The Science Notebook
Gilbert Hydraulics - Part 3

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NOTE:  This book was published in 1920, and while many of the experiments and activitied here may be safely done as written, a few of them may not be considered particularly safe today.  If you try anything here, please understand that you do so at your own risk.  See our Terms of Use.

 Pages 51-75

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 P1 forces water from its cylinder into that of the lower lock; the pressure of this water raises the ram P2 , 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 same 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 apparent 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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