 02/01/2017 S.ThangaKasiRajan, stkrajan@gmail.com 23
  24. 24. Problems 02/01/2017 S.ThangaKasiRajan, stkrajan@gmail.com 24
  25. 25. Problem No 1 Two forces act on a bolt at A. Determine their resultant. 02/01/2017 S.ThangaKasiRajan, stkrajan@gmail.com 25
  26. 26. Problem No 1 (contd…) • Graphical solution - A parallelogram with sides equal to P and Q is drawn to scale. The magnitude and direction of the resultant or of the diagonal to the parallelogram are measured,  35N98 R • Graphical solution - A triangle is drawn with P and Q head-to-tail and to scale. The magnitude and direction of the resultant or of the third side of the triangle are measured,  35N98 R 02/01/2017 S.ThangaKasiRajan, stkrajan@gmail.com 26
  27. 27. Problem No 1 (contd…) • Trigonometric solution - Apply the triangle rule. From the Law of Cosines,          155cosN60N402N60N40 cos2 22 222 BPQQPR A A R Q BA R B Q A      20 04.15 N73.97 N60 155sin sinsin sinsin  N73.97R From the Law of Sines,  04.35 02/01/2017 S.ThangaKasiRajan, stkrajan@gmail.com 27
  28. 28. Problem No 1 (contd…) Sl. No. ƩFx ƩFy 1. + 40 cos20 + 40 sin20 2. + 60 cos 45 + 60 cos 45 80.01 56.10 𝑅 = (Ʃ𝐹𝑥)2+(Ʃ𝐹𝑦)2 𝑅 = (80.01)2+(56.10)2 R = 97.72 N Magnitude of the resultant force Direction of resultant force Ʃ Fx = 80.01 ƩFy=56.01 Ө tan Ө = Ʃ𝐹𝑦 Ʃ𝐹𝑥 tan Ө = ( 56.1 80.01 ) Ө = 35ᴼ 02/01/2017 S.ThangaKasiRajan, stkrajan@gmail.com 28
  29. 29. Problem 2 a) the tension in each of the ropes for  = 45o, b) the value of  for which the tension in rope 2 is minimum. A barge is pulled by two tugboats. If the resultant of the forces exerted by the tugboats is a 25 kN directed along the axis of the barge, determine 02/01/2017 S.ThangaKasiRajan, stkrajan@gmail.com 29
  30. 30. Problem 2 • Graphical solution - Parallelogram Rule with known resultant direction and magnitude, known directions for sides. kN12.8kN18.5 21  TT • Trigonometric solution - Triangle Rule with Law of Sines      105sin kN25 30sin45sin 21 TT kN12.94kN18.3 21  TT 02/01/2017 S.ThangaKasiRajan, stkrajan@gmail.com 30
  31. 31. Problem 2 • The angle for minimum tension in rope 2 is determined by applying the Triangle Rule and observing the effect of variations in . • The minimum tension in rope 2 occurs when T1 and T2 are perpendicular.    30sinkN252T kN12.52 T    30coskN251T kN21.71 T  3090  60 02/01/2017 S.ThangaKasiRajan, stkrajan@gmail.com 31
  32. 32. Problem 3 Knowing that the tension in cable BC is 725-N, determine the resultant of the three forces exerted at point B of beam AB. 02/01/2017 S.ThangaKasiRajan, stkrajan@gmail.com 32
  33. 33. Problem 3 SOLUTION: • Resolve each force into rectangular components. • Calculate the magnitude and direction. 02/01/2017 S.ThangaKasiRajan, stkrajan@gmail.com 33
  34. 34. Problems for Practice Fig 1 Fig 2 A disabled automobile is pulled by means of two ropes shown in fig 1. Determine the Magnitude and direction of Resultant by (a) parallelogram law, (b) Triangle law (c) analytical method Determine the magnitude and direction of resultant force shown in fig 2. 02/01/2017 S.ThangaKasiRajan, stkrajan@gmail.com 34
  35. 35. Equilibrium of a Particle 2 - 35 • When the resultant of all forces acting on a particle is zero, the particle is in equilibrium. • 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. • Particle acted upon by two forces: - equal magnitude - same line of action - opposite sense • Particle acted upon by three or more forces: - graphical solution yields a closed polygon - algebraic solution 00 0     yx FF FR  02/01/2017 S.ThangaKasiRajan, stkrajan@gmail.com
  36. 36. Examples for Equilibrium Cables AB and AC carries the spool of weight 02/01/2017 S.ThangaKasiRajan, stkrajan@gmail.com 36
  37. 37. Problem 4 Given: Sack A weighs 20 N. and geometry is as shown. Find: Forces in the cables and weight of sack B. 1. Apply Equilibrium condition at Point E and solve for the unknowns (TEG & TEC). 2. Repeat this process at C. 02/01/2017 S.ThangaKasiRajan, stkrajan@gmail.com 37
  38. 38. Problem 4 +   Fx = TEG sin 30º – TEC cos 45º = 0 +   Fy = TEG cos 30º – TEC sin 45º – 20 N = 0 Solving these two simultaneous equations for the two unknowns, we get TEC = 38.6 N TEG = 54.6 N Note that the assumed directions for the forces in the two cables EG and EC are tensile in nature. 02/01/2017 S.ThangaKasiRajan, stkrajan@gmail.com 38
  39. 39. Problem 4    Fx = 38.64 cos 45 – (4/5) TCD = 0    Fy = (3/5) TCD + 38.64 sin 45 – WB = 0 Solving the first equation and then the second we get TCD = 34.2 N and WB = 47.8 N . Apply Equilibrium Condition Now move on to the point C and consider equilibrium at C 02/01/2017 S.ThangaKasiRajan, stkrajan@gmail.com 39
  40. 40. Problem 5 It is desired to determine the drag force at a given speed on a prototype sailboat hull. A model is placed in a test channel and three cables are used to align its bow on the channel centerline. For a given speed, the tension is 200-N in cable AB and 300-N in cable AE. Determine the drag force exerted on the hull and the tension in cable AC. SOLUTION: • Choosing the hull as the free body, draw a free-body diagram. • Express the condition for equilibrium for the hull by writing that the sum of all forces must be zero. • Resolve the vector equilibrium equation into two component equations. Solve for the two unknown cable tensions. 02/01/2017 S.ThangaKasiRajan, stkrajan@gmail.com 40
  41. 41. Problem 5 • Express the condition for equilibrium for the hull by writing that the sum of all forces must be zero. 0 DAEACAB FTTTR  SOLUTION: • Choosing the hull as the free body, draw a free-body diagram.   25.60 75.1 m4 m7 tan     56.20 375.0 m4 m1.5 tan   02/01/2017 S.ThangaKasiRajan, stkrajan@gmail.com 41
  42. 42. Problem 5 • Resolve the vector equilibrium equation into two component equations. Solve for the two unknown cable tensions.                jT iFT R iFF iT jTiT jTiTT ji jiT AC DAC DD ACAC ACACAC AB          3009363.021.99 3512.066.173 0 N300 9363.03512.0 56.20cos56.20sin N21.99N66.173 26.60cosN20026.60sinN200          02/01/2017 S.ThangaKasiRajan, stkrajan@gmail.com 42
  43. 43. Problem 5     3009363.021.9900 3512.066.17300     ACy DACx TF FTF N98.35 N214.45   D AC F T     jT iFT R AC DAC    3009363.021.99 3512.066.173 0    This equation is satisfied only if each component of the resultant is equal to zero 02/01/2017 S.ThangaKasiRajan, stkrajan@gmail.com 43
  44. 44. Problem 6 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 a = 25ᴼ and b = 15ᴼ and that the tension in cable CD is 80 N, determine (a) the combined weight of the boatswain’s chair and the sailor, (b) in tension in the support cable ACB. 02/01/2017 S.ThangaKasiRajan, stkrajan@gmail.com 44
  45. 45. Problem 6 02/01/2017 S.ThangaKasiRajan, stkrajan@gmail.com 45
  46. 46. Forces in Space • The vector is contained in the plane OBAC. F  • Resolve into horizontal and vertical components. yh FF sin F  yy FF cos • Resolve into rectangular components hF     sinsin sin cossin cos y hy y hx F FF F FF     02/01/2017 S.ThangaKasiRajan, stkrajan@gmail.com 46
  47. 47. Forces in Space • With the angles between and the axes,F    kji F kjiF kFjFiFF FFFFFF zyx zyx zyx zzyyxx         coscoscos coscoscos coscoscos      • is a unit vector along the line of action of and are the direction cosines for F  F    zyx  cosand,cos,cos 02/01/2017 S.ThangaKasiRajan, stkrajan@gmail.com 47
  48. 48. Problem 6 The tension in the guy wire is 2500 N. Determine: a) components Fx, Fy, Fz of the force acting on the bolt at A, b) the angles x, y, z defining the direction of the force SOLUTION: • Based on the relative locations of the points A and B, determine the unit vector pointing from A towards B. • Apply the unit vector to determine the components of the force acting on A. • Noting that the components of the unit vector are the direction cosines for the vector, calculate the corresponding angles. 02/01/2017 S.ThangaKasiRajan, stkrajan@gmail.com 48
  49. 49. Problem 6 SOLUTION: • Determine the unit vector pointing from A towards B.             m3.94 m30m80m40 m30m80m40 222    AB kjiAB  • Determine the components of the force.         kji kji FF    N795N2120N1060 318.0848.0424.0N2500     kji kji   318.0848.0424.0 3.94 30 3.94 80 3.94 40                      02/01/2017 S.ThangaKasiRajan, stkrajan@gmail.com 49
  50. 50. Problem 6 • Noting that the components of the unit vector are the direction cosines for the vector, calculate the corresponding angles. kji kji zyx   318.0848.0424.0 coscoscos       5.71 0.32 1.115    z y x    02/01/2017 S.ThangaKasiRajan, stkrajan@gmail.com 50
  51. 51. Problem 7 A crate is supported by three cables as shown. Determine the weight of the crate knowing that the tension in cable AB is 3750 N. 02/01/2017 S.ThangaKasiRajan, stkrajan@gmail.com 51
  52. 52. Problem 7 Net Force acting at point A is 02/01/2017 S.ThangaKasiRajan, stkrajan@gmail.com 52
  53. 53. Problem 7 02/01/2017 S.ThangaKasiRajan, stkrajan@gmail.com 53
  54. 54. Problems for practice 02/01/2017 S.ThangaKasiRajan, stkrajan@gmail.com 54
  55. 55. Problems for practice 02/01/2017 S.ThangaKasiRajan, stkrajan@gmail.com 55
  56. 56. References 1. Ferdinand P Beer & E.Russell Johnston “VECTOR MECHANICS FOR ENGINEERS STATICS & Dynamics”, (Ninth Edition) Tata McGraw Hill Education Private Limited, New Delhi. 2. Engineering Mechanics – Statics & Dynamics by S.Nagan, M.S.Palanichamy, Tata McGraw-Hill (2001). 02/01/2017 S.ThangaKasiRajan, stkrajan@gmail.com 56
  57. 57. Thank you Any Queries contact S.Thanga Kasi Rajan Assistant Professor Department of Mechanical Engineering Kamaraj College of Engineering & Technology, Virudhunagar – 626001. Tamil Nadu, India Email : stkrajan@gmail.com 02/01/2017 S.ThangaKasiRajan, stkrajan@gmail.com 57

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