HYDRAULIC
AND PNEUMATIC
ENGINEERING 51
This
is very striking and it is explained as above. If, for example, the
area of the pipe is 1 sq. in. and that of the disk is 500 sq. in. then
1
ft), of water in AB will support a weight of 500 lbs. on the disk.
Similarly
1/2 lb. of water in AB will support 1/2 x 500 = 250 lbs. on the
disk, or 1/4 lbs. of water in AB will support 1/4 x 500 = 125 lbs. on
the
disk, and so on.
EXPERIMENT
No. 27
To
make and
operate a hydrostatic bellows.
Arrange
the apparatus as shown in Fig. 72. Place the book on the empty
observation balloon, and fill the balloon with water until it is
about
half full. Do you observe that a very little water in the tube supports
the weight of one end of the book.
Place
an empty tumbler on the book and fill it with
water. Do you find that a small extra amount of
water in the tube supports the glass of water?
Remove
the tumbler and press down on the book with
your hand. Do you find that to lift water in
the tube you must exert a force much greater
than the weight of this water.
These
experiments are certainly very striking and
they illustrate Pascal's law as follows: The
weight of the extra water in the tube exerts
pressure downward on an area equal to that of
the inside of the tube; this pressure is transmitted
equally and undiminished in all directions by
the water, and is exerted against each equal
area of the inside
52
HYDRAULIC AND PNEUMATIC
ENGINEERING
THE HYDRAULIC PRESS
The
hydraulic press is an application of Pascal's
law and of the hydrostatic bellows. It is used
where great pressure is required, for example,
to compress merchandise, to bend ship plates,
to lift great weights, and so on.
The
press has a force pump with handle P which
operates the small piston A in the small
cylinder C and pumps water from the reservoir L
through the valve d, through the connecting
pipe and valve v, and into the large cylinder D.
The large piston B, or ram as it is called, moves up and down in D.
Both
A and B have collars which prevent the escape of water.
If
now the end of ram B has an area 100 times as great as the end of
A, then each 1 lb. exerted on A exerts a lift of 100 lbs. on B, and so
on.
EXPERIMENT
No. 28
To make
and
operate a hydraulic press.
Arrange
the apparatus as shown in Fig. 74, where the
tin can in the tank represents the ram and where the
balloon represents the collar of the ram. Soap the
plunger to make it slippery.
Open
lower clip, raise the plunger, close lower clip, open
side clip and lower the plunger. Repeat until the
balloon is partly filled with water.
Now
fill the tin can with water and repeat the operations
above.
Do you find that
a
small force on the plunger will lift the
relatively large weight of the tin can full of water?
You have shown here that
on
the hydraulic press a small force moving the
small piston a long distance lifts a great
weight on the large piston a small distance.
HYDRAULIC
AND PNEUMATIC ENGINEERING 53
THE HYDRAULIC ELEVATOR
The
simplest form of hydraulic elevator is illustrated in
Fig. 75. The passenger cage A is securely fastened to the
top of a long ram P which moves up and down in a deep
cylinder C. The elevator is raised by the city water pressure
or, if this pressure is not sufficient, by the pressure
of water pumped into a tank on the roof of the building.
The water enters through the pipe m and through
the three-way valve if, and it leaves through the three-way
valve and the lower pipe.
The
weight of the cage and ram is partly counter-balanced
by the weight shown. When water is admitted to
the cylinder, it exerts pressure upward on the bottom
of the ram and raises the ram and cage; when the
discharge pipe of the cylinder is opened, the cage and
ram descend by their own weight and drive the water
out of the cylinder.
The
operation of the three-way valve is illustrated
in Fig. 76. The lever handle is weighted at the
end and is operated by the cord t, t, c, c, which
passes through the cage. When the operator
pulls the cord up the valve takes the upper position,
water is admitted to the cylinder, and the ram
and cage are raised. When the operator
pulls the cord down, the valve takes
the lower position and connects
the cylinder with the discharge pipe;
From
the cage and ram then descend by their own
weight and in doing so force water from
the cylinder to
the sewer.
54
HYDRAULIC AND PNEUMATIC
ENGINEERING
When
speed is desired, for example in carrying passengers,
the elevator is arranged as shown in Fig. 77.
The plunger or ram P moves in a cylinder C. Both
ram and cylinder carry a number of large
separate pulleys, side by side, around which a
steel cable is passed a number of times and
then attached to the counterpoise weight D.
If,
for example, the steel cable makes 10 loops
around the pulleys there are 20 strands between
the two sets of pulleys. If then the ram
moves 1 foot each strand is lengthened 1 foot
and the counterpoise is pulled down 20 feet.
Since the cable attached to the passenger
cage passes around the pulley of the
counterpoise as shown, each foot the
counterpoise descends raises the cage 2 feet.
Thus if the ram moves 1 foot, the counterpoise
moves 20 feet and the cage, 40 feet. This gives
the passenger cage a speed forty times that of the
ram.
The ram is moved
by
water from the city mains which is controlled
by a three-way valve as described above.
HYDRAULIC AND PNEUMATIC
ENGINEERING 55
EXPERIMENT
No. 29
To make and
operate a hydraulic elevator.
Arrange
the apparatus as shown in Fig. 78. Soap the plunger well to
make it slippery.
Open
side
clip. Is the cage raised? Close side clip. Does it stop?
Open
lower clip and press down gently on the cage. Does it descend? Close
lower clip. Does it stop?
Now
open and close side clip to raise the cage a short distance at a time.
Do you find that you control the elevator perfectly as it rises?
Now
open and close lower clip while you force the cage down a short
distance
at a time. Do you find that you can control the elevator perfectly
as it descends and that you cannot move it down when the clip
is closed?
You have
shown
here how the ram and cage of an elevator are raised
by water pressure and how they descend by their own weight. You
have shown also that you can stop them anywhere, while rising or
descending,
by closing the proper valve.
HYDRAULIC
LIFT LOCKS
CANAL LOCKS
56
HYDRAULIC AND PNEUMATIC ENGINEERING
An
ordinary canal lock, Fig. 79, is used to raise or lower steamers a
few feet to enable them to pass up or down stream, around a rapid, dam
or waterfall. It is simply a short canal with a pair of gates at each
end.
If the steamer is
going up stream, it sails through the lower gates of
the lock; the lower gates are closed behind it; water is admitted to
the
lock until its level is equal to that of the water above the lock; the
upper
gates are then opened, and the steamer sails out of the lock at
the upper level. If the steamer is going down stream the reverse
operation
takes place.
If the
difference in level is considerable but over some distance, a number
of these locks are used, for example, if the difference in level were
80 feet in a distance of two miles, there might be, in the two miles, 4
locks with a difference of level of 20 feet each or 8 locks with a
difference of 10 feet each, and so
on.
When the difference
in
level is great in a short distance, however, a
lift lock must be used.
LIFT
LOCKS
Lift
locks are so
called because the whole lock, with the water in it
and the ship, is lifted vertically from the low level to the high, or
is
lowered vertically from the high level to the low. They are always in
pairs and the weight of one balances the weight of the other.
The
lift lock shown in Fig. 80, is one that it is proposed to build on
a canal between Lake Erie and Lake Ontario. It will take ships 650
feet long and of 30 foot draft, and will lift or lower them through a
vertical
height of 208 feet. The inner side of one will be connected with
the inner side of the other by 56 steel cables which pass over 56
sheaves
of 20 foot diameter. The outer side of each will be connected with
large concrete counterweights by means of steel cables passing over 56
sheaves on each side. The locks will be raised and lowered by means of
electrical power applied to the rims of each sheave. The gates at the
ends of each lock and at the ends of the upper and lower canal will be
opened and closed by being moved down and up vertically. The diagram
shows how the locks will look when one ship is being raised and
another lowered. The building at the right is a plant in which
electrical
power will be developed from the excess water from the upper canal.
A small part only of this power will be used to operate the locks.
HYDRAULIC
AND PNEUMATIC ENGINEERING 57
58
HYDRAULIC AND PNEUMATIC ENGINEERING
HYDRAULIC
LIFT LOCKS
Hydraulic
lift
locks are so called because they
are operated by means of water. Each lock
is a large steel tank securely
attached to the top of a very large
ram which moves up and down in a deep
cylinder. The two cylinders
are connected by a pipe through which
the water flows from one
to the other, the flow being
controlled, or stopped entirely, by means
of a valve.
The
operation of
the locks will be understood from
Fig. 82. If the steamer is going up
stream: it sails into the lock B
which is down and the lock gate is closed;
a little water is admitted to the lock
A which is up, to make it weigh more
than the lower lock B and the steamer;
the valve R is opened; the upper lock
descends and its ram P
1 forces water
from its cylinder into that of the
lower lock; the pressure of this water
raises the ram P
2 , the lower lock and
the steamer, to the upper level; the gates are opened; and the steamer
sails out at the upper
level.
HYDRAULIC
AND PNEUMATIC ENGINEERING 59
If
the steamer is going down stream; it sails into the upper lock and
the gates are closed; water is admitted to the upper lock to make it
weigh more than the lower lock; the valve R is opened; water is forced
from the cylinder of the upper lock to that of the lower as the upper
lock descends and the lower lock rises; the gates are opened; and
the steamer sails out at the lower level.
Note.
You might think that the presence of the steamer in one
lock would
make it weigh more than the other lock, but you will learn in
Experiment
36 that a ship displaces its own weight of water and that therefore
the one lock, plus water, plus steamer, weighs the same as the
other lock plus water.
EXPERIMENT
No. 30
To make and operate
a hydraulic lift lock.
Use
the apparatus shown in Fig. 83. The wide tubes and plungers represent
the cylinders and rams of a real lift lock, and the clip represents
the control valve. The inverted tumblers represent the locks, they
should of course be right side up but you have no
way of fastening them.
Place
a button or pebble on the lower lock to represent
a ship, open the clip and press down on the
upper lock. Is the ship raised?
Lower
a steamer in the same way.
Now
place a steamer in the lower lock and press down
on the upper lock while you open and close the
clip from time to time. Do you find that the plungers
stop as soon as you close the clip?
This
shows how the rams of a real lift lock can be
stopped anywhere by closing the valve R, Fig. 82. Water
is incompressible, as you know from Experiment,
No. 7, and when valve R is closed the rams cannot
move because the water in the cylinders cannot
be compressed and cannot move.
Repeat
this but close the clip only partly.
Do
you find that the plungers can move slowly and
that you can regulate the speed by opening of the
clip more or less?
60
HYDRAULIC AND PNEUMATIC ENGINEERING
This
shows how the rams in a real lift lock can be allowed to move rapidly
or slowly by opening the valve R more or less
In
this experiment you have illustrated the working of a hydraulic lift
lock: you have shown that the downward movement of one ram drives
water into the second cylinder and that the pressure of this water
raises
the ram in the second cylinder; you have shown also that the rams can
be stopped anywhere by closing the valve R or that they can be made
to move very slowly by closing the valve partly.
THE
PRESSURE EXERTED BY WATER
HYDRAULIC
AND PNEUMATIC ENGINEERING 61
A
very astonishing fact is illustrated in Fig. 84, namely that the
pressure
at the nozzles is the same no matter what size and shape the tank
may be and no matter what size and shape the pipe may be, provided
the water level in the tank is at the same distance above the
nozzle in all cases. You will now prove this.
EXPERIMENT
No. 31
To show that the
pressure at a nozzle is independent of the size and
shape of the tank and pipe.
Make
the experiments illustrated in Fig. 85 one after the other using the
same nozzle in all. Are the streams of the same height in all cases if
the water level in the tank is at the same distance above the nozzle?
62
HYDRAULIC AND PNEUMATIC ENGINEERING
You
have shown here that the pressure exerted by water is independent
of the volume of the water but that it depends upon the height of
the water above the nozzle. This is known as the Hydrostatic Paradox
which you will now illustrate.
THE
HYDROSTATIC PARADOX
The
Hydrostatic Paradox is stated as follows : The pressure exerted by
a liquid on any base is
independent
of the volume of the liquid, but depends
only on the area of the base, the depth of the liquid, and the
density
of the liquid.
Note.
The
density of a
liquid is its weight per cubic foot, or per cubic
inch, or per cubic centimeter.
The
hydrostatic paradox is illustrated
by means of the apparatus shown in
Fig. 86. The three tops are of different
sizes and shapes, but they fit a common
base. The bottom of this base is
covered by a sheet of rubber or by a
sheet of corrugated metal. The base
sinks as the pressure increases and
moves the pointer, which indicates
the pressure.
If the
tops
are screwed to the base, one after
the other, and then filled with water
to the same
height,
the pointer indicates the s
ame
pressure in all cases.
The
volume of water in the tops is different in each case, but the pressure
is the same in all. This shows that the pressure exerted by a
liquid is
independent of
the volume of the liquid, provided the
area of
the base,
the
depth
and the
density of
the liquid are the same in all cases.
Another
form of this apparatus is shown in Fig. 87; the three tops fit
the same base, but the bottom is a brass plate AB which is held on by
a cord attached to one arm of a balance (not shown). The plate AB falls
in each case when the water reaches the same height.
The
hydrostatic paradox is also illustrated in 4; the three tubes are of
very different volumes but the water stands at the same height in all.
HYDRAULIC
AND PNEUMATIC ENGINEERING 63
These
experiments show that the pressure a liquid exerts on a given base
is independent of the volume of the liquid, provided the area of the
base, depth of the liquid, and density of the liquid are constant.
EXPERIMENT
No. 32
To illustrate the
hydrostatic paradox.
64
HYDRAULIC AND PNEUMATIC ENGINEERING
Make
the experiments A, B and C, Fig. 88, one after the other. Is the
water in the small tube always at the same height as that in the funnel
or large tube?
Arrange
the
apparatus as in D, Fig. 88. Is the water at the same level
in all cases?
The
funnel
and wide tube, each contain more water than the small tube;
nevertheless, the downward pressure of the water in each is balanced
by the downward pressure of the water in the small tube.
You
have shown here that the pressure exerted by a liquid is independent
of the volume of the liquid, that is, you have illustrated the
hydrostatic
paradox.
EXPLANATION OF THE
HYDROSTATIC PARADOX
The
hydrostatic paradox seems impossible, and that is why it is called
a paradox. It would seem to be self evident that the greater the
volume of water above a base, the greater would be the pressure;
HYDRAULIC
AND PNEUMATIC ENGINEERING 65
and
the less the volume, the less the pressure. You have shown above,
however,
that the pressure on a given base is independent of the volume of
water and that it depends only on the depth.
The
paradox is explained as follows :
In
1, Fig. 89, the base AB is subject to the pressure of the water in the
cylinder above it, and in this case, the pressure is equal to the
weight
of the water.
In 2,
Fig.
89, the same base AB has a much larger volume of water above
it but the pressure is the same as in 1. You will understand why,
if you consider the water outside the dotted lines. This water exerts
a force perpendicular to the sides of the cone, and another force
horizontally
against the water between the dotted lines, see the arrows. Neither
of these forces has any effect downward on the base and therefore
the base is subject only to the weight of the water between the dotted
lines. This weight is the same as in 1 and therefore the pressure on
AB is the same as in 1.
In
3, Fig. 89, the base AB has a much smaller volume above it than in
either 1 or 2, but still the pressure is the same as in 1 and 2. You
will
understand why from your knowledge of Pascal's law. The water above
AB is exerting pressure downward, and according to Pascal's law
this pressure is transmitted equally and undiminished in all
directions.
The pressure per square inch downward on the whole of AB, therefore,
is equal to what it would be if the whole space between the outer
dotted lines were filled with water. This pressure is equal to that in
(1) and this is why the pressure in (3) is equal to that in (1).
HOW
TO CALCULATE THE PRESSURE EXERTED BY
WATER
The
density (weight)
of fresh water is 62 l / 2 lbs. per cubic foot and if
in (1) Fig. 89, the base AB is 1 square foot and the height of the
water
is 10 feet, there are 10 cubic feet of water in the tank and the total
pressure on the bottom is 10 x 62.5 = 625 lbs.
Since
the pressure exerted by water is independent of the volume of the
water and depends only on the area of the base, the height, and the
density
of the water, the pressure on AB in (2) and (3) is 625 lbs., the same
as in (1).
The rule for
calculating the pressure in any case is : Pressure on any
base = area of base in square feet x height of water in feet x density
of water (weight of 1 cubic foot) or, Pressure = area x height x
density.
66
HYDRAULIC AND PNEUMATIC ENGINEERING
In
the example given:
Pressure
= 1 x 10 x 62.5 = 625 lbs. per square foot.
To
find the pressure per square inch, first find the pressure per square
foot
and then divide the result by 144, the number of square inches in 1
square foot. For example, the pressure on 1 square inch of AB in any
of the tanks illustrated is 625 / 144 = 4.34 lbs.
PRESSURE
UNDER WATER
THE
DEPTH BOMB - TORPEDO
- SUBMARINE
THE
DEPTH BOMB
HYDRAULIC
AND PNEUMATIC ENGINEERING 67
The
depth bomb is used by submarine chasers to destroy submarines. It
is a steel cylinder filled with high explosives and equipped with a
trigger
which sets off the explosive at any desired depth under water.
The
trigger is released by means of a small plunger which is exposed to
the pressure of the sea water on the outside and is supported by a
spring
on the inside. The pressure of the water increases as the bomb sinks
and forces the plunger in farther against the spring, but the spring
can be so adjusted that at any desired depth the plunger releases the
trigger and the bomb explodes.
When
the chaser sights a submarine it steams for it and if it is still
above water, attacks it with guns; but if it has submerged, the chaser
steams in circles around the spot where it disappeared and drops or
fires bombs adjusted to explode at different depths.
THE
TORPEDO
The
torpedo is a cigar shaped tube loaded
in the head with high explosives
which are set off by a contact pin.
It is driven by means of a compressed
air motor and is steered by horizontal
and vertical rudders.
We
are interested in the horizontal rudder
particularly at this point. It steers
the torpedo to a depth of 20 feet
under water and keeps it at this
depth. It does this by means of the
pressure of the sea water. The horizontal
rudder is controlled by a piston,
Fig. 92, which is exposed to the
pressure of the sea water on the
68
HYDRAULIC AND PNEUMATIC ENGINEERING
outside
and is supported by a spring on the inside. This piston and its
spring are so adjusted that at 20 feet under water the rudder is
exactly
horizontal, but at a greater or less depth the rudder is so turned as
to bring the torpedo back to a depth of 20 feet.
THE
SUBMARINE
The
submarine must be able to stand enormous pressures when under water
and for this reason it is made in the shape of a cylinder with pointed
ends, because this curved shape enables it to stand greater pressure
than it could if its sides were flat; also it is made of steel because
this is the strongest material available.
You
cannot experiment with the depth bomb, torpedo, and submarine, of
course, but you can make experiments to illustrate the water pressure
under
which they operate. You can show that the pressure under water
increases
with the depth, that it is equal in all directions at any depth, etc.,
and this you will now do.
HYDRAULIC
AND PNEUMATIC ENGINEERING 69
EXPERIMENT
No. 33
To show that the
pressure under water increases with
the depth
and that it is equal in all
directions at any given depth.
This
is usually shown by means of the
apparatus A, Fig. 94. The U shaped
bend of the three tubes contain
mercury to the same depth. Both ends
of the tubes are open. The short ends
point upward, sidewise and downward
respectively. When the short ends of
these tubes are lowered in water, the
mercury shows that the pressure
increases with the depth and is equal
in all
directions at any given depth.
70
HYDRAULIC AND PNEUMATIC ENGINEERING
This
fact is illustrated in another way by means of the apparatus, B
and C, Fig. 94. A thistle tube covered by a sheet of rubber is placed
under
water and the water in the U tube indicates a greater pressure the
greater
the depth. If the thistle tube is turned in all directions at any given
depth, the water in the U tube shows that the pressure is equal in
all directions at this depth.
Illustrate
these facts by means of the apparatus, Fig. 95.
Shove
the funnel straight down (1). Does the pressure increase with the
depth?
Turn the funnel
sidewise (2) and upward at any depth. Is the pressure
equal in all directions at any given depth?
EXPERIMENT
No. 34
To show that water
exerts pressure upward on anything
under its surface
and that the upward
pressure is
equal to the downward pressure at any given depth.
This
is usually shown with the apparatus Fig. 96. If a
glass lamp chimney A, is fitted with a thin ground glass
bottom O which is held over one end by a thread C,
while this end is placed in water, it is found that the
bottom remains on when the thread is
released. This shows that water
exerts pressure upward on anything
under its surface.
If
now water is poured into the
chimney, the bottom
HYDRAULIC
AND PNEUMATIC ENGINEERING 71
remains
on until the level inside the chimney is the same as the level outside
and this is true at any depth. This shows that the pressure upward
at any depth under water is equal to the pressure downward of
the column of water inside the chimney. In other words, it shows that
the pressure upward at any depth under water is equal to the pressure
downward at this depth.
Illustrate
this with the apparatus (1) Fig. 97. Put the stoppered end in
water. Is a fountain produced and does the flow stop when the level
inside is equal to that outside the tube?
Use
the apparatus (2) Fig. 97, hold the rubber sheet on until it is under
water. Does it remain?
Pour
water into the tube. Does the sheet fall off when the level inside
is equal to that outside?
You
have shown here that water exerts pressure upward against anything
under its surface and that the upward pressure is equal to the
downward pressure at any given depth.
HOW
TO CALCULATE THE PRESSURE ON DEPTH BOMB, TORPEDO
AND SUBMARINE
Sea
water is
heavier than fresh water; it weighs 64 lbs. per cubic foot
while fresh water weighs only 62 l / 2 lbs. per cubic foot.
DEPTH
BOMB
A
depth bomb is set to
explode at a depth of 250 feet. If sea water weighs
64 lbs per cubic foot, what is the pressure per sq. in. against the
plunger
at this depth?
Note:
Calculate the pressure per square foot and divide this by
144,
the number of square inches in one square foot.
Area x
depth x density
Pressure
=
---------------------------
144
1 x 250 x 64
Pressure
=
---------------------------
= 111.1 lb. per sq. in.
144
TORPEDO
A
torpedo is set to travel
at a depth of 15 feet under water. What is
the pressure per sq. in. on the steering plunger at this depth?
Area x
depth x density
Pressure
=
---------------------------
144
1 x 15 x 64
Pressure
=
---------------------------
= 6.6 lbs. per sq. in.
144
SUBMARINE
What
is the pressure per square foot on the outside of a submarine at
an average depth of 150 feet in water?
Pressure
=
Area x depth x density
Pressure
=
1 x 150 x 64 = 9600
lbs. per sa. ft.
72
HYDRAULIC AND PNEUMATIC ENGINEERING
BUOYANCY
WHY DOES A
STEEL SHIP FLOAT?
Modern
ships are made of steel, example, the superdreadnaught shown
in Fig. 98, and although steel is over seven times as heavy as water,
bulk for bulk, steel ships float. Why is this?
You
know the answer, at least partly. You know that if a ship were
a solid lump of steel, it would sink. You know also that a ship is
hollow, except for its equipment, and that this hollowness in some way
enables it to float.
The
true reason is that the ship as a whole is lighter than an equal volume
of water.
You will show
in
the following experiments that water exerts a buoyant
force on anything placed in it, and that as a result: things which
are lighter than an equal volume of water float on water; while things
which are heavier than an equal volume of water sink but are lighter
under water than above water.
HYDRAULIC
AND PNEUMATIC ENGINEERING 73
EXPERIMENT
No. 35
To illustrate the
buoyant effect of water.
Find
about your home an empty tin can with a tight lid. Submerge it
partly as in (1) and release it. Does it shoot upward? This buoyant
effect
of the water is due to the upward pressure of the water.
Submerge
it entirely as in (2) and (3) and release it. Does it shoot upward?
This buoyant effect shows that the upward pressure of the water
on the under side of the can is greater than its downward pressure on
the top side.
Fill the
can
with water, submerge and release it. Does it sink? Lift
the full can under water and out of water. Is it much lighter when
under water? It is lighter because the water buoys up part of its
weight.
THE LAW OF
ARCHIMEDES
The
exact law
which applies to the buoyancy of liquids was discovered
by a Greek philosopher Archimedes 200 years before the Christian
era began. It is called the law
of Archimedes and it is as follows: the
buoyant force exerted by a liquid on a body immersed in it, is exactly
equal to the weight of the liquid displaced by the body.
It
is also stated more concisely as follows: a body when placed in
a liquid appears to lose weight equal to the weight of liquid it
displaces.
74
HYDRAULIC AND PNEUMATIC ENGINEERING
The
law of Archimedes is illustrated by means of the apparatus shown
in Fig. 100. The solid cylinder A is so made that it just fits the
cup B, that is, the cylinder has exactly the same volume as the cup.
The
experiment is as follows: The cylinder A is attached to the bottom
of the cup B and both are suspended from one pan of a balance. Weights
are added to the other pan until the cup and cylinder are just
balanced.
If then, a
vessel
of liquid is raised up under the cylinder A until it
is completely submerged, the cup and cylinder appear to lose weight
because
the liquid buoys up the cylinder. If now the cup B is filled with
the liquid, the balance is exactly restored.
Now
the weight of the liquid which fills the cup is equal to that of
the liquid displaced by the cylinder and therefore this experiment
proves
the law of Archimedes, namely, the buoyant force exerted by a
liquid on a body immersed in it is equal to the weight of the liquid
displaced
by the body.
The law of
Archimedes is also illustrated by means of the apparatus shown
in Fig. 101 and by means of a spring balance, not shown.
The
body is first weighed on the spring balance in air, then in the liquid,
and the apparant loss in weight in the liquid is determined.
The
vessel with the spout is then filled with the liquid until it
overflows,
the body is placed in the liquid, and the liquid displaced is weighed.
HYDRAULIC
AND PNEUMATIC ENGINEERING 75
The
apparent loss in weight of the body
is then compared with the weight of
liquid displaced by the body, and it
is found that in every case they are
equal.
You will now make
experiments to illustrate the law of
Archimedes for bodies which float on
water and for bodies which sink in
water, also you will illustrate some
of the applications of this law.
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