Eighth Edition
8
CHAPTER
VECTOR MECHANICS FOR ENGINEERS:
STATICS
Ferdinand P. Beer
E. Russell Johnston, Jr.
Friction
Lecture Notes:
J. Walt Oler
Texas Tech University
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
Eighth
Edition
Vector Mechanics for Engineers: Statics
Contents
Introduction
Square-Threaded Screws
Laws of Dry Friction. Coefficients
of Friction.
Sample Problem 8.3
Angles of Friction
Problems Involving Dry Friction
Sample Problem 8.1
Journal Bearings. Axle Friction.
Thrust Bearings. Disk Friction.
Wheel Friction. Rolling Resistance.
Sample Problem 8. 4
Sample Problem 8.2
Wedges
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Eighth
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Vector Mechanics for Engineers: Statics
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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Vector Mechanics for Engineers: Statics
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
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Vector Mechanics for Engineers: Statics
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
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8-5
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Vector Mechanics for Engineers: Statics
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)
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• Motion,
(Px > Fm)
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Vector Mechanics for Engineers: Statics
Angles of Friction
• It is sometimes convenient to replace normal force
N and friction force F by their resultant R:
• No friction
• No motion
• Motion impending
Fm  s N
tan  s =
=
N
N
tan  s =  s
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• Motion
Fk  k N
tan  k =
=
N
N
tan  k =  k
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Vector Mechanics for Engineers: Statics
Angles of Friction
• Consider block of weight W resting on board with
variable inclination angle q.
• No friction
• No motion
• Motion
impending
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• Motion
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Vector Mechanics for Engineers: Statics
Problems Involving Dry Friction
• All applied forces known
• All applied forces known
• Coefficient of static
friction is known
• Coefficient of static friction • Motion is impending
• Motion is impending
is known
• Determine value of coefficient
• Determine magnitude or
• Determine whether body
of static friction.
direction of one of the
will remain at rest or slide
applied forces
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Eighth
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Vector Mechanics for Engineers: Statics
Sample Problem 8.1
SOLUTION:
• Determine values of friction force
and normal reaction force from plane
required to maintain equilibrium.
• Calculate maximum friction force
and compare with friction force
required for equilibrium. If it is
greater, block will not slide.
A 100 lb force acts as shown on a 300 lb
block placed on an inclined plane. The
coefficients of friction between the block
and plane are s = 0.25 and k = 0.20.
Determine whether the block is in
equilibrium and find the value of the
friction force.
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• If maximum friction force is less
than friction force required for
equilibrium, block will slide.
Calculate kinetic-friction force.
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Eighth
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Vector Mechanics for Engineers: Statics
Sample Problem 8.1
SOLUTION:
• Determine values of friction force and normal
reaction force from plane required to maintain
equilibrium.
 Fx = 0 :
100 lb - 53 (300 lb ) − F = 0
F = −80 lb
 Fy = 0 :
N - 54 (300 lb ) = 0
N = 240 lb
• Calculate maximum friction force and compare
with friction force required for equilibrium. If it is
greater, block will not slide.
Fm = s N
Fm = 0.25(240 lb ) = 60 lb
The block will slide down the plane.
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8 - 11
Eighth
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Vector Mechanics for Engineers: Statics
Sample Problem 8.1
• If maximum friction force is less than friction
force required for equilibrium, block will slide.
Calculate kinetic-friction force.
Factual = Fk =  k N
= 0.20(240 lb)
Factual = 48 lb
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Eighth
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Vector Mechanics for Engineers: Statics
Sample Problem 8.2
SOLUTION:
• When W is placed at minimum x, the
bracket is about to slip and friction
forces in upper and lower collars are at
maximum value.
• Apply conditions for static equilibrium
to find minimum x.
The moveable bracket shown may be
placed at any height on the 3-in.
diameter pipe. If the coefficient of
friction between the pipe and bracket is
0.25, determine the minimum distance
x at which the load can be supported.
Neglect the weight of the bracket.
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Eighth
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Vector Mechanics for Engineers: Statics
Sample Problem 8.3
SOLUTION:
• When W is placed at minimum x, the bracket is about to
slip and friction forces in upper and lower collars are at
maximum value.
FA =  s N A = 0.25N A
FB =  s N B = 0.25N B
• Apply conditions for static equilibrium to find minimum x.
 Fx = 0 : N B − N A = 0
 Fy = 0 : FA + FB − W = 0
0.25N A + 0.25N B − W = 0
0.5 N A = W
NB = N A
N A = N B = 2W
 M B = 0 : N A (6 in.) − FA (3 in.) − W ( x − 1.5 in.) = 0
6 N A − 3(0.25N A ) − W ( x − 1.5) = 0
6(2W ) − 0.75(2W ) − W ( x − 1.5) = 0
x = 12 in.
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Vector Mechanics for Engineers: Statics
Wedges
• Wedges - simple
machines used to raise
heavy loads.
• Block as free-body
 Fx = 0 :
− N1 +  s N 2 = 0
• Force required to lift
block is significantly
less than block weight.
 Fy = 0 :
• Friction prevents wedge
from sliding out.
or
− W −  s N1 + N 2 = 0
 

R1 + R2 + W = 0
• Want to find minimum
force P to raise block.
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• Wedge as free-body
F = 0:
x
−  s N 2 − N 3 ( s cos 6 − sin 6)
+P=0
F = 0:
y
− N 2 + N 3 (cos 6 −  s sin 6) = 0
or
 

P − R2 + R3 = 0
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Vector Mechanics for Engineers: Statics
Sample Problem 8.3
SOLUTION
The position of the machine
block B is adjusted by moving
the wedge A. Knowing that the
coefficient of static friction is
0.35 between all surfaces of
contact, determine the force P
required (a) to raise block B, (b)
to lower block B.
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Vector Mechanics for Engineers: Statics
Sample Problem 8.3
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Vector Mechanics for Engineers: Statics
Sample Problem 8.3
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Vector Mechanics for Engineers: Statics
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.
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Eighth
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Vector Mechanics for Engineers: Statics
Journal Bearings. Axle Friction
• Angle between R and
normal to bearing
surface is the angle of
kinetic friction jk.
M = Rr sin  k
• May treat bearing
reaction as forcecouple system.
 Rr k
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• For graphical solution,
R must be tangent to
circle of friction.
r f = r sin  k
 r k
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Vector Mechanics for Engineers: Statics
Sample Problem 8.5
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.
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
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.
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Vector Mechanics for Engineers: Statics
Sample Problem 8.5
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 j 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 = 611 lb
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8 - 22
Eighth
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Vector Mechanics for Engineers: Statics
Sample Problem 8.5
• 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.)(500lb) − (2.20 in.)P = 0
P = 409 lb
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8 - 23
Eighth
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Vector Mechanics for Engineers: Statics
Sample Problem 8.5
• 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.
sin q =
OE 0.20 in.
=
= 0.0707
OD (2 in.) 2
q = 4.1
From the force triangle,
P = W cot (45 − q ) = (500 lb ) cot 40.9
P = 577 lb
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