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concepts about friction

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  • 1. Engineering Mechanics Friction 1
  • 2. Chapter Outline Theory of Dry Friction Applications Computational Mechanics 2
  • 3. 9.1 Theory of Dry Friction To examine the nature of friction forces:  Place a book on a table & push it with small horizontal force:  Ifthe force you exert is sufficiently small, the book does not move 3
  • 4. 9.1 Theory of Dry Friction  Free-body diagram:  The force W is the book’s weight & N is the total normal force exerted by the table on the surface of the book that is in contact with the table  The force F is the horizontal force you apply & f is the friction force exerted by the table  Because the book is in equilibrium, f = F 4
  • 5. 9.1 Theory of Dry Friction  As long as the book remains in equilibrium, when you increase the force you apply to the book, the friction force must increase correspondingly  When the force you apply becomes too large, the book moves (slips on the table)  After reaching some maximum value, the friction force can no longer maintain the book in equilibrium  Notice that the force you must apply to keep the book moving on the table is smaller than the force required to cause it to slip 5
  • 6. 9.1 Theory of Dry Friction  Ifthe surfaces of the table & the book are magnified sufficiently, they will appear rough:  Friction forces arise in part from the interactions of the roughness or asperities of the contacting surfaces 6
  • 7. 9.1 Theory of Dry Friction  Suppose that we idealize the asperities of the book & table as the mating 2-D “saw-tooth” profiles in Fig. a  As the horizontal force F increases, the book will remain stationary until the force is sufficiently large to cause the book to slide upward as shown in Fig. b 7
  • 8. 9.1 Theory of Dry Friction  The normal force Ci exerted on the ith saw- tooth asperity of the book:  Notice that in this simple model we assume the contacting surfaces on the asperities to be smooth  Denote the sum of the normal forces exerted on the asperities of the book by the table by C = ∑ Ci i 8
  • 9. 9.1 Theory of Dry Friction  Equilibriumequations: ∑ Fx = F − C sin α = 0 ∑ Fy = C cos α − W = 0  Eliminating C from these equations, we obtain the force necessary to cause the book to slip on the table: F = ( tan α ) W  The force necessary to cause the book to slip is proportional to the force pressing the saw-tooth surfaces together (the book’s weight) 9
  • 10. 9.1 Theory of Dry Friction  Also, the angle α is a measure of the roughness of the saw-tooth surfaces:  As α  0, the surfaces become smooth & the force necessary to cause the book to slip approaches zero  As α increases, the roughness increases & the force necessary to cause the book to slip increases 10
  • 11. 9.1 Theory of Dry Friction Coefficients of Friction:  The theory of dry friction or Coulomb friction predicts the maximum friction forces that can be exerted by dry, contacting surfaces that are stationary relative to each other  It also predicts the friction forces exerted by the surfaces when they are in relative motion or sliding 11
  • 12. 9.1 Theory of Dry Friction The Static Coefficient:  The magnitude of the maximum friction force that can be exerted between 2 plane, dry surfaces in contact that are not in motion relative to 1 another is: f = µS N (9.1) where N is the normal component of the contact force between the surfaces & µS is a constant called the coefficient of static friction 12
  • 13. 9.1 Theory of Dry Friction  The value of µS is assumed to depend only on the materials of the contacting surfaces & the conditions (smoothness & degree of contamination by other materials) of the surfaces  Typical values of µS for various materials are shown in Table 9.1  The relatively large range of values for each pair of materials reflects the sensitivity of µS to the conditions of the surfaces 13
  • 14. 9.1 Theory of Dry Friction  Table 9.1: Coefficient of Materials Station Friction, µ S Metal on metal 0.15−0.20 Masonry on 0.60−0.70 masonry Wood on wood 0.25−0.50 Metal on masonry 0.30−0.70 Metal on wood 0.20−0.60 Rubber on concrete 0.50−0.90 14
  • 15. 9.1 Theory of Dry Friction  Returning to the example of the book on the table:  If we know the coefficient of static friction between the book & the table, Eq. (9.1) tells us the largest friction force that the table can exert on the book: F = f = µSN  Also, from the free-body diagram: N = W, so the largest force that will not cause the book to slip is F = µSW 15
  • 16. 9.1 Theory of Dry Friction  Eq. (9.1) determines the magnitude of the maximum friction force but not its direction  The friction force is a maximum & Eq. (9.1) is applicable when 2 surfaces are on the verge of slipping relative to each other  slip is impending & the friction forces resist the impending motion 16
  • 17. 9.1 Theory of Dry Friction  E.g. the lower surface is fixed & slip of the upper surface toward the right is impending:  The friction force on the upper surface resists its impending motion  The friction force on the lower surface is in the opposite direction 17
  • 18. 9.1 Theory of Dry Friction The Kinetic Coefficient:  According to the theory of dry friction, the magnitude of the friction force between 2 plane dry contacting surfaces that are in motion (sliding) relative to each other is: f = µk N (9.2) where N is the normal force between the surfaces & µk is the coefficient of kinetic friction 18
  • 19. 9.1 Theory of Dry Friction  The value of µk is assumed to depend only on the compositions of the surfaces & their conditions  For a given pair of surfaces, its value is generally smaller than that of µS  Once you have caused the book to begin sliding on the table, the friction force: f = µkN = µkW  Therefore,the force you must exert to keep the book in uniform motion is: F = f = µkW 19
  • 20. 9.1 Theory of Dry Friction  When 2 surfaces are sliding relative to each other, the friction forces resist the relative motion:  E.g. the lower surface is fixed & the upper surface is moving to the right  The friction force on the upper surface acts in the direction opposite to its motion  The friction force on the lower surface is in the opposite direction 20
  • 21. 9.1 Theory of Dry Friction Angles of Friction:  Expressing the reaction exerted on a surface due to its contact with another surface in terms of its components parallel & perpendicular to the surface, the friction force f & normal force N  Expressing the reaction in terms of its magnitude R & the angle of friction θ between the reaction & the normal to the surface 21
  • 22. 9.1 Theory of Dry Friction  The forces f & N are related to R & θ by: f = R sin θ (9.3) N = R cosθ (9.4)  The value of θ when slip is impending is called the angle of static friction θS & its value when the surfaces are sliding relative to each other is called the angle of kinetic friction θk 22
  • 23. 9.1 Theory of Dry Friction  By using Eqs. (9.1)—(9.4), we can express the angles of static & kinetic friction in terms of the coefficients of friction: tan θS = µS (9.5) tan θ k = µk (9.6) 23
  • 24. 9.1 Theory of Dry Friction  Summary:  If slip is impending, the magnitude of the friction is given by Eq. (9.1) & the angle of friction by Eq. (9.5)  If the surfaces are sliding relative to each other, the magnitude of the friction force is given by Eq. (9.2) & the angle of friction by Eq. (9.6)  Otherwise, the friction force must be determined from the equilibrium equations 24
  • 25. 9.1 Theory of Dry Friction  The sequence of decisions in evaluating the friction force & angle of friction is summarized in the Fig 9.8: 25
  • 26. Example 9.1 Determining the FrictionForceThe arrangement in Fig. 9.9exerts a horizontal force on thestationary 180-N crate. Thecoefficient of static frictionbetween the crate & the ramp Fig. 9.9is µS = 0.4.(a) If the rope exerts a 90-N force on the crate, what is the friction force exerted on the crate by the ramp?(b) What is the largest force the rope can exert on the crate without causing it to slide up the ramp? 26
  • 27. Example 9.1 Determining the FrictionForceStrategy(a) We can follow the logic in Fig. 9.8 to decide how to evaluate the friction force. The crate is not sliding on the ramp & we don’t know whether slip is impending, so we must determine the friction force by using the equilibrium equations. 27
  • 28. Example 9.1 Determining the FrictionForceStrategy(b) We want to determine the value of the force exerted by the rope that causes the crate to be on the verge of slipping up the ramp. When slip is impending, the magnitude of the friction force is f = µSN & the friction force opposes the impending slip. We can use the equilibrium equations to determine the force exerted by the rope. 28
  • 29. Example 9.1 Determining the FrictionForceSolution(a) Draw the free-body diagram of the crate, showing the force T exerted by the rope, the weight W of the crate & the normal force N & friction force f exerted by the ramp: 29
  • 30. Example 9.1 Determining the FrictionForceSolutionWe can choose the direction of f arbitrarily & oursolution will indicate the actual direction of thefriction force.By aligning the coordinate system with the ramp asshown, we obtain the equilibrium equation: Σ Fx = f + T cos 20° − W cos 20° = 0 30
  • 31. Example 9.1 Determining the FrictionForceSolutionSolving for the friction force, we obtain: f = −T cos 20° + W sin 20° = −(90 N) cos 20° + (180 N) sin 20° = −23.0 NThe minus sign indicates that the direction of thefriction force on the crate is down the ramp. 31
  • 32. Example 9.1 Determining the FrictionForceSolution(b) In this case the friction force f = µSN & it opposes the impending slip. To simplify our solution for T, we align the coordinate system as shown: 32
  • 33. Example 9.1 Determining the FrictionForceSolutionThe equilibrium equations: Σ Fx = T − N sin 20° − µSN cos 20° = 0 Σ Fy = N cos 20° − µSN sin 20° − W= 0Solving the 2nd equilibrium equation for N, weobtain: W N= cos 20° − µS sin 20° 180 N = cos 20° − 0.4 sin 20° = 224 N 33
  • 34. Example 9.1 Determining the FrictionForceSolutionThen, from the 1st equilibrium equation, the force Tis: T = N (sin 20° + µS cos 20°) = 0 = (224 N) (sin 20° + 0.4 cos 20°) = 161 N 34
  • 35. Example 9.1 Determining the FrictionForceCritical Thinking When you use the equilibrium equations to determine a friction force, often you will not know its direction beforehand:  Depending on the value of the force T exerted on the crate by the rope, the friction force exerted on the crate by the ramp can point either up or down the ramp  In drawing the free-body diagram of the crate in (a), we arbitrarily assumed that the friction force pointed up the ramp 35
  • 36. Example 9.1 Determining the FrictionForceCritical Thinking  The negative value obtained from the equilibrium equations, f = −23.0 N, tells us that the force is in the opposite direction, down the ramp In contrast, when you use the equation f = µSN, the friction force must point in the correct direction on the free-body diagram 36
  • 37. Example 9.1 Determining the FrictionForceCritical Thinking  Indrawing the free-body diagram in (b), we wanted to determine the largest force T that would not cause the crate to slide up the ramp, so we assumed that the slip of the crate up the ramp was impending  This told us that the friction force, resisting the impending slip, pointed down the ramp 37
  • 38. Example 9.2 A Friction Problem in 3DimensionsThe 80-kg climber at A in Fig. 9.12 is supported onan icy slope by friends. The tensions in the ropesAB & AC are 130 N & 220 N respectively, the y axisis vertical & the unit vector e = −0.182i + 0.818j +0.545k is perpendicular to the ground where theclimber stands. What minimum coefficient of staticfriction between the climber’s shoes & the groundis necessary to prevent him from slipping? 38
  • 39. Example 9.4 A Friction Problem in 3Dimensions Fig. 9.12 39
  • 40. Example 9.4 A Friction Problem in 3DimensionsStrategyWe know the forces exerted on the climber by the2 ropes & by his weight so we can use equilibriumto determine the force R exerted on him by theground. The components of R normal & parallel tothe ground are the normal & friction forces exertedon him by the ground. By calculating them, we canobtain the minimum necessary coefficient of staticfriction. 40
  • 41. Example 9.4 A Friction Problem in 3DimensionsSolutionDraw the free-body diagram of the climber showingthe forces TAB & TAC exerted by the ropes, the forceR exerted by the ground & his weight: 41
  • 42. Example 9.4 A Friction Problem in 3DimensionsSolutionThe sum of the forces equals zero: R + TAB + TAC – mgj = 0By expressing TAB & TAC in terms of theircomponents, we can solve this equation for thecomponents of R. The force TAB is:  ( 2 − 3) i + ( 2 − 0) j + ( 0 − 4) k  TAB = TAB    ( 2 − 3) 2 + ( 2 − 0 ) 2 + ( 0 − 4) 2    = (130 N ) ( − 0.218i + 0.436 j − 0.873k ) = −28.4i + 56.7 j − 113.5k 42
  • 43. Example 9.4 A Friction Problem in 3DimensionsSolutionAnd the force TAC is:  ( 5 − 3) i + ( 2 − 0 ) j + ( − 1 − 4 ) k  TAC = TAC    ( 5 − 3) 2 + ( 2 − 0 ) 2 + ( − 1 − 4 ) 2    = ( 220 N ) ( 0.348i + 0.348 j − 0.870k ) = 76.6i + 76.6 j − 191.5kSubstituting these expressions into the equilibriumequation & solving for R, we obtain: R = –48.2i + 651.5j + 305.0k (N) 43
  • 44. Example 9.4 A Friction Problem in 3DimensionsSolutionThe normal force on the climber is the componentof R perpendicular to the surface, which is thecomponent of R parallel to the unit vector e:N = (e · R)e = [(−0.182)(−48.2 N) + (0.818)(651.5 N) + (0.545)(305.0 N)]e = −129i + 579j + 386k (N)The magnitude of the normal force is: N = |N| = 708 N. 44
  • 45. Example 9.4 A Friction Problem in 3DimensionsSolutionThe friction force on the climber is the magnitudeof the component of R parallel to the surface: f = |R – N| = 135 NThe minimum coefficient of friction necessary toprevent the climber from slipping is therefore: f 135 N µS = = = 0.191 N 708 N 45
  • 46. Example 9.4 A Friction Problem in 3DimensionsCritical Thinking Notice the role of the unit vector e in this example:  By using equilibrium, we were able to determine the total force R exerted on the climber by the ground  This force consists of components normal & parallel to the ground (the normal & friction forces respectively)  The unit vector e specified the orientation of the ground on which the climber stood & allowed us to determine the normal & friction forces 46
  • 47. 9.2 Applications Belt Friction:  Ifa rope is wrapped around a fixed post, a large force T2 exerted on 1 end can be supported by a relatively small force T1 applied to the other end: 47
  • 48. 9.2 Applications  Considera rope wrapped through an angle β around a fixed cylinder:  Assume that the tension T1 is known  The objective is to determine the largest force T2 that can be applied to the other end of the rope without causing the rope to slip 48
  • 49. 9.2 Applications  Draw the free-body diagram of an element of the rope whose boundaries are at angles α & α + ∆α from the point where the rope comes into contact with the cylinder:  The force T is the tension in the rope at the position defined by the angle α 49
  • 50. 9.2 Applications  The tension in the rope varies with position, because it increases from T1 at α = 0 to T2 at α =β  Therefore, we write the tension in the rope at the position α + ∆α as T + ∆T  The force ∆N is the normal force exerted on the element by the cylinder  Assume that the friction force is equal to its maximum possible value µS∆N, where µS is the coefficient of static friction between the rope & the cylinder 50
  • 51. 9.2 Applications  The equilibrium equations in the directions tangential to & normal to the centerline of the rope are: ∆α  ∆α  ∑ Ftangential = µS∆N + T cos    − ( T + ∆T ) cos   =0  2   2  Δα ∆α  ∑ Fnormal = ∆N − ( T + ∆T ) sin   − T sin     =0 (9.16)  2   2   Eliminating ∆N, we can write the resulting equilibrium equation as:   ∆α  − µ sin  ∆α  ∆T − µ T sin ( ∆α / 2 ) = 0 cos  2     S    2  ∆α S ∆α / 2 51
  • 52. 9.2 Applications the limit of this equation as ∆α  0  Evaluating & observing that: sin ( ∆α / 2 ) →1 ∆α / 2  We obtain: dT − µST = 0 dα  Thisdifferential equation governs the variation of the tension in the rope 52
  • 53. 9.2 Applications  Separating the variables yields: dT = µS dα T  We can now integrate to determine the tension T2 in terms of the tension T1 & the angle β: T2 dT β ∫T1 T = ∫ µ S dα 0  Thus, we obtain the largest force T2 that can be applied without causing the rope to slip when the force on the other end is T1: µSβ T2 = T1e (9.17) 53
  • 54. 9.2 Applications  The angle β in this equation must be expressed in radians  Replacing µS by the coefficient of kinetic friction µk gives the force T2 required to cause the rope to slide at a constant rate  Eq. (9.17) explains why a large force can be supported by a relatively small force when a rope is wrapped around a fixed support:  The force required to cause the rope to slip increases exponentially as a function of the angle through which the rope is wrapped 54
  • 55. 9.2 Applications  Suppose µS = 0.3:  When the rope is wrapped 1 complete turn around the post (β = 2π), the ratio T2/T1 = 6.59  When the rope is warped 4 complete turns around the post (β = 8π), the ratio T2/T1 = 1880 55
  • 56. Example 9.9 Rope Wrapped Around 2CylindersThe 100-N crate in Fig. 9.29 is suspended from arope that passes over 2 fixed cylinders. Thecoefficient of static friction is 0.2 between the rope& the left cylinder & 0.4 between the rope & theright cylinder. What is the smallest force the womancan exert & support the crate? Fig. 9.29 56
  • 57. Example 9.9 Rope Wrapped Around 2CylindersStrategyShe exerts the smallest possible force when slip ofthe rope is impending on both cylinders. Becausewe know the weight of the crate, we can use Eq.(9.17) to determine the tension in the ropebetween the 2 cylinders & then use Eq. (9.17)again to determine the force she exerts. 57
  • 58. Example 9.9 Rope Wrapped Around 2CylindersSolutionThe weight if the crate is W = 100 N. Let T be thetension in the rope between the 2 cylinders. Therope is wrapped around the left cylinder through anangle β = π/2 rad. 58
  • 59. Example 9.9 Rope Wrapped Around 2CylindersSolutionThe tension necessary to prevent the rope fromslipping on the left cylinder is related to W by: µSβ ( 0.2 )( π / 2 ) W = Te = TeSolving for T, we obtain: T = We− ( 0.2 )( π / 2 ) = −(100 N ) e − ( 0.2 )( π / 2 ) = 73.0 N 59
  • 60. Example 9.9 Rope Wrapped Around 2CylindersSolutionThe rope is also wrapped around the right cylinderthrough an angle β = π/2 rad.The force F the woman must exert to prevent therope from slipping on the right cylinder is related toT by: µ β ( 0.4 )( π / 2 ) T = Fe S = FeThe solution for F is: − ( 0.4 )( π / 2 ) − ( 0.4 )( π / 2 ) F = Te = −( 73.0 N ) e = 39.0 N 60
  • 61. Example 9.9 Rope Wrapped Around 2CylindersCritical Thinking To determine the force that would need to be exerted on the rope to cause the crate to begin moving upward:  Assume that the slip of the rope is impending on both cylinders but in the opposite direction to our analysis in this example  For the left cylinder, the tension T necessary for slip of the rope to be impending in the direction that would cause the crate to move upward is: T = We( 0.2 )( π / 2 ) = (100 N ) e( 0.2 )( π / 2 ) = 137 N 61
  • 62. Example 9.9 Rope Wrapped Around 2CylindersCritical Thinking  For the right cylinder, the force F necessary for slip to be impending in the direction that would cause the crate to move upward is: ( 0.4 )( π / 2 ) ( 0.4 )( π / 2 ) F = Te = (137 N ) e = 257 N  Although the young woman would be able to support the stationary crate, she might need to call for help to raise it. 62
  • 63. Design Example 9.10 Belts & PulleysThe pulleys in Fig. 9.30 turn at a constant rate. Thelarge pulley is attached to a fixed support. Thesmall pulley is supported by a smooth horizontalslot & is pulled to the right by the force F = 200 N.The coefficient of static friction between the pulleys& the belt is µS = 0.8, the dimension b = 500 mm &the radii of the pulleys are RA = 200 mm & RB = 100mm. What are the largest values of the couples MA& MB for which the belt will not slip? 63
  • 64. Design Example 9.10 Belts & Pulleys Fig. 9.30 64
  • 65. Design Example 9.10 Belts & PulleysStrategyBy drawing the free-body diagrams of the pulleys,we can use the equilibrium equations to relate thetensions in the belt to MA & MB & obtain a relationbetween the tensions in the belt & the force F.When slip is slipping, the tensions are also relatedby Eq. (9.17). From these equations we candetermine MA & MB. 65
  • 66. Design Example 9.10 Belts & PulleysSolution 66
  • 67. Design Example 9.10 Belts & PulleysSolutionFrom the free-body diagram of the large pulley, weobtain the equilibrium equation: MA = RA (T2 − T1) (1)and from the free-body diagram of the small pulley,we obtain: F = (T1 + T2) cos α (2) MB = RB (T2 − T1) (3) 67
  • 68. Design Example 9.10 Belts & PulleysSolutionThe belt is in contact with the small pulley throughthe angle π − 2α:From the dashed line parallel to the belt, we seethat the angle α satisfies the relation: RA − RB 200 mm − 100 mm sin α = = = 0.2 b 500 mm 68
  • 69. Design Example 9.10 Belts & PulleysSolutionTherefore α = 15° = 0.201 rad.If we assume that the slip is impends between thesmall pulley & the belt, Eq. (9.17) states that: µSβ ( 0.8 )( π − 2α ) T2 = T1e = T1e = 8.95T1We solve this equation together with Eq. (2) for the2 tensions, obtaining T1 = 20.5 N & T2 = 183.6 N.Then from Eqs. (1) & (3), the couples are: MA = 32.6 N-m & MB = 16.3 N-m 69
  • 70. Design Example 9.10 Belts & PulleysSolutionIf we assume that slip impends between the largepulley & the belt, we obtain: MA = 36.3 N-m & MB = 18.1 N-m,so the belt slips on the small pulley at smallervalues of the couples. 70
  • 71. Design Example 9.10 Belts & PulleysDesign Issues Belts & pulleys are used to transfer power in cars & many other types of machines, including printing presses, farming equipment & industrial robots:  Because 2 pulleys of different diameters connected by belt are subjected to different torques & have different rates of rotation, they can be used as a mechanical “transformer” to alter torque or rotation rate 71
  • 72. Design Example 9.10 Belts & PulleysDesign Issues In this example we assumed that the belt was flat but “V-belts” that fit into matching grooves in the pulleys are often used in applications  This configuration keeps the belt in place on the pulleys & also decreases the tendency of the belt to slip  Suppose that a V-belt is wrapped through an angle β around a pulley 72
  • 73. Design Example 9.10 Belts & PulleysDesign Issues  Ifthe tension T1 is known, what is the largest tension T2 that can be applied to other end of the belt without causing it to slip?  Free-body diagram of an element of the belt whose boundaries are at angles α & α + ∆α from the point where the belt comes into contact with the pulley: 73
  • 74. Design Example 9.10 Belts & PulleysDesign Issues  The equilibrium equations in the directions tangential to & normal to the centers of the belt are:  ∆α  − ( T + ∆T ) cos  ∆α  = 0∑ Ftangential = 2µS∆N + T cos  2      2    γ  − ( T + ∆T ) sin  Δα  − T sin  ∆α  = 0 (4) ∑ Fnormal = 2∆N sin  2    2    2    By the same steps leading from Eqs. (9.16) to Eq. (9.17), it can be shown that: µ S β / sin ( γ / 2 ) T2 = T1e (5) 74
  • 75. Design Example 9.10 Belts & PulleysDesign Issues  Thus, using a V-belt effectively increases the coefficient of friction between the belt & the pulley by the factor 1/sin (γ /2)  When it is essential that the belt not slip relative to the pulley, a belt with cogs & a matching pulley or a chain & sprocket wheel can be used  The chains & sprocket wheels in bicycles & motorcycles are examples 75
  • 76. Computational Example 9.11The mass of the block A in Fig. 3.33 is 20 kg & the coefficient of static friction between the block & the floor is µS = 0.3. The spring constant k = 1 kN/m & the spring is unstretched. How far can the slider B be moved to the right without causing the block to slip? Fig. 3.33 76
  • 77. Computational Example 9.11StrategyDraw the free-body diagram of block A assumingthat the slider B is moved a distance x to the right &slip of block A is impending. Then by applying theequilibrium equations, we can obtain an equationfor the distance x corresponding to impending slip. 77
  • 78. Computational Example 9.11SolutionSuppose that moving the slider B a distance x tothe right causing impending slip of the block. Theresulting stretch of the spring is 1 + x 2 − 1 m. 78
  • 79. Computational Example 9.11SolutionThe magnitude of the force exerted on the block bythe spring is: ( FS = k 1 + x 2 − 1 ) (1)From the free-body diagram of the block, weobtain the equilibrium equations:  x  ∑ Fx =  1 + x 2  F S − µS N = 0    1  ∑ Fy =  1 + x 2  F S + N − mg = 0   79
  • 80. Computational Example 9.11SolutionSubstituting Eq. (1) into these 2 equations & theneliminating N, we can write the resulting equationin the form: h( x ) = k ( x + µS ) ( ) 1 + x 2 − 1 − µSmg 1 + x 2 = 0We must obtain the root of this function todetermine the value of x corresponding toimpending slip of the block. 80
  • 81. Computational Example 9.11SolutionFrom the graph of h(x), we estimate that h(x) = 0 atx = 0.43 m. By examining computed results near thisvalue of x (see table), we see that h(x) = 0 & slip isimpending, when x is approximately 0.4284 m. x (m) h(x) 0.4281 −0.1128 0.4282 −0.0777 0.4283 −0.0425 0.4284 −0.0074 0.4285 0.0278 0.4286 0.0629 0.4287 0.0981 81
  • 82. Computational Example 9.11Critical Thinking Many software packages are available that allow you to obtain solutions to nonlinear algebraic equations such as the 1 we obtained in this example Even when you have access to such software, it is a good idea to examine graphical results like those we have presented: 82
  • 83. Computational Example 9.11Critical Thinking  Nonlinear equations sometimes have multiple roots & to insure that you have obtained all the solutions within the range of interest & you have identified the 1 you want  In addition, you can often gain insight by examining the behaviour of an equation over a range of its variables instead of obtaining just 1 solution 83
  • 84. Chapter Summary Dry Friction:  The forces resulting from the contact of 2 plane surfaces can be expressed in terms of the normal force N & friction force f or the magnitude R & angle of friction θ : 84
  • 85. Chapter Summary  If slip is impending, the magnitude of the friction force is: f = µS N (9.1) and its direction opposes the impending slip. The angle of friction equals the angle of static friction θS = arctan (µS)  If the surfaces are sliding, the magnitude of the friction force is: f = µk N (9.2) and its direction opposes the relative motion. The angle of friction equals the angle of kinetic friction θk = arctan (µk) 85
  • 86. Chapter Summary Threads:  The slope α of the thread is related to its pitch p by: p tan α = (9.7) 2π r  The couple required for impending rotation & the axial motion opposite to the direction of F is: M = rF tan (θS + α) (9.9) 86
  • 87. Chapter Summary  The couple required for impending rotation & the axial motion of the shaft in the direction of F is: M = rF tan (θS − α) (9.11)  When θS < α, the shaft will rotate & move in the direction of the force F with no couple applied 87
  • 88. Chapter Summary Journal Bearings:  The couple required for impending slip of the circular shaft is: M = rF sin θS (9.12) where F is the total load on the shaft 88
  • 89. Chapter Summary Thrust Bearings & Clutches:  The couple required to rotate the shaft at a constant rate is: 2 µk F  ro3 − ri3  M=  2 2 (9.13) 3 cos α  ro − ri    89
  • 90. Chapter Summary Belt Friction:  The force T2 required for impending slip in the direction of T2 is: µSβ T2 = T1e (9.17) where β is in radians 90