5. Friction
2142111 Statics,
S i
2011/2
© Department of Mechanical
Engineering Chulalongkorn University
Engineering,
1
Objectives Students must be able to
Course Objective
 Include friction into equilibrium analyses
p
Objectives
j
Chapter
For general dry friction
 Describe the mechanism and characteristics of dry friction,
coefficient 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
pp p
FBDs
 Analyze bodies/structures with friction for unknown
loads/reactions by appropriate FBDs
2
2
Objectives Students must be able to
Course Objective
 Include friction into equilibrium analyses
p
Objectives
j
Chapter
For friction in machines
 Prove, state limitation and apply formula for impending motion in
wedges single continuous threads/screws
wedges,
threads/screws, flat belts
belts,
disks/clutches, pivot/collar/thrust bearings, journal bearings
 Analyze machines with friction
3
3
Contents


Dry friction
 Characteristics,
Characteristics theory
theory, coefficient and angle of friction
Applications
 Wedges
g
 Threads, screws
 Belts
 Disks
Di k and
d Cl
Clutches
t h
 Collar, pivot, journal and thrust bearings
4
4
Dry Friction
Friction


Force of resistance acting on a body which prevents or
retards slipping of the body relative to a surface with
which it is in contact.
 The frictional force acts tangent to the contacting
surface
f
in
i a di
direction
ti opposed
d tto th
the relative
l ti motion
ti 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 fluid
fluid.
 Dry friction occurs between contacting surfaces
without lubricating fluid.
5
5
Dry Friction
Dry Friction Theory #1


Friction forces are tangential forces generated between
contacting surfaces
surfaces.
Friction forces arise in part from the interactions of the
roughnesses or asperities of the contacting surfaces.
6
6
Dry Friction
Equilibrium


Slipping and/or tipping effects
Frictional force F increases with force P.
P
7
7
Dry Friction
Impending Motion
Static Friction Equilibrium

Maximum value of frictional force Fs on the object in
equilibrium is called the limiting static frictional force
force.
8
8
Dry Friction
Impending Motion
Static Friction Coefficient
Fs = μs N
Fs = maximum static frictional force
μs = coefficient of static friction
N = normal force
Fs is the maximum friction forces
that can be exerted by dry
dry, contacting surfaces
that are stationary relative to each other.
9
9
Dry Friction
Impending Motion
Static Friction Angle
g of Static Friction
N
= Rs cos φs
Fs = Rs sin φs = μs N
μs =
Fs R sin φs
=
= tan φs
N R cos φs
φs = tan−1 μs
10
10
Dry Friction
Impending Motion
Static Friction Typical Values
Contact Materials
Metal / ice
Wood / wood
Leather / wood
Leather / metal
Al i
Aluminum
/ Al
Aluminum
i
μs
0.03 – 0.05
0.30 – 0.70
0.20 – 0.50
0.30 – 0.60
1 10 – 1.70
1.10
1 70
How to obtain values of μs ?
11
11
Dry Friction
Frictional Forces

More force is required to start sliding than to keep it
sliding can be explained by the necessity to break these
bonds.
12
12
Dry Friction
Motion
Kinetic Friction Coefficient & Angle
Fk = μk N
Fk = kinetic frictional force
μk = coefficient of kinetic friction
N = normal force
φk = tan−1 μk
13
13
Dry Friction
Motion
Kinetic Friction Angle of Friction
At φ = φs , slip is impending.
At φ = φk , surfa
f ces are slidin
lidi g
relative to each other.
φs ≥ φk
14
14
Dry Friction
Dry Friction Characteristics



Frictional force acts tangentially to the contacting
surfaces opposing the relative or tendency for motion
surfaces,
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:
Slipping:
Very low velocity:
φ = φs
φ = φk
φk ≈ φs
15
15
Dry Friction
Motion


Sliding
 Relative sliding (translation motion) between two
surfaces
Toppling
 Fall over (rotation) about the edge
 Topple,
Topple tipping,
tipping rolling,
rolling tumble
tumble, trip
16
16
Dry Friction
Example Friction 1 #1
Place a sheet of one materials on a flat board. Bond a
sheet 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 μs =
tanα
17
17
Dry Friction
Example Friction 1 #2
The block is about to slip
W
  Fy = 0 
N = W cos α
A
α
N
N − W cos α = 0
  Fx = 0 
−Fs + W sin α = 0
(2) /(1)
μs N = W sin α
μs = tan α
Fs
(1)
(2)
Ans
18
18
Dry Friction
Example Friction 2 #1
Will this crate slide or
topple over?
19
19
Dry Friction
Example Friction 2 #2

The crate is about to slide.
F = Fs = μs N
20
20
Dry Friction
Example Friction 2 #3

The crate is about to topple.
F < Fs (μs N )

The crate is about to slide and
topple.
topple
F = Fs = μs N
21
21
Dry Friction
Example Hibbeler Ex 88-11 #1
The uniform crate has a mass of 20 kg. If a force P = 80 N is
applied to the crate
crate, determine if it remains in equilibrium
equilibrium. The
coefficient of static friction = 0.3.
22
22
Dry Friction
Example Hibbeler Ex 88-11 #2
  Fx = 0 → + 
(80 N)cos 30° − F = 0 → F = 69.3 N
  Fy = 0 ↑ + 
−(80 N) sin30° − 196.2 N + NC = 0 → NC = 236.2 N
  MO = 0 + 
(80 N) sin 30°(0.4 m) − (80 N)cos 30°(0.2 m) + NC ( x ) = 0 →
x = −9.0753 × 10 −3 m = −9.08 mm
x << 0.4
04m
→ Th
The crate
t will
ill nott topple.
t
l
F < μs NC (70.8 N)
→ The crate, is close to but doesn't not slip.
23
23
Dry Friction
Example Hibbeler Ex 88-33 #1
The rod with weight W is about to
slip on rough surfaces at A and B.
B
Find coefficient of static friction.
24
24
Dry Friction
Example Hibbeler Ex 88-33 #2
Impending slip
FA = μs N A
FB = μs NB
3 equilibrium equations
equations,
3 unknowns (N A , NB , μs )
  Fx = 0 → + 
μs N A + μs NB cos 30° − NB sin 30° = 0
(1)
  Fy = 0 ↑ + 
N A − W + NB cos 30° + μs NB sin 30° = 0
(2)
  M A = 0 + 
NB l − W cos 30°(l / 2) = 0
(3)
25
25
Dry Friction
Example
p
Hibbeler Ex 8-3 #3
From (3)
NB = 0.4330
0 4330W
From (1) & (2)
0.2165 μs2 − μs + 0.2165
μs = 0.228
Ans
26
26
Dry Friction
Example Hibbeler Ex 88-44 #1
Find minimum coefficients of static
friction that keep the system in
equilibrium.
27
27
Dry Friction
Example Hibbeler Ex 88-44 #2


Symmetry of lower pipes about
vertical axis passing through O.
Two types of contacts:
 Pipe – pipe
 Pipe – ground
N A = NB , FA = FB ,
FA ≤ μs,p - p N A
FC ≤ μs,p -g NC
28
28
Dry Friction
Example Bedford Ex 9.5 #1
Suppose that α = 10° and the
coefficient of friction between
the surface of the wedge and
the log are μs = 0.22 and μk =
0 20 Neglect the weight of the
0.20.
wedge.
 If the wedge is driven into
th llog att a constant
the
t t rate
t by
b
vertical force F, what are the
magnitude of the normal
f
forces
exerted
t d on the
th log
l b
by
the wedge?
 Will the wedge remain in
place in the log when the
force is removed?
29
29
Dry Friction
Driven at constant rate
Example Bedford Ex 9.5 #2
Symmetry about the center of the wedge
  Fy = 0 ↑ + 
2N sin(α / 2) + 2μk N cos(α / 2) − F = 0
F
N=
2 sin(α / 2) + 2μk (α / 2)
N=
F
2 sin(10
( ° / 2)) + 2(0.20)(10
(
)( ° / 2))
N = 1.75F
Ans
30
30
Dry Friction
On the verge of slipping
Example Bedford Ex 9.5 #3
Symmetry about the center of the wedge
  Fy = 0 ↑ + 
2N sin(α / 2) + 2μs ′N cos(α / 2) = 0
μs ′ = tan(α / 2) = 0.087
0 087
As μs ′ < μs , wedge
g will remain in p
place. Ans
31
31
Dry Friction
Example
p Bedford 9.20 #1
The coefficient of static friction
between the two boxes and
between the lower box and the
inclined surface is μs. What is
th llargestt angle
the
l α for
f which
hi h th
the
lower box will not slip.
32
32
Dry Friction
Example Bedford 9.20 #2
FBD of upper block
  Fy = 0 
N2 − W cos α = 0
N2 = W cos α
FBD of lower block
  Fy = 0 
N2 + W cos α − N1 = 0
N1 = 2W cos α
  Fx = 0 
μs N2 + μs N1 − W sin
i α =0
μs 3W cos α = W sin α
μs = tan α / 3 or α = tan−1(3 μs )
Ans
33
33
Dry Friction
Example Bedford 9.30 #1
The cylinder has weight W. The
coefficient of static friction
between the cylinder and the
floor and between the cylinder
and
d the
th wallll iis μs. What
Wh t is
i the
th
largest couple M that can be
applied to the stationary
cylinder
li d without
ith t causing
i it tto
rotate.
34
34
Dry Friction
Example Bedford 9.30 #2
Impending rotation
  Fy = 0 ↑ + 
μs Nw + Nf − W = 0
  Fx = 0 → + 
  MO = 0 + 
Nw − μs Nf = 0
M − μs (Nw + Ns )r = 0
μsW
W
Nw =
, Nf =
1 + μs2
1 + μs2
(1 + μs )
M = μsWr
1 + μs2
Ans
35
35
Dry Friction
Example Bedford 9.33 #1
The disk of weight W and radius
R is held in equilibrium on the
circular surface by a couple M.
The coefficient of static friction
b t
between
th
the di
disk
k and
d th
the
surface is μs. Show that the
largest value M can have
without
ith t causing
i th
the di
disk
k tto slip
li
is
M=
μs RW
1 + μs2
36
36
Dry Friction
Example
p Bedford 9.33 #2
Impending
p
g slip
p
Fs = μs N
  Fx = 0 + 
Fs − W sin α = 0
  Fy = 0 + 
  MO = 0 + 
N − W cos α = 0
−M + Fs R = 0
μs = tan α , N = W cos α
cos α 1 + tan2 α = 1, tanα =μs
M = μs NR = μsWR cos α
M = μsWR
1 + μs2
37
37
Dry Friction
Example
p Bedford 9.166 #1
Each of the uniform 1-m bars has a mass of 4 kg. The
coefficient of static friction between the bar and the surface at B
is 0.2. If the system is in equilibrium, what is the magnitude of
the friction force exerted on the bar at B.
38
38
Dry Friction
Example
p Bedford 9.166 #2
  Fx = 0 
Ox + Ax = 0
  Fy = 0 
Oy + Ay − 4g = 0
  MO = 0 
Ay (1 m)cos 45°) − Ax (1 m) sin 45°) − (4g N)(0.5 m)cos 45°) = 0
39
39
Dry Friction
Example
p Bedford 9.166 #3
  Fx = 0 
− Ax + F cos 30° − N sin 30° = 0
  Fy = 0 
− Ay − (4g N) + F sin30° + N cos 30° = 0
  MB = 0 
Ay (1 m)cos 45°) + Ax (1 m) sin 45°) + (4g N)(0.5 m)cos 45°) = 0
Ay = 0, Ax = −2g N, Oy = 4g N, Ox = 2g N, N = 438 N, F = 2.63 N Ans
40
40
Dry Friction
Example
p Bedford 9.166 #4
41
41
Dry Friction
Example Bedford 9.163 #1
The coefficient of static friction between the tires of the 1000-kg
tractor 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.
42
42
Dry Friction
Example Bedford 9.163 #2
The crate is about to move.
F = μs,c N
  Fy = 0 
−(450g N) + N = 0
  Fx = 0 
P −F =0
P = 0.3(450g N) = 135g N
43
43
Dry Friction
Example Bedford 9.163 #3
  Fx = 0 
−P + 2F2 = 0
F2 = 67.5g N
  MO = 0 
(0 8 m))F2 − T = 0
(0.8
T = (0.8 m)(135g N) = 1.06 kN ⋅ m Ans
44
44
Dry Friction
Example Bedford 9.165 #1
The 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
incline
before
f
th
the vehicle’s
hi l ’ titires slip?
li ?
45
45
Dry Friction
Example Bedford 9.165 #2
Impending motion:
F2 = μs,t N2
  Fx = 0 
−P cos 20° + 2F2 = 0
  Fy = 0 
2N2 + 2N1 − (900g N) − P sin 20° = 0
  MO = 0 
(900g N)(1 m) + P cos 20°(0.8 m) + P sin 20°(3.7 m) − 2N2 (2.5 m) = 0
P = 563.57g N, 2N2 = 814.75g N, 2F2 = 529.59g N, 2N1 = 278.00g N
46
46
Dry Friction
Example Bedford 9.165 #3
The crate is about to move. F = μs,c N
  Fy = 0 
−W cos 20° + N = 0
  Fx = 0 
P − F − W sin 20° = 0
W = 1.3930
1 3930P N = 7701.1
7701 1 N
m = W / g = 785 kg
Ans
47
47
Dry Friction
Example Bedford 9.165 #4
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 tractor
topples.
48
48
Dry Friction
Example Bedford 9.165 #5
Impending
p
g topple:
pp N1 = 0
  MQ = 0 
P (2.0 m) / 2 − (900g N)(1.5 m) = 0
  Fy = 0 
2N2 − (900g N) − P / 2 = 0
  Fx = 0 
−P / 2 + 2F2 = 0
F2 ≤ μs,t N2 , tires do not slip.
P = 955g N Ans
P = 954.59g N, 2N2 = 1575g N, 2F2 = 675g N
49
49
Machine Friction
Friction in Machines


Friction helps some machines function.
Machines with frictions
 Wedges
g
 Threads, screws
 Belts
 Disks
Di k and
d clutches
l t h
 Bearings
50
50
Machine Friction
Wedges

Wedges exert a large lateral forces from the faces as a
result of small angle
angle.
51
51
Machine Friction
Wedges Analysis #1
The block is about to raise.
The slip of the load and wedge are impending.
F2 = μs N2 , F3 = μs N3
  Fx = 0 → + 
−N3 + N2 sin θ + μs N2 cos θ = 0
  Fy = 0 ↑ + 
N2 cos θ − μs N2 sin θ − μs N3 − W = 0
52
52
Machine Friction
Wedges
g Analysis
y
#2
FBD of the wedge
  Fx = 0 → + 
−N2 sin
i θ − μs N2 cos θ − μs N1 + P = 0
  Fy = 0 ↑ + 
N1 − N2 cos θ + μs N2 sin
i θ =0
(
(
)
)
 1 − μs2 tan θ + 2μs 
W
P=
2
 1 − μs − 2μs tan θ 
53
53
Machine Friction
Threads/Screws

Assumption: the shaft has a single continuous thread.
54
54
Machine Friction
Threads Geometry
θ = tan−1
l
2π r
θ = lead angle
l = pitch of the thread (lead)
r = mean radius
di off th
the th
thread
d
55
55
Machine Friction
Threads Motion
W
W
56
56
Machine Friction
Threads Moving Upwards
  Fx = 0 → + 
S − R sin(
i (θ + φ ) = 0
  Fy = 0 ↑ + 
R cos(θ + φ ) − W = 0
S = W tan(θ + φ ) and M = Sr
M = Wr tan(θ + φ )
On the verge of rotating: φ = φs = tan−1 μs
Uniform upwards motion: φ = φk = tan−1 μk
57
57
Machine Friction
Threads Moving Downwards #1
  Fx = 0 → + 
S ′ − R sin(
i (θ − φ ) = 0
  Fy = 0 ↑ + 
R cos(θ − φ ) − W = 0
S ′ = W tan(θ − φ ) and M ′ = S ′r
M ′ = Wr tan(θ − φ )
On the verge of rotating:
φ = φs = tan−1 μs
Uniform downwards motion: φ = φk = tan−1 μk
58
58
Machine Friction
Threads Moving Downwards #2
M is removed:
W is supported by friction alone,
providing that φ ≥ θ
Self-locking condition
59
59
Machine Friction
Threads Moving Downwards #3
  Fx = 0 → + 
−S ′′ + R sin(
i (φ − θ ) = 0
  Fy = 0 ↑ + 
R cos(φ − θ ) − W = 0
S ′′ = W tan(φ − θ ) and M ′′ = S ′′r
Very rough surface θ < φ:
Reversed motion.
M ′′ = Wr tan(φ − θ )
60
60
Machine Friction
Example Wedges/Threads 1 #1
The mass of block A is 60 kg. Neglect the weight of the 5° wedge.
The 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 mating
groove is 0.2. What couple must be exerted on the threaded shaft
to raise the block A at a constant rate?
61
61
Machine Friction
Example Wedges/Threads 1 #2
Draw FBD off th
D
the mass A and
d th
the wedge
d
Then, follow the same procedure for
wedge analysis:
(
(
)
)
 1 − μk21 tan α + 2μk 1 
W
F=
2
 1 − μk 1 − 2μk 1 tan α 
Substitute for all known values:
F = 667.50 N
WL = 60g N
Ww = 0 N
α = 5°
μk 1 = 0.4 for all
non thread contacts
non-thread
62
62
Machine Friction
Example Wedges/Threads 1 #3
l
r
= 5 mm
= 15 mm
μk 2 = 0.2 for thread/groove
g
φk = arctan( μk 2 ) = 11.310°
θ = arctan(/ 2π r ) = 3.0368°
Draw FBD of the thread
Find M that will push the thread upwards:
M = Fr tan(θ + φk 2 )
M = (667.50
(667 50 N)(0.015
N)(0 015 m) tan(3.0368
tan(3 0368° + 11
11.310
310°)
M = 2.5609 N ⋅ m = 2.56 N ⋅ m
Ans
63
63
Machine Friction
Flat Belts

Involve slippage of flexible cables,
belts and ropes over sheaves and
drums.
64
64
Machine Friction
Flat Belts Analysis #1
Consider line element
  Ft = 0 
dθ
dθ
− (T + dT )cos
=0
2
2
dθ
μ dN = dT cos
2
  Fn = 0 
μ dN + T cos
dθ
dθ
− (T + dT ) sin
=0
2
2
dθ
dθ
dN = 2T sin
+ dT sin
2
2
dN − T sin
65
65
Machine Friction
Flat Belts Analysis #2
μ = μs (motion impending)
μ = μk (ongoing movement)
dθ → 0, cos
dθ
→ 1:
2
μ dN = dT cos
dθ
≈ dT
2
dθ
dθ
dθ → 0, sin
→
:
2
2
dθ
dN = 2T sin
≈ Tdθ
2
dT
= μ dθ
T
→

T2
T1
β
dT
= μ  dθ
0
T
T2 = T1e μβ
→
ln
T2
= μβ
T1
66
66
Machine Friction
Example Flat Belts 1 #1
Friction
F
i i resists
i
the
h
lowering of m:
T2 = mg
T1 = P = mg / 6
A force P = mg/6 is required to
lower the cylinder at a constant
slow speed with the cord
making 1.25 turns around the
fixed shaft. Calculate the
coefficient of friction μs between
bet een
the cord and the shaft.
β = 1.25(2
( π)
T2 = T1e μβ 
mg 2.5πμs
mg =
e
6
μs = 0.228
0 228
Ans
67
67
Machine Friction
Example Flat Belts 2 #1
Calculate the force P on the
h dl off th
handle
the diff
differential
ti l b
band
d
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 between
the band and the flywheel is μ
= 0.40.
68
68
Machine Friction
Example Flat Belts 2 #2
Friction resists the attempt
p of M to rotate
the flywheel in the clockwise direction:
T2 = T1e μβ 
T2 = T1e μ (7π / 6)
  Mcenter = 0 
−(150 N ⋅ m) + (0.15 m)(T2 − T1 ) = 0
T2 − T1 = 1000 N
(1)
(2)
Solve (1) and (2) T1 = 300 N, T2 = 1300 N
69
69
Machine Friction
Example Flat Belts 2 #3
  MO = 0 + 
−(0.15 m)T2 + (0.075 m)(T1 sin30°) + (0.45 m)P = 0
P = 408 N
Ans
70
70
Machine Friction
Disks & Clutches



Use to connect and disconnect two
coaxial rotating shaft
The shaft can support a couple
due to friction forces between the
di k
disks.
Determine the couple by using
coefficient of static friction in the
flat-ended thrust equation.
Draw forces on the plane only
71
71
Machine Friction
Disks & Clutches Analysis #1
dA = 2π r dr
  Fx = 0 
P −  w dA = 0
P=

R
o
p 2π r dr = pπ R
2
2
M = μ PR
3
  M x = 0  M − μ  wr dA = 0
R
M = μ  pr 2π r dr =
o
2
μ pπ R 3
3
72
72
Machine Friction
Disks & Clutches Worn Analysis #1


Distribution of forces on worn plates are not uniformed.
For example, linear distribution
dA = 2π r dr
At r = R,w = 0
At r = 0,, w = p
w = p(R − r ) / R
73
73
Machine Friction
Disks & Clutches Worn Analysis #2
Consider the left plate
  Fx = 0 
P −  w dA = 0
P=

R
o
p (R − r )2π r dr / R =
πp
3
R2
  M x = 0  M − μ  wr dA = 0
R
M = μ  pr (R − r )2π r dr / R
o
1
M = μ PR
2
74
74
Machine Friction
Bearings Pivot Bearings
M=
2
μ PR
3
μ = μs (motion impending)
μ = μk (ongoing movement)
75
75
Machine Friction
Bearings Collar Bearings
Ri3 − Ro3
2
M = μP 2
3
Ri − Ro2
μ = μs (motion impending)
μ = μk (ongoing movement)
76
76
Machine Friction
Bearings Thrust Bearings #1
Ri
 dR 
dA = 2π R ds
d = 2π R 

 cos θ 
A =  dA = 
Ro
Ri
2π R
dR =
cos θ
Ro
dR
π ( Ro2 − Ri2 )
Ri
R
θ
Ro
P
cos θ
77
77
Machine Friction
Bearings Thrust Bearings #2
  Fx = 0 
P − pA cos θ = 0
p=
dR
P
P
=
A cos θ π Ro2 − Ri2
(
Ri
R
θ
Ro
P x
)
  M x = 0 
M −  R( μ p
pdA) = 0
Ro
M = μ R
Ri
P
π ( Ro2 − Ri2 )
2π R
dR
cos θ
2μ P
M=
3 cos θ
 Ro3 − Ri3 
 2
2 
−
R
R
i 
 o
78
78
Machine Friction
Bearings Journal Bearings #1
P
P
φ
79
79
Machine Friction
Bearings Journal Bearings #2
R
R
Rf
  M z = 0 + 
M − P (R sin φ ) = 0
M = PR sin φ
M ≈ PR tan φ
M = PR sin φ ≈ PR μ
80
80
Machine Friction
Example Hibbeler 88-11
11 #1
The 100-mm-diameter pulley
fit loosely
fits
l
l on a 10-mm10
diameter shaft for which the
coefficient of static friction is μs
= 0.4. Determine the minimum
tension T in the belt needed to
(a) raise the 100-kg block and
(b) lower the block. Assume that
no slipping occurs between the
b lt and
belt
d pulley
ll and
d neglect
l t th
the
weight of the pulley.
81
81
Machine Friction
Example
p Hibbeler 8-11 #2
CW rotation
Raise the block
rf = r sin φs = (5 mm) sin(tan−1 0.4) = 1.8570 mm
  M = 0 + 
P2


(981 N)(50 mm + 1.8570 mm − T (50 mm − 1.86 mm) = 0
T = 1056.8
1056 8 N
T = 1.06 kN
Ans
82
82
Machine Friction
Hibbeler Example 88-11
11 #3
CCW rotation
Lower the block
rf = r sin φs = (5 mm) sin(tan−1 0.4) = 1.8570 mm
  M = 0 + 
P2


(981 N)(50 mm − 1.86 mm) − T (50 mm + 1.86 mm) = 0
T = 910.63 N
T = 911 kN
Ans
83
83
Machine Friction
Example Boresi 10.9 #1
84
84
Machine Friction
Example Boresi 10.9 #2
A weight W is lifted at a constant rate
by 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 p
of the screw threads
threads, the mean
radius r of the screw threads, and
the coefficient of kinetic friction μk
of the screw. The friction of the
vertical wall guides is negligible.
For θ = 15
15°, W = 1000 lb,
lb d = 20 in,
in
r = 0.75 in, p = 0.333 in, and μk =
0.15, calculate R.
85
85
Machine Friction
Example Boresi 10.9 #3
  Fy = 0 
2T cos θ − W = 0
W = 2T cos θ
(1)
86
86
Machine Friction
Example Boresi 10.9 #4
  Fy = 0 
Q cos θ − Tc os θ = 0
Q =T
  Fx = 0 
2T sin θ − P = 0
P = 2T sin θ
(1) + (2)
(2)
P = W tan θ
By symmetry, FBD of the left nut is
mirror image of the right nut FBD.
87
87
Machine Friction
Example
p Boresi 10.9 #5
[M = Wr tan(φk + α )]
Rd = 2 Pr tan(φk + α )
R = 2(W tan θ )r tan(tan−1 μk + tan−1
p
2π r
)/d
Substitute numbers in the equation
R = 4.48
4 48 lb
A
Ans
88
88
Machine Friction
Example Hibbeler 88-120
120 (modify) #1
The axle of the pulley fits
l
loosely
l iin a 50-mm-diameter
50
di
t
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 to
turn the pulley counterclockwise
at a constant velocity if the
bl k weighs
block
i h 60 N.
N N
Neglect
l t th
the
weight of the pulley.
89
89
Machine Friction
Case I Journal Bearing
Example Hibbeler 88-120
120 (modify) #2
The shaft is about to rotate
rotate.
φ = tan−1 μk ,b = tan−1 0.3 = 16.699°
rf = r2 sin φ = 7.1837 mm
θ = tan-1 (r1 / r1 ) = 45°
β = sin−1 (rf / 2r1 ) = 2.9117°
  Fy = 0 
−60 + R sin(θ + β ) = 0
  Fx = 0 
T − R cos(θ + β ) = 0
R = 80.850 N, T = 54.192 N
90
90
Machine Friction
Case I Journal Bearing
Example Hibbeler 88-120
120 (modify) #3
If the cord does not slip over the pulley,
the minimum μk ,f 1 is:
T2 = T1e μk ,f β 


60 N = (54.192 N)e
μk ,f 1 = 0.064815
μk ,f 1 ( π / 2)
μk ,b1 < 0.2, the assumption the the
cord does not slip is valid.
T = 54.2
54 2 N
Ans
91
91
Machine Friction
Case II Flexible Belt
Example Hibbeler 88-120
120 (modify) #4
Th belt
The
b l iis about
b
to slide.
lid
T2 = T1e μk ,f β 


60 N = Te0.2(π / 2)
T = 43.824 N
92
92
Machine Friction
Case II Flexible Belt
Example Hibbeler 88-120
120 (modify) #5
If the shaft does not rotate,
  Fy = 0 
−(60 N) + R sin(θ + β ) = 0
  Fx = 0 
T − R cos((θ + β ) = 0
T = 43.824 N, θ = 45°, β = 8.8556°
β = sin−1(rf / 2r1 ) → rf = 21.771 mm
rf = r2 sin φ
→ φ = 60.557°
φ = tan−1 μk ,b1
→ μk ,b1 = 1.77
μk ,b1 > 0.3, the assumption that the
shaft does not rotate is invalid.
93
93
Example Boresi 10
10-86
86 #1
A torque T is applied to pulley A, which drives pulley B. The
t
two
pulleys
ll
h
have equall radii.
dii Th
The ttension
i iin th
the slack
l k side
id off
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
94
Machine Friction
Example
p Boresi 10-86 #2
Given T2 = 2 kN
The belt is about to slip:
T1 = T2e μβ 
T1 = (2 kN)e0.3π = 5
5.1327
1327 kN
T1 = 5.13 kN
Ans
  MO = 0 
−T + (T1 − T2 )r = 0
T = (T1 − T2 )(0.1 m) = 0.31327 kN ⋅ m
As tensions in the belts remain
unchanged between the two pulley,
TB = T = 313 N ⋅ m
Ans
95
95
Machine Friction
Example
p Boresi 10-98 #1
In some applications of collar bearings,
multiple
lti l collars
ll
are used.
d T
Two collars
ll
are used as shown to support the
10,000 lb load (μs = 0.40 and μk = 0.20)


Determine the torque T required to
initiate rotation of shaft S, Assume
that the load is divided equally
between the collars and that the
pressure on each collar is uniform.
Wh are multiple
Why
lti l collars
ll
used?
d?
96
96
Machine Friction
Example Boresi 10
10-98
98 #2
The load is divided equally between the collars,
collars
thus, the load on each collar is 5 kip.
For each collar that is about to rotate:

R23 − R13 
2
T = μs P 2
2 
3
R
−
R
2
1 

2
(1 ft)3 − (0.5 ft)3
= 1555.6 lb ⋅ ft
T = 0.4(5000 lb)
3
3
3
(1 ft) − (0
(0.5
5 ft)
Total required
q
torque
q T = 2M = 3.11 kip
p ⋅ ft
Ans
Less load on each collar, less wear and tear.
Ans
97
97
Review
Concepts #1


Friction is the tangent resistance force which prevents
or 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 together.
Some machine functions by friction. To use the formula,
the physical meanings underlying the friction in machines
must be understood.
98
98