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Friction Friction Problem Situations Physics Montwood High School R. Casao.

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Presentation on theme: "Friction Friction Problem Situations Physics Montwood High School R. Casao."— Presentation transcript:

1 Friction Friction Problem Situations Physics Montwood High School R. Casao

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.

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4 Microscopic Friction Magnified section of a polished steel surface showing surface bumps about 5 x m (500 nm) high, which corresponds to several thousand atomic diameters. Computer graphic from a simulation showing gold atoms (below) adhering to the point of a sharp nickel probe (above) that has been in contact with the gold surface. Surface Roughness Adhesion

5 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.

6 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.

7 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.

8 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.

9 The static frictional force keeps an object from starting to move when a force is applied. The static frictional force has a maximum value, but may take on any value from zero to the maximum, depending on Static Friction what is needed to keep the sum of forces zero.

10 Types of Friction From 0 to the maximum value of the static frictional force F s in the figure, the applied force is resisted by the static frictional force until “breakaway”. Then the sliding (kinetic) frictional force F k is approximately constant.

11 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.

12 Friction vs. Area Question: Why doesn’t friction depend on contact area? The microscopic area of contact between a box and the floor is only a small fraction of the macroscopic area of the box’s bottom surface. If the box is turned on its side, the macroscopic area is increased, but the microscopic area of contact remains the same (because the contact is more distributed). Therefore the frictional force f is independent of contact area.

13 Types of Friction 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.

14 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):

15 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.

16 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 ).

17 As the applied force increases beyond 40 N, the kinetic frictional force remains at 40 N and the 100 N block will accelerate.

18 A Model of Friction Friction

19 Static Friction

20 Kinetic Friction

21 The kinetic frictional force is also independent of the relative speed of the surfaces, and of their area of contact. Kinetic Friction and Speed

22 Rolling Friction

23 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

24 Horizontal Surface: a > O m/s 2

25 If solving for: F x : F f : a:

26 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.

27 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.

28 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.

29 Friction, Cars, & Antilock Brakes The diagram shows forces acting on a car with front-wheel drive. Typically, F n > F n ’ because the engine is over the front wheels. The largest frictional force f s the tire can exert on the road is µ s · Fn. Attempts to make the tire exert a force larger than this causes the tire to “burn rubber” and actually reduces the force, since µ k < µ s. Note that while all points on the rolling tire have the same speed v in the reference frame of the car, in the reference frame of the road the bottom of the tire is at rest, while top is moving forward with a speed of 2 · v. Antilock brakes sense the wheel rotation and “ease off” if it close to stopping, maintaining static friction with the road and allowing better control of steering than if the wheels were locked.

30 Antilock Brakes

31 Example: The Effect of Antilock Brakes A car is traveling at 30 m/s along a horizontal road. The coefficients of friction are m s =0.50 and m k =0.40. (a) What is the braking distance  x a with antilock brakes? (b) What is the braking distance  x b if the brakes lock?

32 Example: A Game of Shuffleboard A cruise-ship passenger uses a shuffleboard cue to push a shuffleboard disk of mass 0.40 kg horizontally along the deck, so that the disk leaves the cue at a speed of 8.5 m/s. The disk then slides a distance of 8.0 m. What is the coefficient of kinetic friction between the disk and deck?

33 Down an Inclined Plane

34 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.

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

36 Down an Inclined Plane If we place an object on an inclined plane and increase the tilt angle  to the point at which the object just begins to slide. What is the relation between  and the static coefficient of friction µ s ?

37 Down an Inclined Plane If the object slides down the incline at constant speed (a = 0 m/s 2 ), the relation between  and the kinetic coefficient of friction µ k :

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

39 Example: A Sliding Coin A hardcover book is resting on a tabletop with its front cover facing upward. You place a coin on the cover and very slowly open the book until the coin starts to slide. The angle  is the angle of the cover just before the coin begins to slide. Find the coefficient of static friction µ s between the coin and book.

40 Example: Dumping a file cabinet A 50.0 kg steel file cabinet is in the back of a dump truck. The truck’s bed, also made of steel, is slowly tilted. What is the size of the static friction force when the truck’s bed is tilted by 20°? At what angle will the file cabinet begin to slide? Steel on dry steel  Free-body diagram

41 Example: Dumping a file cabinet File cabinet will begin to slide when:

42 Non-Parallel Applied Force on Ramp m·gm·g m · g · cos  m · g · sin  fkfk N  If an applied force acts on the box at an angle  above the horizontal, 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(  +  ) 

43 Non-Parallel Applied Force on Ramp m·gm·g m · g · cos  m · g · sin  fkfk N  F A  F A · cos(  +  ) F A · sin (  +  )  If 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(  +  ) + F N = m · g · cos . Remember, F 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, F N = m · g · cos  - F A · sin(  +  ), which means f k =  k ·F N =  k · [ m · g · cos  - F A · sin(  +  )].

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

45 Up an Inclined Plane

46 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.

47 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.

48 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)

49 Pulling an Object on a Flat Surface

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

51 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.

52 Example: Pulling A Sled Two children sitting on a sled at rest in the snow ask you to pull them. You pull on the sled’s rope, which makes an angle of 40° with the horizontal. The children have a combined mass of 45 kg, and the sled has a mass of 5.0 kg. The coefficients of static and kinetic friction are µ s =0.20 and µ k =0.15, and the sled is initially at rest. Find the acceleration of the sled and children if the rope tension is 100 N.

53 Simultaneous Pulling and Pushing an Object on a Flat Surface

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55 Pushing an Object on a Flat Surface

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

57 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.

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

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

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

61 Pulling and Tension m 2 = mass

62 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.

63 Revisiting Tension and Friction

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

65 Revisiting Tension and Friction

66 A block of mass m 2 = 5.0 kg has been adjusted so that the block m 1 = 7.0 kg is just on the verge of sliding. (a)What is the coefficient of static friction m s between the table and the block? Example: A Sliding Block

67 (b) With a slight push, the blocks move with acceleration a. Find a if µ k = 0.54.

68 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

69 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)


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