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**SPH3UW 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

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**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

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

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**Surface Friction... Force of friction acts to oppose relative motion:**

Parallel to surface. Perpendicular to Normal force. j N F i ma fF mg

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**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.

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**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

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**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

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**Lecture 7, Act 1 Solution First draw FBD of the top box: N1 m1**

f = mKN1 = mKm1g T m1g

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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

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**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

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

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**Inclined Plane with Friction:**

Draw free-body diagram: ma KN j N mg i

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**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

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**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

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**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

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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

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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

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**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 )

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**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

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**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

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**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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**(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

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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: ffSN Smg cos M SN i j mg sin MSmg cos M N M mg Stan M

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**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...).

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

fF = FA FA

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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

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**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

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**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

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**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)

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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!

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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

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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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