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WEEK 8: FRICTION THE BEST APPLICATION OF FRICTION Copyright © 2010 Pearson Education South Asia Pte Ltd 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. 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 Copyright © 2010 Pearson Education South Asia Pte Ltd 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. x • 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 The Laws of Dry Friction. Coefficients of Friction x • As P increases, the static-friction force F increases as well until it reaches a maximum value Fm proportional to N. Fm is called limiting static frictional force F N F N m m No motion s s Coefficient of static friction • Further increase in P causes the block to begin to move, & F drops to a smaller kinetic-friction force Fk. Fm F N Fk k k s k Coefficient of kinetic friction > k 0.75 k s Characteristics of Dry Friction Important relationship – you can assume but must be technically valid/sound Theory of Dry Friction Typical Values of μs - you can used/assumed 0.75 k s Contact Materials Coefficient of Static Friction μs 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 Copyright © 2010 Pearson Education South Asia Pte Ltd Statement: for assumption, but must be technically sound/valid • “Assume that the coefficient of friction between wood and wood is µs = 0.7” or any value you put in, but must be realistic based on sound Or understanding of friction • “Assuming that the µk = 0.7µs” Copyright © 2010 Pearson Education South Asia Pte Ltd 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 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. FBD Copyright © 2010 Pearson Education South Asia Pte Ltd Solution ∑ Fx = 0; 80 cos 30o N – F = 0 F = 69.3 N ∑ Fy = 0; -80 sin 30o N + Nc – 196.2 N = O Nc = 236 N Taking moment: ∑ Mo = 0; 80 sin 30o (0.4 m) - 80 cos 30o (0.2 m) + Nc (x) = O Substituting Nc = 236 N, x = - 9.08 mm Copyright © 2010 Pearson Education South Asia Pte Ltd Solution G x Since x is negative, the resultant force Nc 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) Copyright © 2010 Pearson Education South Asia Pte Ltd Solution To find the max frictional force which can be developed at the surface of contact: Fmax = μsNC = 0.3(236N) = 70.8N G x F=69.3 N Since F = 69.3N < 70.8N, the crate will not slip BUT it is close to doing so. Copyright © 2010 Pearson Education South Asia Pte Ltd BACK TO BASIC ax = y, then logay = x 102 = 100, then log10100 = 2 Logaxy = Logax + Logay Loga(x/y) = Logax - Logay Logaxn = nLogax Copyright © 2010 Pearson Education South Asia Pte Ltd BACK TO BASIC ʃ4x7 dx = _4x8_ + c 8 ʃ 2 dx = 2_ ʃ _x-5_ = _2 x-4_ + c 3x5 3 3 * -4 Copyright © 2010 Pearson Education South Asia Pte Ltd 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 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 T2 to pull the belt • Thefefore T2 > T1 • 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 Copyright © 2010 Pearson Education South Asia Pte Ltd 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 F 0; x d d T cos dN (T dT ) cos 0 2 2 d T *1 dN T *1 dT cos 0 2 dN dT 0 When d small Cos(d /2) = 1 dN dT Copyright © 2010 Pearson Education South Asia Pte Ltd F 0; y d d When d small dN (T dT ) sin T sin 0 Sin(d /2) =( d /2) 2 2 d d d dN T dT T 0 2 2 2 d dN T *2 0 0 The product of infinitesimal 2 size dT and d - neglected dN Td 2 Copyright © 2010 Pearson Education South Asia Pte Ltd 8.5 Frictional Forces on Flat Belts • We have dN dT dN Td …….1 ……..2 dT d T T T1 , 0, T T2 , dT T1 T 0 d T In 2 T1 T2 T2 T1e 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 μs = 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 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 T2 in the cord occur at D where T2 = 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 T2 T1e s ; 500 N T1e 0.253 / 4 Pulling F= T1 500 N e 0.253 / 4 500 N 277.4 N 1.80 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 T2 T1e s ; 277.4 We 0.253 / 4 W 153.9 N W 153.9 N m 15.7kg 2 g 9.81m / s Copyright © 2010 Pearson Education South Asia Pte Ltd Sample Problem SOLUTION: • Since angle of contact is smaller, slippage will occur on pulley B first. Determine belt tensions based on pulley B. 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. 8 - 23 • Taking pulley A as a free-body, sum moments about pulley center to determine torque. Sample Problem 8.8 30o 8 - 24 Sample Problem SOLUTION: • Since angle of contact is smaller, slippage will occur on pulley B first. Determine belt tensions based on pulley B. T2 e s T1 T1 2700 N e 0.252 3 1.688 T1 2700 N 1600 N 1.688 • Taking pulley A as free-body, sum moments about pulley center to determine torque. MA 0: M A 200 mm 1600 N 2700 N 0 M A 220 N m 8 - 25 THANK YOU Copyright © 2010 Pearson Education South Asia Pte Ltd