Seely F.B. Analytical Mechanics for Engineers
Подождите немного. Документ загружается.
FORCES
IN
RITE'S
INERTIA GOVERNOR
401
located
in
region
III,
the
moment
7\
becomes
equal
to
(I Mrc)a
and
this
quantity
may
be
positive
or
negative
according
as
7
is
greater
or less
than
Mrc.
With
G located
in
regions
II and
IV,
that
is,
to
the
left
of the
vertical
diameter,
the
moment
of
the
cen-
trifugal
force
and
the
moment
la
are
opposite
in sense.
In
a
gov-
ernor
in
which
the
whole
mass
is
concentrated
near
to
G so
that
7
is
small,
it
is
permissible
to
locate G
in
region
IV;
with
a,
gov-
ernor
of this
type
the
engine
may,
therefore,
rotate
in
either
direc-
tion without
changing
the
governor.
FIG.
403.
In
modern
governors
of the Rite's
type
7
is
made
very large
and
hence
the moment
la
is
very large compared
with
the
moment
Mrca.
Therefore,
in
these
governors,
the mass-center
of
M
may
be located on the circle or even in
region
II if
it is
desirable to do so
in
order
to decrease the
centrifugal
moment
Mrca
and,
in
conse-
quence,
the
size of the
spring
required.
From
Fig. 403,
it will be noted
that
the
moment of
the
cen-
trifugal
force
Mrui
2
is
2Mu>
2
times
the
area of the
triangle
OSG',
hence,
this moment
is
directly proportional
to
the area of
the tri-
angle
OSG.
From
this
fact,
it
appears
that the
size of the
spring
required
varies
directly
as
the
area
of
the
triangle OSG;
and,
to
obtain as
light
a
spring
as
possible,
consistent with
good
regula-
402
FORCE,
MASS,
AND
ACCELERATION
tion,
the
triangle
OSG
should be
made as
small as
possible.
If G
is
kept
in
region I,
this
object
is
accomplished
by
locating
G
either
near to S
at
G'
',
or
near
to at G"
(Fig. 403).
In
the
governors
shown
in
Fig.
404
and
405
it will
be
noted
that
G
is
located
near
to S.
ILLUSTRATIVE
PROBLEM
431. In
Fig.
404
is
shown a Rite's
inertia
governor
designed
for
a
6-h.p.
Nagle
engine.
Experience
shows that from
0.5 to
8 ft.-lb. of
energy per
FIG.
404.
h.p.
should
be
stored
in
the
governor
weight
in
moving
through
its arc
(work
done
in
stretching
the
spring).
In the
design,
5
ft.-lb.
per h.p.
was
assumed.
Find
the inertia
weight
required
and the modulus
of
the
spring
from
the follow-
ing
data:
0=e=3.4in.;
h
=
l.7m.;
mean
speed
=
250
r.p.m.;
1
=
QA in.
=
unstretched
length
of
spring;
Zi
=
ll
in.
=
minimum
length
of
spring;
1-2
=
12.2
in.
=
maximum
length
of
spring;
p
=
3.8
in.
=
moment
arm
of
spring
tension.
Solution. Since
the moment
of the
centrifugal
force
(T
c]
must
equal
the
moment of
the
spring
tension
(T
f)
we
have the
equation,
That
is,
ro>
2
/=
7^X3.8,
FORCES IN
RITE'S INERTIA
GOVERNOR 403
or,
W
.&
(since rf
=
he),
where
T
m
denotes
the mean value
of the
spring
tension.
In
order to determine
the value
of T
m
,
the
work stored
in the
governor
is
equated
to the work done
in
stretching
the
spring. Thus,
_
12.2-11.0
12
'
Hence,
Using
this
value
of
T
m
in
the above
equation,
we
have,
From
which,
w=in\b.
The
length
of the
spring,
when the
governor
is in its
mean
position
is
(12.2
+
11.0)
=11.
6.
in.
Hence the modulus of the
spring
is,
300
11.6-9.4
2.2
300
=
-
=
136 Ib.
perm.
PROBLEM
432.
In
the Rite's
governor
shown
in
Fig.
405,
the
inertia
weight
is 111
Ib.
and
the tension
of the
spring
in
mid-position
is 500
Ib. If e
=
7
in.,
h
=
2.6
in.,
r
=
3
in.,
and
p
=8.
2
in.;
find,
FIG. 405.
404
FORCE,
MASS,
AND
ACCELERATION
(a)
The normal
speed
of
the
engine,
that
is,
the
speed
when the
governor
is
in
mid-position,
and
(6)
The
power
of the
governor
in
ft.-lb.
per
h.p.
if
the
engine
is
rated
at
50
h.p., assuming
that the
spring
stretches from
20
in. to 22
in.
when
the
governor
moves
through
its whole
range.
Ans.
w=268
r.p.m.;
1.67 ft.-lb.
per h.p.
CHAPTER
X
WORK
AND
ENERGY
176.
Introduction.
In the
preceding
chapter
the
relations
between
force,
mass,
and acceleration
were
developed
from
New-
ton's laws
of motion
and
applied
to the
motion of
bodies
under
the
action
of unbalanced
forces. As
already
noted,
the
quantities
involved
directly
in Newton's laws are
force,
mass,
and
acceleration.
But,
acceleration involves the
quantities
velocity, distance,
and
time.
Now,
in
many
problems
in
engineering,
it is
convenient
to use
certain other
quantities,
the
more
important
of which
are:
work,
power, energy,
impulse,
and momentum. The ex-
pression
for each
of these
quantities
is a combination
of
some of
the six
quantities
(force, mass,
acceleration, velocity, distance,
and
time)
which are involved
in
Newton's
laws of
motion.
Thus,
force
and distance combine to measure
work;
force, distance,
and
time combine to measure
power;
mass and
velocity
combine to
measure
momentum and
some forms of
energy;
force and time com-
bine to
measure
impulse,
etc.
Although
the
conceptions
of
these
quantities
are
more or less a
result
of our
experience
with
physical
phenomena,
the exact relations between
them,
as
expressed
in
certain
principles
to be
developed
in
the
following pages,
are
based
on the definite
fundamental laws of
Newton.
The
present chapter
is
devoted to
a discussion of
the
meaning
and
use of
work,
of
energy,
and
of certain
principles
which involve
these two
quantities.
Although
no
fundamental
physical
laws
other
than those of Newton
are used
in
developing
the
principles
of work
and
energy, nevertheless,
the
method
of
analysis
which
makes use of work and
energy, possesses
certain
advantages
over
the
method which
makes
use
directly
of
force,
mass,
and accelera-
tion,
even
in
certain
types
of
problems
which
involve
only rigid
bodies
having
rather
simple
types
of
motion
'such
as
translation,
rotation,
and
plane
motion.
And,
in
dealing
with
non-rigid
bodies
having
unordered
motion,
that
is,
motion
in
which
the
particles
405
406
WORK AND
ENERGY
of
the
mass
system
(body)
do
not
follow
definite
known
paths,
the
.principles
of
energy
are of
particular
importance.
In
fact,
the
study
of
the behavior of
non-rigid
bodies in
general,
such
as
water, steam,
gas,
and air
is
largely
based on
the
principles
of
energy and, hence,
these
principles play
an
important
part
in
hydraulics, thermodynamics,
meteorology,
etc.
1.
WORK
AND
POWER
177. Work Defined.
A
force does work on a
body
if
the
body
on
which
the force acts is
moved so that the
displacement
of the
point
of
application
of
the force
has a
component
in
the
direction
of
the force.
The amount of work
done
by
a force
is the
product
of
the force and the
component
of the
displacement
of
its
applica-
tion
point
in
the direction
of
the
force.
The work done
by
a force
may
also be
expressed
as the
product
of
the
component
of the
force
in
the direction of the
displacement
of its
application
point
and
the
displacement.
The
component
of
the force
in
the
direction
of
the
displacement
of its
application point
is
often called
the
working
component.
And the
component
of the
displacement
in
the
direction
of
the
force is
called
the
effective
displacement.
178.
Algebraic Expressions
for Work Done
by
a Force.
The
mathematical
expression
for
the
work,
w,
done
by
a
force, F,
in a
displacement,
s,
of its
application point
depends
on the
way
the
force varies
during
the
displacement.
Several
important
special
cases are
considered here.
7. The
force is constant
in
magnitude
and
in
direction
and
agrees
in
direction
with the
displacement
as,
for
example,
the force
exerted in
lifting
a
body vertically
upward
with
a uniform
acceler-
ation.
The
amount of
r*
work
done
is,
/?Vw
WORK
DONE BY A FORCE
407
in
which
F
cos
6
is denoted
by
F
t
since
F
cos 6 is
tangent
to the
path
of the
point
of
application.
777. The
force
varies
in
magnitude
but
not
in
direction and the
direction
agrees
with that of
the
displacement
as,
for
example,
the
force exerted
in
compressing
a
helical
spring
or the steam
pres-
sure
against
the
piston
of a steam
engine
after cut
off.
Thus,
in
compressing
a
spring,
the force
corresponding
to
any displace-
ment, s,
is
F
(Fig.
407)
and
this force
may
be assumed
to
remain constant in an infinitesi-
mal
displacement,
ds.
Hence,
according
to case
7,
the work
done
by
the force F in
the
dis-
placement
ds is dw
=
Fds
and,
p 40
the total work
done
on the
spring,
as
F
varies from its initial to its
final
value,
is
w
=
\ *Fds.
A
In
order to evaluate the
integral by
the method
of
calculus,
F
must be
expressed
in
terms of s. That
is,
the manner
in
which F
varies with
s
must be known.
IV.
The force
varies
in
magnitude
and
in
direction
as,
for
FIG.
408.
example,
the
pressure
of
the
connecting
rod on
the
crank-pin
of
an
engine (Fig.
408).
The
expression
for
the work
done
by
the
force
408
WORK
AND ENERGY
is found
by
the same
method
as
was used in
case III
except
that
the
tangential
component
of
the
force must be
used.
Hence,
w
=
This
expression
applies
whether
the
displacement
is
along
a
circular
path
or
not.
But when the
displacement
takes
place
in
a circular
path
the
elemental
displacement
ds is
expressed
by
the
equation
ds
=
rd6.
Whence,
r
rs
re2
F
t
ds=
I
F
t
rd0=
j
Tdd,
in
which
T
is
the
torque
or moment
of
the
force
about the center of
the
circular
path.
And if
the
torque
remains
constant
during
an
angular
displacement,
6=62
61, then,
CQi.
W
=T I
dO=T(8
2
-6i')
=
T'0.
Thus,
in one
revolution,
the work done
by
the
force
F
having
a
moment
T
is w
=
T
7
-
2?r.
And
if n
revolutions occur
per
unit of
time,
then
the work
done
per
unit
of
time
is,
w=T-2irn.
179.
Work Done
by
a
Couple.
Since the
magnitude
of
the
moment
of a
couple
is the
algebraic
sum
of the
moments
of
the
two
forces which constitute the
couple,
and since
the
moment of
a
couple
is the
product
of either force and the
perpendicular
distance
between the
action
lines of
the
forces,
it
follows from
the
above
discussion that the work done
by
a
couple,
having
a
moment
T,
in
a.
displacement
d6,
is
r
w
=
Jei
TdQ,
when
the moment of the
couple
varies.
And the work
done
is,
w=T-6,
when the
moment of
the
couple
is
constant
in
the
displacement.
Hence,
the work done
by
a
couple
having
a
constant moment
is
the
product
of
the moment of the
couple
and the
angular displace-
ment of
the
couple.
WORK
A
SCALAR
QUANTITY
409
180. Work
a Scalar
Quantity. Sign
and Units
of Work.
Work
is a
scalar
quantity.
Thus
the
work
done
by
one force
may
be
added
(algebraically)
to the work
done
by
another
force
regardless
of the directions
of the
forces
or of the
displacements
of their
points
of
application.
And,
the work
done on one
body
of
a
system
may
be added
(algebraically)
to
the work done
on
the
other bodies
of the
system
in
order
to obtain the
total
work
done
on the
system, regardless
of
the
manner
in
which the bodies
move.
The fact that work is a scalar
quantity, frequently
makes
it convenient to use this
quantity
under conditions which make
the use of vector
quantities
difficult or
impossible.
It is
convenient to
regard
the work done
by
a force as
having
sign.
Work
is
positive
when the
working
component
of the
force
and
the
displacement
of its
application point agree
in
sense;
and
work is
negative
when the
working component
and
the
displace-
ment are
opposite
in
sense.
Thus,
a force which
retards the motion
of a
body
does
negative
work
on
the
body.
The
unit of work is the
work
done
by
a unit force
acting through
a unit
distance and
hence, depends
on the units
used
for
force
and
distance.
Thus,
the more common
units for work
in
the
gravi-
tational
(engineers') system
of units are
the
foot-pound,
inch-
pound,
meter-kilogram,
etc. No
one-term
names
are
given
to
the
units
of work
in
the
gravitational
system
of units. The
common
units
of work
in
the absolute
system
of
units are
the
dyne-centi-
meter,
which is called an
erg,
and
the
joule,
which is 10
7
ergs.
For
large
units of work the
horse-power-hour
and
the
kilowatt-hour
are used. For a definition of
these units see
Art. 184.
181.
Graphical Representation
and
Calculation of
Work.
In
calculating
the
work
done
by
a
variable
force,
by
the
calculus
method,
the
working component,
F
t
,
of
the
force must
be
expressed
in
terms
of the
displacement
s. If
it is
impossible
to
express
F
t
in terms of
s,
or
if,
when
possible,
the
expression
for F
t
is
complex
and difficult to
use,
as
in
the
case of
the
tangential
effort on
the
crank-pin
of a steam
engine,
the relation
between
F
t
and s
may
be
expressed
graphically
by
means of
a
graph
or
curve,
and
the work
done
may
be
found
from
the
graphical
diagram
as
follows : If
values
of
F
t
and
s are
plotted
on a
pair
of
rectangular
axes for all
positions
of the
application
point
of
the
force
F,
the
curve
joining
the
plotted
points
is called
a
tangential-force-space
(F
t
-s)
curve
(Fig.
409).
In
most
problems,
only
a
sufficient
number
of
values
of
F
t
are
410 WORK AND
ENERGY
plotted
to make it
possible
to draw a
reliable F
t
-s
curve,
values
of
F
t
being
plotted
more
frequently
when
the
value of
F
t
is
changing
the more
rapidly.
The
work done
by
a
variable
force
F
as
shown
in
the
preceding
article
is
w
=
r*
=
F
t
ds.
A
But,
F
t
ds
repre-
sents an
elemental
part
of the area
(Fig.
409)
between the F
t
-s
curve
and the s-axis.
And,
the total
area
under the
F
t
-s
curve
between
any
two ordinates
corresponding
to
abscissas
si
and
s
2
is,
area
=
Therefore,
the work done
by
a force
in
any
displacement
s
is
represented
by
the area
under the
tangential-force-space
curve
between
the
ordinates
si
and
S2.
This
diagram
is called
a work
diagram.
In
determining
the
amount of
work
represented by
the work
diagram,
the scales
used
in
plotting
the F
t
-s curve
must be
considered.
Thus,
if
ordinates are
plotted
to a scale
of
1 in.
=
50 Ib.
and
abscissas
to
the
scale of 1 in.
=
5
ft.,
then
each
square
inch of
area under
the
F(-s
curve
represents
250
ft.-lb. of work.
Since the
area of
a work dia-
gram
equals
the
product
of the
average
ordinate
and the
base,
the
work done
by
a
force
equals
the
average
value of the
tangential
(working) component
of
the force
and the
length
of the
path
described
by
the
application
point.
The area of the work
diagram may
be
found
by
means of
a
planimeter
or
by dividing
the area into small
strips
and
applying
Simpson's
rule.
Or,
in
some
cases,
less
exact methods
may
be
employed
in
estimating
the area.
ILLUSTRATIVE PROBLEMS
433.
A
helical"
spring (Fig.
410)
having
a
modulus of 200 Ib.
per
in.
is
com-
pressed
s
=
4 in.
by
an axial load. How
much work
is done
by
the
(variable)
load
in
compressing
the
spring?
s=Displacement
FIG. 409.