Objectives: After completing Objectives: After completing these notes, you should be these notes, you should be able to:able to:• Define and calculate the coefficients of Define and calculate the coefficients of

kinetic and static friction, and give the kinetic and static friction, and give the relationship of friction to the normal relationship of friction to the normal force.force.

• Apply the concepts of static and kinetic Apply the concepts of static and kinetic friction to problems involving constant friction to problems involving constant motion or impending motion.motion or impending motion.

• Define and calculate the coefficients of Define and calculate the coefficients of kinetic and static friction, and give the kinetic and static friction, and give the relationship of friction to the normal relationship of friction to the normal force.force.

• Apply the concepts of static and kinetic Apply the concepts of static and kinetic friction to problems involving constant friction to problems involving constant motion or impending motion.motion or impending motion.

Friction ForcesFriction ForcesWhen two surfaces are in contact, friction When two surfaces are in contact, friction forces oppose relative motion or forces oppose relative motion or impending motion.impending motion.

PPFriction forcesFriction forces are are parallel parallel to the surfaces in to the surfaces in contact and contact and opposeoppose motion or impending motion or impending motion.motion.

22 N N

Friction and the Normal Friction and the Normal ForceForce

4 N4 N

The force required to overcome The force required to overcome staticstatic or or kinetic kinetic friction is proportional to the friction is proportional to the

normal force, normal force, FFnn.

Ff = kFnFf = kFnFf ≤ sFn

Ff ≤ sFn

nn

12 12 NN

6 N6 N

nn8 N8 N

4 N4 N

nn

Friction forces are independent of Friction forces are independent of area.area.

44 NN 44 NN

If the total mass pulled is constant, the If the total mass pulled is constant, the same force (4 N) is required to overcome same force (4 N) is required to overcome friction even with twice the area of friction even with twice the area of contact.contact.For this to be true, it is essential that For this to be true, it is essential that ALL other variables be rigidly controlled.ALL other variables be rigidly controlled.

Friction forces are independent of Friction forces are independent of speed.speed.

2 2 NN2 2 NN

The force of kinetic friction is the The force of kinetic friction is the same at same at 5 m/s5 m/s as it is for as it is for 20 m/s20 m/s. . Again, we must assume that there Again, we must assume that there are no chemical or mechanical are no chemical or mechanical changes due to speed.changes due to speed.

5 m/s5 m/s 20 20 m/sm/s

Friction Forces

Ff Fn- coefficient of the surfaces in contact,

Friction is dependent on:

Static friction opposes the intended motion of two surfaces in contact but at rest relative to one another.Kinetic friction opposes motion of two surfaces in contact that are moving relative to one another.

FaPHYSICS

Fs

- normal force, Fn

FaPHYSICS

Fkvelocity

Kinetic friction is less than static friction.

Fs

book pulled

wheel driven

book dragged

surfaces in contact s kleather-soled shoes on wood 0.3 0.2

rubber-soled shoes on wood 0.9 0.7

climbing boots on rock 1.0 0.8

shoes on ice 0.1 0.05

auto tires on dry concrete 1.0 0.8

auto tires on wet concrete 0.7 0.5

auto tires on icy concrete 0.3 0.02

waxed skis on dry snow 0.08 0.04

waxed skis on wet snow 0.14 0.1

wood on wood 0.4 0.2

glass on glass 0.9 0.4

steel on steel - dry 0.6 0.4

steel on steel - greased 0.1 0.05

synovial joints in humans 0.01 0.003

Coefficients of Friction

The Static Friction ForceThe Static Friction Force

In this unit, when we use the following In this unit, when we use the following equation, we refer only to the equation, we refer only to the maximummaximum value of static friction and value of static friction and simply writesimply write::

Ff = sFn Ff = sFn

When an attempt is made to move an When an attempt is made to move an object on a surface, static friction object on a surface, static friction slowly increases to a slowly increases to a MAXIMUM MAXIMUM valuevalue.

nfs

P

WFf ≤ sFn

Constant or Impending Constant or Impending MotionMotion

For motion that is For motion that is impendingimpending and for and for motion at motion at constant constant speed, the speed, the resultant force is zero and resultant force is zero and F = 0F = 0. . (Equilibrium)(Equilibrium)

FFf

F – Ff = 0

Rest

FFf

F – Ff = 0

Constant Speed

Here the Here the weightweight and and normal forcesnormal forces are balanced and do not affect are balanced and do not affect motion.motion.

Friction and AccelerationFriction and Acceleration

When F is greater than the maximum FF the resultant force produces acceleration.

Note that the kinetic friction force Note that the kinetic friction force remains constant even as the velocity remains constant even as the velocity increases or decreases. Kinetic is a increases or decreases. Kinetic is a hard equality.hard equality.

FFF

Changing Speed

This case we will have to use Newton 2nd law

Ff = kFn

a

EXAMPLE 1:EXAMPLE 1: If If kk = 0.3 = 0.3 and and ss = = 0.50.5, what horizontal pull , what horizontal pull PP is is required to just start a required to just start a 250-N250-N block moving?block moving?

1. Draw sketch and free-1. Draw sketch and free-body diagram as body diagram as shown.shown.

2. List givens and label 2. List givens and label what is to be found:what is to be found:

kk = 0.3; = 0.3; ss = 0.5; = 0.5; W = W = 250 250 NN

Find: Find: P = ? P = ? to just to just startstart

3. Recognize for impending motion:3. Recognize for impending motion: P – P – FFfsfs = 0 = 0

Fn

FFffPP

mgmg++

EXAMPLE 1(Cont.):EXAMPLE 1(Cont.): ss = 0.5 = 0.5, , mg = 250 Nmg = 250 N. . Find Find PP to overcome to overcome FFf f (max)(max). Static friction . Static friction applies.applies.

4. To find P we need to 4. To find P we need to know know FFfs fs , which is:, which is:

5. To find5. To find FFnn

::

Fn

FfP

250 N

+

For this case:For this case: P – F P – Ffsfs = = 00

FFff = = s s FFnn FFnn = ? = ?

FFyy = = 00 FFnn – mg = – mg = 00

mgmg= = 250 250 NN

FFnn = = 250 N250 N

(Continued)(Continued)

EXAMPLE 1(Cont.):EXAMPLE 1(Cont.): ss = 0.5 = 0.5, , WW = 250 N = 250 N. Find . Find

PP to overcome to overcome FFf f (max)(max). Now we know . Now we know FFnn = = 250 N250 N..

7. For this case7. For this case: P – Ff = 0

6. Next we find6. Next we find FFff from:from:FFff = = ss F Fnn = = 0.5 (250 0.5 (250

N)N)

P = FP = Fff = = 0.5 (250 N)0.5 (250 N)

P = 125 NP = 125 N

This force (This force (125 N125 N) is needed to ) is needed to just startjust start motion. Next we consider motion. Next we consider P P needed for needed for constant speedconstant speed..

FFnn

FfP

250 N

+

ss = 0.5 = 0.5

EXAMPLE 1(Cont.):EXAMPLE 1(Cont.): If If kk = 0.3 = 0.3 and and ss = = 0.50.5, what horizontal pull , what horizontal pull PP is required to is required to move with move with constant speedconstant speed? ? (Overcoming (Overcoming kinetickinetic friction) friction)

FFyy = m = maayy = = 00FFnn - mg = - mg =

00FFnn = mg = mg

Now: FNow: Fff = = kk F Fnn = = kkmgmg

FFxx = = 0; 0; P - fP - fkk = = 0 0

P = FP = Ff f = = kkmgmg

P = P = (0.3)(250 N)(0.3)(250 N) P = 75.0 NP = 75.0 N

Ff

FFnnP

mg+

kk = 0.3 = 0.3

The Normal Force and The Normal Force and WeightWeight

The normal force is NOT always equal to the weight. The following are examples:

300

P

m

FFnn

mg

Here the normal force is less than weight due to upward component of P.

PFFnn

mg

Here the normal force is equal to only the compo- nent of weight perpendi- cular to the plane.

m

Example 2.Example 2. A force of 100 N drags a A force of 100 N drags a 300-N block by a rope at an angle of 300-N block by a rope at an angle of 404000 above the horizontal surface. If u above the horizontal surface. If uk k = 0.2, what force P will produce = 0.2, what force P will produce constant speed?constant speed?

1. Draw and label a 1. Draw and label a sketch of the sketch of the problem.problem.400

P =100

fk

nW = 300 N

2. Draw free-body 2. Draw free-body diagram.diagram.

The force P is to be The force P is to be replaced by its com- replaced by its com- ponents ponents PPxx and and PPyy..

400

P

W

n

fk

+

WW

PPxxP P cos cos 404000

PPyy

PPyyP P sinsin 404000

Example 2 (Cont.).Example 2 (Cont.). P = 100; P = 100; W = W = 300 N; u300 N; uk k = = 0.2.0.2.

3. Find components 3. Find components of P:of P: 40

0

P

mg

n

fk

+

P P cos cos 404000

P P sinsin 404000

Px = P cos 400 = 0.766PPy = P sin 400 = 0.643P

Px = 0.766P; Py = 0.643PNote: Vertical forces are balanced, and Note: Vertical forces are balanced, and for horizontal forces are unbalanced.for horizontal forces are unbalanced.

0xF 0xF 0yF 0yF

Example 2 (Cont.).Example 2 (Cont.). P = 100; P = 100; W = W = 300 N; u300 N; uk k = = 0.20.2

4. Apply Equilibrium 4. Apply Equilibrium con- ditions to con- ditions to vertical axis. vertical axis.

400

P

300 N

n

fk

+

0.7660.766PP

0.6430.643PP

Fy = 0Fy = 0

PPxx = = 0.7660.766PP PPyy = =

0.643P

nn + + 0.6430.643P – P – 300 N300 N= = 00

[[PPyy and and nn are up are up ((++)])]nn = = 300 N 300 N – –

64.364.3P; P;

n = 223 Nn = 223 N

Solve for Solve for nn

Example 2 (Cont.).Example 2 (Cont.). P = 100; P = 100; W = W = 300 N; u300 N; uk k = 0.2.= 0.2.

5. Apply 5. Apply FFxx = = 0 to con- 0 to con- stant horizontal stant horizontal motion.motion.Fx = 0.766P – fk =

ma

Fx = 0.766P – fk = ma

ffkk = = k k n n == (0.2)(223)(0.2)(223)

76.676.6 – f – fk k = = ma;ma;

400

P

300 N

n

fk

+

0.766P0.766P

0.643P0.643Pn = 223Nn = 223N

76.6-44.7= ma76.6-44.7= ma

ffkk = = (0.2)(223) = 44.7N(0.2)(223) = 44.7N

Example 2 (Cont.).Example 2 (Cont.).P = 100N; P = 100N; W = W = 300 N; u300 N; uk k = 0.2.= 0.2.

400

P

300 N

n

fk

+

0.766P0.766P

0.643P0.643P76.6-44.7= 31.9N = ma76.6-44.7= 31.9N = ma

6.6. Solve for mass.Solve for mass.

300N= mg; m=30kg

31.9N=30 a If If P = P = 100 N, 100 N, the block will the block will

be accelerated.be accelerated.

a = 1.06 m/s2

a = 31.9/30

a = 1.06 m/s2

xxyy

Example 3:Example 3: What push What push P P up the up the incline is needed to move a incline is needed to move a 230-N230-N block up the incline at constant block up the incline at constant speed if speed if kk = 0.3 = 0.3??

606000

Step 1: Step 1: Draw free-body Draw free-body including forces, angles including forces, angles and components.and components.

PP

230 230 NN

fk

n

600

W W cos 60cos 6000

W W sin 60sin 6000

Step 2: Step 2: FFyy = 0 = 0

n – W cos 600 = 0n = (230 N) cos

600

n = 115 Nn = 115 N

WW =230 N =230 N

PP

Example 3 (Cont.):Example 3 (Cont.): FindFind PP to to give move up the incline (W = give move up the incline (W = 230 N).230 N).

600

Step 3. Apply Step 3. Apply FFxx= = 00

xy P

W

fk

n

600

W cos 600

W sin 600

n = 115 NW = 230

N

P - fP - fkk - W - W sin 60sin 6000 = = 00

ffkk = = kknn = 0.2(115 N) = 0.2(115 N)

ffkk = = 2323 N, N, PP = ? = ?P - P - 2323 NN - - (230 N)sin 60(230 N)sin 6000 = = 00

P - P - 2323 NN - - 199 N199 N= = 00 P = 222 NP = 222 N

Summary: Important Points to Summary: Important Points to Consider When Solving Friction Consider When Solving Friction

Problems.Problems.

• The maximum force of static The maximum force of static friction is the force required to friction is the force required to just start just start motion.motion.

s sf nn

fs

P

W

Equilibrium exists at that instant:Equilibrium exists at that instant:

0; 0x yF F 0; 0x yF F

Summary: Important Points (Cont.)Summary: Important Points (Cont.)

• The force of The force of kinetic frictionkinetic friction is that is that force required to maintain force required to maintain constant motionconstant motion..

k kf n

• ffkk does does notnot get larger as the get larger as the speed is increased.speed is increased.

nfk

P

W

Summary: Important Points (Cont.)Summary: Important Points (Cont.)

• Remember the normal force Remember the normal force nn is is not not always equal to the weight of an always equal to the weight of an object.object.

It is necessary to draw the free-body diagram and sum forces to solve for

the correct n value.

300

P

m

n

W

Pn

W

SummarySummary

Static Friction: No relative motion.

Kinetic Friction: Relative motion.

fk = knfk = knfs ≤ snfs ≤ sn

CONCLUSION: Chapter 4BCONCLUSION: Chapter 4BFriction and EquilibriumFriction and Equilibrium