1. Copyright © 2010 Pearson Education South Asia Pte Ltd WEEK 8: FRICTION THE BEST APPLICATION OF FRICTION

2. FRICTION - Introduction For two surfaces in contact, tangential forces, called friction forces, will develop if one attempts to move one relative to the other. However, the friction forces are limited in magnitude and will not prevent motion if sufficiently large forces are applied. There are two types of friction: i. dry or Coulomb friction (C. A Coulomb 1781) and – OUR FOCUS ii. fluid friction - fluid friction applies to lubricated mechanisms.

3. Copyright © 2010 Pearson Education South Asia Pte Ltd Characteristics of Dry Friction Coulomb friction occurs between contacting surfaces of bodies in the absence of a lubricating fluid 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

4. 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 – NOT MOVING. 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. N, a distance x to the right of W to balance the “tipping effect” of P x

5. 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 proportional to N. F m is called limiting static frictional force Further increase in P causes the block to begin to move, & F drops to a smaller kinetic-friction force F k. No motion FmFm FkFk x Coefficient of static friction Coefficient of kinetic friction >

6. Copyright © 2010 Pearson Education South Asia Pte Ltd Characteristics of Dry Friction Theory of Dry Friction Typical Values of μ s - you can used/assumed Contact Materials Coefficient 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 Important relationship – you can assume but must be technically valid/sound

7. Statement : for assumption, but must be technically sound/valid “Assume that the coefficient of friction between wood and wood is µ s = 0.7” Or “Assuming that the µ k = 0.7µ s ” Copyright © 2010 Pearson Education South Asia Pte Ltd or any value you put in, but must be realistic based on sound understanding of friction

8. Example 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. Mass = 20 kg Weight = 20 x 9.81N = 196.2 N W = 196.2 N

9. Copyright © 2010 Pearson Education South Asia Pte Ltd Solution Resultant normal force N C 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. FBD

10. Copyright © 2010 Pearson Education South Asia Pte Ltd ∑ F x = 0; 80 cos 30 o N – F = 0 F = 69.3 N N c = 236 N Taking moment: ∑ M o = 0; 80 sin 30 o (0.4 m) - 80 cos 30 o (0.2 m) + N c (x) = O Substituting N c = 236 N, x = - 9.08 mm Solving Solution ∑ F y = 0; -80 sin 30 o N + N c – 196.2 N = O

11. Copyright © 2010 Pearson Education South Asia Pte Ltd Since x is negative, the resultant force N c acts 9.08 mm (slightly) to the left of the crate ’ s center line. No tipping will occur since x ≤ 0.4m (location of center of gravity, G is: x = 0.4 m; y = 0.2m) Solution x G

12. Copyright © 2010 Pearson Education South Asia Pte Ltd To find the 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 BUT it is close to doing so. Solution x G F=69.3 N

13. BACK TO BASIC a x = y, then log a y = x 10 2 = 100, then log 10 100 = 2 Log a xy = Log a x + Log a y Log a (x/y) = Log a x - Log a y Log a x n = nLog a x Copyright © 2010 Pearson Education South Asia Pte Ltd

14. BACK TO BASIC ʃ4x 7 dx = _ 4x 8 _ + c 8 ʃ 2 dx = 2_ ʃ _ x -5 _ = _2 x -4 _ + c 3x 5 3 3 * -4 Copyright © 2010 Pearson Education South Asia Pte Ltd

15. Frictional Forces on Flat Belts In belt drive and band brake design - it is necessary to determine the frictional forces developed between the belt and contacting surfaces

16. Copyright © 2010 Pearson Education South Asia Pte Ltd Frictional Forces on Flat Belts Consider the flat belt which passes over a fixed curved surface – to find tension T 2 to pull the belt Thefefore T 2 > T 1 Total angle of contact β, coef of friction = Consider FBD of the belt segment in contact with the surface N and F vary both in magnitude and direction

17. Copyright © 2010 Pearson Education South Asia Pte Ltd =1 ____ When small Cos( /2) = 1 =1 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

18. Copyright © 2010 Pearson Education South Asia Pte Ltd =negelct, 0 When small Sin( /2) =( /2) The product of infinitesimal size dT and - neglected

19. Copyright © 2010 Pearson Education South Asia Pte Ltd 8.5 Frictional Forces on Flat Belts We have …….1 ……..2

20. Copyright © 2010 Pearson Education South Asia Pte Ltd Example 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. =F

21. 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 Pulling F=

22. 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

23. 8 - 23 Sample Problem 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 2.7 kN, determine the largest torque which can be exerted by the belt on pulley A. SOLUTION: Since angle of contact is smaller, slippage will occur on pulley B first. Determine belt tensions based on pulley B. Taking pulley A as a free-body, sum moments about pulley center to determine torque.

24. 8 - 24 Sample Problem 8.8 30 o

25. 8 - 25 Sample Problem SOLUTION: Since angle of contact is smaller, slippage will occur on pulley B first. Determine belt tensions based on pulley B. Taking pulley A as free-body, sum moments about pulley center to determine torque.

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