1. Friction Friction Problem Situations

2. Friction Friction F f is a force that resists motion Friction involves objects in contact with each other. Friction must be overcome before motion occurs. Friction is caused by the uneven surfaces of the touching objects. As surfaces are pressed together, they tend to interlock and offer resistance to being moved over each other.

3. Friction Frictional forces are always in the direction that is opposite to the direction of motion or to the net force that produces the motion. Friction acts parallel to the surfaces in contact.

4. Types of Friction Static friction: maximum frictional force between stationary objects. Until some maximum value is reached and motion occurs, the frictional force is whatever force is necessary to prevent motion. Static friction will oppose a force until such time as the object “breaks away” from the surface with which it is in contact. The force that is opposed is that component of an applied force that is parallel to the surface of contact.

5. Types of Friction The magnitude of the static friction force F fs has a maximum value which is given by: where μ s is the coefficient of static friction and F N is the magnitude of the normal force on the body from the surface.

6. Types of Friction Sliding or kinetic friction: frictional force between objects that are sliding with respect to one another. Once enough force has been applied to the object to overcome static friction and get the object to move, the friction changes to sliding (or kinetic) friction. Sliding (kinetic) friction is less than static friction. If the component of the applied force on the object (parallel to the surface) exceeds F fs then the magnitude of the opposing force decreases rapidly to a value F k given by: where μ k is the coefficient of kinetic friction.

7. Types of Friction Here you can see that the applied force is resisted by the static frictional force F fs ( f s in the figure) until “breakaway”. Then the sliding (kinetic) frictional force F k is approximately constant.

8. Types of Friction Static and sliding friction are dependent on: The nature of the surfaces in contact. Rough surfaces tend to produce more friction. The normal force (F n ) pressing the surfaces together; the greater F n is, the more friction there is.

9. Types of Friction

10. Rolling friction: involves one object rolling over a surface or another object. Fluid friction: involves the movement of a fluid over an object (air resistance or drag in water) or the addition of a lubricant (oil, grease, etc.) to change sliding or rolling friction to fluid friction.

11. Coefficient of Friction Coefficient of friction (  ): ratio of the frictional force to the normal force pressing the surfaces together.  has no units. Static: Sliding (kinetic):

12. The maximum frictional force is 50 N. As the applied force increases from 0 N to 50 N, the frictional force also increases from 0 N to 50 N and will be equal to the applied force as it increases.

13. As the applied force increases beyond 50 N, the frictional force remains at 50 N and the 100 N block will accelerate.

14. Once the static frictional force of 50 N has been overcome, only a 40 N force is needed to overcome the 40 N kinetic frictional force and produce constant velocity (a = 0 m/s 2 ).

15. Horizontal Surface – Constant Speed Constant speed: a = O m/s 2. The normal force pressing the surfaces together is the weight; F n = F w

16. Horizontal Surface: a > O m/s 2

17. If solving for: F x : F f : a:

18. Horizontal Surface: Skidding to a Stop or Slowing Down (a < O m/s 2 ) The frictional force is responsible for the negative acceleration. Generally, there is no F x.

19. Horizontal Surface: Skidding to a Stop or Slowing Down (a < O m/s 2 ) Most common use involves finding acceleration with a velocity equation and finding  k : Acceleration will be negative because the speed is decreasing.

20. Horizontal Surface: Skidding to a Stop or Slowing Down (a < O m/s 2 ) The negative sign for acceleration a is dropped because  k is a ratio of forces that does not depend on direction. Maximum stopping distance occurs when the tire is rotating. When this happens, a = -  s ·g. Otherwise, use a = -  k ·g to find the acceleration, then use a velocity equation to find distance, time, or speed.

21. Down an Inclined Plane

22. Resolve F w into F x and F y. The angle of the incline is always equal to the angle between Fw and F y. F w is always the hypotenuse of the right triangle formed by F w, F x, and F y.

23. Down an Inclined Plane The force pressing the surfaces together is NOT F w, but F y ; F n = F y. or

24. Down an Inclined Plane For constant speed (a = 0 m/s 2 ):

25. Down an Inclined Plane To determine the angle of the incline: If moving: If at rest:

26. Non-parallel Applied Force on Ramp mg mg cos  mg sin  fkfk N  Suppose the applied force acts on the box, at an angle  above the horizontal, rather than parallel to the ramp. We must resolve F A into parallel and perpendicular components using the angle  +  (F A cos (  + θ) and F A sin (  + θ)). F A serves to increase acceleration directly and indirectly: directly by F A cos (  + θ) pulling the box down the ramp, and indirectly by F A sin (  + θ) lightening the normal support force with the ramp (thereby reducing friction). F A  F A cos(  +  ) F A sin(  +  )  continued on next slide

27. Non-parallel Applied Force on Ramp mg mg cos  mg sin  fkfk N  F A  F A cos(  +  ) F A sin (  +  )  Because of the perp. comp. of F A, N < mg cos . Assuming F A sin(  +  ) is not big enough to lift the box off the ramp, there is no acceleration in the perpendicular direction. So, F A sin(  +  ) + N = mg cos . Remember, N is what a scale would read if placed under the box, and a scale reads less if a force lifts up on the box. So, N = mg cos  - F A sin(  +  ), which means f k =  k N =  k [mg cos  - F A sin(  +  )]. continued on next slide

28. Non-parallel Applied Force on Ramp mg mg cos  mg sin  fkfk N  F A  F A cos(  +  ) F A sin(  +  )  If the combined force of F A cos(  +  ) + mg sin  is is enough to move the box: F A cos(  +  ) + mg sin  -  k [mg cos  - F A sin(  +  )] = m a

29. Up an Inclined Plane

30. Resolve F w into F x and F y. The angle of the incline is always equal to the angle between F w and F y. F w is always the hypotenuse of the right triangle formed by F w, F x, and F y.

31. Up an Inclined Plane F a is the force that must be applied in the direction of motion. F a must overcome both friction and the x-component of the weight. The force pressing the surfaces together is F y.

32. Up an Inclined Plane For constant speed, a = 0 m/s 2. F a = F x + F f For a > 0 m/s 2. F a = F x + F f + (m·a)

33. Pulling an Object on a Flat Surface

34. The pulling force F is resolved into F x and F y.

35. Pulling an Object on a Flat Surface F n is the force that the ground exerts upward on the mass. F n equals the downward weight F w minus the upward force F y from the pulling force. For constant speed, a = 0 m/s 2.

36. Simultaneous Pulling and Pushing an Object on a Flat Surface

38. Pushing an Object on a Flat Surface

39. The pushing force F is resolved into F x and F y.

40. Pushing an Object on a Flat Surface F n is the force that the ground exerts upward on the mass. F n equals the downward weight F w plus the upward force F y from the pushing force. For constant speed, a = 0 m/s 2.

41. Pulling and Tension The acceleration a of both masses is the same.

42. Pulling and Tension For each mass: Isolate each mass and examine the forces acting on that mass.

43. Pulling and Tension m 1 = mass T 1 may not be a tension, but could be an applied force (F a ) that causes motion.

44. Pulling and Tension m 2 = mass

45. Pulling and Tension This problem can often be solved as a system of equations: See the Solving Simultaneous Equations notes for instructions on how to solve this problem using a TI or Casio calculator.

46. Revisiting Tension and Friction

47. For the hanging mass, m 2 : The acceleration a of both masses is the same. For the mass on the table, m 1 :

48. Revisiting Tension and Friction

49. Normal Force Not Associated with Weight. A normal force can exist that is totally unrelated to the weight of an object. applied force friction weight normal F N = applied force

50. Friction is always parallel to surfaces…. In this case, for the block to remain in position against the wall without moving: the upward frictional force F f has to be equal and opposite to the downward weight F w. The rightward applied force F has to be equal ad opposite to the leftward normal force F N. FFWFW FfFf FNFN (0.20)