Presentation on theme: "SPH3UW Today’s Agenda Friction What is it?"— Presentation transcript:
1 SPH3UW Today’s Agenda Friction What is it? Systematic catagories of forcesHow do we characterize it?Model of frictionStatic & Kinetic friction (kinetic = dynamic in some languages)Some problems involving friction
2 New Topic: Friction What does it do? It opposes relative motion of two objects that touch!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)jNFAPPLIEDimafFRICTIONsome roughness heremg
3 Surface Friction...Friction is caused by the “microscopic” interactions between the two surfaces:
4 Surface Friction... Force of friction acts to oppose relative motion: Parallel to surface.Perpendicular to Normal force.jNFimafFmg
5 Model for Sliding (kinetic) Friction The direction of the frictional force vector is perpendicular to the normal force vector N.The magnitude of the frictional force vector |fF| is proportional to the magnitude of the normal force |N |.|fF| = K | N | ( = K|mg | in the previous example)The “heavier” something is, the greater the friction will be...makes sense!The constant K is called the “coefficient of kinetic friction.”These relations are all useful APPROXIMATIONS to messy reality.
6 Model... Dynamics: i : F KN = ma j : N = mg so F Kmg = ma (this works as long as F is bigger than friction, i.e. the left hand side is positive)jNFimaK mgmg
7 Lecture 7, Act 1 Forces and Motion A box of mass m1 = 1.5 kg is being pulled by a horizontal string having tension T = 90 N. It slides with friction (mk = 0.51) on top of a second box having mass m2 = 3 kg, which in turn slides on a frictionless floor. (T is bigger than Ffriction, too.)What is the acceleration of the second box ?(a) a = 0 m/s2 (b) a = 2.5 m/s2 (c) a = 3.0 m/s2Hint: draw FBDs of both blocks – that’s 2 diagramsslides with friction (mk=0.51 )Tm1a = ?m2slides without friction
8 Lecture 7, Act 1 Solution First draw FBD of the top box: N1 m1 f = mKN1 = mKm1gTm1g
9 Lecture 7, Act 1 SolutionNewtons 3rd law says the force box 2 exerts on box 1 is equal and opposite to the force box 1 exerts on box 2.As we just saw, this force is due to friction:= mKm1gm1f1,2f2,1m2
10 Lecture 7, Act 1 Solution Now consider the FBD of box 2: N2 (contact from…)(friction from…)f2,1 = mkm1gm2(contact from…)(gravity from…)m1gm2g
11 Lecture 7, Act 1 SolutionFinally, solve F = ma in the horizontal direction:mKm1g = m2aa = 2.5 m/s2f2,1 = mKm1gm2
12 Inclined Plane with Friction: Draw free-body diagram:maKNjNmgi
13 Inclined plane... Consider i and j components of FNET = ma : i mg sin KN = maj N = mg cos KNmg sin Kmg cos = mamajNa / g = sin Kcos mgmg cos img sin
14 Static Friction... j N F i fF mg So far we have considered friction acting when the two surfaces move relative to each other- I.e. when they slide..We also know that it acts in when they move together: the ‘static” case.In these cases, the force provided by friction will depend on the OTHER forces on the parts of the system.jNFifFmg
15 Static Friction… (with one surface stationary) Just like in the sliding case except a = 0.i : F fF = 0j : N = mgWhile the block is static: fF FjNFifFmg
16 Static Friction…The maximum possible force that the friction between two objects can provide is fMAX = SN, where s is the “coefficient of static friction.”So fF S N.As one increases F, fF gets bigger until fF = SN and the object starts to move.If an object doesn’t move, it’s static frictionIf an object does move, it’s dynamic frictionjNFifFmg
17 Static Friction...S is discovered by increasing F until the block starts to slide:i : FMAX SN = 0j : N = mgS FMAX / mgjNFMAXiSmgmg
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 ms = 0.4.A rope is attached to the box and pulled at an angle of q = 30o above horizontal with tension T = 40 N.Does the box move?(a) yes (b) no (c) too close to callTqmstatic friction (ms = 0.4 )
19 Lecture 7, Act 2 Solution y x Pick axes & draw FBD of box: Apply FNET = may: N + T sin q - mg = maY = 0NN = mg - T sin q= 80 NTx: T cos q - fFR = maXqThe box will move if T cos q - fFR > 0fFRmmg
20 Lecture 7, Act 2 Solution y x y: N = 80 N x: T cos q - fFR = maX The box will move if T cos q - fFR > 0NT cos q = 34.6 NTfMAX = msNfMAX = msN = (.4)(80N) = 32 NqmSo T cos q > fMAX and the box does movemgNow use dynamic friction: max = Tcosq - mKN
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 (Newton’s 2nd Law along x-axis) Static Friction...The force provided by friction, fF , depends on .fFma = 0 (block is not moving)mg sin ff ijN(Newton’s 2nd Law along x-axis)mg
23 Static Friction...We can find s by increasing the ramp angle until the block slides:mg sin ffIn this case, when it starts to slide:ffSN Smg cos MSNijmg sin MSmg cos MNMmgStan M
24 Additional comments on Friction: Since fF = 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 ³ K for any system (think about it...).
25 Aside: Graph of Frictional force vs Applied force: fF = SN fF = KN fF = FAFA
26 Problem: Box on TruckA box with mass m sits in the back of a truck. The coefficient of static friction between the box and the truck is S.What is the maximum acceleration a that the truck can have without the box slipping?Sma
27 Problem: Box on Truck Draw Free Body Diagram for box: Consider case where fF is max... (i.e. if the acceleration were any larger, the box would slip).NjifF = SNmg
28 Problem: Box on Truck Use FNET = ma for both i and j components i SN = maMAXj N = mgaMAX = S gNjaMAXifF = SNmg
29 Lecture 7, Act 3 Forces and Motion 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?SaFfFfFf(a) (b) (c)
30 Lecture 7, Act 3 SolutionFirst consider the case where the inclined plane is not accelerating.NFfmgmgFfNAll the forces add up to zero!
31 Lecture 7, Act 3 SolutionIf the inclined plane is accelerating, the normal force decreases and the frictional force increases, but the frictional force still points along the plane:NFfamgAll the forces add up to ma!F = maThe answer is (a)mgFfNma
32 Putting on the brakesAnti-lock brakes work by making sure the wheels roll without slipping. This maximizes the frictional force slowing the car since S > K .