29-09-2014

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.

Fm = s N

Further increase in P causes the block to begin

to move as F drops to a smaller kinetic-friction
force Fk.
Fk = k N

The Laws of Dry Friction. Coefficients of Friction

Maximum static-friction force:
Fm = s N
Kinetic-friction force:
Fk = k N

k 0.75 s
Maximum static-friction force and kineticfriction force are:
- proportional to normal force
- dependent on type and condition of
contact surfaces
- independent of contact area

29-09-2014

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 motion
No friction

Motion impending

F
N
tan s = m = s
N
N
tan s = s

Motion

tan k =

Fk k N
=
N
N

tan k = k

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Angles of Friction
Consider block of weight W resting on board with
variable inclination angle .

No friction

No motion

Motion
impending

Motion

Problems Involving Dry Friction

All applied forces known

All applied forces known

Coefficient of static friction

is known

Motion is impending

Determine whether body

will remain at rest or slide

Coefficient of static
friction is known

Motion is impending
Determine value of coefficient
Determine magnitude or
of static friction.
direction of one of the
applied forces

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Problem-1
Determine the maximum angle 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 s.

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

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 k. 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)

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Problem-5
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 s= 0.3. Determine the angle
of inclination (If the ladder and the normal
reaction at B if the ladder is on the verge of
slipping.

Solution

Since the ladder is on the verge of slipping, then

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Problem-6
The 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

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

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

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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

Wedge as free-body

Fx = 0 :

Fx = 0 :

N1 + s N 2 = 0

s N 2 N3 ( s cos 6 sin 6)

Fy = 0 :
W s N1 + N 2 = 0

or
r
r
v
R1 + R2 + W = 0

+P=0

Fy = 0 :
N 2 + N3 (cos 6 s sin 6) = 0

or
r r
r
P R2 + R3 = 0

29-09-2014

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

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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

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Problem-7
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 which the wedge may
have and not pop out of the wood after
being driven by the sledge.

/ 2 = = tan 1
= 2 tan 1 (0.2) = 22.6 0

R
/2

Problem-8
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 s = 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

/2

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The force P required to start the wedge is

found from the equilibrium triangles of the
forces on the load and on the wedge.

2 - 28

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Problem-9
Determine the force P required to
force the 100 wedge under the 90kg uniform crate which rests
against the small stop at A. The
coefficient of friction for all
surfaces is 0.40.

= tan 1 = tan 1 (0.4) = 21.8

90(9.8)N

90(9.81)[cos 50 (0.6) + sin 50 (0.35)]

0.6m

R2 cos( + 50 ) 1.2 = 0

0.35m

R2 = 518 N

50

R2

A
R3

2 - 29

90(9.8)N

P cos10 0 518 sin R1 sin( + 10 0 ) = 0

0.6m

0.35m

P sin 10 0 518 cos + R1 cos( + 10 0 ) = 0

50

R2

R3

On solving
R1 = 471 N and P = 449 N

R2

100
R1

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Problem 13
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.

100 lb

R3

R2
150

R2

150
R1

= tan 1 = tan 1 (0.2) = 11.310

2 - 31

= tan 1 = tan 1 (0.2) = 11.310

a)

Block:

=0

100 lb

R3

100 + R2 cos(150 + 11.310 ) = 0

R2 = 111.6 lb

R2
150

Wedge:

R2

=0
150

R2 sin(150 + 11.310 ) P = 0

R1

P2 = 49.4 lb
y

2 - 32

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b) Rollers removed: Value of R2 from

100 lb body is unchanged.

=0
0

R2
0

R2 sin(15 + 11.31 ) P + R1 sin(11.31 ) = 0

150
R1

R1 is calculated from overall equilibrium as

R1 = 100 / cos(11.310 ) = 102.0 lb

P = 69.4 lb

Problem 10
SOLUTION
Calculate lead angle .
Using block and plane analogy with
impending motion up the plane, calculate
the clamping force with a force triangle.
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 s = 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.

With impending motion down the plane,

calculate the force and torque required to
loosen the clamp.

29-09-2014

tan =

L
2 r

4
= 0.1273
10

= 7.3

tan s = s = 0.30

s = 16.7
Using block and plane analogy with impending
motion up the plane, calculate clamping force with
force triangle.

40 N m
= 8 kN
5 mm
8 kN
W=
tan 24

Q r = 40 N m
tan ( + s ) =

Q=

Q
W

W = 17.97 kN

With impending motion down the plane, calculate

the force and torque required to loosen the clamp.

tan ( s ) =

Q
W

Q = (17.97 kN ) tan 9.4

Q = 2.975 kN

Torque = Q r = (2.975 kN )(5 mm )

)(

= 2.975 103 N 5 10 3 m

Torque = 14.87 N m

29-09-2014

Problem 11
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.
Helix angle

= tan 1 ( 24 / 40 ) = 10.810

Friction angle = tan 1 (0.15) = 8.530

> so screw is not self locking.

+ = 19.34 0 ,

a)

M = Pr tan( )
60 = P(0.02) tan 2.28

= 2.28 0

b)

P = 75.3N

M = Pr tan( + )
60 = P(0.02) tan 19.340
P = 75.3 N

Problem 12
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.

R1
100(9.8) N

R2
100
y
R2

R3

2 - 38

29-09-2014

Angle between R2 and y axis = 2+100

x

R1
100(9.8) N

R1

90-

100(9.8) N

90-

R2

100

R2

R2

= tan 1 = tan 1 (0.4) = 21.800

P

Block:

=0

R3

W cos = R2 cos(2 + 100 )

981 cos 21.800 = R2 cos 53.600

R2 = 1535N

2 - 39

= tan 1 = tan 1 (0.4) = 21.800

Wedge:

=0

R2 cos 36.400 = P cos

1535 cos 36.400 = P cos 21.800
P2 = 1331N
y

R2

x
P

Angle between
R2 and x axis =
90-(2+100)

R3

29-09-2014

Screw:

L
10
=
= 0.1273
2 r 2 (15)mm
tan s = s = 0.25

tan =

s = 20.090

M = P r tan( + )
M = 1331(0.015) tan 20.080
M = 7.30 N

Belt Friction
Consider the flat belt which passes over a
fixed curved surface. The total angle of
belt to surface contact in radians is , and
the coefficient of friction between the
two surfaces is .
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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2 - 44

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Problem 13
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.
A flat belt connects pulley A to pulley B.
The coefficients of friction are s = 0.25
and k = 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.

Problem 13
SOLUTION:
Since angle of contact is smaller, slippage will
occur on pulley B first. Determine belt tensions
based on pulley B.

T2
= e s
T1
T1 =

600 lb
= e 0.25(2 3) = 1.688
T1

600 lb
= 355.4 lb
1.688

Taking pulley A as free-body, sum moments about

pulley center to determine torque.

MA = 0:

M A + (8 in.)(355.4 lb 600 lb ) = 0
M A = 163.1lb ft

29-09-2014

Problem 13
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.

=/3

P
P

P = 180 sin 600 = 155.9lb

P
= e ,
T
= 0.699

155.9
= e / 3
75
180 lb

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

Journal Bearings. Axle Friction

Angle between R and

normal to bearing
surface is the angle of
kinetic friction k.
M = Rr sin k

May treat bearing

reaction as forcecouple system.

For graphical solution,

R must be tangent to
circle of friction.

r f = r sin k
r k

Rr k

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

Problem 14
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.

M = Rr sin ,

sin =

3
M
=
Rr 2(40)(9.81)(0.040 / 2)

= 11.02 0

= tan = 0.1947
r f = r sin = 0.020 sin 11.020 = 0.00382 m = 3.82 mm
2 - 51

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 counterclockwise 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

Sample Problem 8.6

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.
The perpendicular distance from center O of pulley
to line of action of R is
r f = r sin s r s r f (1 in.) 0.20 = 0.20 in.
Summing moments about B,

MB = 0:

(2.20 in.)(500 lb) (1.80 in.)P = 0

P = 611lb

Sample Problem 8.6

Impending motion is counter-clockwise as load is held
stationary with smallest force P. Sum moments about
C to find P.
The perpendicular distance from center O of pulley to
line of action of R is again 0.20 in. Summing
moments about C,

MC = 0 :

(1.80 in.)(500 lb ) (2.20 in.)P = 0

P = 409 lb

29-09-2014

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.

OE
0.20 in.
=
= 0.0707
OD (2 in.) 2
= 4.1

sin =

From the force triangle,

P = W cot (45 ) = (500 lb ) cot 40.9

P = 577 lb