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29-09-2014
1
Friction
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).
• 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.
• 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.
• 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 nonlubricated
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.
• As P increases, the static-friction force F
increases as well until it reaches a maximum
value Fm.
NF sm µ=
• Further increase in P causes the block to begin
to move as F drops to a smaller kinetic-friction
force Fk.
NF kk µ=
The Laws of Dry Friction. Coefficients of Friction
• Maximum static-friction force:
NF sm µ=
• Kinetic-friction force:
sk
kk NF
µµ
µ
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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The Laws of Dry Friction. Coefficients of Friction
• Four situations can occur when a rigid body is in contact with
a horizontal surface:
• No friction,
(Px = 0)
• No motion,
(Px < Fm)
• Motion impending,
(Px = Fm)
• Motion,
(Px > Fm)
Angles of Friction
• It is sometimes convenient to replace normal force
N and friction force F by their resultant R:
• No friction
• Motion impending• No motion• Motion
kk
kkk
N
N
N
F
µφ
µφ
=
==
tan
tan
ss
sms
N
N
N
F
µφ
µφ
=
==
tan
tan
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4
Angles of Friction
• Consider block of weight W resting on board with
variable inclination angle θ.
• No friction • No motion • Motion
impending
• Motion
Problems Involving Dry Friction
• All applied forces known
• Coefficient of static friction
is known
• Determine whether body
will remain at rest or slide
• All applied forces known
• Motion is impending
• Determine value of coefficient
of static friction.
• Coefficient of static
friction is known
• Motion is impending
• Determine magnitude or
direction of one of the
applied forces
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Problem-1
Determine the maximum angle θ which the adjustable
incline may have with the horizontal before the block of
mass m begins to slip. The coefficient of static friction
between the block and the inclined surface is µs.
Problem-2
Determine the range of values which the mass m0 may
have so that the 100-kg block shown in the figure will
neither start moving up the plane nor slip down the plane.
The coefficient of static friction for the contact surface is
0.30.
Case-I
The maximum value of m0 will be given by the
requirement for motion impending up the plane.
Case-II
The maximum value of m0 will be given by the
requirement for motion impending down the plane.
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Problem-3
Determine the magnitude and direction of the friction force
acting on the 100-kG block shown if, first, P = 500 N and,
second, P = 100 N. the coefficient of static friction is 0.20 and
the coefficient of kinetic friction is 0.17. The forces are
applied with the block initially at rest.
A balance of the forces in x and y directions gives
It follows that 242 N of friction cannot be supported. Therefore, equilibrium cannot
exist, and we obtain the correct value of the friction force by using the kinetic coefficient
of friction accompanying the motion down the plane. Hence, the answer is
29-09-2014
7
Problem-4
The homogeneous rectangular block of mass m, width b and
height H is placed on the horizontal surface and subjected to
a horizontal force P which moves the block along the
surface with a constant velocity . The coefficient of kinetic
friction between the block and the surface is µk. Determine
(a) The greatest value which h may have so that the block
will slide down without tipping over
(b) The location of a point C on the bottom face of the
block through which the resultant of the friction and
the normal forces acts if h = H/2.
Solution (a)
OR
Problem-4 continued
Solution (b)
29-09-2014
8
The uniform 10-kg ladder rests against the
smooth wall at B. and the end A rests on the
rough horizontal plane for which the coefficient
of static friction is µs= 0.3. Determine the angle
of inclination θ (If the ladder and the normal
reaction at B if the ladder is on the verge of
slipping.
Solution
Problem-5
Since the ladder is on the verge of slipping, then
29-09-2014
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Problem-6The three flat blocks are positioned on the 300 incline
as shown, and a force P parallel to the incline is applied
to the middle block. The upper block is prevented from
moving by a wire which attaches it to the fixed support.
The coefficient of static friction for each of the three
pairs of mating surfaces is shown. Determine the
maximum value which P may have before any slipping
takes place
For 50 Kg Block
For 40 Kg Block
Thus, 468 N cannot be supported and our initial assumption was wrong.
We conclude, therefore, that slipping occurs first between the 40-kg block and the incline.
With the corrected value F3 = 459 N, equilibrium of the 40-kg block for its impending motion
requires
We will assume arbitrarily that only the 50-kg block slips, so that the 40-kg block remains in
place. Thus, for impending slippage at both surfaces of the 50-kg block, we have
29-09-2014
10
Equilibrium of the 50-kg block gives, finally,
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
µ
µ
or
021 =++ WRRvrr
( )
( ) 06sin6cos
:0
0
6sin6cos
:0
32
32
=°−°+−
=
=+
°−°−−
=
∑
∑
s
y
ss
x
NN
F
P
NN
F
µ
µµ
• Wedge as free-body
or
032 =+− RRPrrr
29-09-2014
11
Frictional Forces on Screws
In most cases, screws are used as fasteners; however, in many types of machines
they are incorporated to transmit power or motion from one part of the machine
to another.
A square-threaded screw is commonly used for the latter purpose, especially
when large forces are applied along its axis.
Screw Thread
29-09-2014
12
A screw is said to be self-locking if it
remains in place under any axial load W
when the moment M is removed. For this
to occur, the direction of the frictional
force must be reversed so that R acts on
the other side of N .
If the screw will unwind
by itself and M’ is the moment
required to prevent unwinding.
M’’ is the moment required to lower the
screw
29-09-2014
13
If the coefficient of friction between the
steel wedge and the moist fibres of the
newly cut stamp is 0.20, determine the
maximum angle α which the wedge may
have and not pop out of the wood after
being driven by the sledge.
01
1
6.22)2.0(tan2
tan2/
==
==−
−
α
µφα
Problem-7
α/2 α/2
R R
The uniform stone has a mass of 500 kg
and is held in the horizontal position using
a wedge at B . If the coefficient of static
friction is µs = 0.3 at the surfaces of
contact, determine the minimum force P
needed to remove the wedge. Assume that
the stone does not slip at A .
FBD
Problem-8
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2 - 28
The force P required to start the wedge is
found from the equilibrium triangles of the
forces on the load and on the wedge.
29-09-2014
15
Problem-9
2 - 29
Determine the force P required to
force the 100 wedge under the 90-
kg uniform crate which rests
against the small stop at A. The
coefficient of friction for all
surfaces is 0.40.
R2R3
A50
90(9.8)N
G0.6m
0.35m
8.21)4.0(tantan 11 === −− µφ
NR
R
518
02.1)5cos(
)]35.0(5sin)6.0(5)[cos81.9(90
2
0
2
00
=
=×+−
+
φ
ϕ
0)10sin(sin51810cos 0
1
0 =+−− φφ RP
0)10cos(cos51810sin 0
1
0 =++− φφ RP
R1 = 471 N and P = 449 N
On solving
R2
P 100
R1
ϕ
R2R3
A50
90(9.8)N
G0.6m
0.35m
ϕ
ϕ
29-09-2014
16
2 - 31
The coefficient of static friction
between the 100-lb body and the
wedge is 0.20. Determine the
magnitude of the force P required to
begin raising the 100-lb body if
(a) rollers of negligible friction are
present under the wedge, as illustrated,
(b) the rollers are removed and the
coefficient of static friction applies at
this surface as well.
Problem 13
R2
R2
150
R1
ϕ
ϕ
P
100 lbR3
011 31.11)2.0(tantan === −− µφ
150
2 - 32
R2
R2
150
R1
ϕ
ϕ
P
100 lbR3
011 31.11)2.0(tantan === −− µφ
a)
y
x
lbR
R
Fy
6.111
0)31.1115cos(100
0
2
00
2
=
=++−
=∑
lbP
PR
Fx
4.49
0)31.1115sin(
0
2
00
2
=
=−+
=∑150
Block:
Wedge:
29-09-2014
17
b) Rollers removed: Value of R2 from
100 –lb body is unchanged.
0)31.11sin()31.1115sin(
0
0
1
00
2 =+−+
=∑RPR
Fx R2
R1
ϕ
ϕ
P
y
150
lbP
lbR
4.69
0.102)31.11cos(/1000
1
=∴
==
R1 is calculated from overall equilibrium as
Problem 10
A clamp is used to hold two pieces of
wood together as shown. The clamp
has a double square thread of mean
diameter equal to 10 mm with a lead
of 4 mm. The coefficient of friction
between threads is µs = 0.30.
If a maximum torque of 40 N*m is
applied in tightening the clamp,
determine (a) the force exerted on the
pieces of wood, and (b) the torque
required to loosen the clamp.
SOLUTION
• Calculate lead angle .
• Using block and plane analogy with
impending motion up the plane, calculate
the clamping force with a force triangle.
• With impending motion down the plane,
calculate the force and torque required to
loosen the clamp.
29-09-2014
18
30.0tan
1273.0 10
4
2tan
==
===
ss
r
L
µφ
ππθ °= 3.7θ
°= 7.16sφ
• Using block and plane analogy with impending
motion up the plane, calculate clamping force with
force triangle.
kN8mm5
mN 40mN 40 =
⋅=⋅= QrQ
( )°
==+24tan
kN8tan W
W
Qsφθ
kN97.17=W
• With impending motion down the plane, calculate
the force and torque required to loosen the clamp.
( ) ( ) °==− 4.9tankN97.17tan QW
Qs θφ
kN975.2=Q
( )( )
( )( )m105N10975.2
mm5kN975.2
33 −××=
== rQTorque
mN87.14 ⋅=Torque
29-09-2014
19
The 40-mm diameter screw has a double
square thread with a pitch of 12 mm and a lead
of 24 mm. the screw and its mating threads in
the fixed block are graphite-lubricated and
have a friction coefficient of 0.15. if a torque
M = 60 N.m is applied to the right handed
portion of the shaft, determine (a) the force P
required to advance the shaft to the right and
(b) the force P which would allow the shaft to
move to the left at a constant speed.
0181.10)40/24(tan == − πθ
01 53.8)15.0(tan == −ϕ
00 28.2,34.19 =−=+ φθφθ
NP
P
M
3.75
28.2tan)02.0(60
)tan(Pr
0
=
=
−= φθ
NP
P
M
3.75
34.19tan)02.0(60
)tan(Pr
0
=
=
+= φθ
Helix angle
Friction angle
Θ > ϕ so screw is not self locking.
a) b)
Problem 11
2 - 38
The vertical position of the 100-kg block is
adjusted by the screw-activated wedge.
Calculate the moment M which must be
applied to the handle of the screw to raise
the block. The single-thread screw has
square threads with a mean diameter of 30
mm and advances 10 mm for each complete
turn. The coefficient of friction for the
screw threads is 0.25, and the coefficient of
friction for all mating surfaces of the block
and wedge is 0.40. Neglect friction at the
ball joint A.
R2
R2
100
R3
ϕ
ϕ
ϕ
P
100(9.8) Nϕ
R1x
y
ϕ
x
Problem 12
29-09-2014
20
2 - 39
R2
R2
100
R3
ϕ
ϕ
ϕ
P
100(9.8) Nϕ
R1x
y
ϕ
x
ϕ90-ϕ
y
R2ϕ
100(9.8) Nϕ
R1x
y
ϕ
90-ϕ
y
Angle between R2 and y axis = 2ϕ+100
NR
R
RW
Fy
1535
60.53cos80.21cos981
)102cos(cos
0
2
0
2
0
0
2
=
=
+=
=∑φφ
Block:
011 80.21)4.0(tantan === −− µφ
R2
R3
ϕ
ϕ
P
ϕ
x
y
ϕ
y
x
Angle between
R2 and x axis =
90-(2ϕ+100)
Wedge:
NP
P
PR
Fx
1331
80.21cos40.36cos1535
cos40.36cos
0
2
00
0
2
=
=
=
=∑φ
011 80.21)4.0(tantan === −− µφ
29-09-2014
21
009.20
25.0tan
1273.0(15)mm 2
10
2tan
=
==
===
s
ss
r
L
φ
µφ
ππθ
Screw:
NM
M
rPM
30.7
08.20tan)015.0(1331
)tan(
0
=
=
+= φθ
Belt Friction
Consider the flat belt which passes over a
fixed curved surface. The total angle of
belt to surface contact in radians is β, and
the coefficient of friction between the
two surfaces is µ.
We wish to determine the tension T2 in
the belt, which is needed to pull the belt
counterclockwise over the surface, and
thereby overcome both the frictional
forces at the surface of contact and the
tension T1 in the other end of the belt.
Obviously, T2 > T1.
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22
2 - 44
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23
Problem 13
A flat belt connects pulley A to pulley B.
The coefficients of friction are µs = 0.25
and µk = 0.20 between both pulleys and
the belt.
Knowing that the maximum allowable
tension in the belt is 600 lb, determine
the largest torque which can be exerted
by the belt on pulley A.
SOLUTION:
• Since angle of contact is smaller,
slippage will occur on pulley B first.
Determine belt tensions based on
pulley B.
• Taking pulley A as a free-body, sum
moments about pulley center to
determine torque.
Problem 13
SOLUTION:
• Since angle of contact is smaller, slippage will
occur on pulley B first. Determine belt tensions
based on pulley B.
( )
lb4.3551.688
lb600
688.1lb600
1
3225.0
11
2
==
===
T
eT
eT
Ts πβµ
• Taking pulley A as free-body, sum moments about
pulley center to determine torque.
( )( ) 0lb600lb4.355in.8:0 =−+=∑ AA MM
ftlb1.163 ⋅=AM
29-09-2014
24
The 180-lb rock climber is lowered over the edge
of the cliff by his two companions, who together
exert a horizontal pull T of 75-lb on the rope.
Compute the coefficient of friction between rope
and the rock.
lbP 9.15560sin180 0 ==
699.0
75
9.155, 3/
=
==
µ
πµβµee
T
P
180 lb
P
P
β=π/3
Problem 13
Journal Bearings. Axle Friction
• Journal bearings provide lateral support to rotating
shafts. Thrust bearings provide axial support
• Frictional resistance of fully lubricated bearings
depends on clearances, speed and lubricant viscosity.
Partially lubricated axles and bearings can be
assumed to be in direct contact along a straight line.
• Forces acting on bearing are weight W of wheels and
shaft, couple M to maintain motion, and reaction R
of the bearing.
• Reaction is vertical and equal in magnitude to W.
• Reaction line of action does not pass through shaft
center O; R is located to the right of O, resulting in
a moment that is balanced by M.
• Physically, contact point is displaced as axle
“climbs” in bearing.
29-09-2014
25
Journal Bearings. Axle Friction
• Angle between R and
normal to bearing
surface is the angle of
kinetic friction ϕk.
k
k
Rr
RrM
µ
φ
≈
= sin
• May treat bearing
reaction as force-
couple system.
• For graphical solution,
R must be tangent to
circle of friction.
k
kf
r
rr
µ
φ
≈
= sin
Wheel Friction. Rolling Resistance
• Point of wheel in contact
with ground has no
relative motion with
respect to ground.
Ideally, no friction.
• Moment M due to frictional
resistance of axle bearing
requires couple produced by
equal and opposite P and F.
Without friction at rim,
wheel would slide.
• Deformations of wheel and
ground cause resultant of
ground reaction to be
applied at B. P is required
to balance moment of W
about B.
Pr = Wb
b = coef of rolling resistance
29-09-2014
26
Problem 14
2 - 51
The two flywheels arre mounted on a common shaft
which is supported by a journal bearing between
them. Each flywheel has a mass of 40 kg and the
diameter of the shaft is 40 mm. if a 3 N.m couple M
on the shaft is required to maintain rotation of the
flywheels and shaft at a constant low speed, compute
(a) the coefficient of friction in the bearing and (b)
the radius rf of the friction circle.
002.11
)2/040.0)(81.9)(40(2
3sin,sin
=
===
φ
φφRr
MRrM
mm82.3m00382.002.11sin020.0sin
1947.0tan
0 ====
==
φ
φµ
rr f
Sample Problem 8.6
A pulley of diameter 4 in. can
rotate about a fixed shaft of
diameter 2 in. The coefficient of
static friction between the pulley
and shaft is 0.20.
Determine:
• the smallest vertical force P
required to start raising a
500 lb load,
• the smallest vertical force P
required to hold the load,
and
• the smallest horizontal force
P required to start raising
the same load.
SOLUTION:
• With the load on the left and force
P on the right, impending motion
is clockwise to raise load. Sum
moments about displaced contact
point B to find P.
• Impending motion is counter-
clockwise as load is held
stationary with smallest force P.
Sum moments about C to find P.
• With the load on the left and force
P acting horizontally to the right,
impending motion is clockwise to
raise load. Utilize a force triangle
to find P.
29-09-2014
27
Sample Problem 8.6SOLUTION:
• With the load on the left and force P on the right,
impending motion is clockwise to raise load. Sum
moments about displaced contact point B to find P.
The perpendicular distance from center O of pulley
to line of action of R is
( ) in.20.020.0in.1sin =≈≈= fssf rrrr µϕ
Summing moments about B,
( )( ) ( ) 0in.80.1lb500in.20.2:0 =−=∑ PM B
lb611=P
Sample Problem 8.6
The perpendicular distance from center O of pulley to
line of action of R is again 0.20 in. Summing
moments about C,
( )( ) ( ) 0in.20.2lb500in.80.1:0 =−=∑ PM C
lb409=P
• Impending motion is counter-clockwise as load is held
stationary with smallest force P. Sum moments about
C to find P.
29-09-2014
28
Sample Problem 8.6• With the load on the left and force P acting
horizontally to the right, impending motion is
clockwise to raise load. Utilize a force triangle to
find P.
Since W, P, and R are not parallel, they must be
concurrent. Line of action of R must pass through
intersection of W and P and be tangent to circle of
friction which has radius rf = 0.20 in.
( )
°=
===
1.4
0707.02in.2
in.20.0sin
θ
θOD
OE
From the force triangle,
( ) ( ) °=−°= 9.40cotlb50045cot θWP
lb577=P
