Pressure in liquid and gas. Atmospheric pressure at different altitudes

a) Clutch of lead cylinders.

Equipment: set of lead cylinders; tripod; kettlebell 2 kg.

It is advisable to have two sets of lead cylinders for the experiment: one to show the uneven and oxidized surfaces that prevent the cylinders from connecting, and the second for the main part of the demonstration. Clean the ends of the cylinders by inserting it into the base and, pressing the knife lightly, give it a few turns. After sharpening, the end of the cylinder should be smooth and shiny.

Having placed both cylinders at their ends, press them towards each other and slightly rotate them around their axis, after which the cylinders should connect. To assess the magnitude of their forces mutual attraction hang the upper cylinder by a hook on a tripod, and hang a 2-kilogram weight from the lower one without jerking. Then, having removed the weight and separated the cylinders with their hands, they demonstrate to the students the unevenness of the ends that have arisen at the coupling points.

CONCLUSION: at a small distance between molecules solid there are forces of mutual attraction that manifest themselves when molecules come into very close contact, and on greater distance Repulsive forces act between molecules.

b) Adhesion of glass to water.

Equipment: two glass plates; water.

Connect the surfaces of the dry plates, make sure there are no attractive forces, drop water between the glasses and try

separate them.

CONCLUSION: there are forces of mutual attraction between the molecules of a liquid and a solid.

Control questions

1. What is the significance of the material being studied for the formation of students’ worldview?

2. What bodies are made of:

a) from molecules;

b) from particles;

c) from atoms?

3. Propose a way to summarize students’ knowledge in the form of a table about the aggregate states of matter and the properties of bodies.

4. Is it true that when heated, all substances expand?

5. Give examples of diffusion used in everyday life and technology.

L A B O R A T O R N A Y WORK No. 2

PRESSURE S T O R D Y X T E L,

LIQUID C O S T E Y AND GAS O V

Experiment No. 1. Pressure of a solid body on a support

Equipment: sand bath, board with nails driven in, 1 kg weight.

Before demonstration, wet sand is poured into the bath and the surface is well leveled. Nails are driven into the corners of a small board. The board with its nail heads is placed on a layer of sand and a weight is placed on top of it. the nails are only slightly pressed into the sand. The board is then turned over on the point of the nails. In this case, the support area of the board is reduced, and under the same force, the nails go significantly deeper into the sand.

CONCLUSION: The result of the force depends on the area of the contacting bodies.

Assignment: Knowing your mass and the area of the shoe, determine how much pressure you produce when walking and standing still. Determine the support area of the boot using a sheet of checkered paper.

EXPERIMENT No. 2. Inflating a rubber ball under the bell of an air pump

Equipment: Komovsky pump, vacuum plate with glass bell, balloon.

This experiment serves as an illustration to explain the mechanism of gas pressure on the walls of the vessel.

IN balloon leave a small amount of air and tie tightly. The ball is placed on the air pump plate so that it does not cover the hole in the discharge tube of the plate, and covered with a glass bell. Connect the plate to the pump and pump out the air. As the air is pumped out, the ball gradually inflates and takes the shape of a ball. Then air is slowly let in under the bell, the opposite phenomenon is observed, and a conclusion is drawn that the gas pressure is equal in all directions.

EXPERIMENT No. 3. Transmission of pressure by gases and liquids.

Equipment: Pascal's ball, vessel with water, bath.

Unscrew the ball from the cylinder and extend the piston with the rod until it stops. Water is poured into the cylinder and the ball is screwed back on. Having placed the device over the bath, slowly push in the piston. They show that the jets from the hole of the ball spray at approximately the same distance. This indicates the same speed of water flow from all holes, that is, the same pressure in all places of the ball.

The experiment is more effective if the jets are illuminated with side light. In this case, they stand out in relief against the black background of the chalkboard.

Draw a conclusion.

EXPERIMENT No. 4. Design and operation of a hydraulic press

ABOUT
equipment: Hydraulic Press.

The educational hydraulic press is widely used in schools. Before the demonstration, the teacher makes a schematic drawing of a hydraulic press with a pressure gauge and a safety valve on the board. First, compare the main parts of the press on the device with the schematic image on the board. By naming the individual parts of the device and their purpose, they tell how a hydraulic press works and how its individual parts interact with each other. The teacher shows on the diagram and in reality an oil tank, a small cylinder with a handle in the form of a lever, valves, a pressure gauge and a large cylinder. Raising and lowering the handle several times and referring to the diagram, he explains the path of oil from the reservoir to the large cylinder.

The device is equipped with two safety devices that protect the device from destruction: a pressure gauge with a red line indicating the maximum permissible pressure in the press, and a safety valve that will automatically open if for some reason the permissible pressure is exceeded.

Reminding students that the pressure in the press cylinders is the same, and the pressure force is proportional to the area of the pistons, they find out how the presses achieve a large gain in strength.

Task: Having measured the diameters of the large and small pistons, the length of the handle arm, calculate the gain in force (that is, how many times the pressure force developed by the large piston, more power small piston pressure).

EXPERIMENT No. 5. Liquid pressure on the bottom and walls of the vessel.

ABOUT
equipment: a plastic bag with water, two dynamometers, tripods, an open pressure gauge, a device for demonstrating the pressure inside the liquid, an aquarium with water.

a) To demonstrate the pressure of the liquid on the walls of the vessel, a plastic bag with water is placed between the tables of dynamometers and the presence of equal pressure on the walls of the bag is shown.

b) To study the dependence of the pressure of a liquid column, it is necessary to show the presence of pressure on the platform located inside the liquid, and to prove that the magnitude of this pressure does not depend on the position of the platform, but changes only on the depth of its immersion.

To demonstrate the operation of the device, connect the capsule nipple via a rubber tube to an open water pressure gauge. Show that the liquid in both elbows of the pressure gauge is at the same level. Then, pressing slightly on the outer surface of the rubber film with a finger, students pay attention to how the liquid levels in the pressure gauge change.

Next, they lower the capsule into an aquarium with water and show that the pressure on the rubber film, observed on the pressure gauge, increases with the depth of immersion. After this, install the device at a certain depth and, using a wire hook, turn the bracket on which the capsule is attached around horizontal axis. Students are reminded that at this level of immersion, the pressure inside the liquid does not depend on the location of the rubber film.

EXPERIMENT No. 6. Hydrostatic paradox.

ABOUT
equipment: Pascal's device, vessel for draining water, colored water.

Device for this experience consists of a base on which a ring-shaped frame with thread is fixed. This frame, open at the top, is covered at the bottom with a thin rubber film, resting on a round plate connected by a lever to an easily movable arrow.

The device comes with three vessels of different shapes and volumes, but with the same base area. Each vessel has a threaded frame with which it is installed on the device.

To demonstrate, first screw a cylindrical vessel into the frame and pour water into it to a height below the top edge by 2-3 cm. The water level in the vessel is noted by a pointer moving on the rod, and the place on the scale where the arrow is installed is noted by a “hussar” that is clearly visible from afar. .

After this, water is poured through the drain tap and instead of a cylindrical vessel, another one is installed, for example, one that expands upward, then a third one. In each experiment, they are convinced that much more or less water has to be taken than for a cylindrical vessel, and the pressure remains the same if the water level rises to the indicated level. This is Pascal's paradox.

EXPERIMENT No. 7. Equilibrium of liquid in communicating vessels.

Equipment: communicating vessels, glass of water, screen.

a) First, an experiment is performed with two communicating vessels in the form of glass tubes connected to each other by a rubber tube, and it is established that a homogeneous liquid is located at the same level. Then they move on to the demonstration device, which consists of 4-5 glass tubes of various shapes, different sections, connected to each other by a horizontal tube of the same cross-section.

To show the equilibrium of a homogeneous liquid in these tubes, slightly colored water is poured through a wide tube. Take so much water that it rises slightly above half the height vertical tubes. A white screen is placed behind the device and students’ attention is drawn to the liquid levels, which in all tubes are located on the same horizontal straight line. If you tilt the device to the right or left, the liquid levels in the tubes will again remain on the same horizontal level.

First, tinted water is poured into the tube to a height of 100-150 mm. Then another liquid that does not mix with water, for example gasoline or kerosene, is poured into one of the branches. The fluids will be installed as shown in the figure.

Task: Using a ruler, mark the boundary of a column of liquid and water and make sure that the ratio of these heights is inversely proportional to the densities of the liquids.

EXPERIMENT No. 8. Archimedes' Law.

Equipment: Archimedes bucket, universal tripod, casting vessel, chemical beaker.

The validity of Archimedes' law is demonstrated using a device consisting of a bucket, a cylinder, which are connected to each other, and a spring dynamometer with a disk pointer. The stretch of the spring is marked with a clamp.

Before the experiment, they show that the capacity of the bucket corresponds to the volume of the cylinder. Assemble the installation according to the drawing. Pay attention to the position of the pointer disk and mark its position with an arrow. Lower the cylinder into a vessel with water. In this case, part of the water is poured out of the vessel. Draw students' attention to the position of the pointer arrow. The spilled water is poured into a bucket. Pay attention to the arrow again. They draw a conclusion.

Lesson

Subject: Transmission of pressure by liquids and gases.
The purpose of the lesson: create conditions for students to acquire new knowledge about pressure in liquids and gases.

Lesson objectives:

Educational:

Introduce students to the concepts of pressure in a liquid, transmission of pressure by liquids.

Developmental:

Develop the ability to analyze, compare, present material competently and logically;

Educational:

Developing accuracy, the ability to listen and be heard;

Lesson type: learning new material

During the classes:

I. Class organization.

II. Updating knowledge.

Frontal survey:

What does the pressure of solids depend on?

How do gases exert pressure?

What does gas pressure depend on?

Do you think liquids exert pressure? How? What does it depend on?

Conclusion: This means that you cannot say for sure that liquids do or do not exert pressure.

Today in lesson we will talk about fluid pressure. Write down the topic of the lesson. Our goal is to prove that fluids exert pressure and how; what does fluid pressure depend on? how can you calculate fluid pressure; how they are taken into account and applied in practice.

III. Learning new knowledge.

So, we put forward a hypothesis: liquids exert pressure. How to test this hypothesis?

Liquid, like all bodies on earth, is affected by gravity. Therefore, each layer of liquid poured into a vessel, with its weight, creates pressure on other layers, which, according to Pascal’s law, is transmitted in all directions. Therefore, there is pressure inside the liquid.

Experience 1. An experiment confirming the existence of pressure in a liquid.

Pour water into the tube; under the weight of the liquid, the rubber film will bend.

Questions:

How long will the rubber film sag?

What happens if you increase the liquid column?

Experiment 2. An experiment confirming the existence of pressure inside a liquid.(a glass tube, the hole of which is closed with a rubber film, a glass of water)

Let's lower the tube with a rubber bottom, into which water is poured, into another, wider vessel with water. We will see that as the tube is lowered, the rubber film gradually straightens. Full straightening of the film shows that the forces acting on it from above and below are equal. Complete straightening of the film occurs when the water levels in the tube and vessel coincide.

Experiment 3. Experience showing that at the same level the pressure of the liquid is the same.(a glass tube with a side hole covered with a rubber film, a glass of water)

Conclusion: experience shows that there is pressure inside a liquid and at the same level it is equal in all directions. With depth, pressure increases.

Experiment 4. An experiment showing that the liquid acts on the walls of the vessel and that the pressure changes with depth. ( a glass of water; three small holes are made in the side surface at different heights from the bottom).

Let's take a tall vessel in which three small holes are made on the side surface at different heights. Let's close them and fill the vessel with water. Then let's open the holes and see what happens.

Conclusion: The pressure inside the liquid is different at different altitudes. It increases with depth.

Experiment 5. An experiment showing the change in fluid pressure with depth.

Take a glass tube and a light disk on a thread. Let's pull the thread to get a vessel with a falling bottom. Immerse the resulting vessel in a jar of water.

Question: Why doesn't the bottom fall off even if the thread is not pulled?

Carefully pour a little colored water into the vessel so that its level is lower than the water level in the jar.

Question: What are we seeing?

Add the remaining water from the glass. The height of the layer of colored water in the vessel will increase.

Question: What are we seeing? Why did the bottom fall out?

Conclusion: Pressure increases with depth.

Exercise: Find the conclusion in the textbook on page, read and write it down in your notebook.

Gases are no different from liquids in this respect. But we must remember that the gas density is hundreds of times less density liquids. The weight of the gas in the vessel is small and its pressure in many cases can be ignored.

IV.Consolidation and clarification of knowledge:

How do liquids and gases transmit pressure?

Why do liquids and gases transmit pressure equally in all directions?

How to prove that the pressure inside a liquid is different levels different, but at the same level in all directions the same?

Why is it that in many cases the pressure of a gas created by its weight is not taken into account?

V.Homework:

VI.Reflection

Complete the sentence Today in class I learned...

A man with and without skis.

A man walks through loose snow with with great difficulty, sinking deeply with every step. But, having put on skis, he can walk without almost falling into it. Why? With or without skis, a person acts on the snow with the same force equal to his weight. However, the effect of this force is different in both cases, because the surface area on which a person presses is different, with skis and without skis. Almost 20 times the surface area of skis more area soles. Therefore, while standing on skis, a person acts on every square centimeter snow surface area with a force 20 times less than standing on the snow without skis.

A student, pinning a newspaper to the board with buttons, acts on each button with equal force. However, a button with a sharper end will go into the wood more easily.

This means that the result of the force depends not only on its modulus, direction and point of application, but also on the area of the surface to which it is applied (perpendicular to which it acts).

This conclusion is confirmed by physical experiments.

Experience. The result of the action of a given force depends on what force acts on a unit surface area.

You need to drive nails into the corners of a small board. First, place the nails driven into the board on the sand with their points up and place a weight on the board. In this case, the nail heads are only slightly pressed into the sand. Then we turn the board over and place the nails on the edge. In this case, the support area is smaller, and under the same force the nails go significantly deeper into the sand.

Experience. Second illustration.

The result of the action of this force depends on what force acts on each unit of surface area.

In the examples considered, the forces acted perpendicular to the surface of the body. The man's weight was perpendicular to the surface of the snow; the force acting on the button is perpendicular to the surface of the board.

Magnitude, equal to the ratio force acting perpendicular to a surface to the area of that surface is called pressure.

To determine the pressure, the force acting perpendicular to the surface must be divided by the surface area:

pressure = force / area.

Let us denote the quantities included in this expression: pressure - p, the force acting on the surface is F and surface area - S.

Then we get the formula:

p = F/S

It is clear that a larger force acting on the same area will produce greater pressure.

The unit of pressure is taken to be the pressure produced by a force of 1 N acting on a surface with an area of 1 m2 perpendicular to this surface.

Unit of pressure - newton per square meter (1 N/m2). In honor of the French scientist Blaise Pascal it's called pascal ( Pa). Thus,

1 Pa = 1 N/m2.

Other units of pressure are also used: hectopascal (hPa) And kilopascal (kPa).

1 kPa = 1000 Pa;

1 hPa = 100 Pa;

1 Pa = 0.001 kPa;

1 Pa = 0.01 hPa.

Let's write down the conditions of the problem and solve it.

Given : m = 45 kg, S = 300 cm 2; p = ?

In SI units: S = 0.03 m2

Solution:

p = F/S,

F = P,

P = g m,

P= 9.8 N · 45 kg ≈ 450 N,

p= 450/0.03 N/m2 = 15000 Pa = 15 kPa

"Answer": p = 15000 Pa = 15 kPa

Ways to reduce and increase pressure.

A heavy crawler tractor produces a pressure on the soil equal to 40 - 50 kPa, i.e. only 2 - 3 times more than the pressure of a boy weighing 45 kg. This is explained by the fact that the weight of the tractor is distributed over a larger area due to the track drive. And we have established that the larger the support area, the less pressure produced by the same force on this support .

Depending on whether low or high pressure is needed, the support area increases or decreases. For example, in order for the soil to withstand the pressure of the building being erected, the area of the lower part of the foundation is increased.

Tires trucks and the landing gear of aircraft is made much wider than that of passenger cars. The tires of cars designed for driving in deserts are made especially wide.

Heavy vehicles, such as a tractor, a tank or a swamp vehicle, having a large support area of the tracks, pass through swampy areas that cannot be passed by a person.

On the other hand, when small area surfaces, a small force can produce a lot of pressure. For example, when pressing a button into a board, we act on it with a force of about 50 N. Since the area of the tip of the button is approximately 1 mm 2, the pressure produced by it is equal to:

p = 50 N / 0.000 001 m 2 = 50,000,000 Pa = 50,000 kPa.

For comparison, this pressure is 1000 times greater than the pressure exerted by a crawler tractor on the soil. You can find many more such examples.

The blades of cutting instruments and the points of piercing instruments (knives, scissors, cutters, saws, needles, etc.) are specially sharpened. The sharpened edge of a sharp blade has a small area, so even a small force creates a lot of pressure, and this tool is easy to work with.

Cutting and piercing devices are also found in living nature: these are teeth, claws, beaks, spines, etc. - all of them are made of hard material, smooth and very sharp.

Pressure

It is known that gas molecules move randomly.

We already know that gases, unlike solids and liquids, fill the entire container in which they are located. For example, a steel cylinder for storing gases, a car tire inner tube or a volleyball. In this case, the gas exerts pressure on the walls, bottom and lid of the cylinder, chamber or any other body in which it is located. Gas pressure is caused by factors other than the pressure of a solid body on the support.

It is known that gas molecules move randomly. As they move, they collide with each other, as well as with the walls of the container containing the gas. There are many molecules in a gas, and therefore the number of their impacts is very large. For example, the number of impacts of air molecules in a room on a surface with an area of \u200b\u200b1 cm 2 in 1 s is expressed as a twenty-three-digit number. Although the impact force of an individual molecule is small, the effect of all molecules on the walls of the vessel is significant - it creates gas pressure.

So, the pressure of the gas on the walls of the vessel (and on the body placed in the gas) is caused by impacts of gas molecules .

Consider the following experiment. Place a rubber ball under the air pump bell. He contains a small amount of air and has irregular shape. Then we pump out the air from under the bell. The shell of the ball, around which the air becomes increasingly rarefied, gradually inflates and takes the shape of a regular ball.

How to explain this experience?

Special durable steel cylinders are used for storing and transporting compressed gas.

In our experiment, moving gas molecules continuously hit the walls of the ball inside and outside. When air is pumped out, the number of molecules in the bell around the shell of the ball decreases. But inside the ball their number does not change. Therefore, the number of impacts of molecules on the outer walls of the shell becomes less than the number of impacts on the inner walls. The ball inflates until the elastic force of its rubber shell becomes equal to the force of gas pressure. The shell of the ball takes the shape of a ball. This shows that gas presses on its walls in all directions equally. In other words, the number of molecular impacts per square centimeter of surface area is the same in all directions. The same pressure in all directions is characteristic of gas and is a consequence of random movement huge number molecules.

Let's try to reduce the volume of gas, but so that its mass remains unchanged. This means that in every cubic centimeter There will be more gas molecules, the gas density will increase. Then the number of impacts of molecules on the walls will increase, i.e., the gas pressure will increase. This can be confirmed by experience.

On the image A shows a glass tube, one end of which is closed with a thin rubber film. A piston is inserted into the tube. When the piston moves in, the volume of air in the tube decreases, i.e. the gas is compressed. The rubber film bends outward, indicating that the air pressure in the tube has increased.

On the contrary, as the volume of the same mass of gas increases, the number of molecules in each cubic centimeter decreases. This will reduce the number of impacts on the walls of the vessel - the gas pressure will become less. Indeed, when the piston is pulled out of the tube, the volume of air increases and the film bends inside the vessel. This indicates a decrease in air pressure in the tube. The same phenomena would be observed if instead of air there were any other gas in the tube.

So, when the volume of a gas decreases, its pressure increases, and when the volume increases, the pressure decreases, provided that the mass and temperature of the gas remain unchanged.

How will the pressure of a gas change if it is heated at a constant volume? It is known that the speed of gas molecules increases when heated. Moving faster, the molecules will hit the walls of the container more often. In addition, each impact of the molecule on the wall will be stronger. As a result, the walls of the vessel will experience greater pressure.

Hence, The higher the gas temperature, the greater the gas pressure in a closed vessel, provided that the gas mass and volume do not change.

From these experiments it can be concluded general conclusion, What The gas pressure increases the more often and harder the molecules hit the walls of the vessel .

To store and transport gases, they are highly compressed. At the same time, their pressure increases, the gases must be enclosed in special, very durable cylinders. Such cylinders, for example, contain compressed air in submarines, oxygen used in metal welding. Of course, we must always remember that gas cylinders cannot be heated, especially when they are filled with gas. Because, as we already understand, an explosion can occur with very unpleasant consequences.

Pascal's law.

Pressure is transmitted to every point in the liquid or gas.

The pressure of the piston is transmitted to each point of the fluid filling the ball.

Now gas.

Unlike solids, individual layers and fine particles liquids and gases can move freely relative to each other in all directions. It is enough, for example, to lightly blow on the surface of the water in a glass to cause the water to move. On a river or lake, the slightest breeze causes ripples to appear.

The mobility of gas and liquid particles explains that the pressure exerted on them is transmitted not only in the direction of the force, but to every point. Let's consider this phenomenon in more detail.

On the image, A depicts a vessel containing gas (or liquid). The particles are evenly distributed throughout the vessel. The vessel is closed by a piston that can move up and down.

By applying some force, we will force the piston to move slightly inward and compress the gas (liquid) located directly below it. Then the particles (molecules) will be located in this place more densely than before (Fig, b). Due to mobility, gas particles will move in all directions. As a result, their arrangement will again become uniform, but more dense than before (Fig. c). Therefore, gas pressure will increase everywhere. This means that additional pressure is transmitted to all particles of gas or liquid. So, if the pressure on the gas (liquid) near the piston itself increases by 1 Pa, then at all points inside gas or liquid, the pressure will become greater than before by the same amount. The pressure on the walls of the vessel, the bottom, and the piston will increase by 1 Pa.

The pressure exerted on a liquid or gas is transmitted to any point equally in all directions .

This statement is called Pascal's law.

Based on Pascal's law, it is easy to explain the following experiments.

The picture shows a hollow ball with small holes in various places. A tube is attached to the ball into which a piston is inserted. If you fill a ball with water and push a piston into the tube, water will flow out of all the holes in the ball. In this experiment, a piston presses on the surface of water in a tube. The water particles located under the piston, compacting, transfer its pressure to other layers that lie deeper. Thus, the pressure of the piston is transmitted to each point of the fluid filling the ball. As a result, part of the water is pushed out of the ball in the form of identical streams flowing out of all holes.

If the ball is filled with smoke, then when the piston is pushed into the tube, equal streams of smoke will begin to come out of all the holes in the ball. This confirms that gases transmit the pressure exerted on them in all directions equally.

Pressure in liquid and gas.

Under the influence of the weight of the liquid, the rubber bottom in the tube will bend.

Liquids, like all bodies on Earth, are affected by gravity. Therefore, each layer of liquid poured into a vessel creates pressure with its weight, which, according to Pascal’s law, is transmitted in all directions. Therefore, there is pressure inside the liquid. This can be verified by experience.

Pour water into a glass tube, the bottom hole of which is closed with a thin rubber film. Under the influence of the weight of the liquid, the bottom of the tube will bend.

Experience shows that the higher the column of water above the rubber film, the more it bends. But every time after the rubber bottom bends, the water in the tube comes to equilibrium (stops), since, in addition to the force of gravity, the elastic force of the stretched rubber film acts on the water.

The forces acting on the rubber film are

are the same on both sides.

Illustration.

The bottom moves away from the cylinder due to the pressure of gravity on it.

The same experiment can be carried out with a tube in which a rubber film covers the side hole, as shown in figure a. Let's immerse this tube with water in another vessel with water, as shown in the figure, b. We will notice that the film will straighten again as soon as the water levels in the tube and the vessel are equal. This means that the forces acting on the rubber film are the same on all sides.

Let's take a vessel whose bottom can fall away. Let's put it in a jar of water. The bottom will be tightly pressed to the edge of the vessel and will not fall off. It is pressed by the force of water pressure directed from bottom to top.

We will carefully pour water into the vessel and watch its bottom. As soon as the water level in the vessel coincides with the water level in the jar, it will fall away from the vessel.

At the moment of separation, a column of liquid in the vessel presses from top to bottom, and pressure from a column of liquid of the same height, but located in the jar, is transmitted from bottom to top to the bottom. Both of these pressures are the same, but the bottom moves away from the cylinder due to the action on it own strength gravity.

Experiments with water were described above, but if you take any other liquid instead of water, the results of the experiment will be the same.

So, experiments show that There is pressure inside the liquid, and at the same level it is equal in all directions. Pressure increases with depth.

Gases are no different from liquids in this respect, because they also have weight. But we must remember that the density of gas is hundreds of times less than the density of liquid. The weight of the gas in the vessel is small, and its “weight” pressure in many cases can be ignored.

Calculation of liquid pressure on the bottom and walls of a vessel.

Let's consider how you can calculate the pressure of a liquid on the bottom and walls of a vessel. Let us first solve the problem for a vessel shaped like a rectangular parallelepiped.

Force F, with which the liquid poured into this vessel presses on its bottom, is equal to the weight P liquid in the container. The weight of a liquid can be determined by knowing its mass m. Mass, as you know, can be calculated using the formula: m = ρ·V. The volume of liquid poured into the vessel we have chosen is easy to calculate. If the height of the column of liquid in a vessel is denoted by the letter h, and the area of the bottom of the vessel S, That V = S h.

Liquid mass m = ρ·V, or m = ρ S h .

The weight of this liquid P = g m, or P = g ρ S h.

Since the weight of a column of liquid is equal to the force with which the liquid presses on the bottom of the vessel, then by dividing the weight P To the square S, we get the fluid pressure p:

p = P/S, or p = g·ρ·S·h/S,

We have obtained a formula for calculating the pressure of the liquid at the bottom of the vessel. From this formula it is clear that the pressure of the liquid at the bottom of the vessel depends only on the density and height of the liquid column.

Therefore, using the formula derived, you can calculate the pressure of the liquid poured into the vessel any shape(strictly speaking, our calculation is only suitable for vessels that have the shape of a straight prism and a cylinder. In physics courses for the institute it was proven that the formula is also true for a vessel free form). In addition, it can be used to calculate the pressure on the walls of the vessel. The pressure inside the liquid, including the pressure from bottom to top, is also calculated using this formula, since the pressure at the same depth is the same in all directions.

When calculating pressure using the formula p = gρh you need density ρ express in kilograms per cubic meter(kg/m 3), and the height of the liquid column h- in meters (m), g= 9.8 N/kg, then the pressure will be expressed in pascals (Pa).

Example. Determine the oil pressure at the bottom of the tank if the height of the oil column is 10 m and its density is 800 kg/m3.

Let's write down the condition of the problem and write it down.

Given :

ρ = 800 kg/m 3

Solution :

p = 9.8 N/kg · 800 kg/m 3 · 10 m ≈ 80,000 Pa ≈ 80 kPa.

Answer : p ≈ 80 kPa.

Communicating vessels.

The figure shows two vessels connected to each other by a rubber tube. Such vessels are called communicating. A watering can, a teapot, a coffee pot are examples of communicating vessels. From experience we know that water poured, for example, into a watering can is always at the same level in the spout and inside.

We often encounter communicating vessels. For example, it could be a teapot, watering can or coffee pot.

The surfaces of a homogeneous liquid are installed at the same level in communicating vessels of any shape.

Liquids of different densities.

The following simple experiment can be done with communicating vessels. At the beginning of the experiment, we clamp the rubber tube in the middle and pour water into one of the tubes. Then we open the clamp, and the water instantly flows into the other tube until the water surfaces in both tubes are at the same level. You can mount one of the handsets on a tripod and raise, lower, or tilt the other different sides. And in this case, as soon as the liquid calms down, its levels in both tubes will be equalized.

In communicating vessels of any shape and cross-section, the surfaces of a homogeneous liquid are set at the same level(provided that the air pressure above the liquid is the same) (Fig. 109).

This can be justified as follows. The liquid is at rest without moving from one vessel to another. This means that the pressure in both vessels at any level is the same. The liquid in both vessels is the same, i.e. it has the same density. Therefore, its heights must be the same. When we lift one container or add liquid to it, the pressure in it increases and the liquid moves into another container until the pressures are balanced.

If a liquid of one density is poured into one of the communicating vessels, and a liquid of another density is poured into the second, then at equilibrium the levels of these liquids will not be the same. And this is understandable. We know that the pressure of the liquid at the bottom of the vessel is directly proportional to the height of the column and the density of the liquid. And in this case, the densities of the liquids will be different.

If the pressures are equal, the height of a column of liquid with a higher density will be less than the height of a column of liquid with a lower density (Fig.).

Experience. How to determine the mass of air.

Air weight. Atmosphere pressure.

The existence of atmospheric pressure.

Atmosphere pressure greater than the pressure of rarefied air in the vessel.

Air, like any body on Earth, is affected by gravity, and therefore air has weight. The weight of air is easy to calculate if you know its mass.

We will show you experimentally how to calculate the mass of air. To do this, you need to take a durable glass ball with a stopper and a rubber tube with a clamp. Let's pump the air out of it, clamp the tube with a clamp and balance it on the scales. Then, opening the clamp on the rubber tube, let air into it. This will upset the balance of the scales. To restore it, you will have to put weights on the other pan of the scale, the mass of which will be equal to the mass of air in the volume of the ball.

Experiments have established that at a temperature of 0 °C and normal atmospheric pressure, the mass of air with a volume of 1 m 3 is equal to 1.29 kg. The weight of this air is easy to calculate:

P = g m, P = 9.8 N/kg 1.29 kg ≈ 13 N.

air shell, surrounding the Earth, called atmosphere (from Greek atmos- steam, air, and sphere- ball).

Atmosphere as shown by flight observations artificial satellites The Earth extends to a height of several thousand kilometers.

Due to the force of gravity, the upper layers of the atmosphere, like ocean water, compress the lower layers. The air layer adjacent directly to the Earth is compressed the most and, according to Pascal's law, transmits the pressure exerted on it in all directions.

As a result earth's surface and the bodies located on it experience the pressure of the entire thickness of the air, or, as is usually said in such cases, experience Atmosphere pressure .

The existence of atmospheric pressure can explain many phenomena that we encounter in life. Let's look at some of them.

The figure shows a glass tube, inside of which there is a piston that fits tightly to the walls of the tube. The end of the tube is lowered into water. If you lift the piston, the water will rise behind it.

This phenomenon is used in water pumps and some other devices.

The figure shows a cylindrical vessel. It is closed with a stopper into which a tube with a tap is inserted. Air is pumped out of the vessel using a pump. The end of the tube is then placed in water. If you now open the tap, water will spray like a fountain into the inside of the vessel. Water enters the vessel because atmospheric pressure is greater than the pressure of rarefied air in the vessel.

Why exists air envelope Earth.

Like all bodies, the gas molecules that make up the Earth's air envelope are attracted to the Earth.

But why then don’t they all fall to the surface of the Earth? How is the Earth's air envelope and its atmosphere preserved? To understand this, we must take into account that gas molecules are in continuous and random motion. But then another question arises: why don’t these molecules fly away into outer space, that is, into space.

In order to completely leave the Earth, a molecule, like spaceship or a rocket, must have a very high speed (not less than 11.2 km/s). This is the so-called second escape velocity. The speed of most molecules in the Earth's air shell is significantly less than this escape velocity. Therefore, most of them are tied to the Earth by gravity, only a negligible number of molecules fly beyond the Earth into space.

The random movement of molecules and the effect of gravity on them result in gas molecules “floating” in space near the Earth, forming an air envelope, or the atmosphere known to us.

Measurements show that air density decreases rapidly with altitude. So, at an altitude of 5.5 km above the Earth, the density of air is 2 times less than its density at the surface of the Earth, at an altitude of 11 km - 4 times less, etc. The higher it is, the rarer the air. And finally, in the uppermost layers (hundreds and thousands of kilometers above the Earth), the atmosphere gradually turns into airless space. The Earth's air envelope does not have a clear boundary.

Strictly speaking, due to the action of gravity, the gas density in any closed vessel is not the same throughout the entire volume of the vessel. At the bottom of the vessel, the gas density is greater than in its upper parts, therefore the pressure in the vessel is not the same. It is larger at the bottom of the vessel than at the top. However, for a gas contained in a vessel, this difference in density and pressure is so small that in many cases it can be completely ignored, just known about it. But for an atmosphere extending over several thousand kilometers, this difference is significant.

Measuring atmospheric pressure. Torricelli's experience.

It is impossible to calculate atmospheric pressure using the formula for calculating the pressure of a liquid column (§ 38). For such a calculation, you need to know the height of the atmosphere and air density. But the atmosphere does not have a definite boundary, and the density of air at different altitudes is different. However, atmospheric pressure can be measured using an experiment proposed in the 17th century by an Italian scientist Evangelista Torricelli , student of Galileo.

Torricelli's experiment consists of the following: a glass tube about 1 m long, sealed at one end, is filled with mercury. Then, tightly closing the second end of the tube, it is turned over and lowered into a cup of mercury, where this end of the tube is opened under the level of mercury. As in any experiment with liquid, part of the mercury is poured into the cup, and part of it remains in the tube. The height of the column of mercury remaining in the tube is approximately 760 mm. There is no air above the mercury inside the tube, there is an airless space, so no gas exerts pressure from above on the column of mercury inside this tube and does not affect the measurements.

Torricelli, who proposed the experiment described above, also gave its explanation. The atmosphere presses on the surface of the mercury in the cup. Mercury is in equilibrium. This means that the pressure in the tube is at the level ahh 1 (see figure) is equal to atmospheric pressure. When atmospheric pressure changes, the height of the mercury column in the tube also changes. As pressure increases, the column lengthens. As the pressure decreases, the mercury column decreases its height.

The pressure in the tube at level aa1 is created by the weight of the mercury column in the tube, since there is no air above the mercury in the upper part of the tube. It follows that atmospheric pressure is equal to the pressure of the mercury column in the tube , i.e.

p atm = p mercury

The higher the atmospheric pressure, the higher the mercury column in Torricelli's experiment. Therefore, in practice, atmospheric pressure can be measured by the height of the mercury column (in millimeters or centimeters). If, for example, the atmospheric pressure is 780 mm Hg. Art. (they say “millimeters of mercury”), this means that the air produces the same pressure as a vertical column of mercury 780 mm high.

Therefore, in this case, the unit of measurement for atmospheric pressure is 1 millimeter of mercury (1 mm Hg). Let's find the relationship between this unit and the unit known to us - pascal(Pa).

The pressure of a mercury column ρ of mercury with a height of 1 mm is equal to:

p = g·ρ·h, p= 9.8 N/kg · 13,600 kg/m 3 · 0.001 m ≈ 133.3 Pa.

So, 1 mmHg. Art. = 133.3 Pa.

Currently, atmospheric pressure is usually measured in hectopascals (1 hPa = 100 Pa). For example, weather reports may announce that the pressure is 1013 hPa, which is the same as 760 mmHg. Art.

Observing the height of the mercury column in the tube every day, Torricelli discovered that this height changes, that is, atmospheric pressure is not constant, it can increase and decrease. Torricelli also noted that atmospheric pressure is associated with changes in weather.

If you attach a vertical scale to the tube of mercury used in Torricelli’s experiment, you get the simplest device - mercury barometer (from Greek baros- heaviness, metreo- I measure). It is used to measure atmospheric pressure.

Barometer - aneroid.

In practice, a metal barometer called a metal barometer is used to measure atmospheric pressure. aneroid (translated from Greek - aneroid). This is what a barometer is called because it contains no mercury.

The appearance of the aneroid is shown in the figure. main part it is a metal box 1 with a wavy (corrugated) surface (see other figure). The air is pumped out of this box, and to prevent atmospheric pressure from crushing the box, its lid 2 is pulled upward by a spring. As atmospheric pressure increases, the lid bends down and tightens the spring. As the pressure decreases, the spring straightens the cap. An indicator arrow 4 is attached to the spring using a transmission mechanism 3, which moves to the right or left when the pressure changes. Under the arrow there is a scale, the divisions of which are marked according to the readings of the mercury barometer. Thus, the number 750, against which the aneroid arrow stands (see figure), shows that in this moment in a mercury barometer, the height of the mercury column is 750 mm.

Therefore, the atmospheric pressure is 750 mmHg. Art. or ≈ 1000 hPa.

The value of atmospheric pressure is very important for predicting the weather for the coming days, since changes in atmospheric pressure are associated with changes in weather. Barometer - necessary device for meteorological observations.

Atmospheric pressure at different altitudes.

In a liquid, pressure, as we know, depends on the density of the liquid and the height of its column. Due to low compressibility, the density of the liquid at different depths is almost the same. Therefore, when calculating pressure, we consider its density constant and take into account only the change in height.

The situation with gases is more complicated. Gases are highly compressible. And the more a gas is compressed, the greater its density, and the greater the pressure it produces. After all, gas pressure is created by the impacts of its molecules on the surface of the body.

The layers of air at the surface of the Earth are compressed by all the overlying layers of air located above them. But the higher the layer of air is from the surface, the weaker it is compressed, the lower its density. Therefore, the less pressure it produces. If, for example, balloon rises above the Earth's surface, the air pressure on the ball becomes less. This happens not only because the height of the air column above it decreases, but also because the density of the air decreases. It is smaller at the top than at the bottom. Therefore, the dependence of air pressure on altitude is more complex than that of liquids.

Observations show that atmospheric pressure in areas at sea level is on average 760 mm Hg. Art.

Atmospheric pressure equal to the pressure of a column of mercury 760 mm high at a temperature of 0 ° C is called normal atmospheric pressure.

Normal atmospheric pressure equals 101,300 Pa = 1013 hPa.

How more height above sea level, the lower the pressure.

With small climbs, on average, for every 12 m of rise, the pressure decreases by 1 mmHg. Art. (or by 1.33 hPa).

Knowing the dependence of pressure on altitude, you can determine the altitude above sea level by changing the barometer readings. Aneroids that have a scale by which height above sea level can be directly measured are called altimeters . They are used in aviation and mountain climbing.

Pressure gauges.

We already know that barometers are used to measure atmospheric pressure. To measure pressures greater or less than atmospheric pressure, it is used pressure gauges (from Greek manos- rare, loose, metreo- I measure). There are pressure gauges liquid And metal.

Let's first consider the device and action open liquid pressure gauge. It consists of a two-legged glass tube into which some liquid is poured. The liquid is installed in both elbows at the same level, since only atmospheric pressure acts on its surface in the vessel elbows.

To understand how such a pressure gauge works, it can be connected by a rubber tube to a round flat box, one side of which is covered with rubber film. If you press your finger on the film, the liquid level in the pressure gauge elbow connected to the box will decrease, and in the other elbow it will increase. What explains this?

When pressing on the film, the air pressure in the box increases. According to Pascal's law, this increase in pressure is also transmitted to the fluid in the pressure gauge elbow that is connected to the box. Therefore, the pressure on the fluid in this elbow will be greater than in the other, where only atmospheric pressure acts on the fluid. Under the force of this excess pressure, the liquid will begin to move. In the elbow with compressed air the liquid will fall, in the other it will rise. The fluid will come to equilibrium (stop) when the excess pressure of the compressed air is balanced by the pressure produced by the excess column of liquid in the other leg of the pressure gauge.

The harder you press on the film, the higher the excess liquid column, the greater its pressure. Hence, the change in pressure can be judged by the height of this excess column.

The figure shows how such a pressure gauge can measure the pressure inside a liquid. The deeper the tube is immersed in the liquid, the greater the difference in the heights of the liquid columns in the pressure gauge elbows becomes., therefore, and more pressure is generated by the fluid.

If you install the device box at some depth inside the liquid and turn it with the film up, sideways and down, the pressure gauge readings will not change. That's how it should be, because at the same level inside a liquid, the pressure is the same in all directions.

The picture shows metal pressure gauge . The main part of such a pressure gauge is a metal tube bent into a pipe 1 , one end of which is closed. The other end of the tube using a tap 4 communicates with the vessel in which the pressure is measured. As the pressure increases, the tube unbends. Movement of its closed end using a lever 5 and serrations 3 transmitted to the arrow 2 , moving near the instrument scale. When the pressure decreases, the tube, due to its elasticity, returns to its previous position, and the arrow returns to the zero division of the scale.

Piston liquid pump.

In the experiment we discussed earlier (§ 40), it was established that the water in the glass tube, under the influence of atmospheric pressure, rose upward behind the piston. This is what the action is based on. piston pumps

The pump is shown schematically in the figure. It consists of a cylinder, inside of which a piston moves up and down, tightly adjacent to the walls of the vessel. 1 . Valves are installed at the bottom of the cylinder and in the piston itself 2 , opening only upwards. When the piston moves upward, water under the influence of atmospheric pressure enters the pipe, lifts the lower valve and moves behind the piston.

As the piston moves downward, the water under the piston presses on the bottom valve and it closes. At the same time, under water pressure, a valve inside the piston opens, and water flows into the space above the piston. At next move As the piston moves upward, the water above it also rises, which pours into the outlet pipe. At the same time, a new portion of water rises behind the piston, which, when the piston is subsequently lowered, will appear above it, and this whole procedure is repeated again and again while the pump is running.

Hydraulic Press.

Pascal's law explains the action hydraulic machine (from Greek hydraulics- water). These are machines whose operation is based on the laws of motion and equilibrium of fluids.

The main part of a hydraulic machine is two cylinders of different diameters, equipped with pistons and a connecting tube. The space under the pistons and the tube are filled with liquid (usually mineral oil). The heights of the liquid columns in both cylinders are the same as long as no forces act on the pistons.

Let us now assume that the forces F 1 and F 2 - forces acting on the pistons, S 1 and S 2 - piston areas. The pressure under the first (small) piston is equal to p 1 = F 1 / S 1, and under the second (large) p 2 = F 2 / S 2. According to Pascal's law, pressure is transmitted equally in all directions by a fluid at rest, i.e. p 1 = p 2 or F 1 / S 1 = F 2 / S 2, from:

F 2 / F 1 = S 2 / S 1 .

Therefore, the strength F 2 so many times more power F 1 , How many times is the area of the large piston greater than the area of the small piston?. For example, if the area of the large piston is 500 cm2, and the small one is 5 cm2, and a force of 100 N acts on the small piston, then a force 100 times greater, that is, 10,000 N, will act on the larger piston.

Thus, with the help of a hydraulic machine, it is possible to balance a larger force with a small force.

Attitude F 1 / F 2 shows the gain in strength. For example, in the example given, the gain in strength is 10,000 N / 100 N = 100.

A hydraulic machine used for pressing (squeezing) is called hydraulic press .

Hydraulic presses are used where greater force is required. For example, for squeezing oil from seeds in oil mills, for pressing plywood, cardboard, hay. In metallurgical plants, hydraulic presses are used to make steel machine shafts, railroad wheels, and many other products. Modern hydraulic presses can develop forces of tens and hundreds of millions of newtons.

The structure of a hydraulic press is shown schematically in the figure. The pressed body 1 (A) is placed on a platform connected to the large piston 2 (B). With the help of a small piston 3 (D), high pressure is created on the liquid. This pressure is transmitted to every point of the fluid filling the cylinders. Therefore, the same pressure acts on the second, larger piston. But since the area of the 2nd (large) piston is greater than the area of the small one, the force acting on it will be greater than the force acting on piston 3 (D). Under the influence of this force, piston 2 (B) will rise. When piston 2 (B) rises, body (A) rests against the stationary upper platform and is compressed. Pressure gauge 4 (M) measures the fluid pressure. Safety valve 5 (P) automatically opens when the fluid pressure exceeds the permissible value.

From small cylinder to large liquid pumped by repeated movements of the small piston 3 (D). This is done as follows. When the small piston (D) rises, valve 6 (K) opens and liquid is sucked into the space under the piston. When the small piston is lowered under the influence of liquid pressure, valve 6 (K) closes, and valve 7 (K") opens, and the liquid flows into the large vessel.

The effect of water and gas on a body immersed in them.

Underwater we can easily lift a stone that is difficult to lift in the air. If you put a cork under water and release it from your hands, it will float up. How can these phenomena be explained?

We know (§ 38) that the liquid presses on the bottom and walls of the vessel. And if some solid body is placed inside the liquid, it will also be subject to pressure, just like the walls of the vessel.

Let us consider the forces that act from the liquid on a body immersed in it. To make it easier to reason, let’s choose a body that has the shape of a parallelepiped with bases parallel to the surface of the liquid (Fig.). Forces acting on side faces bodies are equal in pairs and balance each other. Under the influence of these forces, the body contracts. But the forces acting on the upper and lower edges of the body are not the same. The top edge is pressed by force from above F 1 column of liquid high h 1 . At the level of the lower edge, the pressure produces a column of liquid with a height h 2. This pressure, as we know (§ 37), is transmitted inside the liquid in all directions. Therefore, on bottom edge body from bottom to top with force F 2 presses a column of liquid high h 2. But h 2 more h 1, therefore, the force modulus F 2 more power module F 1 . Therefore, the body is pushed out of the liquid with force F you are t, equal difference strength F 2 - F 1, i.e.

But S·h = V, where V is the volume of the parallelepiped, and ρ f ·V = m f is the mass of liquid in the volume of the parallelepiped. Hence,

F out = g m w = P w,

i.e. buoyant force is equal to the weight of the liquid in the volume of the body immersed in it(buoyant force is equal to the weight of a liquid of the same volume as the volume of the body immersed in it).

The existence of a force pushing a body out of a liquid is easy to detect experimentally.

On the image A shows a body suspended from a spring with an arrow pointer at the end. The arrow marks the tension of the spring on the tripod. When the body is released into the water, the spring contracts (Fig. b). The same contraction of the spring will be obtained if you act on the body from bottom to top with some force, for example, press with your hand (lift).

Therefore, experience confirms that a body in a liquid is acted upon by a force that pushes the body out of the liquid.

As we know, Pascal's law also applies to gases. That's why bodies in gas are subject to a force that pushes them out of the gas. Under the influence of this force, the balloons rise upward. The existence of a force pushing a body out of a gas can also be observed experimentally.

We hang a glass ball or a large flask closed with a stopper from the shortened scale pan. The scales are balanced. Then a wide vessel is placed under the flask (or ball) so that it surrounds the entire flask. The vessel is filled with carbon dioxide, the density of which is more density air (therefore carbon dioxide falls down and fills the vessel, displacing air from it). In this case, the balance of the scales is disturbed. The cup with the suspended flask rises upward (Fig.). A flask immersed in carbon dioxide experiences a greater buoyancy force than the force that acts on it in air.

The force that pushes a body out of a liquid or gas is directed opposite to the force of gravity applied to this body.

Therefore, prolkosmos). This is precisely why in water we sometimes easily lift bodies that we have difficulty holding in the air.

A small bucket and a cylindrical body are suspended from the spring (Fig. a). An arrow on the tripod marks the stretch of the spring. It shows the weight of the body in the air. Having lifted the body, a casting vessel filled with liquid to the level of the casting tube is placed under it. After which the body is completely immersed in the liquid (Fig., b). Wherein part of the liquid, the volume of which is equal to the volume of the body, is poured out from the pouring vessel into the glass. The spring contracts and the spring pointer rises, indicating a decrease in body weight in the fluid. IN in this case In addition to gravity, another force acts on the body, pushing it out of the liquid. If liquid from a glass is poured into the upper bucket (i.e., the liquid that was displaced by the body), then the spring pointer will return to its initial position(Fig., c).

Based on this experience it can be concluded that the force pushing out a body completely immersed in a liquid is equal to the weight of the liquid in the volume of this body . We received the same conclusion in § 48.

If similar experience done to a body immersed in some gas, it would show that the force pushing a body out of a gas is also equal to the weight of the gas taken in the volume of the body .

The force that pushes a body out of a liquid or gas is called Archimedean force , in honor of the scientist Archimedes , who first pointed out its existence and calculated its value.

So, experience has confirmed that the Archimedean (or buoyant) force is equal to the weight of the liquid in the volume of the body, i.e. F A = P f = g m and. The mass of liquid mf displaced by a body can be expressed through its density ρf and the volume of the body Vt immersed in the liquid (since Vf - the volume of liquid displaced by the body is equal to Vt - the volume of the body immersed in the liquid), i.e. m f = ρ f ·V t. Then we get:

F A= g·ρ and · V T

Consequently, the Archimedean force depends on the density of the liquid in which the body is immersed and on the volume of this body. But it does not depend, for example, on the density of the substance of the body immersed in the liquid, since this quantity is not included in the resulting formula.

Let us now determine the weight of a body immersed in a liquid (or gas). Since the two forces acting on the body in this case are directed in opposite sides(gravity is down, and Archimedean force is up), then the weight of the body in the liquid P 1 will be less weight bodies in vacuum P = g m on Archimedean force F A = g m w (where m g - mass of liquid or gas displaced by the body).

Thus, if a body is immersed in a liquid or gas, then it loses as much weight as the liquid or gas it displaced weighs.

Example. Determine the buoyant force acting on a stone with a volume of 1.6 m 3 in sea water.

Let's write down the conditions of the problem and solve it.

When the floating body reaches the surface of the liquid, then with its further upward movement the Archimedean force will decrease. Why? But because the volume of the part of the body immersed in the liquid will decrease, and the Archimedean force is equal to the weight of the liquid in the volume of the part of the body immersed in it.

When the Archimedean force becomes equal to the force of gravity, the body will stop and float on the surface of the liquid, partially immersed in it.

The resulting conclusion can be easily verified experimentally.

Pour water into the drainage vessel to the level of the drainage tube. After this, we will immerse the floating body in the vessel, having previously weighed it in the air. Having descended into water, a body displaces a volume of water equal to the volume of the part of the body immersed in it. Having weighed this water, we find that its weight (Archimedean force) is equal to the force of gravity acting on a floating body, or the weight of this body in the air.

Having done the same experiments with any other bodies floating in different liquids - water, alcohol, salt solution, you can be sure that if a body floats in a liquid, then the weight of the liquid displaced by it equal to weight this body in the air.

It's easy to prove that if the density of a solid solid is greater than the density of a liquid, then the body sinks in such a liquid. A body with a lower density floats in this liquid. A piece of iron, for example, sinks in water but floats in mercury. A body whose density is equal to the density of the liquid remains in equilibrium inside the liquid.

Ice floats on the surface of water because its density is less than the density of water.

The lower the density of the body compared to the density of the liquid, the less part of the body is immersed in the liquid .

At equal densities bodies and liquids a body floats inside a liquid at any depth.

Two immiscible liquids, for example water and kerosene, are located in a vessel in accordance with their densities: in the lower part of the vessel - denser water (ρ = 1000 kg/m3), on top - lighter kerosene (ρ = 800 kg/m3) .

Average density of living organisms inhabiting aquatic environment, differs little from the density of water, so their weight is almost completely balanced by the Archimedean force. Thanks to this, aquatic animals do not need such strong and massive skeletons as terrestrial ones. For the same reason, the trunks of aquatic plants are elastic.

The swim bladder of a fish easily changes its volume. When a fish, with the help of muscles, descends to a greater depth, and the water pressure on it increases, the bubble contracts, the volume of the fish’s body decreases, and it is not pushed up, but floats in the depths. Thus, the fish can within certain limits adjust the depth of your dive. Whales regulate the depth of their dive by decreasing and increasing their lung capacity.

Sailing of ships.

Vessels sailing on rivers, lakes, seas and oceans are built from different materials With different densities. The hull of ships is usually made of steel sheets. All internal fastenings that give ships strength are also made of metals. Used to build ships various materials, having both higher and lower densities compared to water.

How do ships float, take on board and carry large cargo?

An experiment with a floating body (§ 50) showed that the body displaces so much water with its underwater part that the weight of this water is equal to the weight of the body in the air. This is also true for any vessel.

The weight of water displaced by the underwater part of the vessel is equal to the weight of the vessel with the cargo in the air or the force of gravity acting on the vessel with the cargo.

The depth to which a ship is immersed in water is called draft . The maximum permissible draft is marked on the ship's hull with a red line called waterline (from Dutch. water- water).

The weight of water displaced by a vessel when submerged to the waterline is equal to strength gravity acting on a ship with cargo is called the ship's displacement.

Currently, ships with a displacement of 5,000,000 kN (5 × 10 6 kN) or more are being built for the transportation of oil, that is, having a mass of 500,000 tons (5 × 10 5 t) or more together with the cargo.

If we subtract the weight of the vessel itself from the displacement, we get the carrying capacity of this vessel. The carrying capacity shows the weight of the cargo carried by the ship.

Shipbuilding existed back in Ancient Egypt, in Phenicia (it is believed that the Phoenicians were one of the best shipbuilders), Ancient China.

In Russia, shipbuilding originated at the turn of the 17th and 18th centuries. Mostly warships were built, but it was in Russia that the first icebreaker and ships with an engine were built internal combustion, nuclear icebreaker "Arktika".

Aeronautics.

Drawing describing the balloon of the Montgolfier brothers from 1783: “View and exact dimensions of the Balloon Earth“who was the first.” 1786

Since ancient times, people have dreamed of being able to fly above the clouds, swim in air ocean as they sailed on the sea. For aeronautics

At first, they used balloons that were filled with either heated air, hydrogen or helium.

In order for a balloon to rise into the air, it is necessary that the Archimedean force (buoyancy) F A acting on the ball was greater than the force of gravity F heavy, i.e. F A > F heavy

As the ball rises upward, the Archimedean force acting on it decreases ( F A = gρV), since the density upper layers atmosphere is less than that of the Earth's surface. To rise higher, a special ballast (weight) is dropped from the ball and this lightens the ball. Eventually the ball reaches its maximum lifting height. To release the ball from its shell, a portion of the gas is released using a special valve.

In the horizontal direction, a balloon moves only under the influence of wind, which is why it is called balloon (from Greek aer- air, stato- standing). Not so long ago, huge balloons were used to study the upper layers of the atmosphere and stratosphere - stratospheric balloons .

Before we learned how to build large planes for air transportation of passengers and cargo, controlled balloons were used - airships. They have an elongated shape; a gondola with an engine is suspended under the body, which drives the propeller.

The balloon not only rises up on its own, but can also lift some cargo: the cabin, people, instruments. Therefore, in order to find out what kind of load a balloon can lift, it is necessary to determine it lift.

Let, for example, let a balloon with a volume of 40 m 3 filled with helium be launched into the air. The mass of helium filling the shell of the ball will be equal to:
m Ge = ρ Ge V = 0.1890 kg/m 3 40 m 3 = 7.2 kg,
and its weight is:
P Ge = g m Ge; P Ge = 9.8 N/kg · 7.2 kg = 71 N.
The buoyant force (Archimedean) acting on this ball in the air is equal to the weight of air with a volume of 40 m 3, i.e.
F A = g·ρ air V; F A = 9.8 N/kg · 1.3 kg/m3 · 40 m3 = 520 N.

This means that this ball can lift a load weighing 520 N - 71 N = 449 N. This is its lifting force.

A balloon of the same volume, but filled with hydrogen, can lift a load of 479 N. This means that its lifting force is greater than that of a balloon filled with helium. But helium is still more often used, since it does not burn and is therefore safer. Hydrogen is a flammable gas.

It is much easier to lift and lower a ball filled with hot air. To do this, a burner is located under the hole located in the lower part of the ball. Using a gas burner, you can regulate the temperature of the air inside the ball, and therefore its density and buoyant force. To make the ball rise higher, it is enough to heat the air in it more strongly by increasing the burner flame. As the burner flame decreases, the air temperature in the ball decreases and the ball goes down.

You can select a temperature of the ball at which the weight of the ball and the cabin will be equal to the buoyant force. Then the ball will hang in the air, and it will be easy to make observations from it.

As science developed, significant changes occurred in aeronautical technology. It became possible to use new shells for balloons, which became durable, frost-resistant and lightweight.

Advances in the field of radio engineering, electronics, and automation have made it possible to design unmanned balloons. These balloons are used to study air currents, for geographical and biomedical research in the lower layers of the atmosphere.

Liquids, like all bodies on Earth, are affected by gravity. Therefore, a liquid poured into a vessel, with its weight, creates pressure, which, according to Pascal’s law, is transmitted in all directions. Therefore, there is pressure inside the liquid. This can be verified by experience.

Pour water into a glass tube, the bottom hole of which is closed with a thin rubber film. Under the influence of the weight of the liquid, the rubber bottom of the tube will bend (Fig. 93, a).

Experience shows that the higher the column of water above the rubber film, the more it bends (Fig. 93, b). But every time after the rubber bottom bends, the water in the tube comes into equilibrium ( stops), since, in addition to gravity, the elastic force of the rubber film acts on the water.

Let's lower the tube with a rubber bottom, into which water is poured, into another, wider vessel with water (Fig. 93, c). We will see that as the tube is lowered down, the rubber film gradually straightens. Full straightening of the film shows that the forces acting on it from above and below are the same. Complete straightening of the film occurs when the water levels in the tube and vessel coincide.

The same experiment can be carried out with a tube in which a rubber film covers the side hole, as shown in Figure 94, a. If this tube with water is immersed in another vessel with water, as shown in Figure 94, b, then we will again notice that the film will straighten as soon as the water levels in the tube and in the vessel are equal. This means that the forces acting on the rubber film are the same on both sides.

The experiment with a vessel, the bottom of which can fall off, is very clear. Such a vessel is lowered into a jar of water (Fig. 95, a). The bottom is pressed tightly against the edge of the vessel by the force of water pressure from bottom to top. Then water is carefully poured into the vessel. The bottom comes off the vessel when the water level in the vessel coincides with the water level in the jar (Fig. 95.6).

At the moment of separation, the column of liquid in the vessel presses from top to bottom, and the pressure of the same column of liquid, but located in the jar, is transmitted from bottom to top to the bottom. Both of these pressures are the same, but the bottom moves away from the cylinder due to the action of gravity on it.

Experiments with water were described above, but it is easy to conclude that something would have happened if some other liquid had been taken instead of water.

So, experiments show that there is pressure inside the liquid and at the same level it is the same in all directions. With depth, pressure increases.

Questions. 1. How can you show experimentally that the pressure inside a liquid is different at different levels, but at the same level it is the same in all directions? 2. Why is it that in many cases the pressure in a gas created by the weight of that gas is not taken into account?