Seely F.B. Analytical Mechanics for Engineers
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THE
WEDGE
141
By
eliminating
R
3
from
(3)
and
(4)
the
following
equation
is
obtained:
_Pcosl2
Kz
sin
39
(6)
By
equating
values of R
2
in
(5)
and
(6)
the
value of P
may
be
found.
Thus,
P
cos
12
1000 cos
12
Therefore,
sin
39 cos
39
P
=
1000 tan
39 =810
Ib.
PROBLEMS
148.
The
load
of
100,000
Ib.
(Fig.
166)
is
raised
by
applying
forces
P,
P
to the
wedges.
What is the
required
value
of
P if
the
coefficient of
friction
is
0.2 for all surfaces
of
contact?
Ans.
P
=
26,667
Ib.
I
100,000
Ib.
FIG.
166.
149.
Find the
value
of
the
force
P
which must be
applied
to
the
wedge
A
(Fig.
167)
in
order
to raise
a
weight
W carried
by
the block
B,
assuming
that
the
coefficients of
friction
for
all
rubbing
surfaces are
equal.
Ans.
P
=
TFta
160.
Find
the
expression
asked
for in Problem
149
if
friction is
neglected
at
all
surfaces
except
that
between
the block B
and
the
wedge
A.
W
1000
Ib.
1000
Ib.
\ /
W \\
V
FIG.
167.
FIG.
168.
161. Find
the
force P
which
must
be
applied
to
the
wedge
shown
in
Fig.
168 in
order
to
overcome the
1000-lb.
forces
applied
to
blocks A
and B
as
142
FRICTION
shown. Assume
the
coefficient of friction
for
all
rubbing
surfaces
to
be
0.1.
Ans.
467
Ib.
152. What
is
the
value of
P in
Problem
151
if
the
horizontal
surfaces,
on
which
blocks
A
and
B
rest,
are
smooth?
71.
The
Screw.
A
screw
is,
in
effect,
an
inclined
plane
wound
around a
cylinder.
Screws are
made
with
square
threads
and
with
triangular
threads,
|w
but
square
-
threaded
screws,
only,
will
be con-
sidered
here.
Fig.
169(a)
shows
a
jack-screw
with
square
threads
which
is
used in
raising
or
lower-
ing
heavy
loads. The
radius of
the base of
the
thread
is
denoted
by
r\
and
the
outer
radius
by
r<2\
a is
called
the
'pitch
angle,
and
p
is
called the
pitch
of
the
screw.
Let
it
be
required
to find
the
force,
P, which,
when
ap-
plied
at the end of a
lever
of
length
a,
is
just
sufficient
to
raise
the
load
W. The forces
which
hold the screw in
equi-
librium
are: the
force
P;
the
pressure
of the
cap,
C,
on the
head, H;
and
the reaction of the
nut,
B,
on the screw.
The
latter is
distributed
over
the area of
the
threads
in
contact with
the
nut.
If the friction
between
FIG.
169.
the
cap
and
the head
of
the screw
is
neglected,
the
pressure
of
the
cap
on
the head of
the
screw
will be a vertical force
equal
to
W.
The
problem
will
first
be
solved on
this
assumption.
Two
of the
equilibrium
equations
which
apply
to this
type
of force
THE
SCREW
143
system
will
be
sufficient for the solution of the
problem,
namely"
2^
=
and
2^
=
where z is taken as the axis
of the
screw.
The
reaction
between
the nut and the thread
of the
screw
on
an
element
of area
dA will be denoted
by
dR.
This force
may
be
resolved
into
components
dN
normal to
the thread
and
dF
f
par-
allel to
the thread as shown in
Fig.
169(6).
In
taking
moments
about the axis
of the thread it
will be
sufficiently
accurate to con-
sider the moment-arm
of dF' to
be
equal
to the mean radius of the
thread,
J (ri+rz),
which will
be denoted
by
r. The
equilibrium
equations
stated
above, then,
become:
cos a-^dF' sin
a-W
=
Q,
2rdN sin a-2rdF'
cos a
=
0.
Since
dF'
=
jjidN,
these
equations may
be
written
:
cosa2dN-iJLsina2dN-W
=
Q,
.....
(1)
Pa-r
smaZdN-pr
cosa2dN
=
.....
(2)
By
eliminating
2dN
from
(1)
and
(2)
the
equation
obtained
is,
cos a
fj.
sin a
By substituting
tan
<t>
for
n
this
equation
may
be
written in
the
form,
Pa
=
Wr
tan
(0+a).
If the
pitch
angle
a is
large
and
the
angle
of
friction is
small,
the load
W
will
cause the
screw to run
down
unless a
force
is
applied
to
prevent
it.
The
force
P
required
to
hold the
load
is
found
by
a method of
analysis
similar
to that used
above,
the
only
difference
being
that
the
sense
of the
frictional
force
is
reversed. The least value of P
required
to
prevent
the
screw from
running
down is
given
by
the
equation,
If
a
=
4>
in the above
equation,
the force P
reduces to
zero,
that
is,
the load
will
be
held
by
friction alone. If
a
<
0,
a
force
144 FRICTION
is
required (the
sense
of
which
is
opposite
to
that in
the two
cases
considered)
to lower the
load. The
value of the
force
required
to
lower
the load is
given
by
the
equation,
Pa
=
Wr
tan
(<j>a).
In
practice, jack-screws
are
always
made self
-locking,
that
is,
a.
is
made less than
<j>.
The
values of
P
for the
three
cases
just
con-
sidered
will
be
changed
somewhat if friction between the
cap
and
the head of the screw
is considered.
If
the area of contact
between
the
cap
and the head
is a full circle the effective arm of the
fric-
tional resistance is two-thirds
the radius of the circle as will
be
shown in Art.
74.
In
practice,
however,
the area of
contact is
usually
a hollow
circle,
and
it
will be
sufficiently
accurate to
take
the mean
radius as
the effective arm of the frictional
resistance.
If the mean radius is
denoted
by
r' and the coefficient
of friction
between
the
cap
and the
head
of screw
by
/z,
the values of
P
are
given
by
the
following equations:
To
raise
the
load,
Pa=Wr
tan
(0
To
hold
the
load,
Pa=Wr
tan
(a-0)-
To
lower the
load,
Pa=Wr
tan
O-a)+wW.
PROBLEMS
153. The mean
diameter
of the screw
of
a
square-threaded jack-screw
is
1.8 in.
The
pitch
of the
thread is
0.4 in. and
the
coefficient
of friction
for the
screw
and
nut
is
0.12. What
force must
be
applied
at the
end of
a
lever 18
in.
long
to raise a
weight
of
5000
lb.? What force is
required
to
lower the
weight?
Ans.
48.1
lb. 12.3 lb.
154.
A
weight
of 1000
lb.
is
lifted
by
applying
a
couple,
Pd,
to
the
hand wheel of
the
apparatus
shown in
Fig.
170
The
diameter.
d,
of the hand wheel
is 20
in.;
the
mean diameter
of the screw is
1.5
in.;
the
pitch
of
the
thread
is
|
in.;
and the coefficient
of
friction is
0.15.
Find
the
values
of the forces
of
the
couple
when
the value of
6
is
15.
W=1000 lb.
FIG. 170.
JOURNAL
FRICTION
AND
FRICTION
CIRCLE
145
FIG.
171.
166.
The
shaft-straightening
hand
press
shown
in
Fig.
171
is
usecT
for
bending
or
straightening
3
in.
steel
shafts. What
force, P,
applied
at the
end of
a 36-in.
lever
is
required
to
produce
a
pressure,
Q,
of
24,000
Ib.
on
the
shaft?
The
threads
have
a
mean
diameter
of
2 in. and
there
are
four
threads
per
inch.
Con-
sider
friction
between
the screw
and
nut
only,
and
use a
value
of
0.2 for
the coefficient
of
friction.
Ans.
P
=
161
Ib.
72.
Journal
Friction and
the
Friction
Circle.
In
Fig.
172
is
shown
an
axle or
jour-
nal
in
a
bearing,
the diameter
of
which
is
slightly
greater
than
the diameter
of
the axle.
The difference
in the diameters
is
exaggerated
in the
figure
for
the
sake
of clearness.
Since
the
contact
between the
axle and
the
bearing
is
along
a
line,
the reaction
of
the
bearing
on
the axle
is distributed
along
the
line
of
contact
and
may
be
replaced
by
a resultant
force, R,
at
A,
the
midpoint
of the
line of
contact.
If
the
axle rotates
in
the
bearing
or,
if
motion
impends,
the
angle
between the
action
line of
the
reaction, R,
and
the
normal to the
surfaces
at
A
is
the
angle
of friction
</>.
In
the former case
the
angle
is
the
angle
of kinetic
friction
and
in the
latter case it
is
the
angle
of
static
friction. If
the radius of
the
axle
is
denoted
by
r,
a circle
having
a
center
which
coincides
with the
center,
0,
of
the
axle
and
which has a
radius
equal
to
FIG.
172.
r
s
i
n
wm<
De
tangent
to the
action line
of
R. This
circle is
called the
friction
circle.
It
will be
noted
that
when
the
angle
is
small,
sin
</>
is
approximately
equal
to
tan
and
hence also
approximately
equal
to
/*. Therefore,
for
small
values of
JJL
the
radius of the
friction
circle
may
be
taken
as
jur
without
serious
error.
146
FRICTION
The friction
circle is of
importance
in
locating
the
line of
action
of
the
reaction
between an
axle and
its
bearing
since
this
line
is
tangent
to the friction circle
when
motion
occurs or is
impending.
For
example,
in
Fig.
172
the
line of
action
of
the
reaction is
one
of
the
two
tangents
that can
be
drawn from A
to
the
friction
circle.
The direction
of motion
(or
impending motion)
will
determine
which one of
the two
tangents
should
be used.
Thus,
if
the
axle
rotates,
or is about
to
rotate,
in a
clockwise direction
the action
line of the reaction is as
indicated in
Fig.
172.
In
general
the
posi-
tion
of
the
line of
contact
between the
axle
and
bearing
is
not
(6)
FIG. 173.
initially
known.
This
position
can
usually
be
determined,
how-
ever,
by
means
of the
friction circle.
For
example,
consider
the
bell-crank,
shown
in
Fig.
173
(a),
which
is
acted
on
by
a known
force
Q.
Let
it
be
required
to determine
the
magnitude
of
the
force,
P
t
which
must
be
applied along
the line OB
in
order
to
cause
motion
to
impend
in
a clockwise
direction. The
forces
acting
on the bell-crank
are
Q,
P,
and
the
reaction, R,
of
the
bearing.
Since
the
three
forces
are
in
equilibrium
they
must be
concurrent.
Hence if P
and
Q
intersect at
0,
R
must
also
pass
through
0.
Furthermore,
R
must be
tangent
to
the friction
circle.
There
are
JOURNAL
FRICTION
AND
FRICTION CIRCLE
147
only
two
tangents
that
can be drawn
from
to the
friction
circle.
The one
to be used
is indicated
in
the
figure.
Therefore,
the loca-
tion of
the
line of contact between
the
axle and the
bearing
is
at A.
Since
the action
line
of R
is
now
known,
the
magnitudes
of P
and
R
can
easily
be found
by
drawing
the
force
polygon
(Fig.
1736)
for
the three forces.
If
motion
impends
in the
opposite
direction and
it is
required
to find
the least
value
of
the force P
which will
maintain
equilibrium
the other
tangent
would
be used.
ILLUSTRATIVE
PROBLEM
156.
Fig.
174
represents
a
pulley,
having
a diameter of
2
ft.,
mounted on
a
2-in.
axle.
The
coefficient
of friction
between the
axle
and
the
bearing
in
which
it rests
is
0.15. What is
the
least
value
of the force
P
which will
raise
the
200-lb.
weight, assuming
that the
fric-
tion between
the
rope
and
pulley
is
sufficient
to
prevent
slipping? Neglect
the
weight
of
the
pulley.
Solution.
There
are
three forces
act-
ing
on the
pulley:
the
200-lb.
force, P,
and
R,
the reaction of
the
bearing.
Since these
three
forces
are in
equi-
librium
they
must be
parallel
(Art.
48).
Hence R
must be
vertical
and,
since it
must
also
be
tangent
to
the
friction
circle,
its line of
action
is
determined.
The
radius of the friction circle
is
pr
or
0.15
in. The
equations
of
equilibrium
are:
FIG.
174.
?F
V
=
R-P-200
=
0.
SM
=
200
X
12+0.
15R
-12P
=
0.
Eliminating
R from
these
equations
we
have,
P
=
2051b.
If the
force R acted
through
the
center
of
the
axle,
that
is,
if
the
axle
were
smooth,
the
value of
P
would
be
200
Ib.
148
FRICTION
PROBLEM
157.
A
load
of
1000
Ib.
is raised
by
a
wooden
wheel
and
axle as
shown in
Fig.
175.
What
is the
least
value
of the
force P
which
will
raise the
load if
the
coefficient
of
friction
between
the
axle and
the
bearing
is
0.4?
W
FIG.
176.
73.
Pivot
Friction.
In
Fig.
176 is shown
a
pivot
in
a
step
bearing.
The
pivot
will be
assumed
to turn in
its
bearing
and
hence
kinetic
friction, only,
is
involved. In
order
to
determine
the
resisting
moment due to the frictional
forces
at
the end
of a
pivot,
certain
assumptions
will be made as
follows:
1.
The coefficient
of
friction
is
constant over
the
end
of the
pivot.
2. The
pressure
between the
pivot
and the
bearing
at
any
point
(a)
is
constant,
or
(6)
varies
in such
a manner
that
the
wear
of
the
pivot
in
the
direction
of its axis
(axial wear)
is
uniform over the
area
of the
pivot. Or,
as
sometimes
stated,
the
wear,
at
any point,
normal
to the
rubbing
surfaces
(normal
wear)
is
proportional
to the work
of
friction.
Since
the value
of
the
coefficient
of
kinetic friction
varies some-
what
with
the
velocity
of the
rubbing
surfaces,
it is
evident
that
the coefficient
is
not constant over the end
of the
pivot,
for the
velocity
of
any
point
varies as its
distance
from
the axis
of
the
pivot.
The
assumption
made in
(1),
however,
will not
cause
SOLID
FLAT PIVOT
149
serious
error if a mean
value
of
the
coefficient
is
used.
For
a new
pivot
the
assumption
2
(a)
seems
reasonable, although
the
pressure,
probably,
will not remain
constant after wear occurs.
If the wear
normal to
the
pivot
(normal
wear)
at
any
point
of the
pivot
is
proportional
to
the work
done
by
the frictional
force at that
point,
that
is,
to the
product
of
the frictional
force,
F',
and
the distance
traveled
by
the
point,
the
normal
wear
will also
be
proportional
to
the normal
pressure,
since
F' is
equal
to
/JV.
If,
then,
the
normal
pressure
is
constant,
it is evident
that,
for a
time,
the
greater
wear will
occur near the
circumference
of the
pivot.
The effect
of this
greater
wear
at the
edge
of
the
pivot
will be
to increase the
intensity
of
pressure
near the
center and
to decrease
it near the
edge
of the
pivot.
The increased
pressure
near the
center
will
in
turn result
in
increasing
the wear
near the center
and in
decreasing
it near the
edge.
This
process
will
continue
until
a uniform condition has been
established,
after
which,
the
wear will
be uniform
over
the end of
the
pivot.
The
assumption
made
in
2(6), therefore,
seems reasonable
for
pivots
that
have
been in
use for some time.
The
frictional
moment
will now
be
determined
for
several
pivots.
74.
Solid Flat Pivot.
Uniform
Pressure.
In the
flat
pivot
shown in
Fig.
176
the
axial load will
be
denoted
by W,
the radius
of the
pivot by
r,
and
the area of the
pivot
by
A.
The frictional
moment
may
be found
as follows:
W
Pressure
per
unit of area
=
-
2
.
W
W
Pressure on element of
area
=
^dA=
^
pdpdd.
uW
Frictional force on
element
dA
=
~
pdpdd.
irr
2H
U.W
Moment of frictional force on dA
=
-5-
p
2
dpdd.
Total frictional moment
=
~
I
I
P
2
dpdO
_w
r* r
~^Jo
Jo
'
Uniform
Wear.
Since the
wear
at
any point
is
proportional
to
the
distance
traveled
by
that
point,
it is
proportional
to
the
dis-
150
FRICTION
tance, p,
of the
point
from the
axis of the
pivot.
The
wear,
how-
ever,
is also
proportional
to the
friction
force
and
hence,
to
the
pressure,
p.
That
is, pp
is
equal
to
some
constant,
C. The
frictional
moment,
then, may
be
found as
follows:
r
2
"
r
Frictional
moment
=
I
jj.pp
2
dpdd
Jo
Jo
=
C (
I
pdpdd
(1)
But,
the total
pressure
must be
equal
to
W.
Hence,
another
equation containing
C
may
be
found.
Thus,
W
-jrr
ppdpdd
(2)
FIG. 177.
By
eliminating
C from
equations
(1)
and
(2)
the
resulting
expres-
sion
is,
Frictional
moment
=
75.
Hollow
Flat
Pivot or Collar
Bearing.
The determination
of
the frictional
moment for a hollow flat
pivot
(Fig.
177a),
or
collar
bearing
(Fig.
1776),
involves the same
steps
as were used
in
finding