1. Friction 8 Engineering Mechanics: Statics in SI Units, 12e Copyright © 2010 Pearson Education South Asia Pte Ltd

2. Chapter Objectives Introduce the concept of dry friction To present specific applications of frictional force analysis on wedges, screws, belts, and bearings To investigate the concept of rolling resistance

3. Copyright © 2010 Pearson Education South Asia Pte Ltd Chapter Outline 1.Characteristics of Dry Friction 2.Problems Involving Dry Friction 3.Wedges 4.Frictional Forces on Screws 5.Frictional Forces on Flat Belts 6.Frictional Forces on Collar Bearings, Pivot Bearings, and Disks 7.Frictional Forces on Journal Bearings 8.Rolling Resistance

4. Copyright © 2010 Pearson Education South Asia Pte Ltd 8.1 Characteristics of Dry Friction Friction Force that resists the movement of two contacting surfaces that slide relative to one another Acts tangent to the surfaces at points of contact with other body Opposing possible or existing motion of the body relative to points of contact Two types of friction – Fluid and Coulomb Friction

5. Copyright © 2010 Pearson Education South Asia Pte Ltd 8.1 Characteristics of Dry Friction Fluid friction exist when the contacting surface are separated by a film of fluid (gas or liquid) Depends on velocity of the fluid and its ability to resist shear force Coulomb friction occurs between contacting surfaces of bodies in the absence of a lubricating fluid

6. Copyright © 2010 Pearson Education South Asia Pte Ltd 8.1 Characteristics of Dry Friction Theory of Dry Friction Consider the effects caused by pulling horizontally on a block of uniform weight W which is resting on a rough horizontal surface Consider the surfaces of contact to be nonrigid or deformable and other parts of the block to be rigid

7. Copyright © 2010 Pearson Education South Asia Pte Ltd 8.1 Characteristics of Dry Friction Theory of Dry Friction Normal force ∆N n and frictional force ∆F n act along the contact surface For equilibrium, normal forces act upward to balance the block ’ s weight W, frictional forces act to the left to prevent force P from moving the block to the right

8. Copyright © 2010 Pearson Education South Asia Pte Ltd 8.1 Characteristics of Dry Friction Theory of Dry Friction Many microscopic irregularities exist between the two surfaces of floor and the block Reactive forces ∆R n developed at each of the protuberances Each reactive force consist of both a frictional component ∆F n and normal component ∆N n

9. Copyright © 2010 Pearson Education South Asia Pte Ltd 8.1 Characteristics of Dry Friction Theory of Dry Friction Equilibrium Effect of normal and frictional loadings are indicated by their resultant N and F Distribution of ∆F n indicates that F is tangent to the contacting surface, opposite to the direction of P Normal force N is determined from the distribution of ∆N n

10. Copyright © 2010 Pearson Education South Asia Pte Ltd 8.1 Characteristics of Dry Friction Theory of Dry Friction Equilibrium N is directed upward to balance W N acts a distance x to the right of the line of action of W This location coincides with the centroid or the geometric center of the loading diagram in order to balance the “ tipping effect ” caused by P

11. Copyright © 2010 Pearson Education South Asia Pte Ltd 8.1 Characteristics of Dry Friction Theory of Dry Friction Impending Motion As P is slowly increased, F correspondingly increase until it attains a certain maximum value F, called the limiting static frictional force Limiting static frictional force F s is directly proportional to the resultant normal force N F s = μ s N

12. Copyright © 2010 Pearson Education South Asia Pte Ltd 8.1 Characteristics of Dry Friction Theory of Dry Friction Impending Motion Constant of proportionality μ s is known as the coefficient of static friction Angle Φ s that R s makes with N is called the angle of static friction

13. Copyright © 2010 Pearson Education South Asia Pte Ltd 8.1 Characteristics of Dry Friction Theory of Dry Friction Typical Values of μ s Contact MaterialsCoefficient of Static Friction μ s Metal on ice 0.03 – 0.05 Wood on wood 0.30 – 0.70 Leather on wood 0.20 – 0.50 Leather on metal 0.30 – 0.60 Aluminum on aluminum 1.10 – 1.70

14. Copyright © 2010 Pearson Education South Asia Pte Ltd 8.1 Characteristics of Dry Friction Theory of Dry Friction Motion When P is greater than F s, the frictional force is slightly smaller value than F s, called kinetic frictional force The block will not be held in equilibrium (P > F s ) but slide with increasing speed

15. Copyright © 2010 Pearson Education South Asia Pte Ltd 8.1 Characteristics of Dry Friction Theory of Dry Friction Motion The drop from F s (static) to F k (kinetic) can by explained by examining the contacting surfaces When P > F s, P has the capacity to shear off the peaks at the contact surfaces

16. Copyright © 2010 Pearson Education South Asia Pte Ltd 8.1 Characteristics of Dry Friction Theory of Dry Friction Resultant frictional force F k is directly proportional to the magnitude of the resultant normal force N F k = μ k N Constant of proportionality μ k is coefficient of kinetic friction μ k are typically 25% smaller than μ s Resultant R k has a line of action defined by Φ k, angle of kinetic friction

17. Copyright © 2010 Pearson Education South Asia Pte Ltd 8.1 Characteristics of Dry Friction Theory of Dry Friction F is a static frictional force if equilibrium is maintained F is a limiting static frictional force when it reaches a maximum value needed to maintain equilibrium F is termed a kinetic frictional force when sliding occurs at the contacting surface

18. Copyright © 2010 Pearson Education South Asia Pte Ltd 8.1 Characteristics of Dry Friction Characteristics of Dry Friction The frictional force acts tangent to the contacting surfaces The max static frictional force F s is independent of the area of contact The max static frictional force is greater than kinetic frictional force When slipping, the max static frictional force is proportional to the normal force and kinetic frictional force is proportional to the normal force

19. Copyright © 2010 Pearson Education South Asia Pte Ltd 8.2 Problems Involving Dry Friction Types of Friction Problems In all cases, geometry and dimensions are assumed to be known 3 types of mechanics problem involving dry friction - Equilibrium - Impending motion at all points - Impending motion at some points

20. Copyright © 2010 Pearson Education South Asia Pte Ltd 8.2 Problems Involving Dry Friction Types of Friction Problems Equilibrium Total number of unknowns = Total number of available equilibrium equations Frictional forces must satisfy F ≤ μ s N; otherwise, slipping will occur and the body will not remain in equilibrium We must determine the frictional forces at A and C to check for equilibrium

21. Copyright © 2010 Pearson Education South Asia Pte Ltd 8.2 Problems Involving Dry Friction Equilibrium Versus Frictional Equations Frictional force always acts so as to oppose the relative motion or impede the motion of the body over its contacting surface Assume the sense of the frictional force that require F to be an “ equilibrium ” force Correct sense is made after solving the equilibrium equations If F is a negative scalar, the sense of F is the reverse of that assumed

22. Copyright © 2010 Pearson Education South Asia Pte Ltd Example 8.1 The uniform crate has a mass of 20kg. If a force P = 80N is applied on to the crate, determine if it remains in equilibrium. The coefficient of static friction is μ = 0.3.

23. Copyright © 2010 Pearson Education South Asia Pte Ltd Solution Resultant normal force NC act a distance x from the crate ’ s center line in order to counteract the tipping effect caused by P. 3 unknowns to be determined by 3 equations of equilibrium.

24. Copyright © 2010 Pearson Education South Asia Pte Ltd Solving Solution

25. Copyright © 2010 Pearson Education South Asia Pte Ltd Since x is negative, the resultant force acts (slightly) to the left of the crate ’ s center line. No tipping will occur since x ≤ 0.4m Max frictional force which can be developed at the surface of contact F max = μ s N C = 0.3(236N) = 70.8N Since F = 69.3N < 70.8N, the crate will not slip thou it is close to doing so. Solution

26. Copyright © 2010 Pearson Education South Asia Pte Ltd 8.3 Wedges A simple machine used to transform an applied force into much larger forces, directed at approximately right angles to the applied force Used to give small displacements or adjustments to heavy load Consider the wedge used to lift a block of weight W by applying a force P to the wedge

27. Copyright © 2010 Pearson Education South Asia Pte Ltd 8.3 Wedges FBD of the block and the wedge Exclude the weight of the wedge since it is small compared to weight of the block

28. Copyright © 2010 Pearson Education South Asia Pte Ltd Example 8.6 The uniform stone has a mass of 500kg and is held in place in the horizontal position using a wedge at B. if the coefficient of static friction μ s = 0.3, at the surfaces of contact, determine the minimum force P needed to remove the wedge. Is the wedge self-locking? Assume that the stone does not slip at A.

29. Copyright © 2010 Pearson Education South Asia Pte Ltd Solution Minimum force P requires F = μ s N A at the surfaces of contact with the wedge. FBD of the stone and the wedge as below. On the wedge, friction force opposes the motion and on the stone at A, F A ≤ μ s N A, slipping does not occur.

30. Copyright © 2010 Pearson Education South Asia Pte Ltd Solution 5 unknowns, 3 equilibrium equations for the stone and 2 for the wedge.

31. Copyright © 2010 Pearson Education South Asia Pte Ltd Solution Since P is positive, the wedge must be pulled out. If P is zero, the wedge would remain in place (self-locking) and the frictional forces developed at B and C would satisfy F B < μ s N B F C < μ s N C

32. Copyright © 2010 Pearson Education South Asia Pte Ltd 8.4 Frictional Forces on Screws Screws used as fasteners Sometimes used to transmit power or motion from one part of the machine to another A square-ended screw is commonly used for the latter purpose, especially when large forces are applied along its axis A screw is thought as an inclined plane or wedge wrapped around a cylinder

33. Copyright © 2010 Pearson Education South Asia Pte Ltd 8.4 Frictional Forces on Screws A nut initially at A on the screw will move up to B when rotated 360° around the screw This rotation is equivalent to translating the nut up an inclined plane of height l and length 2πr, where r is the mean radius of the head Applying the force equations of equilibrium, we have

34. Copyright © 2010 Pearson Education South Asia Pte Ltd 8.4 Frictional Forces on Screws Downward Screw Motion If the surface of the screw is very slippery, the screw may rotate downward if the magnitude of the moment is reduced to say M ’ < M This causes the effect of M ’ to become S ’ M ’ = Wr tan(θ – Φ)

35. Copyright © 2010 Pearson Education South Asia Pte Ltd Example 8.7 The turnbuckle has a square thread with a mean radius of 5mm and a lead of 2mm. If the coefficient of static friction between the screw and the turnbuckle is μ s = 0.25, determine the moment M that must be applied to draw the end screws closer together. Is the turnbuckle self-locking?

36. Copyright © 2010 Pearson Education South Asia Pte Ltd Solution Since friction at two screws must be overcome, this requires Solving When the moment is removed, the turnbuckle will be self- locking

37. Copyright © 2010 Pearson Education South Asia Pte Ltd 8.5 Frictional Forces on Flat Belts It is necessary to determine the frictional forces developed between the contacting surfaces Consider the flat belt which passes over a fixed curved surface Obviously T 2 > T 1 Consider FBD of the belt segment in contact with the surface N and F vary both in magnitude and direction

38. Copyright © 2010 Pearson Education South Asia Pte Ltd 8.5 Frictional Forces on Flat Belts Consider FBD of an element having a length ds Assuming either impending motion or motion of the belt, the magnitude of the frictional force dF = μ dN Applying equilibrium equations

39. Copyright © 2010 Pearson Education South Asia Pte Ltd 8.5 Frictional Forces on Flat Belts We have

40. Copyright © 2010 Pearson Education South Asia Pte Ltd Example 8.8 The maximum tension that can be developed In the cord is 500N. If the pulley at A is free to rotate and the coefficient of static friction at fixed drums B and C is μ s = 0.25, determine the largest mass of cylinder that can be lifted by the cord. Assume that the force F applied at the end of the cord is directed vertically downward.

41. Copyright © 2010 Pearson Education South Asia Pte Ltd Example 8.8 Weight of W = mg causes the cord to move CCW over the drums at B and C. Max tension T 2 in the cord occur at D where T 2 = 500N For section of the cord passing over the drum at B 180° = π rad, angle of contact between drum and cord β = (135°/180°)π = 3/4π rad

42. Copyright © 2010 Pearson Education South Asia Pte Ltd Example 8.8 For section of the cord passing over the drum at C W < 277.4N

43. Copyright © 2010 Pearson Education South Asia Pte Ltd 8.6 Frictional Forces on Collar Bearings, Pivot Bearings and Disks Pivot and collar bearings are used to support axial load on a rotating shaft Laws of dry friction is applied to determine the moment M needed to turn the shaft when it supports an axial force P

44. Copyright © 2010 Pearson Education South Asia Pte Ltd 8.6 Frictional Forces on Collar Bearings, Pivot Bearings and Disks Frictional Analysis The collar bearing on the shaft is subjected to an axial force P and has a total contact area π(R 2 2 – R 1 2 ) Normal pressure p is considered to be uniformly distributed over this area – a reasonable assumption provided the bearing is new and evenly distributed Since ∑F z = 0, p measured as a force per unit area p = P/π(R 2 2 – R 1 2 )

45. Copyright © 2010 Pearson Education South Asia Pte Ltd Example 8.9 The uniform bar has a total mass m. if it is assumed that the normal pressure acting at the contracting surface varies linearly along the length of the bar, determine the couple moment M required to rotate the bar. Assume that the bar ’ s width a is negligible in comparison to its length l. the coefficient of static friction is equal to μ s.

46. Copyright © 2010 Pearson Education South Asia Pte Ltd Solution FBD of the bar is as shown. Bar has a total weight of W = mg. Intensity w o of the distributed lead at the center (x = 0) is determined from vertical force equilibrium.

47. Copyright © 2010 Pearson Education South Asia Pte Ltd Solution Since w = 0 at x = l/2, for distributed load function, For magnitude of normal force acting on a segment of area having length dx,

48. Copyright © 2010 Pearson Education South Asia Pte Ltd Solution For magnitude of the frictional force acting on the same element of area, For moment created by this force about z axis, Summation of moments by integration,

49. Copyright © 2010 Pearson Education South Asia Pte Ltd 8.7 Frictional Forces on Journal Bearings When a shaft or axle is subjected to lateral loads, a journal bearing is used for support Well-lubricated journal bearings are subjected to the laws of fluid mechanisms When the bearing is not lubricated, analysis of the frictional resistance can be based on the laws of dry friction If the lateral load is P, the bearing reactive force R acting at A is equal and opposite to P

50. Copyright © 2010 Pearson Education South Asia Pte Ltd 8.7 Frictional Forces on Journal Bearings Moment needed to maintain constant rotation of the shaft can be found by the summation of moments about the z axis of the shaft, If the bearing is partially lubricated, μ k is small, μk = tanΦ k ≈ sinΦ k ≈ Φ k Frictional resistance M ≈ Rrμ k

51. Copyright © 2010 Pearson Education South Asia Pte Ltd Example 8.10 The 100mm diameter pulley fits loosely on a 10mm diameter shaft for which the coefficient of static friction is μ s = 0.4. Determine the minimum tension T in the belt needed to (a) raise the 100kg block and (b) lower the block. Assume that no slipping occurs between the belt and the pulley and neglect the weight of the pulley.

52. Copyright © 2010 Pearson Education South Asia Pte Ltd Example 8.10 Part (a) FBD of the pulley is shown. As tension T is increased, the pulley will roll around the shaft to point before motion P 2 impends. Friction circle ’ s radius, r f = r sinΦ s. Using the simplification,

53. Copyright © 2010 Pearson Education South Asia Pte Ltd Example 8.10 Part (a) For radius of friction circle, Therefore,

54. Copyright © 2010 Pearson Education South Asia Pte Ltd Example 8.10 Part (b) When the block is lowered, the resultant force R acting on the shaft passes through the point P 3. Summing moments about this point,

55. Copyright © 2010 Pearson Education South Asia Pte Ltd 8.8 Rolling Resistance A rigid cylinder of weight W rolls at constant velocity along a rigid surface, the normal force is at tangent point of contact Hard material cylinder will compresses the soft surface underneath it

56. Copyright © 2010 Pearson Education South Asia Pte Ltd 8.8 Rolling Resistance We consider the resultant of the entire normal pressure acting on the cylinder N = N d + N r To keep the cylinder in equilibrium, rolling at constant rate, N must be concurrent with the driving force P and the weight W Summation of moment about A, Wa = P (r cosθ) Wa ≈ Pr P ≈ (Wa)/r

57. Copyright © 2010 Pearson Education South Asia Pte Ltd Example 8.11 A 10kg steel wheel has a radius of 100mm and rest on an inclined plans made of wood. If θ is increased so that the wheel begins to roll down the incline with constant velocity when θ = 1.2°, determine the coefficient of rolling resistance.

58. Copyright © 2010 Pearson Education South Asia Pte Ltd Solution FBD of the wheel is as shown. Wheel has impending motion. Normal reaction N acts at point A defined by dimension a. Summing moments about point A, Solving

59. Copyright © 2010 Pearson Education South Asia Pte Ltd QUIZ 1.A friction force always acts _____ to the contact surface. A)NormalB) At 45° C) ParallelD) At the angle of static friction 2. If a block is stationary, then the friction force acting on it is ________. A)   s NB) =  s N C)   s ND) =  k N

60. Copyright © 2010 Pearson Education South Asia Pte Ltd QUIZ 3. A 100 lb box with a wide base is pulled by a force P and  s = 0.4. Which force orientation requires the least force to begin sliding? A) P(A) B) P(B) C) P(C) D) Not determined 4. A ladder is positioned as shown. Please indicate the direction of the friction force on the ladder at B. A)  B)  C) D) P(A) P(B) P(C) 100 lb A B

61. Copyright © 2010 Pearson Education South Asia Pte Ltd QUIZ 5. A wedge allows a ______ force P to lift a _________ weight W. A) (large, large)B) (small, small) C) (small, large)D) (large, small) 6. Considering friction forces and the indicated motion of the belt, how are belt tensions T 1 and T 2 related? A) T 1 > T 2 B) T 1 = T 2 C) T 1 < T 2 D) T 1 = T 2 e 

62. Copyright © 2010 Pearson Education South Asia Pte Ltd QUIZ 7. When determining the force P needed to lift the block of weight W, it is easier to draw a FBD of ______ first. A) The wedge B) The block C) The horizontal groundD) The vertical wall 8. In the analysis of frictional forces on a flat belt, T 2 = T 1 e  . In this equation,  equals ______. A) Angle of contact in deg B) Angle of contact in rad C) Coefficient of static friction D) Coefficient of kinetic friction