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Lecture 8 Static friction and Kinetic or Sliding friction
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Friction l Friction is caused by the “microscopic” interactions between the two surfaces
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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
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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”
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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
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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
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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
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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
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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?
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Coefficients of Friction Material on Material s = static friction k = kinetic friction steel / steel0.60.4 add grease to steel0.10.05 metal / ice0.0220.02 brake lining / iron0.40.3 tire / dry pavement0.90.8 tire / wet pavement0.80.7
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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
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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
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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
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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
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