Seely F.B. Analytical Mechanics for Engineers
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PROBLEMS
IN
EQUILIBRIUM
121
117.
In
the crane
represented
in
Fig.
135
the
weight
of the
post
AB
is
Flon,
the
weight
of the boom
CE
is 0.8
ton
and that of
the
brace
DF
is
1
ton
The
reaction
at
B is
horizontal.
Find
the
reaction at
B
and the horizontal
and
vertical
components
of the
reaction at
A.
Also find the
pin
reaction on the
brace
at
F
and at
I).
Ans.
5
=
5.84
tons;
Z)
=
11.2tons;
F
=
11.9
tons.
118.
In the
crane
represented
in
Fig.
136,
the
weight
of the
post
AB is
1600
lb.,
the
weight
of
the
boom DE is 1200
lb.,
and the
weight
of
the
rod CE
may
be
neglected.
Find the
tension in the
rod
and
the
reaction of the
pin
at
D
on the
post.
6 tons
4
tons
FIG.
135.
FIG
136.
119.
Two
members
AB
and
BC
(Fig.
137)
are
connected
by
a smooth
pin
at
B and
their
lower
ends rest on
a
smooth horizontal
surface,
slipping
on the surface
being
prevented
by
a
cord
which
connects
the
ends A and
C.
The
weight
of AB is 120 lb. and
the
weight
of BC is
180 lb.
The
member
BC
FIG.
137.
122
EQUILIBRIUM
OF FORCE
SYSTEMS
also
carries a concentrated load of 450 Ib. as
shown. Find
the reactions
of
the
surface,
the
tension
in the
cord,
and the
pin
reaction at
B.
Ans.
A
=3721b.
72(7
=
378 Ib. 7
7
=
2341b.
fo
=
3441b.
120. In
Fig.
139
is
shown
a
shear
for
cutting
steel bars in
a
repair
yard.
What
force
P
is
required
to
give
a
pressure
of
70,000
Ib. on the anvil when the
3
ft.-6 in. arm is vertical. Solve
graphically.
121. Two bars
AC
and
BD
(Fig.
138)
are
connected to each other
by
a
smooth
pin
at C and to the floor
by
smooth
pins
at A and
B.
AC
=
CB
=
4
ft.
CD
=
3
ft. Find the
pin pressures
at
A,
B,
and C.
Ans. A
=
1401b.
B
=
109.8
Ib. C
=
1401b.
FIG.
138.
FIG. 139.
122. In
Fig.
140,
when
r is zero,
the
tension
in the
spring,
S,
is
50 Ib.
The
modulus
of
the
spring
is
58
when
r
=
3
in. and
also
find
the
pulls
P,
P
required
to
cause
this
spring
tension.
What
is
the
total
stretch
of
the
spring?
Ans.
S
=
94.1
Ib.
P
=
24.31b.
Stretch
=
1.62
in.
500 Ib.
1
FIG. 141.
PROBLEMS
IN
EQUILIBRIUM
123
123.
The
weights
of
the members
AD,
BC,
and
BD of
the frame
shown in
Fig.
141
are
100
lb.,
50
lb.,
and
75
lb., respectively.
Find
the
pin pressure
at
C
and the
pressure
at
A,
the surface at A
being
smooth.
Ans.
A
=
152
lb.
(7
=
225
lb.
124. Find
by
the
method of
sections,
the stresses
in
the members
CE, DE,
and DF
of
the
Howe
truss shown
in
Fig.
142.
FIG. 142.
124
EQUILIBRIUM
OF
FORCE
SYSTEMS
129. The
framework
shown in
Fig.
147
is
connected
at
the
points.
A,
D,
C,
and
E
by
smooth
pins
and is
supported
by
a smooth
surface at
B. Find
FIG.
146.
the
pin
reactions
at
A, D,
C,
and
E,
and
the
reaction at
B.
Neglect
the
weights
of
the
members.
1000
Ib.
^ | I
FIG. 147.
130. In
the
hydraulic
crane
shown
in
Fig.
148
the
weight
of the
boom
ABC
is 500
Ib.,
as
shown
in the
figure,
and
the
weights
of
the
members AD
and
BD
may
be
neglected.
In
raising
the boom
the
hydraulic
pressure
from
the
cylinder
transmits
a
pressure
P
to the
pin
at
D.
Neglecting
the
frictional
re-
sistance
find the
value of
P
required
to raise the boom
when loaded
as
shown.
PROBLEMS
IN
EQUILIBRIUM
125
Find
also
the stresses
in
BD and
AD and
the
pressures
of
the
rolls
at A and
at
D
against
the vertical
post.
Ans.
P
=
25001b.;
D
=
A=55401b.
DB
=
+6550
Ib.
;
AD=-7601b.
2000 Ib.
131.
The
crank-pin pressures,
PI
and
P,
on the crank shaft
shown
in
Fig
150
are
6000
Ib.
and 4800 Ib.
respectively.
Find the
bearing
reactions
RI
FIG.
149.
and
Rz
(Fig.
1506)
and
the
resisting
moment
Qq
required
for
equilibrium
of
the
shaft.
Ans.
#1
=
31201^
#2
=
19201b.
Qq
=
64,800
Ib.
-in.
132.
Three
rods each of
length
I
rest on
each
other
(Fig.
149)
in
such
a
126
EQUILIBRIUM
OF FORCE SYSTEMS
way
as
to
form a central
equilateral
triangle
each
side
of which
is
f
Z. A
load
P
is
applied
at
one corner
as shown.
Find the reactions
Ri, Rz,
and
R
3
in
terms
of
P.
FIG.
150.
CHAPTER IV
FRICTION
64.
Friction
Defined.
Friction is defined as the
resistance
which
one
body
offers to the motion of a second
body
when
the
second
body
slides or tends to slide over
the former.
The fric-
tional force is
tangent
to the surfaces of contact
of the two bodies
and
always
opposes
motion.
Friction is of
great
importance
in
engineering
practice.
Since
it
always
opposes
motion,
it
is an undesirable
and
expensive
factor
in
the
operation
of
many
machines or machine
elements,
and
in
such cases is reduced as much as
practicable
by
means
of lubricants.
In
other
machines it
becomes a
very
desirable and
useful
element,
as in various forms of
brakes,
friction
drives,
etc. In
fact, many
of our normal
physical
activities,
such as
walking,
would
be
impos-
sible without the aid of friction.
If
the resistance between
two
bodies
prevents
motion of one
body
relative to the
other,
the
resistance
is
called
static
friction,
while the frictional resistance be-
.
tween two
bodies which move
rela-
tive to each
other
is called kinetic
1
friction.
If the friction is
static,
the amount
of friction
developed
is
just
sufficient to maintain
equilib-
rium
with
the other
forces
acting
on the
body.
That
is,
static friction
\R
\|
is
an
adjustable
force
the
magni-
tude of which
is determined from
the
equations
of
equilibrium
for the forces
which
act on the
body.
Thus,
let
Fig.
151
represent
a
body
in
equilibrium
on
a
rough
horizontal
plane
under
the
action
of a
horizontal
force
P,
which
tends
to move the
body,
the
reaction, R,
of
the
plane,
and
the
weight, W,
of the
body.
Let the
reaction
R
be
resolved into
two
components
F and
N
parallel
and
perpendicular,
127
128
FRICTION
respectively,
to the
plane.
The
component
F
along,
or
tangent
to,
the
plane
is the frictional
force
as defined
above.
The
component
N
is
called
the
normal
pressure
and
R
is
called
the total
reaction.
Since the
body
is in
equilibrium,
the
equations
of
equilibrium
must
be
satisfied,
and hence F is
equal
to P. If the
force
P
is
gradually
increased,
F must increase
in the same
ratio in
order to
maintain
a
condition of
equilibrium.
There is
a definite
limit,
however,
to
the amount of
frictional resistance
that
can be
developed,
and
when
the value
of P
exceeds this
limiting
value
motion will
ensue.
The
limiting
or
maximum
value
of
the
frictional force is called
limiting friction
and is denoted
by
F'. Its
value
depends
on
the
normal
pressure
and on the
roughness
of
the
surfaces of
contact.
65.
Coefficient
of
Friction. In
order
to
compare
the frictional
resisting properties
of various
pairs
of materials
or of the same
pair
of
materials under
varying
conditions of their
surfaces
of
contact,
and in order to calculate the
maximum
frictional
force corre-
sponding
to
any
normal
pressure,
a certain
experimental
constant,
called the
coefficient
of friction,
is
used.
The
coefficient
of
static
friction
for
any
two
surfaces is defined
as the
ratio
of the
limiting
friction
to the
corresponding
normal
pressure. Thus,
if the
coefficient of static
friction is
denoted
by
ju,
it
may
be
expressed
as follows :
It is
important
to note that F'
in
the
above
equation
is the
maxi-
mum
friction which
the surfaces
can
develop,
that
is,
the
friction
corresponding
to
impending
motion. Thus
the
maxi-
mum
frictional force which
any
two
surfaces
can
develop
is
equal
to
pN.
The value
of
n
must be
determined
experimentally
and,
as
stated
above,
it
is a constant for
any
two materials
for a definite
condition
of
the surfaces of
contact.
It
varies
considerably,
however,
for
different conditions of the
surfaces,
and
it varies
widely
for different
pairs
of
materials,
as is shown in the
following
table which
gives
the values of the
coefficient of
friction for
dry
surfaces as
determined
by
Morin and others.
ANGLE
OF
FRICTION
129
COEFFICIENT
OF
STATIC FRICTION
Wood
on
wood
0.25
to 0.50
Metal
on
wood
20 to .60
Metal
on metal 15
to .30
Metal on leather 30
to
.60
Wood on leather .25 to .50
Stone
on stone 40
to
.65
Metal
on
stone
30 to
.70
Earth on
earth
25
to
1
.00
If
two
surfaces move relative to each
other,
the
ratio
of
the
friction
developed
to the
corresponding
normal
pressure
is
defined
as the
coefficient of
kinetic
friction.
The
value
of
the
coefficient
of
kinetic
friction for two surfaces
is
influenced
by
more
factors than
is the value of the coefficient
of
static
friction.
A brief
discussion
of the
influencing
factors is
given
in
Art.
68.
For values
of
the
coefficient of
kinetic friction for
various
conditions of
rubbing
surfaces the reader
is referred
to Good-
Mechanics
Applied
to En-
|
w
gmeenng.
66.
Angle
of Friction and the
Fric-
tion Cone. The
angle of
static
friction
for
two
surfaces is defined as
the
angle
between the directions of the total
reaction and
the
normal
pressure
when
motion is
impending.
Thus
in
Fig.
FlQ
152
152,
if
the force P
is
just
large
enough
to
develop
the
limiting
friction,
the
angle
which
R,
the reaction
of the
plane
on the
body,
makes with
the
normal
pressure,
N,
is
the
angle
of static friction and
is denoted
by
<.
Since the
components
of R
} parallel
and
perpendicular
respect-
ively
to the
plane,
are
F'
and
N,
it
is evident
from the
figure
that
tan
=
T=.
But since the ratio
is defined as the
coefficient
N
N
of static
friction, ju,
the
following important
relation
may
be
written :
\L=
tan
<(>,
that
is,
the
coefficient
of
static
friction is
equal
to the
tangent
of the
angle
of
static friction.
If
the two surfaces move
relative to
each other then
the
angle
between the
total
reaction and
the
normal
pressure
is
called the
130 FRICTION
angle
of
kinetic
friction.
Its value is
somewhat
less than
the
angle
of
static
friction,
since
the frictional
force
after
motion
ensues
becomes less
than the
limiting
friction.
The relation
/*
=
tan
also
holds for kinetic
friction,
the
value of
/*
for
kinetic
friction
being
somewhat less
than
for
static
friction. The
angle
of
friction
(for
both static and kinetic
friction)
is
convenient to
use
particu-
larly
in
the
solution of
problems
by graphical
methods.
If
a
body
rests on a
rough
plane
and
certain forces
are
applied
to the
body,
a
convenient
graphical
method
of
determining
whether the
body
will slide
or
not
makes
use of the
cone of friction.
The cone
of friction
for
two
plane
surfaces is
the cone
generated
by
revolving,
about the
normal to
the
plane,
a line
making
with
the normal an
angle
equal
to the
angle
of
friction as
shown
in
Fig.
153. If
the re-
sultant of
all
of
the forces
acting
on
the
body, except
the reaction of the
plane
(total
reaction),
falls
within the cone of
friction,
the
body
will not
slide;
if
the resultant
falls
outside the
cone of
friction,
the
body
will
slide. Thus in
Fig.
153 let
P
represent
the
resultant of
all of the
forces
acting
on the
body except
R,
the reaction of
the
plane.
If
the
angle
which
P makes
with the
FIG. 153.
normal to the
plane
is called
0,
the
force
tending
to move the
body
is
P
sin
0,
and the
normal
pressure
is
P
cos
0. The
body
will
not
slide
if
P
sin
0</iP
cos
0,
that
is,
if
tan
<
/*
or if
tan
<
tan
</>,
or if
0<0.
If >
0,
that
is,
if
P
falls
outside of the
cone of
friction
the
body
will
slide,
and
if
=
<f>
motion
is
impending.
ILLUSTRATIVE
PROBLEM
133.
A
lift
shown
in
Fig.
154
slides
on
a vertical
shaft
having
a
square
cross-section
2
in.
on a side. Find the
greatest
distance, x,
that a
load
W
can
be
placed
from
the
edge
of the
shaft and still
cause the lift to
slide on the
shaft.
Neglect
the
weight
of the lift
and use
0.2 for the
coefficient of
friction.