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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 Static & Kinetic friction (kinetic = dynamic in some languages) l Some problems involving friction

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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

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

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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

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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.”

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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)

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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

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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

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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

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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…)

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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

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

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SPH3UW: Lecture 18, Pg 13 Inclined plane... ij Fa l Consider i and j components of F NET = ma : i j mg N KNKN 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

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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.

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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

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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

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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

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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 )

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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

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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

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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.

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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

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SPH3UW: Lecture 18, Pg 23 Static Friction... We can find s by increasing the ramp angle until the block slides: M mg N SNSN 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

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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...).

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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

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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 SS a

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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

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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

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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 SS a

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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

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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

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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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