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Friction

Friction. Friction caused by irregularities in surfaces. Always acts against the direction of motion. Static Friction – no motion. Force Normal. F applied. Force friction. Wgt. If static equilibrium,. F app = F fr. Static Friction – no motion. Force Normal. F applied. Force

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Friction

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  1. Friction

  2. Friction caused by irregularities in surfaces. Always acts against the direction of motion.

  3. Static Friction – no motion Force Normal Fapplied Force friction Wgt If static equilibrium, Fapp = Ffr

  4. Static Friction – no motion Force Normal Fapplied Force friction Wgt If increase Fapp, Ffr also increases, up to Ffr, max

  5. If increase m, Ffr, max also increases Ffr, max α m On horizontal surface, FN = mg Ffr, maxα FN Force Normal Fapplied Force friction Wgt

  6. Ffr, max is proportional to FN Ffr, max -------- = constant FN = μs μs = coefficient of static friction Ffr≤ μs FN

  7. wood on wood: μs = 0.40 100 N Ffr, max = μs FN = 40 N

  8. Kinetic (moving) Friction Force Normal Forceapplied Force friction weight kinetic friction < static friction If Fapp = Ffr , then speed is constant

  9. Coefficient of kinetic friction Ffr μk = ------ FN

  10. Forceapp 100. kg μk = 0.10 980 N Weight: Force Normal: Force Friction: 980 N 98 N Force applied to keep a constant velocity: 98 N Net force: 0 N

  11. 100. kg 200. N μk = 0.10 102 N Net force: acceleration: 1.02 m/s2

  12. constant speed of 3.0 m/s 50.0 N 100. kg acceleration: net force: friction force: 0 0 50 N

  13. Free Fall and Air Resistance wgt air resistance Fnet = wgt – air resistance

  14. wgt Air resistance increases as speed increases air resistance When air resistance = wgt: Fnet = 0 a = 0 velocity is constant; terminal velocity

  15. Skydiver (total mass M) in free fall, falls for t0 seconds before opening chute. Air resistance = kv Acceleration when chute is opened? Terminal speed?

  16. m1 m2 F μk acceleration? force of tension?

  17. m1 μk m2 acceleration? tension in rope?

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