STATICS. Friction VECTOR MECHANICS FOR ENGINEERS: Eighth Edition CHAPTER. Ferdinand P. Beer E. Russell Johnston, Jr.

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1 Eighth E 8 Friction CHAPTER VECTOR MECHANICS FOR ENGINEERS: STATICS Ferdinand P. Beer E. Russell Johnston, Jr. Lecture Notes: J. Walt Oler Texas Tech University

2 Contents Introduction Laws of Dry Friction. Coefficients of Friction. Angles of Friction Problems Involving Dry Friction Sample Problem 8.1 Sample Problem 8.3 Wedges Square-Threaded Screws Sample Problem 8.5 Journal Bearings. Axle Friction. Thrust Bearings. Disk Friction. Wheel Friction. Rolling Resistance. Sample Problem 8.6 Belt Friction. Sample Problem

3 Introduction In preceding chapters, it was assumed that surfaces in contact were either frictionless (surfaces could move freely with respect to each other) or rough (tangential forces prevent relative motion between surfaces). 8-3

4 Introduction Actually, no perfectly frictionless surface exists. For two surfaces in contact, tangential forces, called friction forces, will develop if one attempts to move one relative to the other. 8-4

5 Introduction However, the friction forces are limited in magnitude and will not prevent motion if sufficiently large forces are applied. The distinction between frictionless and rough is, therefore, a matter of degree. 8-5

6 Introduction There are two types of friction: dry or Coulomb friction and fluid friction. Fluid friction applies to lubricated mechanisms. The present discussion is limited to dry friction between non-lubricated surfaces. 8-6

7 The Laws of Dry Friction. Coefficients of Friction Block of weight W placed on horizontal surface. Forces acting on block are its weight and reaction of surface N. Small horizontal force P applied to block. For block to remain stationary in equilibrium, a horizontal component F of the surface reaction is required. F is a static-friction force. 8-7

8 The Laws of Dry Friction. Coefficients of Friction As P increases, the static-friction force F increases as well until it reaches a maximum value F m. Fm sn Further increase in P causes the block to begin to move as F drops to a smaller kinetic-friction force F k. F N k k 8-8

9 The Laws of Dry Friction. Coefficients of Friction 8-9

10 The Laws of Dry Friction. Coefficients of Friction 8-10

11 The Laws of Dry Friction. Coefficients of Friction Maximum static-friction force: Fm sn Kinetic-friction force: F k N k s Maximum static-friction force and kinetic-friction force are: - proportional to normal force k dependent on type and con of contact surfaces - independent of contact area 8-11

12 The Laws of Dry Friction. Coefficients of Friction Four situations can occur when a rigid body is in contact with a horizontal surface: No friction, (P x = 0) No motion, (P x < F m ) 8-12

13 The Laws of Dry Friction. Coefficients of Friction Four situations can occur when a rigid body is in contact with a horizontal surface: Motion impending, (P x = F m ) Motion, (P x > F m ) 8-13

14 Angles of Friction It is sometimes convenient to replace normal force N and friction force F by their resultant R: Impending motion tan tan s s F N m s s N N Motion impending 8-14

15 Angles of Friction It is sometimes convenient to replace normal force N and friction force F by their resultant R: Motion tan tan k k F N k k k N N 8-15

16 Problems Involving Dry Friction Impending motion at all points of contact (body on verge of slipping) No apparent Impending motion Impending motion at some points of contact 8-16

17 Impending Motion at All Points of Contact Find the smallest angle q for which the 100 N bar can be placed against the wall, without slipping. (Five unknowns). Slipping occurs simultaneously at A and B 8-17

18 For a minute assume there was a roller at B. Then Na=mg and F = (N-W/2)/tanq. F is large if theta is small. A limiting value will reached. At larger theta F will be smaller than (mu NA) and will not slip. 8-18

19 Sample Problem 8.3 The moveable bracket shown may be placed at any height on the 3-cm diameter pipe. If the coefficient of friction between the pipe and bracket is 0.25, determine the minimum distance x at which the load can be supported. Neglect the weight of the bracket. 8-19

20 Sample Problem 8.3 The moveable bracket shown may be placed at any height on the 3-cm diameter pipe. If the coefficient of friction between the pipe and bracket is 0.25, determine the minimum distance x at which the load can be supported. Neglect the weight of the bracket. 8-20

21 Sample Problem

22 SOLUTION: When W is placed at minimum x, the bracket is about to slip and friction forces in upper and lower collars are at maximum value. Apply cons for static equilibrium to find minimum x. 8-22

23 SOLUTION: If x is larger than minimum, by inspection reactions NA and NB are larger (moment due to W is larger which tries to tilt the bracket more creating larger reactions). F A and F B do not then reach their limiting value. Apply cons for static equilibrium to find minimum x. 8-23

24 When W is placed at minimum x, the bracket is about to slip and friction forces in upper and lower collars are at maximum value. F F A B N s N s A B 0.25N 0.25N A B 8-24

25 Apply cons for static equilibrium to find minimum x. x F 0 : N B N 0 A N B N A F y 0 : F A 0.25N 0.5N F A B A W N W B W 0 N A NB 2W 8-25

26 M 0 : N 6 cm F 3 cmw x 1.5 cm 0 B A 6N 6 A 3 A 0.25N A W x 1.5 2W 0.752W W x x 12 cm 8-26

27 If x is larger, then slip will not occur at any of the locations since Na is equal to Nb and coeff. of friction is same. Also, value of W does not matter. It seems to decide the contact and friction is related to reactions for slipping. 8-27

28 ighth Fig. shows an automobile on a road inclined at an angle q with the horizontal. s =0.6, d =0.5. What is the maximum inclination the car can climb at uniform speed. The car is rear wheel drive and has a loaded weight of 3600 lb (1 ib =4.5 N). Assume tires do not slip (and spin) i.e pure rolling at point of contact.

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30

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32 If drive wheel was allowed to slip we replace s by d 8-32

33 Compute the torque needed by the drive wheels to move the car at a uniform speed up an incline where q =15 0. The diameter of tire is 25 in. (1 inch =2.54 cm)

34

35

36 No Apparent Impending Motion Second type of problems: No Apparent Impending Motion 8-36

37 No apparent impending motion ighth 6 unknowns, 6 equations 8-37

38 It is not known whether motion is impending or not All applied forces known Coefficient of static friction is known Determine whether body will remain at rest or slide 8-38

39 Equilibrium Versus Frictional Equations 8-39

40 Equilibrium Versus Frictional Equations Problems where the friction force F is to be an equilibrium force and satisfies the inequality F < μ s N then we can assume the sense of direction of F on the free-body diagram. The correct sense is made known after solving the equations of equilibrium for F. 8-40

41 Equilibrium Versus Frictional Equations This is possible because the equilibrium equations equate to zero the components of vector acting in the same direction. 8-41

42 Equilibrium Versus Frictional Equations However, in cases where the frictional equation F = μ s N is used in the solution of a problem the convenience of assuming the sense of F is lost. Since the frictional equation relates only the magnitude of two perpendicular vectors. Consequently, F must always be shown acting with its correct sense on the free-body diagram. 8-42

43 Sample Problem

44 Sample Problem 8.1 A 100 N force acts as shown on a 300 N block placed on an inclined plane. The coefficients of friction between the block and plane are s = 0.25 and k = Determine whether the block is in equilibrium and find the value of the friction force. 8-44

45 Determine values of friction force and normal reaction force from plane required to maintain equilibrium. Calculate maximum friction force and compare with friction force required for equilibrium. If it (maximum friction force) is greater, block will not slide. 8-45

46 If maximum friction force is less than friction force required for equilibrium, block will slide. Calculate kineticfriction force. 8-46

47 F x 0 : 100 N N F 0 F 80 N F y 0 : N N 5 0 N 240 N 8-47

48 Calculate maximum friction force and compare with friction force required for equilibrium. If it is greater, block will not slide. F N F N 60 N m s m The block will slide down the plane. 8-48

49 8-49

50 Sample Problem 8.1 If maximum friction force is less than friction force required for equilibrium, block will slide. Calculate kineticfriction force. F actual F k k N 240 N F actual 48 N 8-50

51 Impending Motion at Some Points of Contact Determine the horizontal force P needed to cause movement. Weight of each member is 100 N. 8-51

52 Impending Motion at Some Points of Contact From the FBD 7 unknowns. 6 equilibrium equations and one static friction equation 8-52

53 Impending Motion at Some Points of Contact As P increases it will either cause slipping at A and no slipping at C, so that F A = 0.3N A and F c 0.5N c. OR slipping occurs at C and no slipping at A, in which case F c = 0.5N c and F A 0.3N A Calculate P for each case and choose the smaller P. 8-53

54 Beam A B is subjected to a uniform load of 200 N/m and is supported at B by post BC. If the coefficients of static friction at B and C are μ B = 0.2 and μ C = 0.5. determine the force P needed to pull the post out from under the beam. Neglect the weight of the members and the thickness of the beam. 8-54

55 M A =0 N B = 400 N 8-55

56 8-56

57 4 unknowns. So assumption on slipping. 8-57

58 4 unknowns. So assumption on slipping. 8-58

59 8-59

60 Cylinder rolling from Assignment sheet 8-60

61 Thrust Bearings. Disk Friction 8-61

62 Thrust bearings Support the axial thrust of both horizontal as well as vertical shafts Functions are to prevent the shaft from drifting in the axial direction and to transfer thrust loads applied on the shaft Vertical thrust bearings also need to support the weight of the shaft and any components attached to it The moving surface exerted against a thrust bearing may be the area of the end of the shaft or the area of a collar attached at any point to the shaft Sorurce:web.iitd.ac.in/~rribeiro/MEL204170LubricationLec% ppt

63 Vector Types Mechanics of thrust bearings for Engineers: Statics Plain thrust: Consists of a stationary flat bearing surface against which the flat end of a rotating shaft is permitted to bear Flat end of rotor ROTOR Axial movement Bearing surface Sorurce:web.iitd.ac.in/~rribeiro/MEL204170LubricationLec% ppt

64 Consider rotating hollow shaft: 8-64

65 ighth R 2 R A P r A A P r N r F r M k k k

66 ighth 8-66 Thrust Bearings. Disk Friction For full circle of radius R, M k PR R R R R P drd r R R P M k R R k q

67 Clutch and friction The largest torque which can be transmitted by a clutch without slipping 8-67

68 Belt Friction 8-68

69 Belt Friction The normal and frictional forces, acting at different points along the belt vary both in magnitude and direction. Due to this unknown distribution, the analysis of the problem will first require a study of forces acting on a differential element of the belt. 8-69

70 8-70

71 Belt Friction Relate T 1 and T 2 when belt is about to slide to right. 8-71

72 Belt Friction Draw free-body diagram for element of belt 8-72

73 Belt Friction F x q q 0 : T T cos T cos s N Fy q q 0 : N T T sin T sin

74 Belt Friction Combine to eliminate N, divide through by q, T q cos q 2 s T T 2 sin q q 2 2 In the limit as q goes to zero, dt dq s T

75 Belt Friction Separate variables and integrate from q 0 to q T ln 2 or s T 1 T T 2 1 e s 8-75

76 Belt Friction Note: Which is T2 and which is T1? - T2 is the larger force and always opposes friction. 8-76

77 8-77

78 T2 is always greater than T1. T2 represents tension in the part of belt which pulls while T1 is tension in part which resists. Angle of contact is expressed in radians. may be larger than 2, if a rope is wrapped n times around a post = 2n 8-78

79 Sample Problem 8.8 A flat belt connects pulley A to pulley B. The coefficients of friction are s = 0.25 and k = 0.20 between both pulleys and the belt. Knowing that the maximum allowable tension in the belt is 3000 N, determine the largest torque which can be exerted by the belt on pulley A. 8-79

80 Sample Problem

81 Sample Problem 8.8 A flat belt connects pulley A to pulley B. The coefficients of friction are s = 0.25 and k = 0.20 between both pulleys and the belt. Knowing that the maximum allowable tension in the belt is 3000 N, determine the largest torque which can be exerted by the belt on pulley A. 8-81

82 Sample Problem 8.8 Solution Since angle of contact is smaller, slippage will occur on pulley B first. Determine belt tensions based on pulley B. 8-82

83 Sample Problem 8.8 Solution T T 1 2 / T 1 e s 3000 N 3000 N / T / e N

84 Sample Problem 8.8 Taking pulley A as free-body, sum moments about pulley center to determine torque. M 0 : M 0. 2m N 3000N 0 A A M A N m 8-84

85 Sample Problem 8.8 Check that belt does not slip at A. = 240 o = 4/3 rad 4 s 3 s T ln T N ln N s 8-85

86 8-86

87 Angle of wrap 8-87

88 8-88

89 8-89

90 8-90

91 8-91

92 8-92

93 Angle of lap decided by taking tangent at the points of contact. 8-93

94 8-94

95 8-95

96 8-96

97 Wedges Wedges - simple machines used to raise heavy loads. Force required to lift block is significantly less than block weight. Friction prevents wedge from sliding out. Want to find minimum force P to raise block. Block as free-body F N 1 F 0 : N 0 : W N or R x y s s R2 W 0 N

98 8-98

99 8-99

100 8-100

101 8-101

102 8-102

103 8-103

104 8-104

105 8-105

106 8-106

107 8-107

108 8-108

109 8-109

110 8-110

111 8-111

112 8-112

113 Assume impending slip at A and no slip at B 8-113

114 8-114

115 8-115

116 8-116

117 8-117

118 8-118

119 8-119

120 8-120

121 Wedges Wedge as free-body N F s F N x y 2 0 : or P R 2 0 : P N 3 N R3 cos6 sin 6 s cos6 sin s 8-121

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