Physics 211: Lecture 7 - homes.ieu.edu.trhomes.ieu.edu.tr/hozcan/Phys100/Lect07.pdfPhysics 211: Lecture 7, Pg 32 Problem: Putting on the brakes Anti-lock brakes work by making sure

Physics 211: Lecture 7, Pg 1

Physics 211: Lecture 7

Today’s Agenda

Friction

What is it?

How do we characterize it?

Model of friction

Static & Kinetic friction

Some problems involving friction


New Topic: Friction

What does it do?

It opposes relative motion!

How do we characterize this in terms we have learned?

Friction results in a force in the direction opposite to the direction of relative motion!

ma

FAPPLIED

fFRICTION mg

N

i

j


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.

ma

F

fF mg

N

i

j


These relations are all useful APPROXIMATIONS to messy reality.

Model for Sliding 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.”

Suitcase


Model...

Dynamics:

i : F KN = ma

j : N = mg

so F Kmg = ma

ma

F

mg

N

i

j

K 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 (k = 0.51) on top of a second box having mass m2 = 3 kg, which in turn slides on a frictionless floor.

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

m2

T m1 slides with friction (k=0.51)

slides without friction a = ?


Lecture 7, Act 1 Solution

First draw FBD of the top box:

m1

N1

m1g

T f = KN1 = Km1g



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.

m1 f1,2

m2 f2,1

As we just saw, this force is due to friction:

= Km1g



Now consider the FBD of box 2:

m2 f2,1 = km1g

m2g

N2

m1g



Finally, solve F = ma in the horizontal direction:

m2 f2,1 = Km1g

Km1g = m2a gm

ma k

2

1 2sm819510

kg3

kg51..

.

a = 2.5 m/s2


Inclined Plane with Friction:

Draw free-body diagram:

i

j

mg

N

KN ma


Inclined plane...

Consider i and j components of FNET = ma :

i

j

mg

N

KN

ma

i mg sin KN = ma

mg sin

j N = mg cos

mg cos

mg sin Kmg cos = ma

a / g = sin Kcos


Static Friction...

F

mg

N

i

j

fF

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.


Static Friction… (with one surface stationary)

Just like in the sliding case except a = 0.

i : F fF = 0

j : N = mg

F

mg

N

i

j

fF

While the block is static: fF F


Static Friction… (with one surface stationary)

F

mg

N

i

j

fF

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.


Static Friction...

S is discovered by increasing F until the block starts to slide:

i : FMAX SN = 0

j : N = mg

S FMAX / mg

FMAX

mg

N

i

j

Smg

Suitcase



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 = 30o above horizontal with tension T = 40 N.

Does the box move?

(a) yes (b) no (c) too close to call

T

m

static friction (s= 0.4)



Pick axes & draw FBD of box:

T

m

N

mg

y

x

Apply FNET = ma

y: N + T sin - mg = maY = 0

N = mg - T sin = 80 N

x: T cos - fFR = maX

The box will move if T cos - fFR > 0

fFR



T

m

fMAX = sN

N

mg

y

x

x: T cos - fFR = maX

y: N = 80 N

The box will move if T cos - fFR > 0

T cos = 34.6 N

fMAX = sN = (.4)(80N) = 32 N

So T cos > fMAX and the box does move


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.


Static Friction...

mg

N

ma = 0 (block is not moving)

The force provided by friction, fF , depends on .

fF

mg sin ff

(Newton‟s 2nd Law along x-axis)

i

j


Static Friction...

We can find s by increasing the ramp angle until the block slides:

M mg

N

SN

In this case:

mg sin MSmg cos M

Stan M

i

j

mg sin ff

ffSN Smg cos M

Blocks


Additional comments on Friction:

Since fF = N , the force of 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)

By definition, it must be true that S K for any system (think about it...).


Aside:

Graph of Frictional force vs Applied force:

fF

FA

fF = FA

fF = KN

fF = SN


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?

m S

a



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

fF = SN mg

i

j



Use FNET = ma for both i and j components

i SN = maMAX

j N = mg

aMAX = S g N

fF = SN mg

aMAX

i

j



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)

Ff

Ff Ff

S a



First consider the case where the inclined plane is not accelerating.

mg

Ff

N All the forces add up to zero!

mg

N Ff


mg

N Ff


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

All the forces add up to ma!

F = ma

The answer is (a)

mg

Ff

N ma


Problem: 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 .

The driver of a car moving with speed vo slams on the brakes. The coefficient of static friction between the wheels and the road is S . What is the stopping distance D?

ab vo

v = 0

D

Wheel



N

fF = SN mg

Use FNET = ma for both i and j components

i SN = ma

j N = mg

a = S g

a

i

j



As in the last example, find ab = Sg.

Using the kinematic equation: v2 - v02 = 2a( x -x0 )

In our problem: 0 - v02 = 2ab( D )

ab vo

v = 0

D



In our problem: 0 - v02 = 2ab( D )

Solving for D:

Putting in ab = Sg

ab vo

v = 0

D

b

20

a2

vD =

Dv

gs 0

2

2


Recap of today’s lecture

Friction

What is it? (Text: 5-1)

How do we characterize it?

Model of friction.

Static & Kinetic friction. (Text: 5-1)

Some problems involving friction.

Box on truck.

Braking distance. (Example 5-7)

Look at textbook problems Chapter 5: # 26, 28, 34, 49

Documents

Physics 211: Lecture 7 - homes.ieu.edu.trhomes.ieu.edu.tr/hozcan/Phys100/Lect07.pdfPhysics 211: Lecture 7, Pg 32 Problem: Putting on the brakes Anti-lock brakes work by making sure