Formula for calculating the pressure of a solid body. Academy of Entertaining Sciences
7. The task of calculating the pressure of solids
Task: A machine weighing 12,000 N has a support area of 2.5 m2. Determine the pressure of the machine on the foundation.
Given:
P=12000 N
S=2.5 m2
p-?
Solution :
p=P/S
=> p=P/S
p=12000 N/2.5 m2=4.8 kPa
Answer. p=4.8 kPa
Task: A box weighing 960 N exerts a pressure of 5 kPa on the support. What is the area of support of the box?
Given:
P=960 N
p=5 kPa
S-?
SI
5*103Pa
Solution :
p=F/S
=> p=P/S
=> S=P/p
S=960 N/5*103 Pa=0.192 m2
Answer. S \u003d 0.192 m 2
Task: A two-axle trailer with a load has a mass of 2.5 tons. Calculate the pressure exerted by the trailer on the road if the area of contact of each wheel with the road is 125 cm2.
Given:
m=2.5 t
S=125 cm2
2 axles;
4 wheels
g=10 N/kg
p-?
SI
2.5*103kg
125*10-4m2
Solution :
p=F/S
F=m*g
S=4S to
=> p=m*g/4Sk
p=2.5*103kg*10N/kg/4*125*10-4m2=5*105Pa
Answer. p= 5*10 5 Pa
A boy weighing 48 kg exerts pressure on a support. Calculate how much pressure he exerts if the total area of his soles is 320 cm 2 .
After analyzing the condition, we write it in short form, indicating the weight of the boy and the area of his soles (Fig. 1). Then, in a separate column, we write in the SI system those quantities that are given in the condition in non-systemic units. The mass of the boy is given in the SI system, but the area, expressed in square centimeters, should be expressed in square meters:
320 cm 2 \u003d 320 ∙ (0.01 m) 2 \u003d 320 0.0001 m 2 \u003d 0.032 m 2.
Rice. one. Brief condition tasks number 1
To find the pressure, we need the force with which the boy acts on the support, divided by the area of \u200b\u200bthe support:
We do not know the value of the force, but the condition of the problem includes the mass of the boy. The force with which it acts on the support is its weight. Assuming that the boy is stationary, we can assume that his weight is equal to the force of gravity, which is equal to the product of the boy's mass and acceleration free fall
Now we can combine both formulas into one final one. To do this, instead of the force F, we substitute the productmg from the second formula into the first formula. Then the calculation formula will look like:
The next step is to check the dimension of the result. Units of mass [m] = kg, units of gravitational acceleration [g] = N/kg, units of area [S] = m 2. Then
Finally, let's substitute the numerical data from the problem statement into the final formula:
Don't forget to write down your answer. In the answer, we can use multiples
Answer: p = 15 kPa.
(If you write = 15,000 Pa in your answer, then it will also be correct.)
Complete Solution in the final form it will look like this (Fig. 2):
Rice. 2. Complete solution of problem No. 1
2. Task number 2The bar acts on the support with a force of 200 N, while it exerts a pressure of 4 kPa. What is the area of the bar support?
Let's write a short condition and express the pressure in the SI system (4 kPa = 4000 Pa) (Fig. 3).
Rice. 3. Brief condition of problem No. 2
The value of the surface area is included in the formula known to us for calculating pressure.
From this formula, we need to express the area of \u200b\u200bthe support. Let's remember the mathematical rules. Force F is the dividend, support area S is the divisor, pressure p is the quotient. To find an unknown divisor, you need to divide the dividend by the quotient. We'll get:
Let's check the dimension of the result. The area must be expressed in square meters.
When checking, we replaced pascals with newtons per square meter, and the fractional line with a division sign. Recall that division of fractions is replaced by multiplication. In this case, the fraction, which is a divisor, is turned over, that is, its numerator and denominator are reversed. After that, the newton in the numerator (before the fraction) and the newton in the denominator of the fraction are reduced, and square meters remain.
Note that the dimension check is a very important step in solving the problem, as it allows you to detect errors that were accidentally made when performing mathematical transformations.
After checking the dimension of the result, we will calculate numerical value area, substituting the data from the brief condition:
Let's not forget to record the answer.
Answer: S \u003d 0.05 m 2.
A complete solution to the problem will look like this (Fig. 4):
Fig 4. Complete solution of problem No. 2
Solid body pressure/Article for 7th grade students/
§ Content :
1. What is pressure?
2. Ways to increase and decrease pressure.
3. pressure in nature.
4. pressure in technology.
5. Solving problems for calculating pressure.
6. Experimental tasks.
7. Just interesting things.
1.What is pressure?Imagine that you are going skiing. Skis glide through the snow, leaving a very shallow track. What happens if you take off your skis? Of course, you immediately fall into the snow. Let's see why this is happening. Weight, i.e. the force with which a person presses on the snow, remained the same. What has changed? Only the area of support (compare the soles of boots and skis). This means that it can be assumed that the result of the action of a force depends not only on the force itself - the point of application, direction, modulus - but also on the area of contact.
To test this, let's do an experiment. Take a foam sponge and a bar of soap. Put the soap on the sponge with the largest side. Pay attention to the deformation of the sponge. Now turn the soap on edge. What changed? Now we can conclude: the result of the action of a force depends on the force itself and on the area of its influence. Therefore, you need physical quantity that takes into account both factors. This value is called pressure. The ratio of the force F to the surface area S, provided that the force acts perpendicular to the surface, is called pressure.
p = F/S
The pressure units are calculated by the formula: 1 N/sq.m = 1 Pa (Pascal). The unit of measurement is named after the famous scientist Blaise Pascal. In addition to the basic units, prefixes are also used:
1 kPa = 1000 Pa, 1 MPa = 1,000,000 Pa
Think about whether they use the prefixes "milli", "micro"? Why?
2. Ways to increase and decrease pressure.First, let's answer the question: what is it for? Have you seen what traces heavy vehicles and tractors leave on the ground? Such deep ruts arise just because of high pressure. So, in such cases, it must be reduced. Since pressure depends on force and area, it can be changed by changing these values.
Why increase the pressure? Try cutting bread with a dull knife. What is the difference between a dull knife and a sharp one? Of course, the blade area and the pressure created. Therefore, all cutting and piercing tools must be very sharp.
3. Pressure in wildlife. 4. Engineering pressurePressure must be taken into account in mechanical engineering, architecture, and transport. We have already talked about machines that deform the soil. They cause irreparable damage to the environment. For example, during the development of the Far North, caterpillar tractors destroyed huge areas of reindeer moss - the main food for deer, which negatively affected their population. To avoid this, it is necessary to reduce the pressure, i.e. either reduce the pressure force or increase the area. Reducing strength is difficult: to do this, you need to reduce mass by using lighter materials. But these substances are either fragile or very expensive. Therefore, it is most often used to increase the area. This can be done in different ways: using caterpillars on tractors, increasing the diameter of tires, using paired wheels. How the tires are inflated is also of great importance, because the contact area also depends on this. Caterpillars significantly reduce pressure ( see table), increasing the permeability of the mechanism, but at the same time severely damage the upper layers of the soil. It is very important to take pressure into account in architecture and construction. The foundation of the building is used to relieve pressure. Since ancient times, hollow columns have been used in construction. Having sufficient strength, they are much lighter than solid ones, and therefore the pressure generated is also less.
Mechanism
Pressure, in kPa
Caterpillar tractors (marsh) with wide caterpillars
20 -30
Crawler tractors
40 -50
Car wheels
230 -300
Railway car wheels on rails
300 000
5. Solving problems for calculating pressure.§ 1) There is a brick on the floor with dimensions: height -5 cm, width - 10 cm, length - 20 cm. Its mass is 2 kg. What pressure does the brick exert on the floor, being in three different positions?
§ 2) What is the length of the skis if a person weighing 80 kg standing on them exerts a pressure of 2.5 kPa on the snow? Ski width 8 cm.
§ 3) What pressure does a caterpillar tractor exert on the soil if the mass of the tractor is 3.2 tons and the area of one caterpillar is 0.8 sq.m?
6.Experimental tasks.§ 1) Determine the pressure of a glass of tea on the table. Will the pressure change if you drink tea? How many times?
§ 2) How many times will the pressure of the physics textbook on the table change if it is placed on its edge? and if the physics textbook is replaced by history?
7. Just interesting tasks.§ 1) The scout must cross the river along thin ice. Come up with a device that reduces the risk of crossing.
§ 2) Why are the rails not laid directly on the ground?
§ 3) Why is it easier to accidentally cut yourself with a sharp razor than with a knife?
§ 4) They pressed a wooden wall with a force of 200 N, first with a palm, then with an awl with the same force. The forces are equal in magnitude, why is the result different?
Release 16In the video lesson of physics from the Academy entertaining sciences Professor Daniil Edisonovich will introduce young viewers to a new physical quantity that serves to measure pressure - Pascal. After watching the program, you will learn the importance of the area of support of a solid body, how not to fall through ice or snow, and also get acquainted with the formula for the pressure of solid bodies.
Solid body pressure formula
As you probably remember from the last program, weight is the force with which the body presses on the support. Why is it that the same person, walking in the snow in boots, falls through, but not when skiing? To understand this issue, Professor Daniil Edisonovich will teach you the formula for the pressure of solids. The tractor weighs much more than the car, and does not get stuck in loose soil. At the same time, a light vehicle hitting such soil is likely to get stuck and have to be pulled out by a tractor. The result of a force acting on a surface depends not only on the magnitude of this force, but also on the area to which this force is applied. When a person steps into the snow, the weight of his body is distributed over the area of his feet. And if a person is wearing skis, then the weight is distributed over their area, which is much more area feet. Since the application area has become larger, a person will not fall into the snow. Pressure is a scalar physical quantity equal to the ratio of the pressure force applied to a given surface to the area of this surface. To determine the pressure, it is necessary to divide the force acting perpendicular to the surface by the area of \u200b\u200bthis surface. The formula for the pressure of solids is written as follows: p \u003d F / S, where p is the pressure, F is the pressure force, S is the area of \u200b\u200bthe support. The unit of pressure is the pressure produced by a force of 1 newton acting on a surface of 1 m2 perpendicular to this surface. Pressure is measured in pascals. Thus, according to the formula for the pressure of solids, 1 pascal is equal to 1 newton per square meter. Between the force of pressure and pressure there is a directly proportional relationship, that is, the greater the force, the greater the pressure and vice versa, the smaller the force, the less pressure. If we talk about the dependence of pressure on the area of \u200b\u200bthe support, then there is an inversely proportional relationship, that is, the larger the area of \u200b\u200bthe support, the less pressure and vice versa, the smaller the contact area of the bodies, the greater the pressure. The pressure value has great importance not only in human life, but also in the life of animals. For example, a hare exerting a pressure of 1.2 kPa can relatively easily run away from a wolf that exerts a pressure of 12 kPa on loose snow, but will not escape from it on solid ground.
Man on skis, and without them.
On loose snow a man walks with with great difficulty sinking deep with every step. But, having put on skis, he can walk, almost without falling into it. Why? On skis or without skis, a person acts on the snow with the same force equal to his own weight. However, the effect of this force in both cases is different, because the surface area on which the person presses is different, with and without skis. The surface area of the ski is almost 20 times the area of the sole. Therefore, standing on skis, a person acts on every square centimeter of the snow surface area with a force 20 times less than standing on snow without skis.
The student, pinning a newspaper to the board with buttons, acts on each button with the same force. However, a button with a sharper end is easier to enter into the tree.
This means that the result of the action of a 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 this force depends on what force acts per unit area of the surface.
Nails must be driven into the corners of a small board. First, we set the nails driven into the board on the sand with their points up and put a weight on the board. In this case, the nail heads are only slightly pressed into the sand. Then turn the board over and put the nails on the tip. In this case, the area of support is smaller, and under the action of the same force, the nails go deep into the sand.
An experience. Second illustration.
The result of the action of this force depends on what force acts on each unit of surface area.
In the considered examples, the forces acted perpendicular to the surface of the body. The person's weight was perpendicular to the surface of the snow; the force acting on the button is perpendicular to the surface of the board.
The value equal to the ratio of the force acting perpendicular to the surface to the area of \u200b\u200bthis surface is called pressure.
To determine the pressure, it is necessary to divide the force acting perpendicular to the surface by the surface area:
pressure = force / area.
Let us denote the quantities included in this expression: pressure - p, the force acting on the surface, - F and the 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 more pressure.
The pressure unit is taken as the pressure that produces a force of 1 N acting on a surface of 1 m 2 perpendicular to this surface.
Unit of pressure - newton per square meter(1 N / m 2). In honor of the French scientist Blaise Pascal it's called pascal Pa). In this way,
1 Pa = 1 N / m 2.
Other pressure units 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 condition of the problem and solve it.
Given : m = 45 kg, S = 300 cm 2; p = ?
In SI units: S = 0.03 m 2
Solution:
p = F/S,
F = P,
P = g m,
P= 9.8 N 45 kg ≈ 450 N,
p\u003d 450 / 0.03 N / m 2 \u003d 15000 Pa \u003d 15 kPa
"Answer": p = 15000 Pa = 15 kPa
Ways to reduce and increase pressure.
A heavy caterpillar tractor produces a pressure on the soil equal to 40-50 kPa, that is, only 2-3 times more than the pressure of a boy weighing 45 kg. This is because the weight of the tractor is distributed over a larger area due to the caterpillar drive. And we have established that the larger the area of the support, the less pressure produced by the same force on this support .
Depending on whether you need to get a small or a large pressure, the area of \u200b\u200bsupport increases or decreases. For example, in order for the soil to withstand the pressure of a building being erected, the area of \u200b\u200bthe lower part of the foundation is increased.
Truck tires and aircraft chassis are made much wider than passenger cars. Particularly wide tires are made for cars designed to travel in deserts.
Heavy machines, like a tractor, a tank or a swamp, having a large bearing area of the tracks, pass through swampy terrain that a person cannot pass through.
On the other hand, with a small surface area, a large pressure can be generated with a small force. For example, pressing a button into a board, we act on it with a force of about 50 N. Since the area of the button tip is approximately 1 mm 2, the pressure produced by it is equal to:
p \u003d 50 N / 0.000001 m 2 \u003d 50,000,000 Pa \u003d 50,000 kPa.
For comparison, this pressure is 1000 times more than the pressure exerted by a caterpillar tractor on the soil. Many more such examples can be found.
The blade of cutting and piercing tools (knives, scissors, cutters, saws, needles, etc.) is specially sharpened. The sharpened edge of a sharp blade has a small area, so even a small force creates a lot of pressure, and it is easy to work with such a tool.
Cutting and piercing devices are also found in wildlife: these are teeth, claws, beaks, spikes, etc. - they are all 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 vessel in which they are located. For example, a steel cylinder for storing gases, a car tire 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 due to other causes than pressure solid body on a support.
It is known that gas molecules move randomly. During their movement, they collide with each other, as well as with the walls of the vessel in which the gas is located. There are many molecules in the 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 of 1 cm 2 in 1 s is expressed as a twenty-three-digit number. Although the impact force of an individual molecule is small, the action of all molecules on the walls of the vessel is significant - it creates gas pressure.
So, gas pressure on the walls of the vessel (and on the body placed in the gas) is caused by impacts of gas molecules .
Consider the following experience. Place a rubber ball under the air pump bell. It contains a small amount of air and has irregular shape. Then we pump out the air from under the bell with a pump. The shell of the ball, around which the air becomes more and more rarefied, gradually swells and takes the form of a regular ball.
How to explain this experience?
Special durable steel cylinders are used for storage and transportation of compressed gas.
In our experiment, moving gas molecules continuously hit the walls of the ball inside and out. 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 balloon is inflated until the force of elasticity of its rubber shell becomes equal to the pressure force of the gas. The shell of the ball takes the shape of a ball. This shows that gas presses on its walls equally in all directions. 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 a gas and is a consequence of the random movement of a huge number of molecules.
Let's try to reduce the volume of gas, but so that its mass remains unchanged. This means that in each cubic centimeter of gas there will be more molecules, the density of the gas 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 A glass tube is shown, one end of which is covered with a thin rubber film. A piston is inserted into the tube. When the piston is pushed in, the volume of air in the tube decreases, i.e., the gas is compressed. The rubber film bulges outward, indicating that the air pressure in the tube has increased.
On the contrary, with an increase in the volume of the same mass of gas, the number of molecules in each cubic centimeter decreases. This will reduce the number of impacts on the walls of the vessel - the pressure of the gas will become less. Indeed, when the piston is pulled out of the tube, the volume of air increases, 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 in the tube there would be any other gas.
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 does the pressure of a gas change when it is heated at a constant volume? It is known that the speed of movement of gas molecules increases when heated. Moving faster, the molecules will hit the walls of the vessel 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 more pressure.
Consequently, The pressure of a gas in a closed vessel is greater the higher the temperature of the gas, provided that the mass of the gas and the volume do not change.
From these experiments it can be concluded that the pressure of the gas is greater, the more often and stronger the molecules hit the walls of the vessel .
For storage and transportation of gases, they are highly compressed. At the same time, their pressure increases, 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 each point of the liquid or gas.
The pressure of the piston is transmitted to each point of the liquid filling the ball.
Now gas.
Unlike solids, individual layers and small particles of liquid and gas 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. Ripples appear on a river or lake at the slightest breeze.
The mobility of gas and liquid particles explains that the pressure produced on them is transmitted not only in the direction of the force, but at every point. Let's consider this phenomenon in more detail.
On the image, a a vessel containing a gas (or liquid) is depicted. 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, let's make the piston move a little inward and compress the gas (liquid) directly below it. Then the particles (molecules) will be located in this place more densely than before (Fig., b). Due to the mobility of the gas particles will move in all directions. As a result, their arrangement will again become uniform, but more dense than before (Fig. c). Therefore, the pressure of the gas will increase everywhere. This means that additional pressure is transferred to all particles of a 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 pressure will be greater than before by the same amount. The pressure on the walls of the vessel, and on the bottom, and on 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 figure shows a hollow sphere with small holes in various places. A tube is attached to the ball, into which a piston is inserted. If you draw water into the ball and push the piston into the tube, then water will flow from all the holes in the ball. In this experiment, the piston presses on the surface of the water in the tube. The water particles under the piston, condensing, transfer its pressure to other layers lying deeper. Thus, the pressure of the piston is transmitted to each point of the liquid filling the ball. As a result, part of the water is pushed out of the ball in the form of identical streams flowing from all holes.
If the ball is filled with smoke, then when the piston is pushed into the tube, identical streams of smoke will begin to come out of all the holes in the ball. This confirms that and gases transmit the pressure produced on them equally in all directions.
Pressure in liquid and gas.
Under the weight of the liquid, the rubber bottom in the tube will sag.
Liquids, like all bodies on Earth, are affected by the force of 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 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 sags. But every time after the rubber bottom sags, the water in the tube comes to equilibrium (stops), because, in addition to gravity, the elastic force of the stretched rubber film acts on the water.
Forces acting on the rubber film |
are the same on both sides. |
Illustration.
The bottom moves away from the cylinder due to the pressure on it due to gravity.
Let's lower a 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 out. Full straightening of the film shows that the forces acting on it from above and below are equal. Full 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 closes the side opening, as shown in figure a. Immerse this tube of water into another vessel of water, as shown in the figure, b. We will notice that the film straightens again as soon as the water levels in the tube and vessel are equal. This means that the forces acting on the rubber film are the same from all sides.
Take a vessel whose bottom can fall off. Let's put it in a jar of water. In this case, 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 the bottom up.
We will carefully pour water into the vessel and watch its bottom. As soon as the level of water in the vessel coincides with the level of water in the jar, it will fall away from the vessel.
At the moment of detachment, a column of liquid in the vessel presses down on the bottom, and pressure is transmitted from bottom to top to the bottom of a column of liquid of the same height, but located in the jar. Both these pressures are the same, but the bottom moves away from the cylinder due to the action of its own gravity on it.
The experiments with water were described above, but if we take any other liquid instead of water, the results of the experiment will be the same.
So, experiments show that inside the liquid there is pressure, and at the same level it is the same in all directions. Pressure increases with depth.
Gases do not differ in this respect from liquids, because they also have weight. But we must remember that the density of a gas is hundreds of times less than the density of a liquid. The weight of the gas in the vessel is small, and in many cases its "weight" pressure can be ignored.
Calculation of liquid pressure on the bottom and walls of the vessel.
Calculation of liquid pressure on the bottom and walls of the vessel.
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 having the shape of a rectangular parallelepiped.
Strength F, with which the liquid poured into this vessel presses on its bottom, is equal to the weight P the liquid in the vessel. The weight of a liquid can be determined by knowing its mass. m. Mass, as you know, can be calculated by 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 liquid column in the vessel is denoted by the letter h, and the area of the bottom of the vessel S, then V = S h.
Liquid mass m = ρ V, or m = ρ S h .
The weight of this fluid P = gm, or P = g ρ S h.
Since the weight of the liquid column is equal to the force with which the liquid presses on the bottom of the vessel, then, 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 a liquid on the bottom of a vessel. From this formula it can be seen that the pressure of a liquid at the bottom of a vessel depends only on the density and height of the liquid column.
Therefore, according to the derived formula, it is possible to calculate the pressure of the liquid poured into the vessel any form(Strictly speaking, our calculation is only suitable for vessels having the shape of a straight prism and a cylinder. In physics courses for the institute, it was proved that the formula is also true for a vessel of arbitrary shape). In addition, it can be used to calculate the pressure on the walls of the vessel. The pressure inside the fluid, including 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 = gph need density ρ expressed in kilograms per cubic meter (kg / m 3), and the height of the liquid column h- in meters (m), g\u003d 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/m 3 .
Let's write down the condition of the problem and write it down.
Given :
ρ \u003d 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.
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. We know from experience that water poured, for example, into a watering can, always stands at the same level in the spout and inside.
Communicating vessels are common to us. For example, it can be a teapot, a watering can or a coffee pot. |
The surfaces of a homogeneous liquid are installed at the same level in communicating vessels of any shape. |
Liquids of various densities. |
With communicating vessels, the following simple experiment can be done. 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 fix one of the tubes in a tripod, and raise, lower or tilt the other in different directions. And in this case, as soon as the liquid calms down, its levels in both tubes will equalize.
In communicating vessels of any shape and section, the surfaces of a homogeneous liquid are set at the same level(provided that the air pressure over 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 pressures in both vessels are the same at any level. The liquid in both vessels is the same, that is, it has the same density. Therefore, its heights must also be the same. When we raise one vessel or add liquid to it, the pressure in it increases and the liquid moves into another vessel until the pressures are balanced.
If a liquid of one density is poured into one of the communicating vessels, and 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 a liquid on the bottom of a 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.
With equal pressures, the height of a liquid column with a higher density will be less than the height of a liquid column with a lower density (Fig.).
An experience. How to determine the mass of air.
Air weight. Atmosphere pressure.
existence of atmospheric pressure.
Atmospheric pressure is greater than the pressure of rarefied air in a vessel.
The force of gravity acts on the air, as well as on any body located on the Earth, and, therefore, the air has weight. The weight of air is easy to calculate, knowing its mass.
We will show by experience how to calculate the mass of air. To do this, take a strong glass ball with a cork and a rubber tube with a clamp. We pump air out of it with a pump, clamp the tube with a clamp and balance it on the scales. Then, opening the clamp on the rubber tube, let air into it. In this case, the balance of the scales will be disturbed. To restore it, you will have to put weights on the other pan of scales, 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 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.
The air envelope that surrounds the earth is called atmosphere (from Greek. atmosphere steam, air, and sphere- ball).
Atmosphere as shown by flight observations artificial satellites Earth, extends to a height of several thousand kilometers.
Due to the action 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, transfers the pressure produced on it in all directions.
As a result earth's surface and the bodies 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 be explained by many phenomena that we encounter in life. Let's consider some of them.
The figure shows a glass tube, inside which there is a piston that fits snugly against the walls of the tube. The end of the tube is dipped in water. If you raise the piston, then 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 cork into which a tube with a tap is inserted. Air is pumped out of the vessel by a pump. The end of the tube is then placed in water. If you now open the tap, then the water will splash into the inside of the vessel in a fountain. Water enters the vessel because the atmospheric pressure is greater than the pressure of rarefied air in the vessel.
Why does the air shell of the Earth exist.
Like all bodies, the molecules of gases that make up the air envelope of the Earth are attracted to the Earth.
But why, then, do they not all fall to the surface of the Earth? How is the air shell of the Earth, its atmosphere, preserved? To understand this, we must take into account that the molecules of gases are in continuous and random motion. But then another question arises: why these molecules do not fly away into the world space, that is, into space.
In order to completely leave the Earth, the molecule, like spaceship or a rocket, must have a very high speed (at least 11.2 km / s). This so-called second escape velocity. The speed of most molecules in the Earth's air envelope is much less than this cosmic speed. 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 the fact that gas molecules "float" in space near the Earth, forming an air shell, or the atmosphere known to us.
Measurements show that air density decreases rapidly with altitude. So, at a height of 5.5 km above the Earth, the air density is 2 times less than its density at the Earth's surface, at a height of 11 km - 4 times less, etc. The higher, 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 air shell of the Earth does not have a clear boundary.
Strictly speaking, due to the action of gravity, the density of the gas in any closed vessel is not the same throughout the entire volume of the vessel. At the bottom of the vessel, the density of the gas is greater than in its upper parts, and 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 the gas contained in the vessel, this difference in density and pressure is so small that in many cases it can be completely ignored, just be aware of it. But for an atmosphere extending over several thousand kilometers, the difference is significant.
Measurement of atmospheric pressure. The Torricelli 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 the density of the air. But the atmosphere does not have a definite boundary, and the air density at different heights is different. However, atmospheric pressure can be measured using an experiment proposed in the 17th century by an Italian scientist. Evangelista Torricelli a student of Galileo.
Torricelli's experiment is as follows: 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 with mercury, where this end of the tube is opened under the level of mercury. As in any liquid experiment, part of the mercury is poured into the cup, and part of it remains in the tube. The height of the mercury column 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 mercury column inside this tube and does not affect the measurements.
Torricelli, who proposed the experience described above, also gave his explanation. The atmosphere presses on the surface of the mercury in the cup. Mercury is in balance. This means that the pressure in the tube is aa 1 (see figure) is equal to atmospheric pressure. When atmospheric pressure changes, the height of the mercury column in the tube also changes. As the pressure increases, the column lengthens. As the pressure decreases, the mercury column decreases in height.
The pressure in the tube at the 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. Hence it follows that atmospheric pressure is equal to the pressure of the mercury column in the tube , i.e.
p atm = p mercury.
The greater 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, 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 produces.
Therefore, in this case, 1 millimeter of mercury (1 mm Hg) is taken as the unit of atmospheric pressure. 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:
p = g ρ h, p\u003d 9.8 N / kg 13,600 kg / m 3 0.001 m ≈ 133.3 Pa.
So, 1 mm Hg. 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 daily the height of the mercury column in the tube, Torricelli discovered that this height changes, that is, atmospheric pressure is not constant, it can increase and decrease. Torricelli also noticed that atmospheric pressure is related to changes in the weather.
If you attach a vertical scale to the mercury tube used in Torricelli's experiment, you get the simplest device - mercury barometer (from Greek. baros- heaviness, metreo- measure). It is used to measure atmospheric pressure.
Barometer - aneroid.
In practice, a metal barometer is used to measure atmospheric pressure, called aneroid (translated from Greek - aneroid). The barometer is called so because it does not contain mercury.
The appearance of the aneroid is shown in the figure. Its main part is a metal box 1 with a wavy (corrugated) surface (see other fig.). Air is pumped out of this box, and so that atmospheric pressure does not crush the box, its cover 2 is pulled up by a spring. As atmospheric pressure increases, the lid flexes downward and tensions the spring. When the pressure decreases, the spring straightens the cover. An arrow-pointer 4 is attached to the spring by means of a transmission mechanism 3, which moves to the right or left when the pressure changes. A scale is fixed under the arrow, the divisions of which are marked according to the indications of a mercury barometer. So, the number 750, against which the aneroid needle stands (see Fig.), shows that at the given moment in the mercury barometer the height of the mercury column is 750 mm.
Therefore, atmospheric pressure is 750 mm Hg. 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 the weather. A barometer is a necessary instrument for meteorological observations.
Atmospheric pressure at various altitudes.
In a liquid, the pressure, as we know, depends on the density of the liquid and the height of its column. Due to the low compressibility, the density of the liquid at different depths is almost the same. Therefore, when calculating the pressure, we consider its density to be constant and take into account only the change in height.
The situation is more complicated with gases. Gases are highly compressible. And the more the gas is compressed, the greater its density, and the greater the pressure it produces. After all, the pressure of a gas is created by the impact of its molecules on the surface of the body.
The layers of air near the surface of the Earth are compressed by all the overlying layers of air above them. But the higher the layer of air from the surface, the weaker it is compressed, the lower its density. Hence, the less pressure it produces. If, for example, a balloon rises above the surface of the Earth, then the air pressure on the balloon becomes less. This happens not only because the height of the air column above it decreases, but also because the air density decreases. It is smaller at the top than at the bottom. Therefore, the dependence of air pressure on altitude is more complicated than that of liquids.
Observations show that atmospheric pressure in areas lying at sea level is on average 760 mm Hg. Art.
Atmospheric pressure equal to the pressure of a mercury column 760 mm high at a temperature of 0 ° C is called normal atmospheric pressure..
normal atmospheric pressure equals 101 300 Pa = 1013 hPa.
The higher the altitude, the lower the pressure.
With small rises, on average, for every 12 m of rise, the pressure decreases by 1 mm Hg. Art. (or 1.33 hPa).
Knowing the dependence of pressure on altitude, it is possible to determine the height above sea level by changing the readings of the barometer. Aneroids having a scale on which you can directly measure the height above sea level are called altimeters . They are used in aviation and when climbing mountains.
Pressure gauges.
We already know that barometers are used to measure atmospheric pressure. To measure pressures greater or less than atmospheric pressure, the pressure gauges (from Greek. manos- rare, inconspicuous metreo- measure). Pressure gauges are liquid and metal.
Consider first the device and action open liquid manometer. It consists of a two-legged glass tube into which some liquid is poured. The liquid is installed in both knees at the same level, since only atmospheric pressure acts on its surface in the knees of the vessel.
To understand how such a pressure gauge works, it can be connected with a rubber tube to a round flat box, one side of which is covered with a rubber film. If you press your finger on the film, then the liquid level in the manometer knee connected in the box will decrease, and in the other knee it will increase. What explains this?
Pressing on the film increases the air pressure in the box. According to Pascal's law, this increase in pressure is transferred to the liquid in that knee of the pressure gauge, which is attached to the box. Therefore, the pressure on the liquid in this knee will be greater than in the other, where only atmospheric pressure acts on the liquid. Under the force of this excess pressure, the liquid will begin to move. In the knee with compressed air, the liquid will fall, in the other it will rise. The liquid will come to equilibrium (stop) when the excess pressure of the compressed air is balanced by the pressure that the excess liquid column produces in the other leg of the pressure gauge.
The stronger the pressure on the film, the higher the excess liquid column, the greater its pressure. Consequently, 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 manometer knees becomes., so, therefore, and fluid produces more pressure.
If you install the device box at some depth inside the liquid and turn it with a film up, sideways and down, then the pressure gauge readings will not change. That's the way it should be, because at the same level inside a liquid, the pressure is the same in all directions.
The picture shows metal manometer . 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 with a tap 4 communicates with the vessel in which the pressure is measured. As pressure increases, the tube flexes. Movement of its closed end with a lever 5 and gears 3 passed to the shooter 2 moving around the scale of the instrument. When the pressure decreases, the tube, due to its elasticity, returns to its previous position, and the arrow returns to zero division of the scale.
Piston liquid pump.
In the experiment we considered earlier (§ 40), it was found that water in a glass tube, under the action of atmospheric pressure, rose up behind the piston. This action is based piston pumps.
The pump is shown schematically in the figure. It consists of a cylinder, inside which goes up and down, tightly adhering to the walls of the vessel, the piston 1 . Valves are installed in the lower part of the cylinder and in the piston itself. 2 opening only upwards. When the piston moves upwards, water enters the pipe under the action of atmospheric pressure, lifts the bottom valve and moves behind the piston.
When the piston moves down, the water under the piston presses on the bottom valve, and it closes. At the same time, the pressure of the water opens the valve inside the piston, and the water flows into the space above the piston. With the next movement of the piston upwards, the water above it also rises in the place with it, 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 be above it, and this whole procedure is repeated again and again while the pump is running.
Hydraulic Press.
Pascal's law allows you to explain the action hydraulic machine (from Greek. hydraulicos- water). These are machines whose operation is based on the laws of motion and equilibrium of liquids.
The main part of the 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 there are no forces acting 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 - areas of pistons. The pressure under the first (small) piston is p 1 = F 1 / S 1 , and under the second (large) p 2 = F 2 / S 2. According to Pascal's law, the pressure of a fluid at rest is transmitted equally in all directions, i.e. p 1 = p 2 or F 1 / S 1 = F 2 / S 2 , from where:
F 2 / F 1 = S 2 / S 1 .
Therefore, the strength F 2 so much more power F 1 , how many times the area of the large piston is greater than the area of the small piston. For example, if the area of the large piston is 500 cm 2, and the small one is 5 cm 2, and a force of 100 N acts on the small piston, then a force 100 times greater will act on the larger piston, that is, 10,000 N.
Thus, with the help of a hydraulic machine, it is possible to balance a large force with a small force.
Attitude F 1 / F 2 shows the gain in strength. For example, in the example above, the gain in force is 10,000 N / 100 N = 100.
The hydraulic machine used for pressing (squeezing) is called hydraulic press .
Hydraulic presses are used where a lot of power is required. For example, for squeezing oil from seeds at oil mills, for pressing plywood, cardboard, hay. In iron and steel works, hydraulic presses are used to make steel machine shafts, railway wheels, and many other products. Modern hydraulic presses can develop a force of tens and hundreds of millions of newtons.
The device of the hydraulic press is shown schematically in the figure. The body to be pressed 1 (A) is placed on a platform connected to a large piston 2 (B). The small piston 3 (D) creates a large pressure on the liquid. This pressure is transmitted to every point of the fluid filling the cylinders. Therefore, the same pressure acts on the second, large piston. But since the area of the 2nd (large) piston is larger than the area of the small one, then the force acting on it will be greater than the force acting on piston 3 (D). Under this force, piston 2 (B) will rise. When piston 2 (B) rises, the body (A) rests against the fixed upper platform and is compressed. The pressure gauge 4 (M) measures the fluid pressure. Safety valve 5 (P) automatically opens when the liquid pressure exceeds the allowable value.
From a small cylinder to a large liquid is pumped by repeated movements of the small piston 3 (D). This is done in the following way. When the small piston (D) is lifted, valve 6 (K) opens and liquid is sucked into the space under the piston. When the small piston is lowered under the action of liquid pressure, valve 6 (K) closes, and valve 7 (K") opens, and the liquid passes into a large vessel.
The action of water and gas on a body immersed in them.
Under water, we can easily lift a stone that can hardly be lifted in the air. If you submerge the cork under water and release it from your hands, it will float. 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, then it will also be subjected to pressure, like the walls of the vessel.
Consider the forces that act from the side of the liquid on the body immersed in it. To make it easier to reason, we choose a body that has the shape of a parallelepiped with bases parallel to the surface of the liquid (Fig.). The forces acting on the side faces of the body are equal in pairs and balance each other. Under the influence of these forces, the body is compressed. But the forces acting on the upper and lower faces of the body are not the same. On the upper face presses from above with force F 1 column of liquid tall h one . At the level of the lower face, the pressure produces a liquid column with a height h 2. This pressure, as we know (§ 37), is transmitted inside the liquid in all directions. Therefore, on the lower face of the body from the bottom up with a force F 2 presses a liquid column high h 2. But h 2 more h 1 , hence the modulus of force F 2 more power modules F one . Therefore, the body is pushed out of the liquid with a force F you are t, equal difference forces F 2 - F 1 , i.e.
But S·h = V, where V is the volume of the parallelepiped, and ρ W ·V = m W is the mass of fluid in the volume of the parallelepiped. Consequently, F vyt \u003d g m well \u003d P well, i.e. buoyant force is equal to the weight of the liquid in the volume of the body immersed in it(The 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 that pushes a body out of a liquid is easy to discover 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 the bottom up with some force, for example, press it with your hand (raise it). Therefore, experience confirms that a force acting on a body in a fluid pushes the body out of the fluid. For gases, as we know, Pascal's law also applies. That's why bodies in the gas are subjected to a force pushing them out of the gas. Under the influence of this force, the balloons rise up. 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 cork to a 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 greater than the density of air (therefore carbon dioxide descends and fills the vessel, displacing air from it). In this case, the balance of the scales is disturbed. A cup with a suspended flask rises up (Fig.). A flask immersed in carbon dioxide experiences a greater buoyant force than that which 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, prolcosmos). This explains why in the water we sometimes easily lift bodies that we can hardly keep in the air. A small bucket and a cylindrical body are suspended from the spring (Fig., a). The arrow on the tripod marks the extension of the spring. It shows the weight of the body in the air. Having lifted the body, a drain vessel is placed under it, filled with liquid to the level of the drain tube. After that, 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 a pouring vessel into a glass. The spring contracts and the pointer of the spring rises to indicate the decrease in the weight of the body in the fluid. In this case, in addition to the force of gravity, another force acts on the body, pushing it out of the fluid. If the liquid from the glass is poured into the upper bucket (i.e., the one 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 that pushes a body completely immersed in a liquid is equal to the weight of the liquid in the volume of this body . We reached the same conclusion in § 48. If a similar experiment were done with a body immersed in some gas, it would show that the force pushing the body out of the 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 to its existence and calculated its significance. So, experience has confirmed that the Archimedean (or buoyant) force is equal to the weight of the fluid in the volume of the body, i.e. F A = P f = g m and. The mass of liquid m f , displaced by the body, can be expressed in terms of its density ρ w and the volume of the body V t immersed in the liquid (since V l - the volume of the liquid displaced by the body is equal to V t - the volume of the body immersed in the liquid), i.e. m W = ρ W V t. Then we get: F A= g ρ and · V t Therefore, 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 a body immersed in a 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 directions (gravity is down, and the Archimedean force is up), then the weight of the body in fluid P 1 will be less than the weight of the body in vacuum P = gm to the Archimedean force F A = g m w (where m w is the mass of liquid or gas displaced by the body). In this way, if a body is immersed in a liquid or gas, then it loses in its weight as much as the liquid or gas displaced by it 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 condition 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 is easy to verify experimentally. Pour water into the drain vessel up to the level of the drain pipe. After that, let's immerse the floating body into the vessel, having previously weighed it in the air. Having descended into the water, the 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 air. Having done the same experiments with any other bodies floating in different liquids - in water, alcohol, salt solution, you can make sure that if a body floats in a liquid, then the weight of the liquid displaced by it is equal to the weight of this body in air. It is 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. The body, on the other hand, 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 that of water. The lower the density of the body compared to the density of the liquid, the minority body is immersed in liquid . With equal densities of the body and liquid, the body floats inside the 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 / m 3), on top - lighter kerosene (ρ = 800 kg / m 3) . The 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 the fish descends to a great depth with the help of muscles, and the water pressure on it increases, the bubble contracts, the volume of the fish's body decreases, and it does not push upwards, but swims in the depths. Thus, the fish can, within certain limits, regulate the depth of its dive. Whales regulate their diving depth by contracting and expanding their lung capacity. Sailing ships.Ships floating 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 fasteners that give ships strength are also made of metals. For the construction of ships, various materials are used, which, compared with water, have both higher and lower densities. How do ships float, take on board and carry large loads? An experiment with a floating body (§ 50) showed that the body displaces so much water with its underwater part that this water is equal in weight to the weight of the body in air. This is also true for any ship. The weight of water displaced by the underwater part of the ship is equal to the weight of the ship with cargo in the air or the force of gravity acting on the ship with cargo. The depth to which a ship is submerged in water is called draft . The deepest allowable draft is marked on the ship's hull with a red line called waterline (from Dutch. water- water). The weight of water displaced by the vessel when immersed to the waterline, equal to the force of gravity acting on the vessel with cargo, is called the displacement of the vessel. At present, ships with a displacement of 5,000,000 kN (5 10 6 kN) and more are being built for the transportation of oil, i.e., having a mass of 500,000 tons (5 10 5 t) and more together with the cargo. If we subtract the weight of the ship itself from the displacement, then we get the carrying capacity of this ship. Carrying capacity shows the weight of the cargo carried by the ship. Shipbuilding has existed since Ancient Egypt, in Phoenicia (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. Mainly warships were built, but it was in Russia that the first icebreaker, ships with an internal combustion engine, and the nuclear icebreaker Arktika were built. Aeronautics.Drawing describing the ball of the Montgolfier brothers in 1783: “View and exact dimensions of the Balloon Earth"Which was the first." 1786 Since ancient times, people have dreamed of being able to fly above the clouds, to swim in the ocean of air, as they sailed on the sea. For aeronautics At first, balloons were used, which were filled either with heated air, or with 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 more than gravity F heavy, i.e. F A > F heavy As the ball rises, the Archimedean force acting on it decreases ( F A = gρV), since the density of the upper 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 lift height. To lower the ball, part of the gas is released from its shell using a special valve. In the horizontal direction, the balloon moves only under the influence of the wind, so it is called balloon (from Greek air- air, stato- standing). Not so long ago, huge balloons were used to study the upper layers of the atmosphere, the stratosphere - stratostats . Before they learned how to build large aircraft for transporting passengers and cargo by air, 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 by itself, but can also lift some cargo: a cabin, people, instruments. Therefore, in order to find out what kind of load a balloon can lift, it is necessary to determine it. lifting force. Let, for example, 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: 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 still, helium is used more often, since it does not burn and is therefore safer. Hydrogen is a combustible gas. It is much easier to raise and lower a balloon filled with hot air. For this, a burner is located under the hole located in the lower part of the ball. Using a gas burner, you can control the temperature of the air inside the ball, which means its density and buoyancy. In order for the ball to rise higher, it is enough to heat the air in it more strongly, increasing the flame of the burner. When the burner flame decreases, the temperature of the air in the ball decreases, and the ball goes down. It is possible to choose such a temperature of the ball at which the weight of the ball and the cabin will be equal to the buoyancy force. Then the ball will hang in the air, and it will be easy to make observations from it. As science developed, there were also significant changes in aeronautical technology. It became possible to use new shells for balloons, which became durable, frost-resistant and light. Achievements in the field of radio engineering, electronics, automation 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. Pressure is a physical quantity that plays a special role in nature and human life. This phenomenon, imperceptible to the eye, not only affects the condition environment, but also very well felt by everyone. Let's figure out what it is, what types of it exist and how to find the pressure (formula) in different environments. What is called pressure in physics and chemistryThis term refers to an important thermodynamic quantity, which is expressed as the ratio of the perpendicularly exerted pressure force to the surface area on which it acts. This phenomenon does not depend on the size of the system in which it operates, and therefore refers to intensive quantities. In a state of equilibrium, the pressure is the same for all points in the system. In physics and chemistry, it is denoted by the letter "P", which is short for Latin name term - pressūra. If a we are talking about the osmotic pressure of a liquid (the balance between the pressure inside and outside the cell), the letter "P" is used. Pressure unitsAccording to the standards of the International SI system, the physical phenomenon under consideration is measured in pascals (in Cyrillic - Pa, in Latin - Ra). Based on the pressure formula, it turns out that one Pa is equal to one N (newton - divided by one square meter (a unit of area). However, in practice, it is rather difficult to use pascals, since this unit is very small. In this regard, in addition to the standards of the SI system, this value can be measured in a different way. Below are its most famous analogues. Most of them are widely used in the former USSR.
General pressure formula (7th grade physics)From the definition of a given physical quantity, one can determine the method of finding it. It looks like the photo below.
In it, F is force, and S is area. In other words, the formula for finding pressure is its force divided by the surface area on which it acts. It can also be written as follows: P = mg / S or P = pVg / S. Thus, this physical quantity is related to other thermodynamic variables: volume and mass. For pressure, the following principle applies: the smaller the space affected by the force, the greater the amount of pressing force it has. If, however, the area increases (with the same force) - the desired value decreases. Hydrostatic pressure formulaDifferent aggregate states of substances provide for the presence of their properties that are different from each other. Based on this, the methods for determining P in them will also be different. For example, the formula for water pressure (hydrostatic) looks like this: P = pgh. It also applies to gases. At the same time, it cannot be used to calculate atmospheric pressure, due to the difference in altitudes and air densities. In this formula, p is the density, g is the gravitational acceleration, and h is the height. Based on this, the deeper the object or object sinks, the higher the pressure exerted on it inside the liquid (gas).
The variant under consideration is an adaptation of the classical example P = F / S. If we recall that the force is equal to the derivative of the mass by the free fall velocity (F = mg), and the mass of the liquid is the derivative of the volume by the density (m = pV), then the pressure formula can be written as P = pVg / S. In this case, the volume is area multiplied by height (V = Sh). If you insert this data, it turns out that the area in the numerator and denominator can be reduced and the output is the above formula: P \u003d pgh. Considering the pressure in liquids, it is worth remembering that, unlike solids, the curvature of the surface layer is often possible in them. And this, in turn, contributes to the formation of additional pressure. For such situations, a slightly different pressure formula is used: P \u003d P 0 + 2QH. In this case, P 0 is the pressure of a non-curved layer, and Q is the liquid tension surface. H is the average curvature of the surface, which is determined by Laplace's Law: H \u003d ½ (1 / R 1 + 1 / R 2). The components R 1 and R 2 are the radii of the main curvature. Partial pressure and its formulaAlthough the P = pgh method is applicable to both liquids and gases, it is better to calculate the pressure in the latter in a slightly different way. The fact is that in nature, as a rule, absolutely pure substances are not very common, because mixtures predominate in it. And this applies not only to liquids, but also to gases. And as you know, each of these components exerts a different pressure, called partial pressure. It's pretty easy to define. It is equal to the sum of the pressure of each component of the mixture under consideration (ideal gas). It follows from this that the partial pressure formula looks like this: P \u003d P 1 + P 2 + P 3 ... and so on, according to the number of constituent components.
There are often cases when it is necessary to determine the air pressure. However, some mistakenly carry out calculations only with oxygen according to the scheme P = pgh. But air is a mixture of different gases. It contains nitrogen, argon, oxygen and other substances. Based on the current situation, the air pressure formula is the sum of the pressures of all its components. So, you should take the aforementioned P \u003d P 1 + P 2 + P 3 ... The most common instruments for measuring pressureDespite the fact that it is not difficult to calculate the thermodynamic quantity under consideration using the above formulas, sometimes there is simply no time to carry out the calculation. After all, you must always take into account numerous nuances. Therefore, for convenience, a number of devices have been developed over several centuries to do this instead of people. In fact, almost all devices of this kind are varieties of a pressure gauge (it helps to determine the pressure in gases and liquids). However, they differ in design, accuracy and scope.
Types of pressureConsidering the pressure, the formula for finding it and its variations for different substances, it is worth learning about the varieties of this quantity. There are five of them.
AbsoluteThis is the name of the total pressure under which a substance or object is located, without taking into account the influence of other gaseous components of the atmosphere. It is measured in pascals and is the sum of excess and atmospheric pressure. It is also the difference between barometric and vacuum types. It is calculated by the formula P = P 2 + P 3 or P = P 2 - P 4. The reference point for absolute pressure under the conditions of the planet Earth is taken as the pressure inside the container from which air is removed (that is, classical vacuum). Only this type of pressure is used in most thermodynamic formulas. barometricThis term refers to the pressure of the atmosphere (gravity) on all objects and objects found in it, including the surface of the Earth itself. Most people also know it under the name atmospheric. It is referred to and its value varies with the place and time of measurement, as well as weather conditions and being above / below sea level. The value of barometric pressure is equal to the modulus of the force of the atmosphere per unit area along the normal to it. In a stable atmosphere, the value of this physical phenomenon equal to the weight of a column of air on a base with an area equal to one. The norm of barometric pressure is 101,325 Pa (760 mm Hg at 0 degrees Celsius). Moreover, the higher the object is from the surface of the Earth, the lower the air pressure on it becomes. Every 8 km it decreases by 100 Pa.
Thanks to this property, in the mountains, water in kettles boils much faster than at home on the stove. The fact is that pressure affects the boiling point: with its decrease, the latter decreases. And vice versa. The work of such kitchen appliances as a pressure cooker and an autoclave is built on this property. The increase in pressure inside them contributes to the formation of higher temperatures in the dishes than in ordinary pans on the stove.
The barometric altitude formula is used to calculate atmospheric pressure. It looks like the photo below.
P is the desired value at height, P 0 is the density of air near the surface, g is the free fall acceleration, h is the height above the Earth, m is the molar mass of the gas, t is the temperature of the system, r is the universal gas constant 8.3144598 J⁄ ( mol x K), and e is the Eclair number, equal to 2.71828. Often in the above formula for atmospheric pressure, instead of R, K is used - Boltzmann's constant. The universal gas constant is often expressed in terms of its product by the Avogadro number. It is more convenient for calculations when the number of particles is given in moles. When making calculations, it is always worth taking into account the possibility of changes in air temperature due to a change in the meteorological situation or when climbing above sea level, as well as geographical latitude.
Gauge and vacuumThe difference between atmospheric and measured ambient pressure is called overpressure. Depending on the result, the name of the value changes. If it is positive, it is called gauge pressure. If the result obtained is with a minus sign, it is called a vacuum gauge. It is worth remembering that it cannot be more than barometric. differentialThis value is the pressure difference at different measuring points. As a rule, it is used to determine the pressure drop on any equipment. This is especially true in the oil industry. Having figured out what kind of thermodynamic quantity is called pressure and with the help of what formulas it is found, we can conclude that this phenomenon is very important, and therefore knowledge about it will never be superfluous.  |
