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5. Friction

2142111 S i 2011/22142111 Statics, 2011/2

© Department of Mechanical Engineering Chulalongkorn UniversityEngineering, Chulalongkorn University

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Objectives Students must be able toObjectives Students must be able to

Course ObjectiveInclude friction into equilibrium analyses Include friction into equilibrium analyses

Chapter Objectivesp jFor general dry friction Describe the mechanism and characteristics of dry friction,

coefficient of friction and angle of frictioncoefficient of friction and angle of friction Specify and determine the limiting conditions for sliding, toppling,

free rolling and driven wheels under dry friction by appropriate FBDs

Specify the limiting conditions for rear/front wheel drives by appropriate FBDs

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pp p Analyze bodies/structures with friction for unknown

loads/reactions by appropriate FBDs

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Objectives Students must be able toObjectives Students must be able to

Course ObjectiveInclude friction into equilibrium analyses Include friction into equilibrium analyses

Chapter Objectivesp jFor friction in machines Prove, state limitation and apply formula for impending motion in

wedges single continuous threads/screws flat beltswedges, single continuous threads/screws, flat belts, disks/clutches, pivot/collar/thrust bearings, journal bearings

Analyze machines with friction

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ContentsContents Dry friction

Characteristics theory coefficient and angle of friction Characteristics, theory, coefficient and angle of friction Applications

Wedgesg Threads, screws Belts

Di k d Cl t h Disks and Clutches Collar, pivot, journal and thrust bearings

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FrictionDry Friction

Friction Force of resistance acting on a body which prevents or

retards slipping of the body relative to a surface withretards slipping of the body relative to a surface with which it is in contact. The frictional force acts tangent to the contacting

f i di ti d t th l ti tisurface in a direction opposed to the relative motion or tendency for motion of one surface against another

Two types of friction Fluid friction exists when the contacting surfaces are

separated by a film of fluidseparated by a film of fluid. Dry friction occurs between contacting surfaces

without lubricating fluid.

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Dry Friction Theory #1

Dry Friction

Dry Friction Theory #1

Friction forces are tangential forces generated between contacting surfacescontacting surfaces.

Friction forces arise in part from the interactions of the roughnesses or asperities of the contacting surfaces.

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EquilibriumDry Friction

Equilibrium Slipping and/or tipping effects

Frictional force F increases with force P Frictional force F increases with force P.

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Static Friction Equilibrium

Impending MotionDry Friction

Static Friction Equilibrium

Maximum value of frictional force Fs on the object in equilibrium is called the limiting static frictional forceequilibrium is called the limiting static frictional force.

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Static Friction Coefficient

Dry Friction Impending Motion

Static Friction Coefficient

F Nμ=s sF Nμ=

maximum static frictional forcecoefficient of static friction

s

s

Fμ

==

normal forceN =

is the friction forces that can be exerted by dry contacting surfaces

maximumsF

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that can be exerted by dry, contacting surfaces that are relative tstation o each ot rary he .

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Static Friction Angle of Static Friction


g

N R φcossin

sin

s s

s s s s

N RF R N

F R

φφ μ

φ

== =

sin tancos

s ss s

s

F RN R

φμ φφ

= = =

φ μ−= 1tans sφ μs s

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Static Friction Typical Values


Static Friction Typical Values

Contact Materials μs

Metal / ice 0.03 – 0.05Wood / wood 0.30 – 0.70Leather / wood 0.20 – 0.50Leather / metal 0.30 – 0.60Al i / Al i 1 10 1 70Aluminum / Aluminum 1.10 – 1.70

sHow to obtain values of ?μ

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sμ

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Frictional ForcesDry Friction

Frictional Forces More force is required to start sliding than to keep it

sliding can be explained by the necessity to break thesesliding can be explained by the necessity to break these bonds.

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Kinetic Friction Coefficient & Angle

MotionDry Friction

Kinetic Friction Coefficient & Angle

μ=F Nμ=k kF N

kinetic frictional forcekF =coefficient of kinetic frictionnormal force

k

k

Nμ =

=

φ μ−= 1tanφ μ= tank k

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Kinetic Friction Angle of Friction

Dry Friction Motion

Kinetic Friction Angle of Friction

At , is impending.At f

sliplidi

sφ φφ φ

=φ φ≥s k

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At , surfa slidinces are relative to eac

gh other.

kφ φ=φ φs k

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Dry Friction Characteristics

Dry Friction

Dry Friction Characteristics

Frictional force acts tangentially to the contacting surfaces opposing the relative or tendency for motionsurfaces, opposing the relative or tendency for motion.

Fs is independent of the area of contact, provided that the normal pressure is not very low nor great enough for deformation of the surfaces.

In equilibrium:Impending slipping: φ = φs

Slipping: φ = φk

Very low velocity: φk ≈ φs

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MotionDry Friction

SlidingRelative sliding (translation motion) between two Relative sliding (translation motion) between two surfaces

Toppling Fall over (rotation) about the edge

Topple tipping rolling tumble trip Topple, tipping, rolling, tumble, trip

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Example Friction 1 #1

Dry Friction


Place a sheet of one materials on a flat board. Bond a sheet of other material to block A Place the block onto thesheet of other material to block A. Place the block onto the board, and slowly increase the tilt a of the board until the block slips. Show that the coefficient of static friction between the two materials is related to the angle a by μ =between the two materials is related to the angle a by μs = tanα

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Dry Friction


α = − = The block is about to slip

0 cos 0yF N WW

αα

=

= − + =

cos (1)

0 sin 0

y

x s

N W

F F Wα

A

μ αμ α

=

=

sin (2)

(2) /(1) tan Ans

x s

s

s

N WN Fs

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Dry Friction


Will this crate slide or topple over?topple over?

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Dry Friction


The crate is about to slide.

s sF F Nμ= =

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Dry Friction


The crate is about to topple.

( )s sF F Nμ<

The crate is about to slide and toppletopple.

s sF F Nμ= =

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Example Hibbeler Ex 8-1 #1

Dry Friction

Example Hibbeler Ex 8 1 #1

The uniform crate has a mass of 20 kg. If a force P = 80 N is applied to the crate determine if it remains in equilibrium Theapplied to the crate, determine if it remains in equilibrium. The coefficient of static friction = 0.3.

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Dry Friction


= → + ° − = → =

0

(80 N)cos30 0 69.3 NxF

F F = ↑ + − ° − + = → = 0

(80 N)sin30 196.2 N 0 236.2 Ny

C C

F

N N

= + ° − ° + = →

0

(80 N)sin30 (0.4 m) (80 N)cos30 (0.2 m) ( ) 0 O

C

M

N x

−= − × = −39.0753 10 m 9.08 mm

0 4 Th t ill

x

t t l

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<< →0.4 m The crate willxμ< →

not topple. (70.8 N) The crate, is close to but doesn't not slip.s CF N

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Dry Friction


The rod with weight W is about to slip on rough surfaces at A and Bslip on rough surfaces at A and B. Find coefficient of static friction.

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Dry Friction


Impending slipF N

3 equilibrium equations

A s A

B s B

F NF N

μμ

==

3 equilibrium equations,3 unknowns ( , , )A B sN N μ

0 cos30 sin30 0 (1)

0 cos30 sin30 0 (2)x s A s B B

y A B s B

F N N N

F N W N N

μ μ

μ

= → + + ° − ° = = ↑ + − + ° + ° =

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0 cos30 ( / 2) 0 (3)A BM N l W l = + − ° =

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Dry Friction

p

=From (3) 0 4330N W

μ μμ

=

− +=

2

From (3) 0.4330

From (1) & (2) 0.2165 0.21650.228 Ans

B

s s

s

N W

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μ 0.228 Anss

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Dry Friction


Find minimum coefficients of static friction that keep the system infriction that keep the system in equilibrium.

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Dry Friction


Symmetry of lower pipes about Symmetry of lower pipes about vertical axis passing through O.

Two types of contacts: Pipe – pipe Pipe – ground

-

, ,A B A B

A s p p A

N N F FF Nμ

= =≤

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, -

, -

A s p p A

C s p g CF N

μμ≤

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Example Bedford Ex 9.5 #1

Dry Friction


Suppose that α = 10° and the coefficient of friction betweencoefficient of friction between the surface of the wedge and the log are μs = 0.22 and μk = 0 20 Neglect the weight of the0.20. Neglect the weight of the wedge. If the wedge is driven into

th l t t t t bthe log at a constant rate by vertical force F, what are the magnitude of the normal f t d th l bforces exerted on the log by the wedge?

Will the wedge remain in

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place in the log when the force is removed?

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Example Bedford Ex 9.5 #2Driven at constant rateDry Friction


= ↑ + Symmetry about the center of the wedge

0 yF

α μ α

α μ α

+ − =

=+

2 sin( / 2) 2 cos( / 2) 0

2sin( / 2) 2 ( / 2)

kN N FFN

α μ α+

=° + °

2sin( / 2) 2 ( / 2)

2sin(10 / 2) 2(0.20)(10 / 2)

k

FN

=

( ) ( )( )

1.75 AnsN F

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Example Bedford Ex 9.5 #3On the verge of slippingDry Friction


↑ + Symmetry about the center of the wedge

0F

α μ αμ α

′

= ↑ + + =

= =

0

2 sin( / 2) 2 cos( / 2) 0tan( / 2) 0 087

y

s

F

N Nμ α

μ μ

′

′

= =

<

tan( / 2) 0.087

As , wedge will remain in place. Ans

s

s sμ μ g ps s

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Example Bedford 9.20 #1

Dry Friction

pThe coefficient of static friction between the two boxes andbetween the two boxes and between the lower box and the inclined surface is μs. What is th l t l f hi h ththe largest angle α for which the lower box will not slip.

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Dry Friction


α = − = 2

FBD of upper block

0 cos 0yF N W

α=

2 cosFBD of lower block

0 0

N W

F N W Nα

α

= + − = =

2 1

1

0 cos 0

2 cos

0 i 0

yF N W N

N W

F N N Wμ μ α

μ α α−

= + − = =

2 1

1

0 sin 0

3 cos sin

tan / 3 or tan (3 ) Ans

x s s

s

F N N W

W W

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μ α α μ= = 1tan / 3 or tan (3 ) Anss s

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Dry Friction


The cylinder has weight W. The coefficient of static frictioncoefficient of static friction between the cylinder and the floor and between the cylinder

d th ll i Wh t i thand the wall is μs. What is the largest couple M that can be applied to the stationary

li d ith t i it tcylinder without causing it to rotate.

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Dry Friction


μ ↑ + + Impending rotation

0 0F N N Wμ

μ

= ↑ + + − = = → + − =

0 0

0 0y s w f

x w s f

F N N W

F N N

μ = + − + = 0 ( ) 0O s w sM M N N r

μsW WN Nμ

μ μμμ

= =+ +

+=

2 2, 1 1

(1 ) Ans

sw f

s s

s

WN N

M Wr

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μμ

=+ 2 Ans

1ss

M Wr

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Dry Friction


The disk of weight W and radius R is held in equilibrium on theR is held in equilibrium on the circular surface by a couple M. The coefficient of static friction b t th di k d thbetween the disk and the surface is μs. Show that the largest value M can have

ith t i th di k t liwithout causing the disk to slip is

sRWM

μ21

s

s

Mμ

μ=

+

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Dry Friction

p

Impending slip s sF Nμ=p g p

0 sin 0

0 cos 0

s s

x sF F W

F N W

μ

α

α

= + − = = + − =

0 cos 0

0 0y

O s

F N W

M M F R

α + = + − + =

2

tan , cos

cos 1 tan 1, tan =s N Wμ α α

α α α μ

= =

+ = scos 1 tan 1, tan

cos s sM NR WR

α α α μ

μ μ α

+

= =

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21s sWM Rμ μ= +

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Dry Friction

pEach of the uniform 1-m bars has a mass of 4 kg. The coefficient of static friction between the bar and the surface at Bcoefficient of static friction between the bar and the surface at Bis 0.2. If the system is in equilibrium, what is the magnitude of the friction force exerted on the bar at B.

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Dry Friction

p

= 0xF + =

=

0

0 x x

y

O A

F + − =

= 4 0

0

y

y y

O

O A g

M

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° − ° − ° =

(1 m)cos 45 ) (1 m)sin45 ) (4 N)(0.5 m)cos 45 ) 0y xA A g

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Dry Friction

p

= − + ° − ° =

0

cos30 sin30 0x

x

F

A F N

= − − + ° + ° =

0

(4 N) sin30 cos30 0y

y

F

A g F N

= ° + ° + ° =

0

(1 m)cos 45 ) (1 m)sin45 ) (4 N)(0.5 m)cos 45 ) 0B

y x

M

A A g

40= = − = = = =0, 2 N, 4 N, 2 N, 438 N, 2.63 N Ansy x y xA A g O g O g N F

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Dry Friction

p

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Dry Friction


The coefficient of static friction between the tires of the 1000-kg tractor and the ground and between the 450-kg crate and thetractor and the ground and between the 450-kg crate and the ground are 0.8 and 0.3, respectively. Starting from the rest, what torque must the tractor’s engine exert on the rear wheels to cause the crate to move? The front wheels can turn freely.the crate to move? The front wheels can turn freely.

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Dry Friction


μ= ,

The crate is about to move.

s cF N

= − + =

0

(450 N) 0yF

g N =

− = 0

0xF

P F

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= =0.3(450 N) 135 NP g g

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Dry Friction


2

0

2 0xF

P F

= − + =

2 67.5 NF g=

= 0

(0 8 ) 0OM

F T

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− == = ⋅

2(0.8 m) 0(0.8 m)(135 N) 1.06 kN m Ans

F TT g

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Dry Friction

Example Bedford 9.165 #1The mass of the vehicle is 900 kg, it has rear-wheel drive, and the coefficient of static friction between its tires and the surface is 0.65. The coefficient of static friction between the crate and the surface is 0.4. If the vehicle attempts to pull the crate up the incline, what is the largest value of the mass of the crate for which it will slip up the i li b f th hi l ’ ti li ?incline before the vehicle’s tires slip?

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Dry Friction


μ=2 , 2

Impending motion:

s tF N

= − ° + =

2

0

cos20 2 0xF

P F

= + − − ° =

2 1

0

2 2 (900 N) sin20 0yF

N N g P

= + ° + ° − =

2

0

(900 N)(1 m) cos20 (0.8 m) sin20 (3.7 m) 2 (2.5 m) 0OM

g P P N

46= = = =2 2 1563.57 N, 2 814.75 N, 2 529.59 N, 2 2P g N g F g N 78.00 Ng

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Dry Friction


μ=

= ,The crate is about to move.

0 s c

y

F N

F

− ° + =

= cos20 0

0x

W N

F

− − ° =sin20 0

1 3930 N 7701 1 N

P F W

W P= == =

1.3930 N 7701.1 N/ 785 kg Ans

W Pm W g

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Dry Friction


What is the maximum tension in the cord that the tractor may tow a crate over a ramp with 45° angle of incline before the tractora crate over a ramp with 45° angle of incline before the tractor topples.

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Dry Friction


=1Impending topple: 0N

=

− =

1p g pp

0

(2.0 m) / 2 (900 N)(1.5 m) 0

QM

P g =

− − =

(2.0 m) / 2 (900 N)(1.5 m) 0

0

2 (900 N) / 2 0

y

P g

F

N g P =

=

+ =

22 (900 N) / 2 0

0

/ 2 2 0

x

N g P

F

P F

μ≤

=2 , 2, tires do not slip.

955 N Anss tF N

P g

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− + == = =

2

2 2

/ 2 2 0954.59 N, 2 1575 N, 2 675 NP F

P g N g F g

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Friction in MachinesMachine Friction

Friction in Machines Friction helps some machines function.

Machines with frictions Wedgesg Threads, screws Belts

Di k d l t h Disks and clutches Bearings

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WedgesMachine Friction

Wedges Wedges exert a large lateral forces from the faces as a

result of small angleresult of small angle.

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Wedges Analysis #1

Machine Friction

Wedges Analysis #1

The block is about to raise. The slip of the load and wedge are impending.

2 2 3 3,

0

s sF N F N

F

μ μ= =

→ + 3 2 2

0

sin cos 0

0

x

s

F

N N N

F

θ μ θ

= → + − + + =

↑ +

2 2 3

0

cos sin 0y

s s

F

N N N Wθ μ θ μ

= ↑ + − − − =

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Wedges Analysis #2

Machine Friction

g y

FBD of the wedge

0

i 0xF

N N N P

= → + 2 2 1sin cos 0

0

i 0

s s

y

N N N P

F

N N N

θ μ θ μ− − − + =

= ↑ + 1 2 2cos sin 0sN N Nθ μ θ− + =

( )( )

2

2

1 tan 2s sP Wμ θ μ − +

=

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( )21 2 tans sμ μ θ

− −

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Threads/ScrewsMachine Friction

Threads/Screws Assumption: the shaft has a single continuous thread.

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Threads Geometry

Machine Friction

Threads Geometry

1tan2

lr

θπ

−=lead anglepitch of the thread (lead)

di f th th dlθ =

=

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mean radius of the threadr =

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Threads Motion

Machine Friction

Threads Motion

W W

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Threads Moving Upwards

Machine Friction

Threads Moving Upwards

θ φ

= → + 0

i ( ) 0xF

S R θ φ

θ φ

− + =

= ↑ + +

sin( ) 0

0

cos( ) 0y

S R

F

R Wθ φ

θ φ

+ − =

= + =

cos( ) 0

tan( ) and

R W

S W M Srθ φ+tan( ) and S W M Sr

θ φ= +tan( )M Wr

1On the verge of rotating: tans sφ φ μ−= =

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1Uniform upwards motion: tank kφ φ μ−= =

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Threads Moving Downwards #1

Machine Friction


θ φ

= → + ′ 0

i ( ) 0xF

S R θ φ

θ φ

′ − − =

= ↑ + sin( ) 0

0

cos( ) 0y

S R

F

R Wθ φ

θ φ

− − =

′ ′ ′= − =

cos( ) 0

tan( ) and

R W

S W M S rθ φtan( ) and S W M S r

θ φ′ = −tan( )M Wr

1On the verge of rotating: tans sφ φ μ−= =

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1Uniform downwards motion: tank kφ φ μ−= =

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Machine Friction


is removed:Mis supported by friction alone,

providing that W

φ θ≥

Self-locking condition

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Machine Friction


φ θ

= → + ′′ 0

i ( ) 0xF

S R φ θ

φ θ

′′− + − =

= ↑ + sin( ) 0

0

cos( ) 0y

S R

F

R Wφ θ

φ θ

− − =

′′ ′′ ′′= − =

cos( ) 0

tan( ) and

R W

S W M S r

Very rough surface :Reversed motion.

θ φ<

φ θtan( ) and S W M S r

φ θ′′ = −tan( )M Wr

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Example Wedges/Threads 1 #1

Machine Friction


The mass of block A is 60 kg. Neglect the weight of the 5° wedge. The coefficient of kinetic friction between the contacting surfacesThe coefficient of kinetic friction between the contacting surfaces of the block A, the wedge, the table, and the wall is 0.4. The pitch of the threaded shaft is 5 mm, the mean radius is 15 mm, the coefficient of kinetic friction between the thread and the matingcoefficient of kinetic friction between the thread and the mating groove is 0.2. What couple must be exerted on the threaded shaft to raise the block A at a constant rate?

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Machine Friction


D FBD f th d th dADraw FBD of the mass and the wedgeThen, follow the same procedure for wedge analysis:

A

( )( )

21 1

21 1

wedge analysis:

1 tan 2

1 2 tank k

k k

F Wμ α μ

μ μ α

− + =

− − ( )1 11 2 tan

Substitute for all known values:667.50 N

k k

F

μ μ α

=

60 N0 N

L

w

W gW

==

1

50.4 for all

non thread contactsk

αμ

= °=

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non-thread contacts

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Machine Friction


5 mml

2

5 mm15 mm0.2 for thread/groovek

lrμ

===2

2

garctan( ) 11.310arctan(/ 2 ) 3.0368

k

k k

r

μφ μθ π

= = °= = °

Draw FBD of the thread

θ φ= +° + °

2

Find that will push the thread upwards:tan( )

(667 50 N)(0 015 m) tan(3 0368 11 310 )k

MM FrM

63

= ° + °= ⋅ = ⋅

(667.50 N)(0.015 m) tan(3.0368 11.310 )2.5609 N m 2.56 N m Ans

MM

64

Flat BeltsMachine Friction

Flat Belts Involve slippage of flexible cables,

belts and ropes over sheaves andbelts and ropes over sheaves and drums.

64

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Flat Belts Analysis #1

Machine Friction


Consider line elementConsider line element

0 tF

d dθ θ

= cos ( )cos 0

2 2

cos

d ddN T T dT

ddN dT

θ θμ

θμ

+ − + =

= cos2

0 n

dN dT

F

d d

μ

θ θ

= sin ( )sin 0

2 2

2 sin sin

d ddN T T dT

d ddN T dT

θ θ

θ θ

− − + =

+

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2 sin sin2 2

dN T dT= +

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Machine Friction

Flat Belts Analysis #2(motion impending)(ongoing movement)

sμ μμ μ

==

0, cos 1:2

dd θθ → →

(ongoing movement)kμ μ=

cos2

ddN dT dT

d d

θμ

θ θ

= ≈

0, sin :2 2

2 sin

d dd

ddN T Td

θ θθ

θ θ

→ →

= ≈2 sin2

T

dN T Td

TdT dT β

θ= ≈2 1T T eμβ=

66

2

1

20

1

lnT

T

TdT dTd dT T T

βμ θ μ θ μβ= → = → =

67

Example Flat Belts 1 #1

Machine Friction


F i i i h

=

Friction resists the lowering of :mT mg

β π

== ==

2

1 / 61.25(2 )

T mgT P mg

A force P = mg/6 is required to lower the cylinder at a constant slow speed with the cord

μβ

β

= 2 1

( )

T T eslow speed with the cord making 1.25 turns around the fixed shaft. Calculate the coefficient of friction bet een

πμ

μ

= 2.5

60 228 Ans

smgmg e

67

coefficient of friction μs between the cord and the shaft.

μ = 0.228 Anss

68


Machine Friction


Calculate the force P on the h dl f th diff ti l b dhandle of the differential band brake that will prevent the flywheel from turning on its shaft to which the torque M = 150 N·m is applied. The coefficient of friction betweencoefficient of friction between the band and the flywheel is μ= 0.40.

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69


Machine Friction


Friction resists the attempt of to rotate M

μβ μ π = = (7 / 6)

2 1 2 1

pthe flywheel in the clockwise direction:

(1)T T e T T e = − ⋅ + − =

− = 2 1

2 1

0 (150 N m) (0.15 m)( ) 0

1000 N (2)centerM T T

T T

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= =2 1

1 2Solve (1) and (2) 300 N, 1T T 300 N

70


Machine Friction


= + 0 OM

− + ° + ==

2 1(0.15 m) (0.075 m)( sin30 ) (0.45 m) 0408 N Ans

T T PP

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Disks & ClutchesMachine Friction

Disks & Clutches Use to connect and disconnect two

coaxial rotating shaftcoaxial rotating shaft The shaft can support a couple

due to friction forces between the di kdisks.

Determine the couple by using coefficient of static friction in the flat-ended thrust equation.

71Draw forces on the plane only

72

Disks & Clutches Analysis #1

Machine Friction

Disks & Clutches Analysis #1

2dA r drπ=

μ= 2M PR

2

0 0

2

x

R

F P w dA

P p r dr p Rπ π

= − = μ

3M PR

2

0 0o

x

P p r dr p R

M M wr dA

π π

μ

= =

= − =

72

322

3R

oM pr r dr p Rμ π μ π= =

73

Disks & Clutches Worn Analysis #1

Machine Friction


Distribution of forces on worn plates are not uniformed. For example, linear distribution

2dA r drπ

At

2At , 0

0,

dA r drr R wr w p

π== == =

73

,( ) /

pw p R r R= −

74


Machine Friction


Consider the left plate

2

0 0

( )2 /

x

R

F P w dA

pP p R r r dr R Rππ

= − =

= − =

( )2 /3

0 0

o

x

P p R r r dr R R

M M wr dA

π

μ

= − =

= − =

( )2 /R

oM pr R r r dr Rμ π= −

μ= 12

M PR

74

2

75

Bearings Pivot Bearings

Machine Friction

Bearings Pivot Bearings

μ= 23

M PR3

(motion impending)μ μ= (motion impending)(ongoing movement)

s

k

μ μμ μ

==

75

76

Bearings Collar Bearings

Machine Friction

Bearings Collar Bearings

μ −=

3 3

2 2

23

i oR RM P

R R−3 i oR R

(motion impending)μ μ= (motion impending)(ongoing movement)

s

k

μ μμ μ

==

76

77

Bearings Thrust Bearings #1

Machine Friction


RRi Ro

2 2 dRdA R d R dR

( )2 2

2 2cos

2oR o i

dRdA R ds R

R RRA dA dR

π πθππ

= =

−

PRoRi R

θ

( )cos cosiR

A dA dRθ θ

= = =

77

78


Machine Friction


= 0F

PRoRi

dR

Rθ xθ

= − = 0

cos 0xF

P pAP P

( )θ π= =

−2 2cos o i

P PpA R R

μ

=

− =

0

( ) 0

xM

M R pdA

( )

μ

πμθπ

=−

2 2

( )

2cos

o

i

R

Ro i

p

P RM R dRR R

μθ −

= −

3 3

2 2

23cos

o i

o i

R RPMR R

78

( )

79

Bearings Journal Bearings #1

Machine Friction


P P

φ

79

80


Machine Friction


0

( sin ) 0zM

M P R φ

= + − =

R RRf

sintan

PMM R

PRφφ

=≈

R R

φ μ= ≈sinM PR PR

80

81

Example Hibbeler 8-11 #1

Machine Friction

Example Hibbeler 8 11 #1

The 100-mm-diameter pulley fit l l 10fits loosely on a 10-mm-diameter shaft for which the coefficient of static friction is μss= 0.4. Determine the minimum tension T in the belt needed to (a) raise the 100-kg block and(a) raise the 100-kg block and (b) lower the block. Assume that no slipping occurs between the b lt d ll d l t thbelt and pulley and neglect the weight of the pulley.

81

82

Example Hibbeler 8-11 #2CW rotation

Machine Friction

p

Raise the blockφ −= = =

= +

1

Raise the blocksin (5 mm)sin(tan 0.4) 1.8570 mm

0f sr r

M = + + − − =

=

2

0

(981 N)(50 mm 1.8570 mm (50 mm 1.86 mm) 01056 8 N

PM

TT

82

==

1056.8 N1.06 kN Ans

TT

83

Hibbeler Example 8-11 #3CCW rotation

Machine Friction

Hibbeler Example 8 11 #3

φ −= = =

1

Lower the blocksin (5 mm)sin(tan 0.4) 1.8570 mmf sr r

= + − − + =

2

0

(981 N)(50 mm 1.86 mm) (50 mm 1.86 mm) 0P

M

T

83

==

910.63 N911 kN Ans

TT

84

Example Boresi 10.9 #1

Machine Friction


84

85


Machine Friction


A weight W is lifted at a constant rate by applying a couple Rd to theby applying a couple Rd to the compound jackscrew. Derive a formula for R in terms of

the angle θ, the weight W, the length d of the lever rod, the pitch pof the screw threads the meanof the screw threads, the mean radius r of the screw threads, and the coefficient of kinetic friction μkof the screw. The friction of the vertical wall guides is negligible.

For θ = 15° W = 1000 lb d = 20 in

85

For θ 15 , W 1000 lb, d 20 in, r = 0.75 in, p = 0.333 in, and μk = 0.15, calculate R.

86


Machine Friction


0 2 cos 0yF T Wθ = − =

86

2 cos (1)W T θ=

87


Machine Friction


0 cos os 0yF Q Tc

Q T

θ θ = − = =

0 2 sin 0

2 sin (2)xF T P

P T

θ

θ

= − = =

(1) (2) tanP W θ+ =

By symmetry, FBD of the left nut is mirror image of the right nut FBD.

87

88


Machine Friction

p

[ ]Pr

φ αφ α

= +

= +

tan( )2 tan( )

kM WrRd Pr φ α

θ μπ

− −

= +

= +1 1

2 tan( )

2( tan ) tan(tan tan ) /2

k

k

RdpR W r d

r

Substitute numbers in the equation4 48 lb AR

88

= 4.48 lb AnsR

89

Example Hibbeler 8-120 (modify) #1

Machine Friction

Example Hibbeler 8 120 (modify) #1

The axle of the pulley fits l l i 50 di tloosely in a 50-mm-diameter pinhole. If μk = 0.30 between the pinhole and the pulley axle and 0.20 between the pulley and the cord, determine the minimum tension T required tominimum tension T required to turn the pulley counterclockwise at a constant velocity if the bl k i h 60 N N l t thblock weighs 60 N. Neglect the weight of the pulley.

89

90

Example Hibbeler 8-120 (modify) #2Case I Journal Bearing

Machine Friction


The shaft is about to rotate1 1

,

2

The shaft is about to rotate.tan tan 0.3 16.699

sin 7.1837 mmk b

fr rφ μ

φ

− −= = = °

= =2-1

1 1

11

tan ( / ) 45

sin ( / 2 ) 2.9117

f

f

r r

r r

φθ

β −

= = °

= = °1( )

0 60 sin( ) 0

f

yF R

β

θ β = − + + = 0 cos( ) 0

80.850 N, 54.192 NxF T R

R T

θ β = − + =

= =

90

91

Example Hibbeler 8-120 (modify) #3Case I Journal Bearing

Machine Friction


If the cord does not slip over the pulley,

μ β

μ

= ,

, 1

2 1

the minimum is:k f

k f

T T eμ π

μ

==

, 1( / 2)

, 1

60 N (54.192 N)0.064815

k f

k f

e

μ <, 1 0.2, the assumption the thek b

=

cord does not slip is valid.

54 2 N AnsT

91

= 54.2 N AnsT

92

Example Hibbeler 8-120 (modify) #4Case II Flexible Belt

Machine Friction


Th b l i b lidμ β =

,2 1

The belt is about to slide.k fT T e

π==

0.2( / 2)60 N43.824 N

TeT

92

93

Example Hibbeler 8-120 (modify) #5Case II Flexible Belt

Machine Friction


If the shaft does not rotate,

θ β

θ β

= − + + = = − + =

0 (60 N) sin( ) 0

0 cos( ) 0y

x

F R

F T R β

θ β

= = ° = ° ( )

43.824 N, 45 , 8.8556x

T

βφ φ

−= → == → = °

11

2

sin ( / 2 ) 21.771 mmsin 60.557

f f

f

r r rr r

φ μ μμ

−= → =

>

1, 1 , 1

, 1

tan 1.770.3, the assumpt

k b k b

k b ion that the

93

shaft does not rotate is invalid.

94

Example Boresi 10-86 #1Example Boresi 10 86 #1

A torque T is applied to pulley A, which drives pulley B. The t ll h l dii Th t i i th l k id ftwo pulleys have equal radii. The tension in the slack side of the belt is 2 kN. The coefficient of friction between the belt and the pulleys is 0.30. What is the maximum possible torque that can be applied

to pulley A?? What torque is transmitted to pulley B?

94

95

Example Boresi 10-86 #2

Machine Friction

p=2Given 2 kN

The belt is about to slip:T

μβ

π

= = =

1 2

0.3

The belt is about to slip:

(2 kN) 5 1327 kN

T T e

T e= ==

1

1

(2 kN) 5.1327 kN5.13 kN Ans

T eT

= − + − =

1 2

0

( ) 0OM

T T T r= − = ⋅1 2( )(0.1 m) 0.31327 kN m

As tensions in the belts remain T T T

95

unchanged between the two pu= = ⋅

lley, 313 N m AnsBT T

96


Machine Friction

pIn some applications of collar bearings,

lti l ll d T llmultiple collars are used. Two collars are used as shown to support the 10,000 lb load (μs = 0.40 and μk = 0.20)s k

Determine the torque T required to initiate rotation of shaft S, Assume that the load is divided equallythat the load is divided equally between the collars and that the pressure on each collar is uniform.Wh lti l ll d? Why are multiple collars used?

96

97


Machine Friction

Example Boresi 10 98 #2

The load is divided equally between the collarsThe load is divided equally between the collars,thus, the load on each collar is 5 kip.

μ −

= 3 32 12 2

For each collar that is about to rotate:

2s

R RT Pμ − −=

2 22 1

3 3

3

3

2 (1 ft) (0.5 ft)0.4(5000 lb)3 (1 ft) (0 5

s R R

T = ⋅3 1555.6 lb ftft)−3 (1 ft) (0.5

= = ⋅

ft)

Total required torque 2 3.11 kip ft AnsT M

97

q q pLess load on each collar, less wear and tear. Ans

98

Concepts #1

Review

Concepts #1

Friction is the tangent resistance force which prevents or retards slipping of the body relative to a surface withor retards slipping of the body relative to a surface with which it is in contact. Maximum friction occurs that the verge of impending motion which can be slipping or toppling or both togethertoppling or both together.

Some machine functions by friction. To use the formula, the physical meanings underlying the friction in machines must be understood.

98

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