eighth edition - apl100apl100.wdfiles.com/local--files/notes/friction revised.pdf · eighth edition...

VECTOR MECHANICS FOR ENGINEERS:

STATICS

Eighth Edition

Ferdinand P. Beer

E. Russell Johnston, Jr.

Lecture Notes:

J. Walt Oler

Texas Tech University

CHAPTER

© 2007 The McGraw-Hill Companies, Inc. All rights reserved.

8 Friction


Vector Mechanics for Engineers: Statics

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Contents

Introduction

Laws of Dry Friction. Coefficients

of Friction.

Angles of Friction

Problems Involving Dry Friction

Sample Problem 8.1

Sample Problem 8.3

Wedges

Square-Threaded Screws

Sample Problem 8.5

Journal Bearings. Axle Friction.

Thrust Bearings. Disk Friction.

Wheel Friction. Rolling Resistance.

Sample Problem 8.6

Belt Friction.

Sample Problem 8.8



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Introduction

• In preceding chapters, it was assumed that

surfaces in contact were either frictionless

(surfaces could move freely with respect to

each other) or rough (tangential forces

prevent relative motion between surfaces).



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Introduction

• Actually, no perfectly frictionless surface

exists. For two surfaces in contact,

tangential forces, called friction forces, will

develop if one attempts to move one relative

to the other.



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Introduction

• However, the friction forces are limited in

magnitude and will not prevent motion if

sufficiently large forces are applied.

• The distinction between frictionless and

rough is, therefore, a matter of degree.



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Introduction

• There are two types of friction: dry or

Coulomb friction and fluid friction. Fluid

friction applies to lubricated mechanisms.

• The present discussion is limited to dry

friction between non-lubricated surfaces.



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The Laws of Dry Friction. Coefficients of Friction

• Block of weight W placed on

horizontal surface. Forces acting on

block are its weight and reaction of

surface N.

• Small horizontal force P applied to

block. For block to remain stationary,

in equilibrium, a horizontal

component F of the surface reaction

is required. F is a static-friction

force.



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• As P increases, the static-friction

force F increases as well until it

reaches a maximum value Fm.

NF sm

• Further increase in P causes the block to

begin to move as F drops to a smaller

kinetic-friction force Fk.

NF kk



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• Maximum static-friction force:

NF sm

• Kinetic-friction force:

sk

kk NF

75.0

• Maximum static-friction force and kinetic-friction force

are:

- proportional to normal force

- dependent on type and condition of contact surfaces

- independent of contact area



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• Four situations can occur when a rigid body is in contact with a

horizontal surface:

• No friction,

(Px = 0)

• No motion,

(Px < Fm)



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• Four situations can occur when a rigid body is in contact with a

horizontal surface:

• Motion impending,

(Px = Fm)

• Motion,

(Px > Fm)



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Angles of Friction

• It is sometimes convenient to replace normal force N and friction

force F by their resultant R:

• Impending

motion

ss

sms

N

N

N

F

tan

tan

• Motion impending



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Angles of Friction

• It is sometimes convenient to replace normal force N and friction

force F by their resultant R:

• Motion

kk

kkk

N

N

N

F

tan

tan



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Problems Involving Dry Friction

Impending motion at all points of contact

(body on verge of slipping)

No apparent Impending motion

Impending motion at some points of contact



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Impending Motion at All Points of Contact

8 - 17

Find the smallest angle q for which the 100 N

bar can be placed against the wall, without

slipping. (Five unknowns). Slipping occurs

simultaneously at A and B



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For a minute assume there was a roller at B. Then Na=mg and F

= (N-W/2)/tanq. F is large if theta is small. A limiting value will

reached. At larger theta F will be smaller than (mu NA) and will

not slip.



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Sample Problem 8.3

The moveable bracket shown

may be placed at any height

on the 3-cm diameter pipe.

If the coefficient of friction

between the pipe and bracket

is 0.25, determine the

minimum distance x at

which the load can be

supported. Neglect the

weight of the bracket.



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Sample Problem 8.3

The moveable bracket shown may be

placed at any height on the 3-cm

diameter pipe. If the coefficient of

friction between the pipe and bracket is

0.25, determine the minimum distance x

at which the load can be supported.

Neglect the weight of the bracket.



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Sample Problem 8.3



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SOLUTION:

•When W is placed at

minimum x, the bracket

is about to slip and

friction forces in upper

and lower collars are at

maximum value.

• Apply conditions for static equilibrium to find minimum x.



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SOLUTION:

• If x is larger than

minimum, by inspection

reactions NA and NB are

larger (moment due to W is

larger which tries to tilt the

bracket more creating larger

reactions). FA and FB do not

then reach their limiting

value. • Apply conditions for static equilibrium to find minimum x.



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When W is placed at minimum x, the

bracket is about to slip and friction forces

in upper and lower collars are at maximum

value.

BBsB

AAsA

NNF

NNF

25.0

25.0



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• Apply conditions for static equilibrium to find minimum x.

:0 xF 0 AB NNAB NN

:0 yF

WN

WNN

WFF

A

BA

BA

5.0

025.025.0

0

WNN BA 2



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:0 BM

05.1275.026

05.125.036

0cm5.1cm3cm6

xWWW

xWNN

xWFN

AA

AA

cm12x



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If x is larger, then slip will not occur at

any of the locations since Na is equal to

Nb and coeff. of friction is same.

Also, value of W does not matter. It

seems to decide the contact and friction is

related to reactions for slipping.



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Fig. shows an automobile on a road inclined at

an angle q with the horizontal. s =0.6, d

=0.5.

What is the maximum inclination the car can

climb at uniform speed. The car is rear wheel

drive and has a loaded weight of 3600 lb (1 ib

=4.5 N).

Assume tires do not slip (and spin) i.e pure

rolling at point of contact.



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•If drive wheel was allowed to slip we

replace s by d

8 - 32



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•Compute the torque needed by the drive

wheels to move the car at a uniform

speed up an incline where q =150. The

diameter of tire is 25 in. (1 inch =2.54

cm)



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No Apparent Impending Motion

8 - 36

Second type of problems: No Apparent

Impending Motion



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No apparent impending motion

6 unknowns, 6 equations



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It is not known whether motion is impending or not

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• All applied forces known

• Coefficient of static friction is known

• Determine whether body will remain at rest or

slide



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Equilibrium Versus Frictional Equations

8 - 39



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Problems where the friction force F is to

be an equilibrium force and satisfies the

inequality F < μsN then we can assume the

sense of direction of F on the free-body

diagram.

The correct sense is made known after

solving the equations of equilibrium for F.



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This is possible because the equilibrium

equations equate to zero the components

of vector acting in the same direction.



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However, in cases where the frictional

equation F = μsN is used in the solution of

a problem the convenience of assuming

the sense of F is lost.

Since the frictional equation relates only

the magnitude of two perpendicular

vectors. Consequently, F must always be

shown acting with its correct sense on the

free-body diagram.



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Sample Problem 8.1



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Sample Problem 8.1

A 100 N force acts as shown on a 300 N

block placed on an inclined plane. The

coefficients of friction between the block

and plane are s = 0.25 and k = 0.20.

Determine whether the block is in

equilibrium and find the value of the

friction force.



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•Determine values of friction force and

normal reaction force from plane required

to maintain equilibrium.

•Calculate maximum friction force and

compare with friction force required for

equilibrium. If it (maximum friction

force) is greater, block will not slide.



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•If maximum friction force is less than

friction force required for equilibrium,

block will slide. Calculate kinetic-

friction force.



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:0 xF

0N 300 - N 10053 F

N 80F

:0 yF

0N 300 - 54 N

N 240N



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•Calculate maximum friction force and

compare with friction force required

for equilibrium. If it is greater, block

will not slide.

m s mF N F 0.25 240 N 60 N

The block will slide down the plane.



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Sample Problem 8.1

•If maximum friction force is less than

friction force required for equilibrium,

block will slide. Calculate kinetic-

friction force.

N 240200

N

.

FF kkactual

N 48actualF



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Impending Motion at Some Points of Contact

8 - 51

Determine the horizontal force P needed to

cause movement. Weight of each member is

100 N.



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8 - 52

From the FBD 7 unknowns.

6 equilibrium equations and one static

friction equation



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Impending Motion at Some Points of Contact

8 - 53

As P increases it will either cause slipping

at A and no slipping at C, so that FA =

0.3NA and Fc ≤ 0.5Nc.

OR

slipping occurs at C and no slipping at

A, in which case Fc = 0.5Nc and FA ≤ 0.3NA

Calculate ‘P’ for each case and choose

the smaller P.



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Beam A B is subjected to a uniform load of 200 N/m and is

supported at B by post BC. If the coefficients of static

friction at B and C are μB = 0.2 and μC= 0.5. determine the

force P needed to pull the post out from under the beam.

Neglect the weight of the members and the thickness of the

beam.



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∑ MA=0 → NB = 400 N



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4 unknowns. So assumption on

slipping.



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4 unknowns. So assumption on slipping.



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Cylinder rolling from Assignment sheet



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Thrust Bearings. Disk Friction



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n Thrust bearings

Support the axial thrust of both horizontal as well as vertical shafts

Functions are to prevent the shaft from drifting in the axial direction and to transfer thrust loads applied on the shaft

Vertical thrust bearings also need to support the weight of the shaft and any components attached to it

The moving surface exerted against a thrust bearing may be the area of the end of the shaft or the area of a collar attached at any point to the shaft

Sorurce:web.iitd.ac.in/~rribeiro/MEL204170LubricationLec%20150211.ppt



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Plain thrust: Consists of a stationary flat bearing surface against which the flat

end of a rotating shaft is permitted to bear

ROTOR

Bearing surface

Flat end of rotor

Axial movement

Sorurce:web.iitd.ac.in/~rribeiro/MEL204170LubricationLec%20150211.ppt



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Consider rotating hollow shaft:



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21

22 RR

APr

AA

PrNrFrM

k

kk



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Thrust Bearings. Disk Friction

For full circle of radius R,

PRM k32

21

22

31

32

32

2

0

2

21

22

2

1

RR

RRP

drdrRR

PM

k

R

R

k

q



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Clutch and friction

8 - 67

The largest torque which can be

transmitted by a clutch without

slipping



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Belt Friction



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The normal and frictional

forces, acting at different

points along the belt vary

both in magnitude and

direction.

Due to this unknown

distribution, the analysis of

the problem will first

require a study of forces

acting on a differential

element of the belt.

Belt Friction



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Belt Friction

• Relate T1 and T2 when belt is about to slide to

right.



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Belt Friction

Draw free-body diagram for element of belt



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Belt Friction

02

cos2

cos:0

NTTTF sx qq

02

sin2

sin:0

qq

TTTNFy



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Belt Friction

•Combine to eliminate N, divide

through by q,

2

2sin

22cos

q

q

q

q

TT

Ts

•In the limit as q goes to zero,

0 Td

dTs

q



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Belt Friction

•Separate variables and integrate

from qq to0

se

T

T

T

Ts

1

2

1

2 orln



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Belt Friction

Note: Which is T2 and which is T1?

- T2 is the larger force and always

opposes friction.



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T2 is always greater than T1. T2

represents tension in the part of belt which

pulls while T1 is tension in part which

resists.

Angle of contact is expressed in radians.

may be larger than 2, if a rope is

wrapped ‘n’ times around a post

= 2n



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Sample Problem 8.8

A flat belt connects pulley A

to pulley B. The coefficients

of friction are s = 0.25 and k

= 0.20 between both pulleys

and the belt.

Knowing that the maximum

allowable tension in the belt is

3000 N, determine the largest

torque which can be exerted

by the belt on pulley A.



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Sample Problem 8.8



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Sample Problem 8.8

A flat belt connects pulley A to pulley B.

The coefficients of friction are s = 0.25

and k = 0.20 between both pulleys and

the belt.

Knowing that the maximum allowable

tension in the belt is 3000 N, determine

the largest torque which can be exerted by

the belt on pulley A.



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Sample Problem 8.8 Solution

8 - 82

•Since angle of contact is

smaller, slippage will

occur on pulley B first.

Determine belt tensions

based on pulley B.



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Sample Problem 8.8 Solution

8 - 83

N317776881N3000

6881N3000

1

32250

112

../T

.eT/eT/T .s



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Sample Problem 8.8

8 - 84

•Taking pulley A as free-body, sum

moments about pulley center to

determine torque. 0N3000N31777m200 ..M:M AA

mN5244 .MA



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Sample Problem 8.8

8 - 85

•Check that belt does not slip at A. =

240o = 4/3 rad

5240125052403

4

5240N31777

N3000

1

2

...

..

lnT

Tln

ss

s



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Angle of wrap

8 - 87



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Angle of lap decided by

taking tangent at the points

of contact.



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Wedges

• Wedges - simple machines used to

raise heavy loads.

• Force required to lift block is

significantly less than block weight.

• Friction prevents wedge from sliding

out.

• Want to find minimum force P to

raise block.

• Block as free-body

0

:0

0

:0

21

21

NNW

F

NN

F

s

y

s

x

or

021 WRR



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Assume impending slip at A and

no slip at B



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Wedges

06sin6cos

:0

0

6sin6cos

:0

32

32

s

y

ss

x

NN

F

P

NN

F

• Wedge as free-body

or

032 RRP

eighth edition - apl100apl100.wdfiles.com/local--files/notes/friction revised.pdf · eighth edition...

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