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6 Friction

6 Friction. Objectives Students must be able to. Utilize theory of dry friction Describe theory of dry friction Describe physical meanings of frictional effects Describe and differentiate between static and kinetic coefficients of friction Describe the angles of frictions

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6 Friction

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  1. 6 Friction

  2. ObjectivesStudents must be able to • Utilize theory of dry friction • Describe theory of dry friction • Describe physical meanings of frictional effects • Describe and differentiate between static and kinetic coefficients of friction • Describe the angles of frictions • Add friction into the analyses of objects and structures in equilibrium

  3. Objectives Students must be able to • Describe and analyze machines with frictions • Wedges • Threads, screws • Belts • Disks and clutches • Collar, pivot, thrust and journal bearings • Outline rolling resistance • Describe the physical meanings of rolling resistance • Differentiate between frictions and rolling resistance

  4. Topic in textbook We will study this Part first. • Section A: Frictional Phenomena • Characteristics, theory, coefficient of friction, angle of friction • Section B: Applications • Wedges • Screws • Journal Bearings • Thrust Bearings; Disk friction • Flexible Belts • Rolling Resistance

  5. Dry Friction Force of resistance acting on a body which prevents or retards slipping of the body relative to a surface with which it is in contact. Friction exists? • roughnesses of the contacting surfaces. Magnitude: friction’s magnitude limitation will be discussed later • Direction:tangent to the contacting surface and • opposed to the relative motion or tendency for motion Line of Action (Point of application): contact surface

  6. Friction Model Equilibrium y x W In equilibrium a/2 a/2 • Frictional force F FBD is correct? modeling P h F The DN at right side is supporting force more than its left side. x N N The object is toppling (not in equilibrium) If x > a/2 ? increases with force P Slipping and/or Tipping Effect toppling Slipping • The application point (x) of N x-limit F-limit

  7. Motion a/2 a/2 P h • Slipping / Sliding • Relative sliding (translation motion) between two surfaces • Toppling / Tippling • Fall over (rotation) about the edge • Topple, tipping, rolling, tumble, trip F x N N

  8. Static friction (no motion) F Kinetic friction (motion) F=P P Experiment for determining Friction mg P P m FBD F N “impending motion”’ (on the verge of motion) • Object at rest (no motion) • : coefficient of static friction Fk = kN • Object with motion (steady state) • : coefficient of kinetic friction Fs:max = sN : constanton 2 certain contacting surfaces

  9. Angle of Friction Not depend on N not depend on P,v,a • k = arctan(k) = angle of kinetic friction (object in motion) mg P F  N R • s = arctan(s) =angle of (max) static friction (object at rest)

  10. Dry Friction Dry Friction Characteristics • Frictional force acts tangentially to the contacting surfaces, opposing the relative or tendency for motion. • Fs is independent of the area of contact, provided that the normal pressure is not very low nor great enough for deformation of the surfaces. • In equilibrium: Impending slipping:f = fs Slipping:f = fk Very low velocity:fk» fs

  11. Dry Friction Static Friction Typical Values Impending Motion Contact Materials μs Metal / ice 0.03 – 0.05 Wood / wood 0.30 – 0.70 Leather / wood 0.20 – 0.50 Leather / metal 0.30 – 0.60 Aluminum /Aluminum 1.10 – 1.70

  12. Sample 6/1 Determine the maximum angle which the adjustable incline may have before the block of mass m begins to slip. The coefficient of static friction between the block and the inclined surface is s. y x W=mg F N “Impending Slip”: H/2 x Ans (for slipping) Three force member Possibility of toppling? W=mg 3 eq. , 3 unknowns

  13. 6/125 A uniform block of mass m is at rest on an incline z. Determine the maximum force P>0 that can be applied to the block in the direction shown before slipping begins. The coefficient of static friction between the block and the incline is z y x q mg y P N q At this max P, object is about to move at which direction?

  14. Dry Friction Example Friction 2 #1 Will this crate slide or topple over?

  15. 1st Eq. 4 Unknowns: P, F, N, x 2nd Eq. Two possibilities 3rd Eq. 1) about to slip 2) about to tip 4th Eq. 4th Eq. It’s time consuming, Better to know it exactly Check condition: Check condition:

  16. The block of mass m is homogeneous, moving at a constant velocity. The coefficient of kinetic friction is • Determine • a) the greatest value that h may have so that the block will not tip over • b) The location of point C on the bottom face of the block through with the resultant of the friction and normal force act if h = H/2.

  17. The block of mass m is homogeneous, moving at a constant velocity. The coefficient of kinetic friction is • Determine • a) the greatest value that h may have so that the block will not tip over Moving With no acceleration Solution 1 = ? on the verge of tipping Moving The box is in Equilibrium (no acceleration) Three force member in Statics P h=? Concurrent at one point OR parallel mg F F N N Ans on the verge of tipping

  18. The block of mass m is homogeneous. • The block is moving at a constant velocity. • Determine • b) The location of point C on the bottom face of the block through with the resultant of the friction and normal force act if h = H/2. Moving with constant velocity Solution 1 =H/2 C =? N Moving The box is in Equilibrium (no acceleration) Three force member P Concurrent at one point OR parallel h=H/2 mg F x Ans N

  19. Inequality: hard to deal Dry Friction’s Problem static friction + or Equilibrium eq. kinetic friction fricitional eq. Friction • Kinetic motion is known. Moving at constant vel. + Kinetic friction Equilibrium eq.

  20. + Static friction not assuming “impending motion” In static equilibrium friction can support equilibrium Static equilibrium? No. of unknown must = No. of Equilibrium eq. Assume: static equilibrium. Equilibrium eq. Solve for F (and also N) using only equilibrium eq. Only equilibrium eq. can be used to determine unknown values. Assumption Checking Equilibrium eq. Friction is determined by equilibrium eq’s. Static friction If O.K. However, F must <= mN Assumption is not true! Motion occurs. F= kN Static friction

  21. total unknown < equilibrium + frictional eq. + Equilibrium eq. Static friction Find min q to hold equilibrium Find min P to initiate the motion Impending motions occur at both point in the same time. impending motions do not occur in both point at the same time. several possibilities to slip Both can be used together to determine unknown values. total unknown = equilibrium + frictional eq. Assumption needs to be make and that assumption should be checked later. Equilibrium eq. static friction

  22. 500N and 100N Equilibrium? Sample 6/3 Determine the friction force acting on the block shown if P = 500N and P = 100N. The block is initially at rest. Assume: Body in equilibrium Equilibrium state is unknown (not assuming the impending motion) 2 Eq , 2 unknown impending motion OK ! Friction must be enough to maintain equilibrium P=500N Assumption Checking: Correct? P=100N Contradict! Object not in Equilibrium Not valid! still valid!

  23. Dry Friction Example HibbelerEx 8-1 #1 The uniform crate has a mass of 20 kg. If a force P = 80 N is applied to the crate, determine whether it remains in equilibrium. The coefficient of static friction = 0.3. Equilibrium State isnotknown

  24. Example HibbelerEx 8-1 #2 Assume: Equilibrium |x| < 0.4 m (physical boundary of toppling)

  25. Dry Friction Example BedfordEx 9.5 #1 Suppose that a = 10° and the coefficient of friction between the surface of the wedge and the log are ms = 0.22 and mk = 0.20. Neglect the weight of the wedge. • If the wedge is driven into the log at a constant rateby vertical force F, what are the magnitude of the normal forces exerted on the log by the wedge? • Will the wedge remain in place in the log when the force is removed?

  26. = 10° ms = 0.22 and mk = 0.20. • Neglect the weight of the wedge. Moving down at constant rate symmetric/mirror concept

  27. = 10° ms = 0.22 and mk = 0.20. • Neglect the weight of the wedge. • Will the wedge remain in place in the log when the force is removed? Think of minimum friction coefficient that still “self-lock” the wedge Impending Motion (On the verge of slipping) symmetric/mirror concept

  28. Dry Friction Example Bedford9.20 #1 The coefficient of static friction between the two boxes and between the lower box and the inclined surface is ms. What is the largest angle a for which the lower box will not slip.

  29. The coefficient of static friction between the two boxes and between the lower box and the inclined surface is ms. What is the largest angle a for which the lower box will not slip. Impending Motion at same time ?

  30. Bedford9.30 (ex) The cylinder has weight W. The coefficient of static friction between the cylinder and the floor and between the cylinder and the wall is ms. What is the largest couple Mthat can be applied to the stationary cylinder without causing it to rotate. Impending Motion at same time Impending rotation

  31. Bedford9.33 (Ex) The disk of weight W and radius R is held in equilibrium on the circular surface by a couple M. The coefficient of static friction between the disk and the surface is ms. Find that the largest value M can have without causing the disk to slip.

  32. Dry Friction Example Bedford9.166 #1 Each of the uniform 1-m bars has a mass of 4 kg. The coefficient of static friction between the bar and the surface at B is 0.2. If the system is in equilibrium, what is the magnitude of the friction force exerted on the bar at B.

  33. Dry Friction Example Bedford9.166 #2

  34. Dry Friction Example Bedford9.166 #3

  35. Dry Friction Example Bedford9.166 #4 symmetry?

  36. Dry Friction Example HibbelerEx 8-3 #1 The rod with weight W is about to slip on rough surfaces at A and B. Find coefficient of static friction. Direction of N?

  37. Dry Friction Example HibbelerEx 8-3 #2

  38. Dry Friction ExampleHibbeler Ex 8-3 #3

  39. 6/9 For a 20 jaw opening, what is the minimum coefficient of static friction between the jaws and the tube which will enable tongs to grip the tube with out slipping Solution 1 y x

  40. 6/9 For a 20 jaw opening, what is the minimum coefficient of static friction between the jaws and the tube which will enable tongs to grip the tube with out slipping Solution 2 y The object is in equilibrium Two force systems x

  41. T x y x y P y x Sample 6/5 Find the maximum value which P may have before any slipping takes place. ? Possible ways to slip 1) Middle object is going to move lonely. 2) Middle + buttom object is going to move together.

  42. T x y x y P y x Check the Assumption Friction F3 can support the equilibrium of 40-kg object Assume: 1) Middle object is going to move lonely. Impending motion 5 unknowns 3 equilibrium eq 2 friction eq Impossible, the assumption is wrong

  43. T x y x y P y x F2 can support the 50-kg and 40-kg as one body Assume: 2) Middle + bottom object is going to move together. Impending motion OK Ans

  44. T x y x y P y x Assume: ? 3) 40-kg object moves lonely --– possible? Equation without unknown (little chance to be true)

  45. 8-61 The uniform 50-N slender rod rests on the top center of the 20-N block. Determine the largest couple moment M which can be applied to the rod without causing motion of the rod. + y x Motion Possibility? 1. Slip at C, B 2. Slip at C, A 3. Slip at A, B ? 4. other… x

  46. 8-61 The uniform 50-N slender rod rests on the top center of the 20-N block. Determine the largest couple moment M which can be applied to the rod without causing motion of the rod. + y x Assume 2. Slip at C, A 8 unknowns, 6+2 equation. Don’t solve 8 eq paralleling! Solve only 6 eq. excluding Assumption Checking x

  47. 8-56 The uniform 50-N slender rod rests on the top center of the 20-N block. Determine the largest couple moment M which can be applied to the rod without causing motion of the rod. + y x Assume 2. Slip at C, A Block is tipping = 0.12 m = 62.5 N = 9.5 N-m = 42.5 N = 25 N Assumption Checking = 7.5 N Correct?

  48. 8-61 The uniform 50-N slender rod rests on the top center of the 20-N block. Determine the largest couple moment M which can be applied to the rod without causing motion of the rod. + y x Possibility? 1. Slip at C, B 2. Slip at C, A 3. Slip at A, B ? 4. C-Slip, x-limit 4. other… 5. B-Slip, x-limit x 6. A-Slip, x-limit

  49. 8-61 The uniform 50-N slender rod rests on the top center of the 20-N block. Determine the largest couple moment M which can be applied to the rod without causing motion of the rod. + y x Assume 4. C-Slip, x-limit 8 unknowns, 6+2 equation. Don’t solve 8 eq paralleling! Solve only 6 eq. excluding Assumption Checking x

  50. 8-61 The uniform 50-N slender rod rests on the top center of the 20-N block. Determine the largest couple moment M which can be applied to the rod without causing motion of the rod. + y x Assume 4. C-Slip, x-limit = 63.64 N = 5.78 N-m = 21.21 N Ans 21.21 <=0.6*43.64=26.18 = 43.64 N Assumption Checking 21.21 <=0.4*63.64=25.46 = 6.36 N

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