UNIT - V FRICTION

Friction Dry friction (Coulomb friction) Friction forces oppose the tendency of contacting surfaces to slip one relative to the other. Dry friction- the tangential component of the contact force.

Friction Forces have to be concurrent in order to be in equilibrium.

Friction forces as a function of P Maximum value of friction forces is called the limiting value of static friction. This condition is also called the impending motion.

Tip vs. slip P2>P1 F1 F2 Body Tips

The magnitude of the friction forces Fmax=msN Where ms is the coefficient of static friction. It is independent of normal forces and area of contact. F<msN The general case for equilibrium condition Once the body starts to slip F=mkN mk is the coefficient of dynamic friction mk < ms

The Resultant of the friction and normal forces R= N2+F2 tan q= F/N At the point of impending motion: R= N2+F2max = N2+(msN)2 = N 1+ ms2 tan fs= F/N= msN/N= ms fs is the angle of static friction q < fs

Gravity forces in inclined surfaces For equilibrium: q < fs

Example Two blocks with masses mA = 20 kg and mB = 80 kg are connected with a flexible cable that passes over a frictionless pulley as shown in Fig. P9-4. The coefficient of friction between the blocks is 0.25. If motion of the blocks is impending, determine the coefficient of friction between block B and the inclined surface and the tension in the cable between the two blocks.

Solution Free body diagram for block A Fy = An - 196.14 cos(35) = 0 An = 160.67 N For impending motion of the block A: (Af = ms.An) +  Fx = -T + 0.25(160.67) + 196.14 sin 35 = 0 T = 152.67 N

Block B weight: WB = mB.g = 80 (9.807) = 784.56 N Free body diagram for block B: Fy = Bn - 160.67 - 784.56 cos 35 = 0 Bn = 803.34 N Fx = - Bf - 40.17 - 152.67 + 784.56 sin 35 = 0 Bf = 257.17 N = Bf/Bn= 257.17/803.34 = 0.320

Determine the horizontal force P required to start moving the 250-lb block shown in Fig. P9-1 up the inclined surface. The coefficient of friction between the inclined surface and the block is s = 0.30. P = 265.29 lb  265.3b

Solution For impending motion: Ff = Fn = 0.30Fn   Ff = Fn = 0.30Fn Free body diagram for the block: Fy = Fn cos 30 - Ff sin 30 -W = Fn cos 30 - 0.30Fn sin 30 - 250 = 0 Fn = 349.15 lb Fx = P - Fn sin 30 - Ff cos 30 = P - 349.15 sin 30 - 0.30 (349.15) cos 30 = 0 P = 265.29 lb  265.3b

The masses of blocks A and B of Fig The masses of blocks A and B of Fig. P9-36 are mA = 40 kg and mB = 85 kg. If the coefficient of friction is 0.25 for both surfaces, determine the force P required to cause impending motion of block B. P  935 N

Solution WA = mAg = 40 (9.807) = 392.28 N From a free-body diagram for the block A when motion is impending: Af = Af (max) = A An = 0.25 An Fy = An cos 45 + Af sin 45 - WA = An cos 45 + 0.25 An sin 45 - 392.28 = 0 An= 443.81 N Af= 0.25 An = 110.95 N

WB = mBg = 85 (9.807) = 833.60 N From a free-body diagram for the block B when motion is impending: Bf = Bf (max) = Bn = 0.25Bn Fx = P cos 20-WB sin 45-Af -Bf = P cos 20-833.60 sin 45-110.95-0.25Bn = 0 Fy = P sin 20 - WB cos 45 - An + Bn = P sin 20 - 833.60 cos 45 - 443.81 + Bn = 0 P  935 N

Workers are pulling a 400 lb crate up an incline as shown in Fig. P9-3 Workers are pulling a 400 lb crate up an incline as shown in Fig. P9-3. The coefficient of friction between the crate and the surface is 0.20, and the rope on which the workers are pulling is horizontal. (a) Determine the force P that the workers must exert to start sliding the crate up the incline. (b) If one of the workers lets go of the rope for a moment, determine the minimum force the other workers must exert to keep the crate from sliding back down the incline. P=197.8 lb P=25.8lb

A 120 lb girl is walking up a 48-lb uniform beam as shown in Fig. P9-21. Determine how far up the beam the girl can walk before the beam starts to slip if (a)The coefficient of friction is 0.20 at all surfaces (b)The coefficient of friction at the bottom end of the beam is increased to 0.40 by placing a piece of rubber between the beam and the floor. x = 3.21 ft x = 6.26 ft

Solution (a) For a free-body diagram for the beam when motion is impending: Ar = An = 0.2An Bf= Bn = 0.2Bn f = tan-16/8 = 36.87 Fh = Bn sin 36.87 - 0.2Bn cos 36.87 - 0.2An = 0 +  Fy = Bn cos 36.87 + 0.2Bn sin 36.87 + An - 120 - 48 = 0 Solving yields: An = 118.46 lb; Bn = 53.85 lb + MA = 48(6) cos 36.87 + 120(x) cos 36.87 - 53.85(10) = 0 x = 3.209 ft  3.21 ft

For a free-body diagram for the beam when motion is impending: Ar = An = 0.4An Bf= Bn = 0.2Bn +  Fh = Bn sin 36.87 - 0.2Bn cos 36.87 - 0.4An = 0 +  Fy = Bn cos 36.87 + 0.2Bn sin 36.87 + An - 120 - 48 = 0 Solving yields:An = 91.49 lb; Bn = 83.17 lb + MA = 48(6) cos 36.87 + 120(x) cos 36.87 - 83.17(10) = 0 x = 6.264 ft  6.26 ft