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8/12/2019 force of friction
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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.
8/12/2019 force of friction
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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
8/12/2019 force of friction
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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
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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
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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 =
From the freebody of the link 3,
==+= 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
From the freebody of the link 2,
4
C
B
3
50 N
A
B
2
NFf
FCxFCy
FCx
FCy
FBy
FBx
FBy
FBx
FAx
FAy
x
y
T
8/12/2019 force of friction
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==+= 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=
8/12/2019 force of friction
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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
8/12/2019 force of friction
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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
8/12/2019 force of friction
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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.
8/12/2019 force of friction
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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
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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
8/12/2019 force of friction
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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
8/12/2019 force of friction
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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)
Substituting EQ 3 into 1 gives:
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
8/12/2019 force of friction
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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
8/12/2019 force of friction
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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)
Substituting EQ 3 into 1 gives:
010045 =+ NcosFF (4)
A
BC
1
23
4
100 N45x
y
D
F
5
6
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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