Calculating friction
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The document discusses the calculations of frictional forces, including maximum static friction and kinetic friction, influenced by various factors such as material smoothness and the normal force. It provides examples of friction coefficients for different material combinations and detailed calculations for a 4 kg concrete slab and a wooden sled being pulled on different surfaces. Additionally, it describes the forces necessary to maintain constant velocity when pulling the sled on a slope.
Calculating friction
- 2. fs max & fk are affected by
- 3. fs max & fk are affected by Smoothness
- 4. fs max & fk are affected by Smoothness affected by materials
- 7. fs max & fk are affected by Smoothness Normal force
- 8. fs max & fk are affected by Smoothness Normal force
- 9. fs max & fk are affected by µs or µk Smoothness Normal force
- 10. Materials in contact µs µk Steel on Teflon 0,04 0,04 Wood on snow 0,08 0,06 Metal on metal (lubricated) 0,10 0,07 Wood on wood 0,7 0,4 Steel on steel 0,75 0,57 Glass on glass 0,9 0,4 Rubber on concrete 0,9 0,7
- 11. fs max & fk are affected by µs or µk Smoothness N Normal force
- 12. fs max & fk are affected by µs or µk Smoothness N Normal force
- 13. fs max = µs N
- 14. fk = µ k N
- 15. Maximum static friction Fstatic friction max = µs N normal force surface exerts on object; depends on object’s weight, downward forces on object and surface’s slope; if no slope and no extra forces, = weight coefficent of static friction; depends on the materials of the rubbing surfaces
- 16. Kinetic friction Fkinetic friction = µk N normal force surface exerts on object; depends on object’s weight, downward forces on object and surface’s slope; if no slope and no extra forces, = weight coefficent of kinetic friction; depends on the materials of the rubbing surfaces
- 17. Question 1 A 4kg concrete slab is pulled horizontally along a piece of rubber. Calculate the frictional force: a) just before the slab starts to move (maximum static friction) b) as the slab moves along (kinetic friction) Rubber-concrete: µs = 0,9 µd = 0,7
- 18. mass = 4 kg
- 19. mass = 4 kg weight = 40 N
- 21. Normal Weight
- 22. 40 N 40 N
- 23. a) fs max = µs N fs max = 0,9 • 40 N fs max = 36 N
- 24. b) fk = µ k N fk = 0,7 • 40 N fk = 28 N
- 25. Question 2 A 4 kg wooden sled is pulled up a snow-covered slope which is at 20° to the horizontal.
- 26. Question 2 A 4 kg wooden sled is pulled up a snow-covered slope which is at 20° to the horizontal. Calculate the frictional force: a) just before the sled starts to move (maximum static friction) b) as the sled moves along (kinetic friction) Wood-snow: µs = 0,08 µd = 0,06
- 27. mass = 4 kg 20°
- 28. mass = 4 kg weight = 40 N 20°
- 29. Weight
- 30. Wy W
- 31. N Wy W
- 32. N Wy 20° W 20°
- 33. N Wy Wy = W • cos 20° 20° W 20°
- 34. N Wy Wy = W • cos 20° Wy = 40 N • cos 20° 20° W 20°
- 35. N Wy Wy = W • cos 20° Wy = 40 N • cos 20° Wy = 37,59 N 20° W 20°
- 37. Materials in contact µs µk Steel on Teflon 0,04 0,04 Wood on snow 0,08 0,06 Metal on metal (lubricated) 0,10 0,07 Wood on wood 0,7 0,4 Steel on steel 0,75 0,57 Glass on glass 0,9 0,4 Rubber on concrete 0,9 0,7
- 38. a) fs max = µs N fs max = 0,08 • 37,59 N fs max = 3 N
- 39. b) fk = µ k N fk = 0,06 • 37,59 N fk = 2,26 N
- 40. fs max & fk are affected by µs or µk Smoothness N Normal force
- 41. Question 2c A 4 kg wooden sled is pulled up a snow-covered slope which is at 20° to the horizontal. c) How hard must a person pull parallel to the slope upward for the sled to move at a constant velocity?
- 42. Question 2c A 4 kg wooden sled is pulled up a snow-covered slope which is at 20° to the horizontal. c) How hard must a person pull parallel to the slope upward for the sled to move at a constant velocity? (to balance the forces of: - kinetic friction and - the component of the sled’s weight down the slope)
- 43. For CONSTANT VELOCITY: Fdown slope Fup slope
- 44. For CONSTANT VELOCITY: Fdown slope Fup slope
- 45. N Wy W
- 46. W Component of W down slope
- 47. W
- 48. Wx W
- 49. Wx W Wx
- 50. Wy Wx Wx W
- 51. Wx 20° Wx = W • sin 20° W Wx
- 52. Wx 20° Wx = W • sin 20° Wx = 40 • sin 20° 40 N Wx
- 53. Wx 20° 40 N Wx Wx = W • sin 20° Wx = 40 • sin 20° Wx = 13,68 N down the slope
- 54. 13,68 N 20° 40 N
- 57. 13,68 N f
- 58. b) fk = µ k N fk = 0,06 • 37,59 N fk = 2,26 N
- 59. 13,68 N fk = 2,26 N
- 60. Fdown slope = 15,94 N 13,68 N fk = 2,26 N
- 61. For CONSTANT Fdown slope = 15,94 N VELOCITY: 13,68 N Fup slope = 15,94 N fk = 2,26 N