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SPH3UW: Lecture 18, Pg 1 SPH3UW Today’s Agenda l Friction è What is it? è Systematic catagories of forces è How do we characterize it? è Model of friction.

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Presentation on theme: "SPH3UW: Lecture 18, Pg 1 SPH3UW Today’s Agenda l Friction è What is it? è Systematic catagories of forces è How do we characterize it? è Model of friction."— Presentation transcript:

1 SPH3UW: Lecture 18, Pg 1 SPH3UW Today’s Agenda l Friction è What is it? è Systematic catagories of forces è How do we characterize it? è Model of friction  Static & Kinetic friction (kinetic = dynamic in some languages) l Some problems involving friction

2 SPH3UW: Lecture 18, Pg 2 New Topic: Friction l What does it do? è It opposes relative motion of two objects that touch! l How do we characterize this in terms we have learned (forces)?  Friction results in a force in the direction opposite to the direction of relative motion (kinetic friction, static – impending mot) amaama F F APPLIED gmggmg N i j f f FRICTION some roughness here

3 SPH3UW: Lecture 18, Pg 3 Surface Friction... l Friction is caused by the “microscopic” interactions between the two surfaces:

4 SPH3UW: Lecture 18, Pg 4 Surface Friction... l Force of friction acts to oppose relative motion: è Parallel to surface. N è Perpendicular to Normal force. amaama F ffFffF gmggmg N i j

5 SPH3UW: Lecture 18, Pg 5 These relations are all useful APPROXIMATIONS to messy reality. Model for Sliding (kinetic) Friction N l The direction of the frictional force vector is perpendicular to the normal force vector N. f N l The magnitude of the frictional force vector |f F | is proportional to the magnitude of the normal force |N |. fN g  |f F | =  K | N | ( =  K  |  mg | in the previous example) è The “heavier” something is, the greater the friction will be...makes sense! The constant  K is called the “coefficient of kinetic friction.”

6 SPH3UW: Lecture 18, Pg 6 Model... l Dynamics: i :F  K N = ma j :N = mg soF  K mg = ma amaama F gmggmg N i j  K mg (this works as long as F is bigger than friction, i.e. the left hand side is positive)

7 SPH3UW: Lecture 18, Pg 7 Lecture 7, Act 1 Forces and Motion A box of mass m 1 = 1.5 kg is being pulled by a horizontal string having tension T = 90 N. It slides with friction (  k = 0.51) on top of a second box having mass m 2 = 3 kg, which in turn slides on a frictionless floor. (T is bigger than F friction, too.) è What is the acceleration of the second box ? (a) a = 0 m/s 2 (b) a = 2.5 m/s 2 (c) a = 3.0 m/s 2 m2m2m2m2 T m1m1m1m1  slides with friction (  k =0.51  ) slides without friction a = ? Hint: draw FBDs of both blocks – that’s 2 diagrams

8 SPH3UW: Lecture 18, Pg 8 Lecture 7, Act 1 Solution l First draw FBD of the top box: m1m1 N1N1 m1gm1g T f =  K N 1 =  K m 1 g

9 SPH3UW: Lecture 18, Pg 9 Lecture 7, Act 1 Solution l Newtons 3rd law says the force box 2 exerts on box 1 is equal and opposite to the force box 1 exerts on box 2. m1m1 f f 1,2 m2m2 f f 2,1 l As we just saw, this force is due to friction: =  K m 1 g

10 SPH3UW: Lecture 18, Pg 10 Lecture 7, Act 1 Solution l Now consider the FBD of box 2: m2m2 f f 2,1 =  k m 1 g m2gm2g N2N2 m1gm1g (gravity from…) (contact from…) (friction from…)

11 SPH3UW: Lecture 18, Pg 11 Lecture 7, Act 1 Solution l Finally, solve F = ma in the horizontal direction: m2m2 f f 2,1 =  K m 1 g  K m 1 g = m 2 a a = 2.5 m/s 2

12 SPH3UW: Lecture 18, Pg 12 Inclined Plane with Friction: l Draw free-body diagram:  i j mg N KNKN ma 

13 SPH3UW: Lecture 18, Pg 13 Inclined plane... ij Fa l Consider i and j components of F NET = ma :  i j mg N  KNKN ma i i mg sin  K N = ma mg sin  j j N = mg cos  mg cos  mg sin  K mg cos  = ma a / g = sin  K cos 

14 SPH3UW: Lecture 18, Pg 14 Static Friction... F gmggmg N i j fFfF l So far we have considered friction acting when the two surfaces move relative to each other- I.e. when they slide.. è We also know that it acts in when they move together: the ‘static” case. l In these cases, the force provided by friction will depend on the OTHER forces on the parts of the system.

15 SPH3UW: Lecture 18, Pg 15 Static Friction… (with one surface stationary) l Just like in the sliding case except a = 0. i :F  f F = 0 j :N = mg F gmggmg N i j fFfF While the block is static: f F  F

16 SPH3UW: Lecture 18, Pg 16 Static Friction… F gmggmg N i j fFfF The maximum possible force that the friction between two objects can provide is f MAX =  S N, where  s is the “coefficient of static friction.”  So f F   S N.  As one increases F, f F gets bigger until f F =  S N and the object starts to move. è If an object doesn’t move, it’s static friction è If an object does move, it’s dynamic friction

17 SPH3UW: Lecture 18, Pg 17 Static Friction... F  S is discovered by increasing F until the block starts to slide: i :F MAX  S N = 0 j :N = mg  S  F MAX / mg F F MAX gmggmg N i j  S mg

18 SPH3UW: Lecture 18, Pg 18 Lecture 7, Act 2 Forces and Motion A box of mass m =10.21 kg is at rest on a floor. The coefficient of static friction between the floor and the box is  s = 0.4. A rope is attached to the box and pulled at an angle of  = 30 o above horizontal with tension T = 40 N. è Does the box move? (a) yes (b) no (c) too close to call T m  static friction (  s  = 0.4  ) 

19 SPH3UW: Lecture 18, Pg 19 Lecture 7, Act 2 Solution l Pick axes & draw FBD of box: T m  N mg y x l Apply F NET = ma y: N + T sin  - mg = ma Y = 0 N = mg - T sin  = 80 N x: T cos  - f FR = ma X The box will move if T cos  - f FR > 0 f FR

20 SPH3UW: Lecture 18, Pg 20 Lecture 7, Act 2 Solution T m f MAX =  s N N mg y x x: T cos  - f FR = ma X y: N = 80 N The box will move if T cos  - f FR > 0 T cos  = 34.6 N f MAX =  s N = (.4)(80N) = 32 N So T cos  > f MAX and the box does move  Now use dynamic friction: ma x = Tcos  -  K N

21 SPH3UW: Lecture 18, Pg 21 Static Friction: We can also consider  S on an inclined plane. In this case, the force provided by friction will depend on the angle  of the plane. 

22 SPH3UW: Lecture 18, Pg 22 Static Friction...  mg N ma = 0 (block is not moving) The force provided by friction, f F, depends on .  fFfF mg sin  f f  (Newton’s 2nd Law along x-axis) ij

23 SPH3UW: Lecture 18, Pg 23 Static Friction... We can find  s by increasing the ramp angle until the block slides:  M mg N SNSN In this case, when it starts to slide:  mg sin  M  S mg cos  M   S  tan  M  ij mg sin  f f   f f  S N  S mg cos  M

24 SPH3UW: Lecture 18, Pg 24 Additional comments on Friction: Since f F =  N, kinetic friction “does not” depend on the area of the surfaces in contact. (This is a surprisingly good rule of thumb, but not an exact relation. Do you see why??) By definition, it must be true that  S   K for any system (think about it...).

25 SPH3UW: Lecture 18, Pg 25 Aside: l Graph of Frictional force vs Applied force: fFfF FAFA f F = F A f F =  K N f F =  S N

26 SPH3UW: Lecture 18, Pg 26 Problem: Box on Truck A box with mass m sits in the back of a truck. The coefficient of static friction between the box and the truck is  S. a è What is the maximum acceleration a that the truck can have without the box slipping? m SS a

27 SPH3UW: Lecture 18, Pg 27 Problem: Box on Truck l Draw Free Body Diagram for box: è Consider case where f F is max... (i.e. if the acceleration were any larger, the box would slip). N f F =  S N mg i j

28 SPH3UW: Lecture 18, Pg 28 Problem: Box on Truck ij l Use F NET = ma for both i and j components  i  i  S N = ma MAX è j è j N = mg a MAX =  S g N f F =  S N mg a MAX i j

29 SPH3UW: Lecture 18, Pg 29 Lecture 7, Act 3 Forces and Motion l An inclined plane is accelerating with constant acceleration a. A box resting on the plane is held in place by static friction. What is the direction of the static frictional force? (a) (b) (c) FfFf FfFf FfFf SS a

30 SPH3UW: Lecture 18, Pg 30 Lecture 7, Act 3 Solution l First consider the case where the inclined plane is not accelerating. mg FfFf N l All the forces add up to zero! mg N FfFf

31 SPH3UW: Lecture 18, Pg 31 mg N FfFf Lecture 7, Act 3 Solution l If the inclined plane is accelerating, the normal force decreases and the frictional force increases, but the frictional force still points along the plane: a l All the forces add up to ma! è F = ma è The answer is (a) mg FfFf N mama

32 SPH3UW: Lecture 18, Pg 32 Putting on the brakes Anti-lock brakes work by making sure the wheels roll without slipping. This maximizes the frictional force slowing the car since  S >  K.


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