chapter 4b. friction and equilibrium · chapter 4b. friction and equilibrium a powerpoint...

Chapter 4B. Friction and Chapter 4B. Friction and EquilibriumEquilibrium

A PowerPoint Presentation by

Paul E. Tippens, Professor of Physics

Southern Polytechnic State University

A PowerPoint Presentation byA PowerPoint Presentation by

Paul E. Tippens, Professor of PhysicsPaul E. Tippens, Professor of Physics

Southern Polytechnic State UniversitySouthern Polytechnic State University

©

2007

Equilibrium: Until motion begins, all forces on the mower are balanced. Friction in wheel bearings and on the ground oppose the lateral motion.

Objectives: After completing this Objectives: After completing this module, you should be able to:module, you should be able to:

•

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

•

Apply the concepts of static and kinetic friction to problems involving constant 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 force.relationship of friction to the normal 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 forces When two surfaces are in contact, friction forces oppose relative motion or impending motion.oppose relative motion or impending motion.

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

Static Friction:Static Friction: No No relative motion.relative motion.

Kinetic Friction:Kinetic Friction: Relative motionRelative motion.

22

NN

Friction and the Normal ForceFriction and the Normal Force

4 N4 N

The force required to overcome The force required to overcome staticstatic or or kinetic kinetic friction is proportional to the normal force, friction is proportional to the normal force, nn.

fk

= k

nfk

= k

nfs

= s

nfs

= s

n

nn

12 N12 N

6 N6 N

nn8 N8 N

4 N4 N

nn

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

44

NN 44

NN

If the total mass pulled is constant, the same If the total mass pulled is constant, the same force (4 N) is required to overcome friction force (4 N) is required to overcome friction even with twice the area of contact.even with twice the area of contact.

For this to be true, it is essential that ALL For this to be true, it is essential that ALL other variables be rigidly controlled.other variables be rigidly controlled.

Friction forces are independent of Friction forces are independent of temperature, provided no chemical or temperature, provided no chemical or

structural variations occur.structural variations occur.

44

NN 4 N4 N

Heat can sometimes cause surfaces to become Heat can sometimes cause surfaces to become deformed or sticky. In such cases, temperature deformed or sticky. In such cases, temperature can be a factor.can be a factor.

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

2 2 NN2 2 NN

The force of kinetic friction is the same at The force of kinetic friction is the same at 5 m/s5 m/s

as it is for as it is for 20 m/s20 m/s. Again, we must . Again, we must

assume that there are no chemical or assume that there are no chemical or mechanical changes due to speed.mechanical changes due to speed.

5 m/s5 m/s 20 m/s20 m/s

The Static Friction ForceThe Static Friction Force

In this module, when we use the following In this module, when we use the following equation, we refer only to the equation, we refer only to the maximummaximum

value of static friction and simply writevalue of static friction and simply write::

fs

= s

nfs

= s

n

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

s sf nn

fsP

W

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

Pfs

P – fs

= 0

Rest

Pfk

P – fk

= 0

Constant Speed

Here the Here the weightweight

and and normal forcesnormal forces

are are balanced and do not affect motion.balanced and do not affect motion.

Friction and AccelerationFriction and Acceleration

When P is greater than the maximum fs the resultant force produces acceleration.

Note that the kinetic friction force remains Note that the kinetic friction force remains constant even as the velocity increases.constant even as the velocity increases.

Pfk

Constant Speed

This case will be discussed in a later chapter.

fk

= k

n

a

EXAMPLE 1:EXAMPLE 1:

If If kk

= 0.3= 0.3

and and ss

= 0.5= 0.5, , what horizontal pull what horizontal pull PP is required to is required to just start a just start a 250250--NN

block moving?block moving?

1. Draw sketch and free1. Draw sketch and free-- body diagram as shown.body diagram as 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 N250 N

Find: Find: P = ? P = ? to just startto just start

3. Recognize for impending motion:3. Recognize for impending motion:

P P –– ff

ss

= 0= 0

nff

ss

PP

WW++

EXAMPLE 1(Cont.):EXAMPLE 1(Cont.): ss

= 0.5= 0.5, , W = 250 NW = 250 N. Find . Find PP

to overcome to overcome ff

ss

(max)(max). Static friction applies.. Static friction applies.

4. To find P we need to 4. To find P we need to know know ff

ss

, which is:, which is:

5. To find5. To find nn::

nfs

P

250 N

+

For this case:For this case: P P –– ff

ss

= 0= 0

ff

ss

= = ss

nn n = ?n = ?

FF

yy

= = 00 nn –– W = W = 00

WW = = 250 N250 N n = n = 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 ff

ss

(max)(max). Now we know . Now we know nn = 250 N= 250 N..

7. For this case7. For this case: P – fs

= 0

6. Next we find6. Next we find ff

ss

from:from:

ff

ss

= = ss

n n = = 0.5 (250 N)0.5 (250 N)

P = P = ff

ss

= = 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. motion. Next we consider Next we consider P P needed for constant speedneeded for constant speed..

nfs

P

250 N

+

ss

= 0.5= 0.5

EXAMPLE 1(Cont.):EXAMPLE 1(Cont.):

If If kk

= 0.3= 0.3

and and ss

= 0.5= 0.5, , what horizontal pull what horizontal pull PP

is required to move with is required to move with

constant speedconstant speed? (Overcoming ? (Overcoming kinetickinetic

friction)friction)

FF

yy

= m= maayy

= 0= 0

nn -- W = 0W = 0 nn = W= WNow: Now: ff

kk

= = kk

nn = = kk

WW

FF

xx

= = 0; 0; P P -- ff

kk

= = 0 0

P = P = ff

kk

= = kk

WW

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

fk

nP

mg+

kk

= 0.3= 0.3

The Normal Force and WeightThe Normal Force and WeightThe normal force is NOT always equal to the weight. The following are examples:

300

P

m

n

W

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

PnW

Here the normal force is equal to only the compo-

nent of weight perpendi- cular to the plane.

Review of FreeReview of Free--body Diagrams:body Diagrams:

For Friction Problems:

•

Read problem; draw and label sketch.

•

Construct force diagram for each object, vectors at origin of x,y axes. Choose x or y axis along motion or impending motion.

•

Dot in rectangles and label x and y components

opposite and adjacent to angles.

•

Label all components; choose positive direction.

For Friction Problems:For Friction Problems:

••

Read problem; draw and label sketch.Read problem; draw and label sketch.

••

Construct force diagram for each object, Construct force diagram for each object, vectors at origin of x,y axes. Choose x or y vectors at origin of x,y axes. Choose x or y axis along motion or impending motion.axis along motion or impending motion.

••

Dot in rectangles and label x and y compoDot in rectangles and label x and y compo-- nentsnents

opposite and adjacent to angles.opposite and adjacent to angles.

••

Label all components; choose positive Label all components; choose positive direction.direction.

For Friction in Equilibrium:For Friction in Equilibrium:

•

Read, draw and label problem.

•

Draw free-body diagram for each body.

•

Choose x or y-axis along motion or impending motion and choose direction of motion as positive.

•

Identify the normal force and write one of following:

fs

= s

n or fk

= k

n

•

For equilibrium, we write for each axis:

Fx

= 0 Fy

= 0

•

Solve for unknown quantities.

••

Read, draw and label problem.Read, draw and label problem.

••

Draw freeDraw free--body diagram for each body.body diagram for each body.

••

Choose x or yChoose x or y--axis along motion or impending axis along motion or impending motion and choose direction of motion as positive.motion and choose direction of motion as positive.

••

Identify the normal force and write one of Identify the normal force and write one of following:following:

ff

ss

= = ss

nn or or ff

kk

= = kk

nn

••

For equilibrium, we write for each axis:For equilibrium, we write for each axis:

FF

xx

= = 0 0 FF

yy

= = 00

••

Solve for unknown quantities.Solve for unknown quantities.

m

Example 2.Example 2.

A force of 60 N drags a 300A force of 60 N drags a 300--N N block by a rope at an angle of 40block by a rope at an angle of 4000

above the above the

horizontal surface. If horizontal surface. If uu

kk

= 0.2, what force P = 0.2, what force P will produce constant speed?will produce constant speed?

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

400

P = ?

fkn

W = 300 N

2. Draw free2. Draw free--body diagram.body diagram.

The force P is to be The force P is to be replaced by its comreplaced by its com--

ponentsponents

PPxx

and and PPyy

..

400

P

W

n

fk+

WW

PPxxP P coscos

404000

PPyy

PPyyP P sinsin

404000

Example 2 (Cont.).Example 2 (Cont.).

P = ?; P = ?; W = W = 300 N; 300 N; uu

kk

= 0.2.= 0.2.

3. Find components of P:3. Find components of P:400

P

mg

n

fk+

P P coscos

404000

P P sinsin

404000

Px

= P cos

400 = 0.766P

Py

= P sin 400 = 0.643P

Px

= 0.766P; Py

= 0.643P

Note: Vertical forces are balanced, and for Note: Vertical forces are balanced, and for constant speed, horizontal forces are balanced.constant speed, horizontal forces are balanced.

0xF 0yF


P = ?; P = ?; W = W = 300 N; 300 N; uu

kk

= 0.2.= 0.2.

4. Apply Equilibrium con4. Apply Equilibrium con-- ditionsditions to vertical axis. to 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 [[PP

yy

and and nn are up (are up (++)])]nn = = 300 N 300 N –– 0.6430.643P; P;

n = 300 N –

0.643Pn = 300 N –

0.643P

Solve for Solve for nn

in terms of in terms of PP


P = ?; P = ?; W = W = 300 N; 300 N; uu

kk

= 0.2.= 0.2.

5. Apply 5. Apply FF

xx

= = 0 to con0 to con-- stantstant

horizontal motion.horizontal motion.

Fx

= 0.766P – fk

= 0Fx

= 0.766P – fk

= 0

ff

kk

= = kk

n n == (0.2)(300 N (0.2)(300 N --

0.6430.643PP))

0.7660.766P P –– ff

kk

= = 0;0;

400

P

300 N

n

fk+

0.766P0.766P

0.643P0.643Pn = 300 N –

0.643Pn = 300 N –

0.643P

0.766P – (60 N –

0.129P) =

00.766P – (60 N –

0.129P) =

0

ff

kk

= = (0.2)(300 N (0.2)(300 N --

0.6430.643PP) = 60 N ) = 60 N ––

0.1290.129PP


P = ?; P = ?; W = W = 300 N; 300 N; uu

kk

= 0.2.= 0.2.

400

P

300 N

nfk

+

0.766P0.766P

0.643P0.643P0.766P – (60 N –

0.129P )=00.766P – (60 N –

0.129P )=0

6.6. Solve for unknown P.Solve for unknown P.

0.766P – 60 N + 0.129P =0

0.766P + 0.129P = 60 N If If P = P = 67 N, the 67 N, the block will be block will be dragged at a dragged at a

constant speed.constant speed.

P = 67.0 N

0.766P + 0.129P = 60 N

0.895P = 60 N

P = 67.0 N

xxyy

Example 3:Example 3:

What push What push P P up the incline is up the incline is needed to move a needed to move a 230230--NN

block up the block up the

incline at constant speed if incline at constant speed if kk

= 0.3= 0.3??

606000

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

PP

230 N230 N

fk

n

600W W cos 60cos 6000

W W sin 60sin 6000

Step 2: Step 2: FF

yy

= 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 give to give move up the incline (W = 230 N).move up the incline (W = 230 N).

600

Step 3. Apply Step 3. Apply FF

xx

= = 00

xy P

W

fk

n

600W cos

600

W sin 600

n = 115 N W = 230 N

P P -- ff

kk

-- W W sin 60sin 6000

= = 00

ff

kk

= = kk

nn

= 0.2(115 N)= 0.2(115 N)

ff

kk

= = 2323

N, N, PP = ?= ?

P P -- 2323

NN

-- (230 (230 N)sinN)sin

606000

= 0= 0

P P -- 2323

NN

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

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

••

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

s sf nn

fsP

W

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

0; 0x yF F

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

••

The force of The force of kinetic frictionkinetic friction

is that force is that force required to maintain required to maintain constant motionconstant motion..

k kf n••

Equilibrium exists if speed is constant, Equilibrium exists if speed is constant, but but ff

kk

does does notnot

get larger as the get larger as the speed is increased.speed is increased.

0; 0x yF F

nfk

P

W


••

Choose an Choose an xx or or yy--axis along the direction axis along the direction of motion or impending motion.of motion or impending motion.

fk

nnPP

WW++

kk

= 0.3= 0.3

The The FF will be will be zero zero along the along the xx--axisaxis

and and

along the along the yy--axisaxis..

0; 0x yF F

In this figure, we have:In this figure, we have:


••

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

PnW 0; 0x yF F

SummarySummary

Static Friction: No relative motion.

Kinetic Friction: Relative motion.

fk

= k

nfk

= k

nfs

≤

s

nfs

≤

s

n

Procedure for solution of equilibrium problems is the same for each case:

0 0x yF F

CONCLUSION: Chapter 4BCONCLUSION: Chapter 4B Friction and EquilibriumFriction and Equilibrium

Chapter 4B. Friction and EquilibriumSlide Number 2Objectives: After completing this module, you should be able to:Friction ForcesFriction and the Normal ForceFriction forces are independent of area.Friction forces are independent of temperature, provided no chemical or structural variations occur.Friction forces are independent of speed.The Static Friction ForceConstant or Impending MotionFriction and AccelerationEXAMPLE 1: If mk = 0.3 and ms = 0.5, what horizontal pull P is required to just start a 250-N block moving?EXAMPLE 1(Cont.): ms = 0.5, W = 250 N. Find P to overcome fs (max). Static friction applies.EXAMPLE 1(Cont.): ms = 0.5, W = 250 N. Find P to overcome fs (max). Now we know n = 250 N.EXAMPLE 1(Cont.): If mk = 0.3 and ms = 0.5, what horizontal pull P is required to move with constant speed? (Overcoming kinetic friction)The Normal Force and WeightReview of Free-body Diagrams:For Friction in Equilibrium:Example 2. A force of 60 N drags a 300-N block by a rope at an angle of 400 above the horizontal surface. If uk = 0.2, what force P will produce constant speed?Example 2 (Cont.). P = ?; W = 300 N; uk = 0.2.Example 2 (Cont.). P = ?; W = 300 N; uk = 0.2.Example 2 (Cont.). P = ?; W = 300 N; uk = 0.2.Example 2 (Cont.). P = ?; W = 300 N; uk = 0.2.Example 3: What push P up the incline is needed to move a 230-N block up the incline at constant speed if mk = 0.3? Example 3 (Cont.): Find P to give move up the incline (W = 230 N).Summary: Important Points to Consider When Solving Friction Problems.Summary: Important Points (Cont.)Summary: Important Points (Cont.)Summary: Important Points (Cont.)SummaryCONCLUSION: Chapter 4B�Friction and Equilibrium

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