Chapter 4B. Friction and Equilibrium A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University A PowerPoint

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

A PowerPoint Presentation byA PowerPoint Presentation by

Paul E. Tippens, Professor of Paul E. Tippens, Professor of PhysicsPhysics

Southern Polytechnic State Southern Polytechnic State UniversityUniversity

A PowerPoint Presentation byA PowerPoint Presentation by

Paul E. Tippens, Professor of Paul E. Tippens, Professor of PhysicsPhysics

Southern Polytechnic State Southern Polytechnic State UniversityUniversity© 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 Objectives: After completing this module, you should be this module, 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.

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

Kinetic Friction:Kinetic Friction: Relative motionRelative 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, nn.

fk = knfk = knfs = snfs = sn

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 temperature, provided no temperature, provided no

chemical or structural variations chemical or structural variations occur.occur.

44 NN 4 N4 N

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

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

The Static Friction ForceThe Static Friction Force

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

fs = snfs = sn

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.

s sf nn

fs

P

W

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)

Pfs

P – fs = 0

Rest

Pfk

P – fk = 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 P is greater than the maximum fs 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.increases.

Pfk

Constant Speed

This case will be discussed in a later chapter.

fk = kn

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 – fP – fss = 0= 0

nffss

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 ffs s (max)(max). Static . Static friction applies.friction applies.

4. To find P we need to 4. To find P we need to know know ffs s , which is:, which is:

5. To find5. To find

nn::

nfs

P

250 N

+

For this case:For this case: P – f P – fss = 0= 0

ffss = = ssnn n = ?n = ?

FFyy = = 00 nn – W = – W = 00

WW = = 250 N250 N n = n = 250 250 NN

(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 ffs s (max)(max). Now we know . Now we know nn = = 250 N250 N..

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

6. Next we find6. Next we find ffss from:from:ffss = = ssn n = = 0.5 (250 0.5 (250 N)N)

P = fP = fss = = 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..

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 = = 0.50.5, 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 = = 00

nn - W = 0 - W = 0 nn = W = W

Now: fNow: fkk = = kkn n = = kkWW

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

P = fP = fk k = = kkWW

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 The Normal Force and WeightWeight

The 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.

Pn

W

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

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

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 vectors at origin of x,y axes. Choose x or y axis along motion or impending motion.y axis along motion or impending motion.

• Dot in rectangles and label x and y Dot in rectangles and label x and y compo-nents opposite and adjacent to compo-nents opposite and adjacent to angles.angles.

• Label all components; choose positive Label all components; choose positive direction.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 vectors at origin of x,y axes. Choose x or y axis along motion or impending motion.y axis along motion or impending motion.

• Dot in rectangles and label x and y Dot in rectangles and label x and y compo-nents opposite and adjacent to compo-nents opposite and adjacent to angles.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.Read, draw and label problem.

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

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

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

ffss = = ssnn or f or fkk = = kknn

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

FFxx = = 0 0 FFyy = = 00

• Solve for unknown quantities.Solve for unknown quantities.

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

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

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

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

ffss = = ssnn or f or fkk = = kknn

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

FFxx = = 0 0 FFyy = = 00

• Solve for unknown quantities.Solve for unknown quantities.

m

Example 2.Example 2. A force of 60 N drags a A force of 60 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 = ?

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 components 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 = ?; P = ?; 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 constant speed, horizontal forces are for constant speed, horizontal forces are balanced.balanced.

0xF 0xF 0yF 0yF


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 – –

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


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

0

Fx = 0.766P – fk = 0

ffkk = = k k n n == (0.2)(300 N - 0.643(0.2)(300 N - 0.643PP))

0.7660.766P – fP – fk k = = 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

ffkk = = (0.2)(300 N - 0.643(0.2)(300 N - 0.643PP) = 60 N – ) = 60 N – 0.1290.129PP


400

P

300 N

n

fk

+

0.766P0.766P

0.643P0.643P0.766P – (60 N – 0.129P )=00.766P – (60 N – 0.129P )=06.6. 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 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

• Equilibrium exists if speed is Equilibrium exists if speed is constant, but constant, but ffkk does does notnot get get larger as the speed is increased.larger as the speed is increased.

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

nfk

P

W


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

fk

nnPP

WW++

kk = 0.3 = 0.3

The The FF will be will be zero zero along the along the x-x-axisaxis and along the and along the y-axisy-axis..

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

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


• 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 0; 0x yF F 0; 0x yF F

SummarySummary

Static Friction: No relative motion.

Kinetic Friction: Relative motion.

fk = knfk = knfs ≤ snfs ≤ sn

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

0 0x yF F 0 0x yF F

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

Documents

Chapter 4B. Friction and Equilibrium A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University A PowerPoint