force of friction

8/12/2019 force of friction

1/16

FRICTION FORCE MODELS

When two bodies are in contact with each other, with or without relative motion between,

then fiction forces come into account. This forces result in a loss of energy which is dissipated

in the form of heat. Friction between mating parts will cause wear. In some types of machines

and processes we desire to minimize the retarding effect of friction forces. Examples are

bearings of all types, power screws, gears, fluid flow in pipes and ect. In other situations we

wish to maximize the use of friction, as in brakes, clutches, belt drives and wedges. Ordinary

walking depends on friction between the shoe and ground. Friction force are present

throughout the nature and exist to a considerable extend in all machines no matter how

accurately or carefully lubricated.

Dry Friction

Dry friction is encountered when the unlubricated surface of two solids are in contact under a

condition of sliding or tendency to slide. This type of friction is also called Coulombs

friction.

Fluid Friction

Fluid friction is developed when adjacent layers in a fluid are moving at different velocities. It

may be either viscous or turbulent.

Internal Friction

Solid damping or hysteretic damping is caused by internal friction or hysteresis, when a solid

is deformed.


2/16

Example:The Withworth quick return mechanism is acted upon by a 10 Nm external torque

acting on crank BC in CW direction. Calculate the amount of torque required on crank AB at

the instant when crank AB starts rotating CCW, If the coefficient of friction between the

mating surfaces of the prismatic joints is 0,3.

AB = AC = 10 cm.

Solution:

By definition friction force always opposes the motion, in order to put the friction force in

proper direction we need velocity analysis. The velocity of the crank pin B as a point on AB

link is easily found, so that B will be used as a reference point for determining the relative

velocity of link 3 and link 4. The relative velocity equation may now be written;

4/343 VVVvvv

+=

3Vv

is normal to AB link and direction is dictated by problem definition.

4Vv

is normal to BC link.

4/3Vv

is along BC link.

A

B

C60

Text= 10 N.mT

1

2

3

4

A

T


3/16

We now complete the sketch of the velocity diagram shown below.

Vector 4Vv

and 4/3Vv

are shown with their proper sense

such that the head to tail sum of 4Vv

and 4/3Vv

equals 3Vv

.

Friction force direction on link 3 should be opposite sense

of 4/3Vv

. Then the freebody diagram of the link can now

be drawn.

From the freebody of the link 4, taking the moment about C, Nv

becomes;

NAB

TNNABT extext 100

1,0

10* ====

vv

V4

V3

V3/4

OV

Text= 10 N.m

A

T

A

T

B FB=

N

FB'=Ff

N

Ff

C

Fc=N

Fc'=Ff

FA'

FA

B

4

2 d1

d2

FB'=F

f N

Ff

3


4/16

then friction force fF , BF , and BF' become;

NNFf 30100*3,0* ===

v

NFB 100=

NFB

30' =

Using the frebody diagram of the link 2, and taking the moment about A, torque becomes;

NFdNdTf 48,730*086,0100*049,0** 21 =+=+= , CCW

Example: Onto point D of the mechanism shown, a vertical force of 50 N is acting

downwards. Calculate the magnitude of the external motor torque on crank AB, acting in the

CCW direction at the instant when crank just starts rotating CCW, if the coefficient of friction

at the sliding joint is 0,2.

AB = BD = 5 cm. AC = 10 cm.

Solution:

If the crank just starts rotating CCW, link 4 moves leftwards. Since, friction force always

opposes the motion, direction of the friction force is rightwards.

A

1

4

B

C

2 3

50 N

60


5/16

We now complete the sketch of the freebody diagram shown below.

From the freebody of the link 4,

=== CxfCxfx FFFFF 0;0

=== NFNFF CyCyy 0;0

NFf *;0 =


==+= CxBxCxBxx FFFFF 0;0

+=== 50050;0 CyByCyByy FFFFF

=+= 030sin**30cos**30cos**50;0 ooo BCFBCFBDM CxCyB

030sin*0866,0*30cos*0866,0*30cos*05,0*50 =+ ooo

CxCy FF


4

C

B

3

50 N

A

B

2

NFf

FCxFCy

FCx

FCy

FBy

FBx

FBy

FBx

FAx

FAy

x

y

T


6/16

==+= AxBxAxBxx FFFFF 0;0

==+= AyByAyByy FFFFF 0;0

=+= 060cos**60sin**;0 TABFABFM ByBxA oo

TFF ByBx =+ oo

60cos*05,0*60sin*05,0*

Solving the above equations simultaneously, we obtain;

NN 63,32=

NFCx

527,6=

NFCy 63,32=

NFBx 527,6=

NFBy 63,82=

CCWNmT 348,2=


7/16

Example:An external torque of 100 N-m is acting on link 2 of the mechanism shown in

CCW direction. Calculate the magnitude and direction of the external force required on link 4

at the instant when the link just starts moving rightwards if the coefficient of friction at the

sliding joint at B is 0.3 .

AB = AC = 10 cm.

Solution:

By definition friction force always opposes the motion, in order to put the friction force in

proper direction we need velocity analysis. The velocity of the crank pin B as a point on AB

link is easily found, so that B will be used as a reference point for determining the relative

velocity of link 2 and link 3 The relative velocity equation may now be written;

34 VVvv

=

3/232 VVVvvv

+=

2Vv

is normal to AB link and direction is dictated by problem definition.

3Vv

is along BC link.

3/2Vv

is along AB link.

We now complete the sketch of the velocity diagram shown below.

A

1

3F

60

50

2

4

T


8/16

Vector 3Vv

and 3/2Vv

are shown with their proper sense such that

the head to tail sum of 3Vv

and 3/2Vv

equals 2Vv

. Friction force

direction on link 3 should be opposite sense of 3/2Vv

. Then the

freebody diagram of the link can now be drawn.

From the freebody of the link 2, taking the moment about A, Nv

becomes;

==+= AxBxAxBxx FFFFF 0;0

NAB

TNNABT extext 1,1730

0578,0

100* ====

vv

Then friction force fF becomes;

NNFf 03.5191,1730*3,0* ===

v

Using the freebody diagram of the link 4;

=+= 030cos*60cos*;0 FNFF fx oo

NF 83,1757= , rightwards

V2/3

V3

V2

Ov

3F

4

A

2

B

C

B

100 N-m

FC

TCN

Ff

N

Ff

FA

F'A

x

y


9/16

Example:In the figure a mechanism is shown with the appropriate dimensions. A horizontal

100 N force is acting on link 4 leftward. Calculate the magnitude of the external torque on

crank AB, at the instant when the crank just starts rotating CCW, if the coefficient of friction

at the sliding joint is 0,3.

AC=10 cm, CB=BD=4 cm, AF=2 cm

14

2

3

A

C

B

D

=45

F

100 N

Solution: Link 3 is two force member. When the crank just starts rotating CCW, link 4 tends

to move leftward. So, direction of the friction force would be rightward.


10/16

4

23

A

C

B

D

=45

F

100 N

T

dFB

FA

FB

C

B

A

FC

FCFf

NA

TA

From FBD of the 4 link;

0451000 ==++= ocosFF;F Cfx (1)

0300 =+= NsinFN;F Cy (2)

Af NF = (3)

Substituting EQ 3 into 1 gives:

010045 =+AC

NcosF (4)

Then solving EQ 2 and 4 for CF

N.sincos

FC 79108

4545

100=

+=

From link 3;

BC FF =

From link 2;

CWm.N..*.

F*ABT B 677791081000

570=== ANS


11/16

Example: Onto link 4 of the mechanism shown, a force of 50 N is acting downward.

Calculate the magnitude of the external motor torque on crank AB, at the instant when crank

just starts rotating CW, if the coefficient of friction at the sliding joint is 0,2.

=60 degrees. AB=5 cm, AC=10 cm, BD=15 cm,

A 4

B

C

2

3

50 N

D

B

C

FC

FC

FB

FB

FA

Ff

N

T

x

y

+

Link 3 is two force member, Link 4 is four force member

From freebody diagram of the link 4,

0300300 =+=+= cos*FNcos*FF;F CCfx

030500 =+= Nsin*F;F Cy

N.Fsincos

FcosF

sin*F CCC

C 0513503030

030

3050 ==

=+

CWNm..*.F*ABTB

62600513050 === ANSWER


12/16

Example:In the figure a mechanism is shown with the appropriate dimensions. A horizontal

100 N force is acting on link 4 leftward. Calculate the magnitude and direction of the

external torque on crank AB , at the instant when the crank just starts rotating CW, if the

coefficient of friction at the sliding joint is 0,3.

AB = 5 cm, AC = 10 cm,

A

1

4

B

2 3

100 N60


13/16

A 4

B

C

2

3

100 N

60

B

C

N

Ff

FC

FC

FB

FB

FA30

T

x

y

+

From FBD of 4 link;

0100300 == fCx FcosF;F (1)

0300 =+= NsinF;F Cy (2)

NFf = (3)


010030 = NcosFC (4)

Then solving EQ 2 and 4 for CF

o

330661393030

100=

= N.

sincosF

C

From link 3;

BC FF =

From link 2;

CWm.N..*F*AB B 9836661391000

50===T

v


14/16

3- (25%) Consider the mechanism given in question 1. Calculate the magnitude of the

external motor torque on crank AB, at the instant when crank just starts rotating CCW, if the

coefficient of friction at the sliding joint is 0,2.

A

D(6cm, 3cm)

B

C

1

2

3 4100 N

120o

x

y

It is obvious that link 2 is two force member. Freebody diagram would be as shown below.

D

C

3

4

100 N

FD

FC

FC

FC=FBB

C4

Ff30

o

From fourth link freebody diagram:

=+= 0100300 cos*FF;F Cfx (1)

== 0300 sin*FF;F CDy (2)

00260300165030

0

4 =

=.*sin*F.*cos*F

;M

CC

D

(3)

Df FF = (4)

We have 4 unknown and 4 equation. From equation 1,2

and 4; N.sincosFC 51033030

100=

+=

. Tork becomes;

CWNm..*.F*ABC

592510302502 === ANSWER

A

B

2120

o

x

y

FC

2

FC


15/16

2- (20%) Consider the mechanism

given in question 1. Calculate the

magnitude of the external motor

torque on link 2, at the instant

when crank just starts rotating

CCW, if the coefficient of friction

between ground and link 4 is 0,2.Do not forget to complete the

freebody diagrams of the links

given below.

A

B C

2

3D

F

5

6

C

C

FF

FD

FD

FAT

4

100 N

N

FF

Ff

From freebody diagram of the link 4;

0451000 =++= ocosFF;F Ffx (1)

0450 == sinFN;F Fy (2)

NFf = (3)


010045 =+ NcosFF (4)

A

BC

1

23

4

100 N45x

y

D

F

5

6


16/16

Then solving EQ 2 and 4 for FF

N.sincos

FF

871174545

100=

+=

From link 3;

DF FF =

From link 2;CWm.N..*F*ADT D 125487117

1000

35=== ANS

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force of friction