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Friction

Friction. Friction is a force that always opposes motion. It’s direction is always parallel to the surface. 2 Types of Friction. Static Friction also called “start-up” friction only applies to objects at rest always more than moving friction. 2 Types of Friction. Kinetic Friction

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Friction

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  1. Friction

  2. Friction is a force that always opposes motion. It’s direction is always parallel to the surface

  3. 2 Types of Friction • Static Friction • also called “start-up” friction • only applies to objects at rest • always more than moving friction

  4. 2 Types of Friction • Kinetic Friction • Also called “moving” friction • there are several types • sliding friction • rolling friction

  5. Forces Acting on an Object FN FA Ff FW

  6. Forces Acting on an Object FN FA Ff FW If FA = Ff there will be no motion or it will have constant motion

  7. Forces Acting on an Object FN FA Ff FW If FA > Ff The object will accelerate in the direction of FA

  8. Friction • In order to calculate the force of friction, the weight of the object and the surface has to be taken into account. • This is done using the coefficient of Friction

  9. Coefficient of Friction • The coefficient of friction is a ratio of the force of friction to the normal force • It is a measure of how much friction a surface has • The symbol for coefficient of Friction is µ • It has no units

  10. Coefficient of Friction • The formula is: • µ = Ff • FN • or Ff = µFN • FN = FW = (m)(g) • Ff = µ(m)(g)

  11. Friction Problem FN FA Ff FW A box is stationary and has a mass of 10 kg. A force of 60 N is applied to it. Calculate it’s acceleration. Static µ = 0.40 Kinetic µ = 0.30

  12. Friction Problem FN FA Ff FW This is a Fnet problem. The formula is: Fnet = FA - Ff or (m)(a) = FA - Ff

  13. Friction Problem FN FA Ff FW You first need to calculate the Force of Friction. Ff = µ FN ( remember FN = FW = (m)(g))

  14. Friction Problem FN FA Ff FW You first need to calculate the Force of Friction. Ff = µ FN ( remember FN = FW = (m)(g) Next you need to use the correct µ. Static µ = 0.40 Kinetic µ = 0.30

  15. Friction Problem FN FA Ff FW (use static µ) Ff = µ FN Ff = (0.4)(10 kg)(9.8 m/s2) Ff = 39.2 N

  16. Friction Problem FN FA Ff FW Fnet = FA - Ff (10 kg)(a) = 60 N - 39.2 N a = 20 .8 N 10 kg a = 2.08 m/s2

  17. Friction Problem #2 FN FA Ff FW Assume the same box is now moving at a constant speed. What is the applied force ? (This is not an Fnet problem since there is no acceleration) Static µ = 0.40 Kinetic µ = 0.30

  18. Friction Problem #2 FN FA Ff FW You first need to calculate the Force of Friction. Ff = µ FN ( remember FN = FW = (m)(g))

  19. Friction Problem #2 FN FA Ff FW (Use kinetic µ) Ff = µ FN Ff = (0.3)(10 kg)(9.8 m/s2) Ff = 29.4 N

  20. Friction Problem #2 FN FA Ff FW (Use kinetic µ) Ff = µ FN Ff = (0.3)(10 kg)(9.8 m/s2) Ff = 29.4 N Since box is moving at a constant speed, FA=Ff FA = 29.4 N

  21. Friction Problem #3 FN FA Ff FW Now that the box is moving, calculate the acceleration if 60N of force is applied. (This is an Fnet problem) Static µ = 0.40 Kinetic µ = 0.30

  22. Friction Problem #3 FN FA Ff FW You first need to calculate the Force of Friction. Ff = µ FN ( remember FN = FW = (m)(g)

  23. Friction Problem #3 FN FA Ff FW (Use kinetic µ) Ff = µ FN Ff = (0.3)(10 kg)(9.8 m/s2) Ff = 29.4 N

  24. Friction Problem #3 FN FA Ff FW Fnet = FA - Ff (10 kg)(a) = 60 N - 29.4 N a = 30.6 N 10 kg a = 3.06 m/s2

  25. Summary • if FA = Ff box doesn’t move • if FA > Ff box overcomes static friciton and accelerates. • if FA = Ff box moves at constant speed • if FA > Ff box will accelerate (+) • if FA < Ff box will accelerate (-) • only an Fnet can cause acceleation

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