Lecture 8 Static friction and Kinetic or Sliding friction.

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Lecture 8 Static friction and Kinetic or Sliding friction

Friction l Friction is caused by the “microscopic” interactions between the two surfaces

Friction... l Force of friction acts to oppose motion:  Parallel to a surface N  Perpendicular to a Normal force. mama F ffFffF gmggmg N i j

Static and Kinetic Friction l Friction exists between objects and its behavior has been modeled. At Static Equilibrium: A block, mass m, with a horizontal force F applied, N Direction: A force vector  to the normal force vector N and the vector is opposite to the direction of acceleration if  were 0. Magnitude: f is proportional to the applied forces such that f s ≤  s N   s called the “coefficient of static friction”

Friction: Static friction Static equilibrium: A block with a horizontal force F applied, As F increases so does f s F m1m1 FBD fsfs N mg  F x = 0 = -F + f s  f s = F  F y = 0 = - N + mg  N = mg

Static friction, at maximum (just before slipping) Equilibrium: A block, mass m, with a horizontal force F applied, N Direction: A force vector  to the normal force vector N and the vector is opposite to the direction of acceleration if  were 0. Magnitude: f S is proportional to the magnitude of N f s =  s N F m fsfs N mg

Kinetic or Sliding friction (f k < f s ) Dynamic equilibrium, moving but acceleration is still zero As F increases f k remains nearly constant (but now there acceleration is acceleration) F m1m1 FBD fkfk N mg  F x = 0 = -F + f k  f k = F  F y = 0 = - N + mg  N = mg v f k =  k N

Sliding Friction: Quantitatively N l Direction: A force vector  to the normal force vector N and the vector is opposite to the velocity. fN l Magnitude: f k is proportional to the magnitude of N fN g  f k =  k N ( =  K  mg in the previous example) The constant  k is called the “coefficient of kinetic friction” Logic dictates that  S >  K for any system

Static Friction with a bicycle wheel l You are pedaling hard and the bicycle is speeding up. What is the direction of the frictional force? l You are breaking and the bicycle is slowing down What is the direction of the frictional force?

Coefficients of Friction Material on Material  s = static friction  k = kinetic friction steel / steel add grease to steel metal / ice brake lining / iron tire / dry pavement tire / wet pavement0.80.7

An experiment Two blocks are connected on the table as shown. The table has unknown static and kinetic friction coefficients. Design an experiment to find  S Static equilibrium: Set m 2 and add mass to m 1 to reach the breaking point. Requires two FBDs : m1m1 m2m2 m2gm2g N m1gm1g T T Mass 2  F x = 0 = -T + f s = -T +  S N  F y = 0 = N – m 2 g fSfS Mass 1  F y = 0 = T – m 1 g T = m 1 g =  S m 2 g   S = m 1 /m 2

A 2 nd experiment Two blocks are connected on the table as shown. The table has unknown static and kinetic friction coefficients. Design an experiment to find  K. Dynamic equilibrium: Set m 2 and adjust m 1 to find place when a = 0 and v ≠ 0 Requires two FBDs m1m1 m2m2 m2gm2g N m1gm1g T T Mass 2  F x = 0 = -T + f f = -T +  k N  F y = 0 = N – m 2 g fkfk Mass 1  F y = 0 = T – m 1 g T = m 1 g =  k m 2 g   k = m 1 /m 2

An experiment (with a ≠ 0) Two blocks are connected on the table as shown. The table has unknown static and kinetic friction coefficients. Design an experiment to find  K. Non-equilibrium: Set m 2 and adjust m 1 to find regime where a ≠ 0 Requires two FBDs T Mass 2  F x = m 2 a = -T + f k = -T +  k N  F y = 0 = N – m 2 g m1m1 m2m2 m2gm2g N m1gm1g T fkfk Mass 1  F y = m 1 a = T – m 1 g T = m 1 g + m 1 a =  k m 2 g – m 2 a   k = (m 1 (g + a)+m 2 a)/m 2 g

Inclined plane with “Normal” and Frictional Forces Weight of block is mg Normal Force Friction Force Sliding Down “Normal” means perpendicular Note: If frictional Force = Normal Force  (coefficient of friction) F friction =  F normal =  mg sin  then zero acceleration 1.At first the velocity is v up along the slide 2.Can we draw a velocity time plot? 3.What the acceleration versus time?acceleration   v mg sin  f k Sliding Up