In this section we will consider a law of nature that applies only to liquids and gases and does not apply to solids.
Let's mentally imagine that inside the liquid at a given point there is a small platform. The liquid produces pressure on this area. It is important that the fluid pressure on this small area does not depend on the orientation of the area. To prove the validity of this statement, we will use the so-called solidification principle. According to this principle, any volume of liquid or gas in the static case, when the elements of the liquid do not move relative to each other, can be considered as solid and apply the equilibrium conditions of a rigid body to this volume.
Let us select a small volume in the liquid in the form of a long triangular prism (Fig. 9.23, a), one of the faces of which (the OBCD face) is located horizontally. The areas of the prism bases will be considered small compared to the area of the side faces. The volume of the prism will be small, and therefore the force of gravity acting on this prism will be small. This force can be neglected in comparison with the pressure forces acting on the edge of the prism1.
1 Surface area is proportional to the square linear dimensions bodies, and the volume is a cube. Therefore, for a small prism, the force of gravity, proportional to the volume, can always be neglected in comparison with the force of pressure, proportional to the surface area.
Figure 9.23, b shows the cross section of the prism. The forces Flt F2, F3 act on the lateral faces of the prism. We do not take into account the pressure forces on the bases of the prism, since they are balanced. Then according to the equilibrium condition
Fi + F2 + F3 = o.
The vectors of these forces form a triangle similar to triangle AOB, since the angles in these two triangles are respectively equal (Fig. 9.23, c). From the similarity of triangles it follows that
?i = = ї±
OA OB AB"
Let us multiply the denominators of these fractions by OD, BC and KA, respectively (OD = BC = KA):
F1 F2 F3
OA OD OB BC AV KA"
From Figure 9.23a it is clear that the denominator of each fraction is equal to the area of the corresponding lateral face of the prism. Denoting the areas of these prism faces as S2, S3, we obtain:
F±==F_2=F3 S2 "3
or
Рі=Рг=Рз- (9.6.1)
So, the pressure in a stationary liquid (or gas) does not depend on the orientation of the area inside the liquid.
According to formula (9.5.3), the pressure is the same at all points lying at a given level. This pressure on the underlying layers of liquid is created by a column of liquid of height h. Therefore, we can conclude that the pressure of the upper layers of liquid on the layers of liquid located below them is transmitted by the underlying layers equally in all directions.
But pressure on the liquid can be applied external forces, for example using a piston. Taking this into account, we arrive at Pascal's law: the pressure produced by external forces on a fluid at rest is transmitted equally by the fluid in all directions.
In this formulation, Pascal's law remains true for the general case, that is, for the case when we take gravity into account. If the force of gravity creates a pressure inside a fluid at rest, depending on the depth of immersion, then the applied
New external (surface) forces increase the pressure at each point of the liquid by the same amount.
Rice. 9.24
Pascal's law can be confirmed experimentally. If, for example, you fill a metal ball with several holes in it with water, and then compress the water with a piston, then identical jets of water will spray out of all the holes (Fig. 9.24, a). Pascal's law is also valid for gases (Fig. 9.24, b). Hydrostatic paradox
Let's take three vessels of different shapes (Fig. 9.25). Water weighing 3 N is poured into vessel A, water weighing 3 N is poured into vessel B, and weighing 1 N is poured into vessel C. The water level in all three vessels is at a height of 0.1 m. The bottom area of each vessel is 20 cm2 = 0.002 m2. Using the formula p = pgh, we find that the pressure at the bottom of each vessel is 1000 Pa. Knowing the pressure, we use the formula F = pS to find that the pressure force on the bottom of the vessel in all three cases is equal to 2 N. It can’t be, you say. How can water weighing 1 N in the third vessel create a force of 2 N on the bottom? This position, which seems to contradict common sense, is known as the “hydrostatic paradox”, or “Pascal’s paradox”.
Trying to solve the riddle of the hydrostatic paradox, Pascal placed vessels similar to those shown in Figure 9.25 on special scales that made it possible to measure the pressure force on the bottom of each vessel (Figure 9.26, a, b, c). The bottom of the vessel, standing on the scales, was not rigidly connected to the vessel, and the vessel itself was fixed motionless on a special stand. The readings of the scales confirmed the calculations. Thus, contrary to common sense, the force of pressure on the bottom of the vessel does not depend on the shape of the vessel, but depends only on the height of the liquid column, its density and the area of the bottom.
This experience leads to the idea that with the proper shape of the vessel it is possible, with the help of a very small amount of liquid, to - 300 cm3
100 cm3
V)
10 cm
shhhhhhhh,
A)
200 cm3
10 cm
ъшшшяшШЯШ, b)
Rice. 9.26 create very large pressure forces on the bottom. Pascal attached a tube with a cross-sectional area of 1 cm to a tightly sealed barrel and poured water into it to a height of 4 m (weight of water in the tube P = mg = 4 N). The resulting pressure forces tore the barrel (Fig. 9.27). Taking the area of the bottom of the barrel to be 7500 cm2, we obtain a pressure force on the bottom of 30,000 N, and this enormous force is caused by just one mug of water (400 cm3) poured into the tube.
How to explain Pascal's paradox? The force of gravity creates pressure inside a liquid at rest, which, according to Pascal’s law, is transmitted to both the bottom and the walls of the vessel. If a liquid presses on the bottom and walls of a vessel, then the walls of the vessel also produce pressure on the liquid (Newton’s third law).
If the walls of the vessel are vertical (Fig. 9.28, a), then the pressure forces of the vessel walls on the liquid are directed horizontally. Consequently, these forces do not have a vertical component. Therefore, the force of liquid pressure on the bottom of the vessel is equal in this case to the weight of the liquid in the vessel. If the vessel expands upward (Fig. 9.28, b) or narrows (Fig. 9.28, c), then the pressure force of the vessel walls on the liquid has a vertical component, directed upward in the first case, and downward in the second. Therefore, in a vessel expanding upward, the pressure force on the bottom is equal to the difference between the weight of the liquid and the vertical component of the pressure force Fig. 9.27 walls. Therefore, the pressure force on
Rice. 9.28
the bottom in this case is less than the weight of the liquid. In a vessel that tapers upward, on the contrary, the pressure force on the bottom is equal to the sum of the weight of the liquid and the vertical component of the pressure force of the walls on the liquid. Now the force of pressure on the bottom is greater than the weight of the liquid.
Of course, if you put various vessels on the scales without a separable bottom and not fixed on stands, then the scale readings will be different (2 N, 3 N and 1 N, if the mass of the vessels can be neglected). In this case, the vertical component of the forces of liquid pressure on the side surface will be added to the force of liquid pressure on the bottom in an expanding vessel. In a narrowing vessel, the corresponding component of the pressure forces will be subtracted from the pressure force on the bottom.
Hydraulic Press
Pascal's law allows us to explain the action of a device common in technology - a hydraulic press.
A hydraulic press consists of two cylinders of different diameters, equipped with pistons and connected by a tube (Fig. 9.29). The space under the pistons and the tube are filled with liquid (mineral oil). Let us denote the area of the first piston by S1, and the area of the second by S2. Let us apply force F2 to the second piston. Let's find what force F2 must be applied to the first piston to maintain equilibrium.
According to Pascal's law, the pressure at all points of the liquid must be the same (we neglect the effect of gravity on the liquid). But the pressure under the first piston is equal
Fi
-x-, and under the second.
shhhhhhhh,: Fig. 9.29 Therefore,
shhhhhhhhh, Fig. 9.30
i 2
2s:
і
(9.6.2)
F^F,
Hence the force module Fy is the same number of times greater than the force module F2, how many times the area of the first piston is greater than the area of the second. Thus, with the help of a hydraulic press, it is possible, by means of a small force applied to a piston of a small cross-section, to obtain enormous forces acting on a piston of a large cross-section. The hydraulic press principle is used in hydraulic jacks to lift heavy loads.
Thanks to Pascal's law, paradoxical situations are possible when a mug of water added to a barrel leads to its rupture. The same Pascal's law underlies the design of hydraulic presses.
A vessel with water is installed on the edge of the board (Fig. 9.30). Will the balance be disrupted if a plank is placed on the surface of the water and a weight is placed on it so that the plank and the weight float on the surface of the water not in the middle of the vessel?
○ 47. Flight range
A body thrown horizontally has a greater flight range the greater the height it is thrown (all other things being equal). In the well-known experiment on the liquid pressure of the vessel wall (Fig. 26), the flight duration of water jets does not increase with height, but decreases. Explain this apparent contradiction.
○ 48. Pascal's experiment.
The rupture of the barrel in Pascal's experiment (Fig. 27) represents a paradox, since the only force acting here - the gravity of the water in the tube - is obviously insufficient for this; To break a barrel, a force is required that is significantly greater than the weight of the barrel together with water. Where does this additional enormous power come from?
● 49. Once again about Pascal’s experiment
In Pascal's famous experiment (see problem No. 48), the pressure in a barrel of water is created by the weight of a column of water in the tube. If you double the force acting on a body, the pressure will also double. Consequently, if instead of one tube with water we take two (Fig. 28), then the water pressure on the walls of the barrel should double.
Let's assemble the installation shown in Figure 28. The pressure gauge indicating the pressure exerted on the liquid does not change its readings when replacing one tube with two. What is the error in reasoning?
○ 50. Pascal's paradox.
An EADCBF vessel with an attached CD bottom is lowered into a tank of water (Fig. 29). Water in volume ABCD has a mass of 2.5 kg, which means she weighs 24.5 n. If you place a narrow cylinder weighing 25 n, then it doesn’t come off, but if you pour 2.5 kg water, then it comes off. Explain the paradox.
○ 51. Another Pascal paradox.
Pascal's law is formulated as follows: if pressure is applied to any part of the surface of a liquid enclosed in a vessel closed on all sides, then it is transmitted through the liquid equally in all directions. In accordance with this, the pressures on areas A and B (Fig. 30), located in the upper and lower parts of the vessel, must be the same. For if there was more pressure on one of them than on the other, then, according to Pascal’s law, the excess pressure would be completely transferred to the other area and, as a result, the pressures would be equal.
But, on the other hand, it is known that the pressure at any point of a heavy and calm liquid is equal to the weight of a column of liquid, the height of which BC is equal to the immersion depth of the point, and the base is equal to unity. Consequently, the fluid will produce pressure on site B, but not on site A. How can these two conflicting conclusions be resolved?
● 52. Perpetual hydrostatic engine.
A cylinder filled with liquid has a piston, the shape of which is shown in Figure 31. A certain force acts on the piston on the left. Since the piston area on the right is larger than on the left, we can expect a greater force of pressure on the water. And if so, then it seems that more work can be obtained by moving the piston slightly to the left. What is the error in such reasoning?
● 53. Law of communicating vessels.
Identical communicating vessels A and B contain room water (Fig. 32). Tap K was closed and the water in vessel B was heated, as a result of which its level increased slightly. Will water flow from one container to another if you open the tap?
Solution. One student said that increasing the height of the liquid column in vessel B will cause an increase in pressure at the level of the connecting tube. Therefore, liquid will flow from vessel B to vessel A if tap K is opened
Another student argued that after heating the water in vessel B, the pressure in both vessels did not change, since the weight of the water and the base area of the bottom of the vessel did not change.
The third student argued that the water pressure in vessel B at the level of the connecting tube will decrease after it is heated. And if you open tap K, then water will flow from vessel A to vessel B. This happens because vessel B expands upward, and the increase in the height of the water level in it will not be inversely proportional to the decrease in density caused by heating the water. Which student is right?
● 54. Perpetual hydrodynamic engine.
Big round iron pipe, curved in the form of a ring with a gap between points A and B (Fig. 33), half buried in the ground and half hanging above the ground. A mill wheel, consisting of a series of blades mounted on an axle, is placed inside the underground part of the pipe at point B. If 2-3 barrels of water are immediately poured into this pipe, in its left half, near point B, then (according to the author of the project) this the water will accelerate through the pipe, reach point A, fall down again, etc., simultaneously setting the mill wheel into rapid rotational motion. The only thing that is supposedly necessary for the further uninterrupted operation of the mill is to periodically add water to the pipe to replace the water that has evaporated. What's wrong with the project? How would the water actually move in the pipe?
○ 55. What holds water in a glass?
A full glass of water taken at a temperature of 20°C is covered with a sheet of paper and, holding the sheet, the glass is turned over. Then the hand that was holding the sheet is removed. Water does not pour out. It is held in place by atmospheric pressure.
Let's repeat the same experiment without a sheet of paper. Water pours out of a glass. But atmospheric pressure also existed in this case, but the result was different. How to explain this paradox?
○ 56. How much does a barometric tube weigh?
A thin-walled cylindrical barometric tube with mercury is suspended from a pan of equal-arm scales (Fig. 34). The end of the tube is immersed only to an insignificant depth in a cup of mercury. To balance the scales, you need to place a weight on another pan of the scales, the weight of which is equal to the sum of the weights of the tube and the mercury column in it.
But the mercury in the tube (if we neglect the friction of the mercury on the walls) with its weight presses on the mercury in the cup, and not on the tube. How to resolve this contradiction?
● 57. "Eternal" pump.
The following project of an “eternal” pump was proposed. An iron tank with tap A and pipe B (hose) lowered into the river is placed on the river bank (Fig. 35). In order to operate the device, you need to pump air out of the tank through tap A and fill it with water. If you then turn off the air pump and open tap A, then water (according to the inventor) due to its gravity will flow from the tap, and atmospheric pressure through tube B will supply more and more amounts of water to the tank. What's wrong with the project?
● 58. Another project of an “eternal” pump.
A vessel having the shape shown in Figure 36 is filled with water. The radius of hole A is equal to the radius of tube B. What happens if plug A is removed? Neglect the phenomena of capillarity.
Solution. Under the influence of weight, liquid will flow out of hole A. At the same time, new quantities of liquid will flow into the expanded part C through tube B.
Experience refutes this decision. What is the error in reasoning?
○ 59. Why didn’t water flow through the pipe?
Rubber hose with an internal diameter of 15-20 mm, wound on a drum with a diameter of 300 mm(Fig. 37). One end of the hose is lowered into the bucket, and the other is raised above the drum by about 1 m. There is no water in the hose. There are no taps or clamps on it. If you insert a funnel into the upper end of the hose and start pouring water into it from the water supply, then water will not flow out of the lower end of the hose. But the air pressure at the upper and lower ends of the hose is the same. How to explain this paradox?
● 60. Are barometer readings the same indoors and outdoors?
One of the students argued that the barometer would show higher pressure outdoors than indoors. Another believed that the barometer readings would be the same. When they noticed the barometer readings in the physics room on the fourth floor and in the school courtyard, the pressure in the courtyard turned out to be greater than in the room. Does this mean that the first student was right?
● 61. Fountain of Heron.
The ancient physicist Heroes of Alexandria proposed an original design for the fountain (Fig. 38). First, vessel L is filled with water and a certain amount of water is poured into vessel C. Will there be an eternal cycle of water in the system: vessel C, tube 1, vessel B, tube 2, vessel A, tube 3, jet 4, vessel Sit. d.? Since the fountain is an active system (which is easy to verify experimentally by constructing it from cans), is the law of conservation of energy violated: the water in jet 4 is raised by more high level, what was it in vessel C?
○ 62. Why doesn't rubber compress?
With the tap open (Fig. 39), a rubber ball C is inflated through tube B. It inflates and the rubber stretches. Close the tap and disconnect the pump. However, air does not come out of tube B, although it should be pushed out, compressed, by the elastic rubber shell of the ball. Explain the paradox.
● 63. Water pressure at the bottom of the vessel.
A glass containing a stone floats in a vessel of water. How will the water level in the vessel change if a stone is removed from the glass and lowered into the vessel?
Solution. When transferring a stone into a vessel, the total weight of the water, stone and glass does not change. The area of the bottom of the vessel also does not change. Consequently, the pressure of the three-body system on the bottom of the vessel should not change. But the pressure of a column of water on the bottom of the vessel is з = ρр, where ρ is the density of water, g is the acceleration of gravity, h is the height of the column. Therefore, the water level in the vessel should not change.
Let's carry out the appropriate experiment and make sure that the water level in the vessel decreases. What is wrong with the decision?
A cast iron cylinder with a fairly wide base floats in a vessel with mercury, and water is poured on top (Fig. 40). Determine the buoyant force acting on the cylinder.
Solution. The lower part of the cylinder is subject to a buoyant force equal to the weight of mercury in the volume ABKE. The upper part is subject to a buoyancy force equal to the weight of water in the EKSM volume. Therefore, the force that pushes the cylinder is equal to the weight of mercury in the volume ABKE plus the weight of water in the volume EKSM.
On the other hand, water exerts some pressure on mercury, which is transmitted to the lower base of the cylinder ABL. Therefore, the buoyant force acting on the lower part of the cylinder AB is greater than the weight of mercury in the volume ABKE. Along with this, the force with which the water acts on the upper part of the cylinder will not be buoyant, but immersive. Consequently, the force that pushes the cylinder is equal to the difference in the pressure forces on its upper and lower bases. However, calculations show that the buoyant force is equal to the weight of mercury in the volume of ABKE plus the weight of water in the volume of EKSM. Which solution is scientifically more rigorous?
○ 65. Position of the center of gravity of a floating block.
A cast iron bar floats in a vessel with mercury. Will the position of the center of gravity of the block change in relation to the level of mercury if water is poured into the vessel (see problem No. 64)?
Solution. Water presses on the block from above and from the sides. The forces of pressure on the block from the sides are balanced, and the force of pressure on the block from above should lower the position of the center of gravity of the block relative to the level of mercury.
Let us carry out the corresponding experiment and find that the block in mercury does not sink, but floats up a little. What is wrong with the solution presented above?
○ 66. When is a body in stable equilibrium?
It is known that the balance of a body is more stable the lower its center of gravity is located. Figure 41 shows two positions of a floating ice floe. Position b is undoubtedly more stable, but the center of gravity of the ice floe is higher than in the position A. How to resolve this contradiction?
○ 67. How did the energy transition take place?
A piece of wood placed at the bottom of a vessel with water, floating up, acquired kinetic energy. According to the law of conservation, energy cannot arise “out of nothing.” What body transferred the energy to the piece of wood?
○ 68. Is the law of conservation of energy violated?
It is usually believed that a siphon can be used to transfer liquid from an upper vessel to a lower one. If two vessels A and B are placed in a large vessel with water, the first containing kerosene, and the second containing water (Fig. 42), and connect them with a tube, then the kerosene will overflow from the lower vessel A into the upper vessel B.
Thus, kerosene, floating, increases its potential energy in relation to the Earth. Doesn't this experience contradict the law of conservation of energy?
○ 69. Terrestrial and lunar hydrometers.
Two students argued. One said that the astronauts would have to change the scale (increase the divisions by 6 times) of the earth's hydrometer when they have to use it on the Moon, since the gravity on it is 6 times less than on Earth.
Another argued that terrestrial hydrometers can be used on any planet, because if the weight of the hydrometer itself changes by a certain number of times, then the weight of the water displaced by it changes by the same number of times. Which one is right?
● 70. Why did the cuvette overturn?
A cuvette with water stands on a block (Fig. 43). A box with a weight floats on the water. The cuvette is in equilibrium.
If you take the weight out of the box and place it on the bottom of the cuvette under the place where the box was floating, the balance will be disrupted (Fig. 44), although the weight of the left side of the cuvette does not seem to have changed. Explain the error in reasoning.
● 71. What kind of cargo is needed?
If in the installation shown in Figure 45 the thread AB is burned out, then the body P, having a volume 100 cm 3, is completely immersed in water and remains hanging on the DIA thread. In this case, the balance of the scales is disturbed. On which scale and what additional weight should you put to restore balance?
Solution. In accordance with Archimedes' law, body P, immersed in water, will be pushed upward with force 0.98 n. Therefore, the weight of the tripod and the weight of the body P, reduced by the weight of the water displaced by it, will act on the right cup of the weight.
Therefore, to restore the balance of the scales, it is necessary to place a weight with a mass on the right side of the cup 100 g.
However, experience shows that it is necessary to place a weight with a mass of 200 g. What is the error in the solution?
● 72. Which part of the vessel is heavier?
The vessel ABCD (Fig. 46), symmetrical with respect to the vertical plane OK, is filled with water and rests on the edge of a fixed prism. A piece of aluminum weighing 0.5 kg, and to the left - a piece of lead weighing 0.4 kg. Which part of the vessel will overtighten?
Solution. The vessel is a complex equal-armed lever. Since the weight of a piece of aluminum is greater than that of lead, the right side of the vessel where the piece of aluminum lies will be pulled.
Experience, however, refutes this conclusion. What is the error in the solution?
An hourglass is placed in a tall glass cylindrical vessel, water is poured to the very top and the lid is closed (Fig. 47). The clock floats up right under the lid. The cylinder is then turned over. The watch does not float (Fig. 48), although it is surrounded by water and the buoyant force is greater than the weight of the watch. After a certain period of time, when a certain amount of sand is poured into the lower compartment, the watch will begin to slowly float up. Thus, the flow of sand from the upper compartment of the watch to the lower one affects its buoyancy. But the clock is hermetically sealed and its weight does not change due to the flow of sand. How to explain this paradox?
● 74. How to avoid overload?
A spacecraft takes off from Earth with an acceleration several times greater than the acceleration due to gravity. Therefore, the astronaut in the spacecraft is exposed to overload (a force that presses a person to the support).
To avoid overload, it is proposed to place the astronaut in a chamber with water (the density of water is approximately equal to the density of the human body). The authors of the project believed that a person, being in water, becomes weightless and, therefore, completely gets rid of the effects of both natural and artificial gravity (overload). What is wrong with this conclusion?
● 75. A simple project for a perpetual motion machine.
Let's consider one of the projects of a perpetual motion machine. A shaft (Fig. 49) is inserted into the cutout of the wall AB of the liquid tank, the axis O of which lies in the plane of the wall AB.
The shaft covers the entire cutout, so that liquid does not spill out; the shaft can rotate on its axis. According to Archimedes' law, half of the shaft immersed in liquid is subject to a lifting force, which, according to the inventor, should cause the shaft to rotate counterclockwise. This rotation would have to go on forever. What's wrong with the project?
● 76. Leonard's perpetual motion machine.
The Swiss G. Leonard in 1865 proposed the following project for a perpetual motion machine. An endless chain of tin floats passes with the right half through vessel B with water (Fig. 50). According to the author, the floats, trying to float, will rotate the wheel C, through which this chain is thrown, counterclockwise. What's wrong with the project?
● 77. Perpetual motion machine from the time of Leonardo da Vinci.
In the 15th century a perpetual motion machine project was proposed, based on Archimedes' law 1. The design of this perpetual motion machine has a wheel with seven hinged weights (Fig. 51). The inventor immersed one third of the wheel in water, reasonably assuming that the weight of this part of the wheel and the loads would decrease according to the well-known law of Archimedes and the wheel would begin to rotate. What's wrong with the project?
1 (The blueprint for the project was discovered in the notes and sketches of the famous Italian artist and scientist Leonardo da Vinci. It is reliably known that he was not involved in the invention of a perpetual motion machine. Apparently, this drawing came to him for conclusion from some Italian inventor.)
● 78. Perpetual motion machine by V. Congreve.
The English artilleryman and engineer William Congreve designed a perpetual motion machine consisting of a triangular prism with rollers K, M, H at the corners and jaws stretched around the prism (Fig. 52). The whole thing is partially submerged in water. The inventor believed that the weight of sponge A would increase due to the absorbed water. As a result, the balance will be disrupted and the tape with the jaws will move. Then sponge B absorbs water, replacing sponge A, the tape turns again, and so on endlessly. To increase the difference between the weight of the sponges emerging from the water above roller K and plunging into the water at roller M (i.e., to more reliably ensure movement), the author provided for squeezing water out of the sponges above roller K using weights P attached to the sponges. But... the engine didn't work. What's wrong with the project?
● 79. Air resistance force.
The ball moves in the air, having a current speed v(Fig. 53). Since the force of air resistance is proportional to the square of the speed, it can be represented in the form F=kv 2, Where k- proportionality coefficient.
On the other hand, decomposing the speed v into the horizontal and vertical components, we get: v 1 = vcos60° and v 2 = vsin60°. Therefore, F 1 = kv 1 2 = kv 2 cos 2 60° and F 2 = kv 2 2 = kv 2 sin 2 6O°, where F 1 and F 2 are the resistance forces caused by the components v 1 and v 2. Thus, the total drag force is:
which does not coincide with the expression F = kv 2. How to resolve this contradiction?
○ 80. Do clouds fall?
All bodies fall to the ground. Clouds are made up of small droplets of water, which means the clouds must fall to the ground.
However, no one was able to observe that the cloud, descending, ever reached the ground. How to resolve this paradox?
○ 81. How to shoot from a flying airplane.
When testing a rocket mounted at the tail of an aircraft to protect it from attack from behind, a surprising paradox was discovered. When a projectile was fired, it first moved away from the plane, and then turned around and caught up with the plane. How can this phenomenon be explained?
Get lost! Get lost!
He's had bad luck all his life. As a child, an unexplained illness nearly ended his life. Fate spared him, but not for long. In his youth, sudden paralysis made him crippled - his legs refused to work, he could barely move. But all the more immeasurable is his feat in science. Overcoming physical suffering, he worked with tenacity and ecstasy, characteristic only of a brilliant thinker,
At the age of 16, Blaise Pascal became no less famous a mathematician than his contemporaries such as Fermat and Descartes. At the age of 18 he invented counting machine- the predecessor of the adding machine and the great-grandmother of the computer.
The time came when he invaded that area of \u200b\u200bknowledge in which the great Galileo failed. He began with a discrepancy between the mass of water poured into a vessel and the force with which this mass presses on the bottom. Wanting to obtain a visual proof of the “hydrostatic paradox,” Pascal performs an experiment called “Pascal’s barrel.”
According to his instructions, strong oak barrel filled to the brim with water and tightly closed the lid. The end of a vertical glass tube of such length was inserted into a small hole in the lid that its end was at the level of the second floor.
Going out onto the balcony, Pascal began filling the tube with water (Fig. 2). He had not even managed to pour out a dozen glasses when suddenly, to the amazement of the onlookers who surrounded the barrel, the barrel burst with a crash. She was torn apart by an incomprehensible force.
Pascal is convinced: yes, the force that breaks the barrel does not at all depend on the amount of water in the tube. It's all about the height to which the tube was filled. Next, the amazing property of water manifests itself - to transmit the pressure created on its surface (in the barrel) throughout the entire volume, to each point of the wall or bottom of the barrel.
So he comes to the discovery of the law that received his name, the name of Blaise Pascal: “Pressure applied to the surface of a liquid is transferred to each of its particles without changing its original value.”
On the surface of the water in the barrel under the lid, this pressure is P = ρgh, where ρ is the density of water; g - free fall acceleration; h is the height of the water column in the tube. By multiplying the resulting pressure by the diametrical cross-sectional area of the barrel (S = DH), we get the force that crushed its strong oak walls:
P= ρg(h+H/2)(DH)
If we take the height of the water in the tube to be 4 m (second floor balcony), the diameter of the barrel is 0.8 m and the height of the barrel is 0.8 m, then no matter how small the amount of water in the tube is, the force breaking the barrel will be 27.6 kN.
Already relying on the law he discovered, Pascal obtains a corollary: “If a full vessel, closed on all sides, has two holes, one of which is 100 times larger than the other, then, placing in each hole a piston corresponding to this hole, the person pressing a small piston will create a force equal to the force of 100 people pressing on a piston with an area 100 times larger." Thus, Pascal substantiated the possibility of obtaining arbitrarily large forces from arbitrarily small ones using a liquid. It is difficult to overestimate the importance of this consequence for modern mechanical engineering. It led to the creation of superpresses with a pressure of (65-75) * 10 7 Pa. It formed the basis of the hydraulic drive, which in turn led to the emergence of hydraulic automation that controls modern jetliners, spaceships, computer-controlled machines, powerful dump trucks, mining combines, excavators...
But what about Pascal himself? Did he foresee that his law would mark an era in technological progress?
Suddenly Pascal stopped all research activities and, leaving Paris, settled in a cell in the Port-Royal monastery. He cut off all ties with people of science, renounced everything that just yesterday constituted the meaning of his existence and completely devoted himself to religion. If the great Galileo, even the most cruel tortures in the dungeons of the Inquisition, did not force him to change science, then Pascal did it himself, without any coercion.
He ended his days dressed in hair shirt with a Bible on his lap. He mortified his flesh in order to atone for the most terrible, from the point of view of religion, sin - curiosity, passion for knowledge. And he died when he was only 39 years old.
But why did he recant? Perhaps he was frightened by his truly anti-divine discoveries, which promised the world such power, in comparison with which the divine power paled, or he lacked that single step from ignorance to knowledge that Archimedes was able to take, and which would have allowed him to reveal the paradoxical properties of water. In the bright chronicle of the history of science, the tragedy of Blaise Pascal became the only dark spot.
The newest book of facts. Volume 3 [Physics, chemistry and technology. History and archaeology. Miscellaneous] Kondrashov Anatoly Pavlovich
What is the hydrostatic paradox?
The hydrostatic paradox lies in the fact that the weight of a liquid poured into a vessel may differ from the pressure it exerts on the bottom of the vessel. Thus, in vessels that expand upward, the pressure force on the bottom is less than the weight of the liquid, and in vessels that narrow, it is greater. In a cylindrical vessel both forces are equal. If the same liquid is poured to the same height into vessels of different shapes, but with the same bottom area, then, despite the different weight of the poured liquid, the pressure force on the bottom is the same for all vessels and is equal to the weight of the liquid in a cylindrical vessel. This follows from the fact that the pressure of a fluid at rest depends only on the depth below the free surface and on the density of the fluid. The hydrostatic paradox is explained as follows. Since hydrostatic pressure is always normal to the walls of the vessel, the pressure force on the inclined walls has a vertical component, which compensates for the weight of the volume of liquid that is excessive against the cylinder in a vessel expanding upward and the weight of the volume of liquid that is missing against the cylinder in a vessel that narrows upward. The hydrostatic paradox was discovered by the French physicist Blaise Pascal (1623–1662).
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