SPH4U: Lecture 7 Today’s Agenda Friction What is it? Systematic catagories of forces How do we characterize it? Model of friction Static & Kinetic friction (kinetic = dynamic in some languages) Some problems involving friction

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) j N FAPPLIED i ma fFRICTION some roughness here mg

Surface Friction... Friction is caused by the “microscopic” interactions between the two surfaces:

Surface Friction... Force of friction acts to oppose relative motion: Parallel to surface. Perpendicular to Normal force. j N F i ma fF mg

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.

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) j N F i ma K mg mg

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/s2 Hint: draw FBDs of both blocks – that’s 2 diagrams slides with friction (mk=0.51 ) T m1 a = ? m2 slides without friction

Lecture 7, Act 1 Solution First draw FBD of the top box: N1 m1 f = mKN1 = mKm1g T m1g

Lecture 7, Act 1 Solution 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. As we just saw, this force is due to friction: = mKm1g m1 f1,2 f2,1 m2

Lecture 7, Act 1 Solution Now consider the FBD of box 2: N2 (contact from…) (friction from…) f2,1 = mkm1g m2 (contact from…) (gravity from…) m1g m2g

Lecture 7, Act 1 Solution Finally, solve F = ma in the horizontal direction: mKm1g = m2a a = 2.5 m/s2 f2,1 = mKm1g m2

Inclined Plane with Friction: Draw free-body diagram: ma KN j N  mg  i

Inclined plane... Consider i and j components of FNET = ma : i mg sin KN = ma j N = mg cos  KN mg sin Kmg cos  = ma ma j N  a / g = sin Kcos   mg mg cos  i mg sin 

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. j N F i fF mg

Static Friction… (with one surface stationary) Just like in the sliding case except a = 0. i : F fF = 0 j : N = mg While the block is static: fF F j N F i fF mg

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 friction If an object does move, it’s dynamic friction j N F i fF mg

Static Friction... S is discovered by increasing F until the block starts to slide: i : FMAX SN = 0 j : N = mg S FMAX / mg j N FMAX i Smg mg

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 call T q m static friction (ms = 0.4 )

Lecture 7, Act 2 Solution y x Pick axes & draw FBD of box: Apply FNET = ma y: N + T sin q - mg = maY = 0 N N = mg - T sin q = 80 N T x: T cos q - fFR = maX q The box will move if T cos q - fFR > 0 fFR m mg

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 > 0 N T cos q = 34.6 N T fMAX = msN fMAX = msN = (.4)(80N) = 32 N q m So T cos q > fMAX and the box does move mg Now use dynamic friction: max = Tcosq - mKN

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. 

(Newton’s 2nd Law along x-axis) Static Friction... The force provided by friction, fF , depends on . fF ma = 0 (block is not moving) mg sin ff  i j N  (Newton’s 2nd Law along x-axis) mg 

Static Friction... We can find s by increasing the ramp angle until the block slides: mg sin ff In this case, when it starts to slide: ffSN  Smg cos M SN i j mg sin MSmg cos M N M mg  Stan M

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...).

Model for Surface Friction The direction of the frictional force vector fF is perpendicular to the normal force vector N, in the direction opposing relative motion of the two surfaces. Kinetic (sliding): The magnitude of the frictional force vector is proportional to the magnitude of the normal force N. fF = KN It moves, but it heats up the surface it moves on! Static: The frictional force balances the net applied forces such that the object doesn’t move. The maximum possible static frictional force is proportional to N. fF  SN and as long as this is true, then fF = fA in opposite direction It doesn’t move!

Aside: Graph of Frictional force vs Applied force: fF = SN fF = KN fF = FA FA

Example Determine the acceleration of the 2.0kg Block in the Diagram below (given that the coefficient of friction is 0.5). First we note that since all three blocks are connected, they all have the same acceleration magnitude (just different directions). Obviously the 3.0 kg mass beats the 1.0 kg mass so, it will dictate the direction of movement.

Solution Determine the acceleration of the 2.0kg Block in the Diagram below (given that the coefficient of friction is 0.5). Let’s draw a free body diagram, treating the three objects as one (since connected), and using the right direction as positive. 6 kg

Solution Therefore the acceleration of the 2.0 kg block is about 1.6 m/s2 to the right.

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. What is the maximum acceleration a that the truck can have without the box slipping? S m a

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). N j i fF = SN mg

Problem: Box on Truck Use FNET = ma for both i and j components i SN = maMAX j N = mg aMAX = S g N j aMAX i fF = SN mg

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? S a Ff Ff Ff (a) (b) (c)

Lecture 7, Act 3 Solution First consider the case where the inclined plane is not accelerating. N Ff mg mg Ff N All the forces add up to zero!

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

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 .