SPH3UW: Lecture 18, Pg 1 SPH3UW Today’s Agenda l Friction è What is it? è Systematic catagories...

SPH3UW: Lecture 18, Pg 1

SPH3UWSPH3UW

Today’s AgendaToday’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: New Topic: FrictionFriction

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)

maaFFAPPLIED

mgg

NN

ii

j j

ffFRICTION some roughness here


Surface Friction...Surface Friction...

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


Surface Friction...Surface Friction...

Force of friction acts to oppose relative motion: Parallel to surface. Perpendicular to NNormal force.

maaFF

ffF mgg

NN

ii

j j


These relations are all useful APPROXIMATIONS to messy reality.

Model for Sliding Model for Sliding (kinetic)(kinetic) Friction Friction

The direction of the frictional force vector is perpendicular to the normal force vector NN.

The magnitude of the frictional force vector |ffF| is proportional to the magnitude of the normal force |N N |.

|ffF| = K | N N | ( = K|mg g | 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.”


Model...Model...

Dynamics:

i : F KN = ma

j : N = mg

so F Kmg = ma

maaFF

mgg

NN

ii

j j

K mg

(this works as long as F is bigger than friction, i.e. the left hand side is positive)


Lecture 7, Lecture 7, Act 1Act 1Forces and MotionForces 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. (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

mm2 2

T mm11slides with friction (k=0.51)

slides without frictiona = ?

Hint: draw FBDs of both blocks – that’s 2 diagrams


Lecture 7, Lecture 7, Act 1Act 1SolutionSolution

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 ff1,2

m2 ff2,1

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

= Km1g



Now consider the FBD of box 2:

m2 ff2,1 = km1g

m2g

N2

m1g(gravity from…)(contact from…)

(contact from…)

(friction from…)



Finally, solve F = ma in the horizontal direction:

m2 ff2,1 = Km1g

Km1g = m2ag

mm

a k2

1 2sm819510kg3

kg51..

.

a = 2.5 m/s2


Inclined Plane with Friction:Inclined Plane with Friction:

Draw free-body diagram:

ii

jj

mgN

KNma


Inclined plane...Inclined plane...

Consider ii and j j components of FFNET = maa :

ii

jj

mg

N

KN

ma

i i mg sin KN = ma

mg sin

jj N = mg cos

mg cos

mg sin Kmg cos = ma

a / g = sin Kcos


Static Friction...Static Friction...

FF

mgg

NN

ii

j 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…Static Friction…(with one surface stationary)(with one surface stationary)

Just like in the sliding case except a = 0.

i : F fF = 0

j : N = mg

FF

mgg

NN

ii

j j

fF

While the block is static: fF F


Static Friction…Static Friction…

FF

mgg

NN

ii

j 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. If an object doesn’t move, it’s static friction If an object does move, it’s dynamic friction



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

i : FMAX SN = 0

j : N = mg

S FMAX / mg

FFMAX

mgg

NN

ii

j j

Smg



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

mfMAX = 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

Now use dynamic friction: max = Tcos - KN


Static Friction: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.



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)

ii

jj



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

M mg

N

SN

In this case, when it starts to slide:

mg sin MSmg cos M

Stan Mii

jj

mg sin ff

ffSN Smg cos M


Additional comments on Friction: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 K for any system (think about it...).


Aside:Aside:

Graph of Frictional force vs Applied force:

fF

FA

fF = FA

fF = KN

fF = SN


Problem: Box on TruckProblem: 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 aa that the truck can have without the box slipping?

m S

aa



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

ii

jj



Use FNET = ma for both ii and jj components ii SN = maMAX

jj N = mg

aMAX = S g N

fF = SN mg

aMAX

ii

jj



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 aa



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

mg

Ff

N All the forces add up to zero!

mg

NFf


mg

NFf


If the inclined plane is accelerating, the normal force decreases and the frictional force increases, but the frictional force still points along the plane:

aa

All the forces add up to ma! F = ma The answer is (a)

mg

Ff

Nma


Putting on the brakesPutting 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 .

SPH3UW: Lecture 18, Pg 1 SPH3UW Today’s Agenda l Friction è What is it? è Systematic catagories...

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