Buoyancy is a buoyant force acting on a body immersed in a liquid (or gas) and directed opposite to the force of gravity. In general cases, the buoyancy force can be calculated using the formula: F b = V s × D × g, where F b is the buoyancy force; V s is the volume of the body part immersed in the liquid; D is the density of the liquid in which the body is immersed; g – gravity.

Steps

Calculation by formula

Find the volume of the body part immersed in the liquid (submerged volume). The buoyant force is directly proportional to the volume of the part of the body immersed in the liquid. In other words, the more a body sinks, the greater the buoyant force. This means that even sinking bodies are subject to buoyancy. The immersed volume should be measured in m3.

• For bodies that are completely immersed in a liquid, the immersed volume is equal to the volume of the body. For bodies floating in a liquid, the immersed volume is equal to the volume of the part of the body hidden under the surface of the liquid.
• As an example, consider a ball floating in water. If the diameter of the ball is 1 m, and the surface of the water reaches the middle of the ball (that is, it is half immersed in water), then the immersed volume of the ball is equal to its volume divided by 2. The volume of the ball is calculated by the formula V = (4/3)π( radius) 3 = (4/3)π(0.5) 3 = 0.524 m 3. Immersed volume: 0.524/2 = 0.262 m3.

1. Find the density of the liquid (in kg/m3) into which the body is immersed. Density is the ratio of the mass of a body to the volume occupied by that body. If two bodies have the same volume, then the mass of the body with higher density will be greater. As a rule, the greater the density of the liquid in which the body is immersed, the greater the buoyant force. The density of a liquid can be found on the Internet or in various reference books.

   • In our example, the ball floats in water. The density of water is approximately 1000 kg/m3 .
   • The densities of many other liquids can be found.

2. Find the force of gravity (or any other force acting vertically downward on the body). It doesn't matter whether a body floats or sinks, gravity always acts on it. IN natural conditions the force of gravity (more precisely, the force of gravity acting on a body weighing 1 kg) is approximately equal to 9.81 N/kg. However, if there are other forces acting on the body, for example, centrifugal force, such forces must be taken into account and the resulting force directed vertically downward must be calculated.

   • In our example, we are dealing with a conventional stationary system, so the only force acting on the ball is gravity, equal to 9.81 N/kg.
   • However, if a ball floats in a container of water that rotates around a certain point, then a centrifugal force will act on the ball, which does not allow the ball and water to splash out and which must be taken into account in the calculations.

3. If you have the immersed volume of the body (in m3), the density of the liquid (in kg/m3) and the force of gravity (or any other force directed vertically downwards), then you can calculate the buoyant force. To do this, simply multiply the above values ​​and you will find the buoyant force (in N).

   • In our example: F b = V s × D × g. F b = 0.262 m 3 × 1000 kg/m 3 × 9.81 N/kg = 2570 N.

4. Find out whether the body will float or sink. Using the above formula, you can calculate the buoyancy force. But by doing more calculations, you can determine whether the body will float or sink. To do this, find the buoyant force for the entire body (that is, in the calculations use the entire volume of the body, not the immersed volume), and then find the force of gravity using the formula G = (body mass) * (9.81 m/s 2). If the buoyant force is greater than the force of gravity, then the body will float; if the force of gravity is greater than the buoyant force, then the body will sink. If the forces are equal, then the body has “neutral buoyancy”.

   • For example, consider a 20 kg log (cylindrical) with a diameter of 0.75 m and a height of 1.25 m, immersed in water.
     • Find the volume of the log (in our example, the volume of the cylinder) using the formula V = π(radius) 2 (height) = π(0.375) 2 (1.25) = 0.55 m 3 .
     • Next, calculate the buoyant force: F b = 0.55 m 3 × 1000 kg/m 3 × 9.81 N/kg = 5395.5 N.
     • Now find the force of gravity: G = (20 kg)(9.81 m/s2) = 196.2 N. This value is much less than value buoyant force, so the log will float.

5. Use the calculations described above for a body immersed in gas. Remember that bodies can float not only in liquids, but also in gases, which may well push out some bodies, despite the very low density of gases (think about a balloon filled with helium; the density of helium is less than the density of air, so a balloon with helium flies (floats) ) in the air).

   Setting up the experiment

   1. Place a small cup in the bucket. In this simple experiment we will show that a body immersed in a liquid is subject to a buoyant force, since the body pushes out a volume of liquid equal to the immersed volume of the body. We will also demonstrate how to find the buoyant force through experiment. Start by placing a small cup in a bucket (or pan).

   2. Fill the cup with water (to the brim). Be careful! If the water in the cup spills into the bucket, throw the water away and start again.

      • For the sake of experiment, let's assume that the density of water is 1000 kg/m3 (unless you are using salt water or other liquid).
      • Use a pipette to fill the cup to the brim.

   3. Get a small item that will fit in the cup and won't be damaged by water. Find the mass of this body (in kilograms; to do this, weigh the body on a scale and convert the value in grams to kilograms). Then slowly lower the object into the cup of water (that is, submerge your body in the water, but do not submerge your fingers). You will see that some water has spilled from the cup into the bucket.

      • In this experiment, we will lower a toy car weighing 0.05 kg into a cup of water. We don't need the volume of this car to calculate the buoyancy force.
   4. ), and then multiply the volume of displaced water by the density of water (1000 kg/m3).
      • In our example, the toy car sank, displacing about two tablespoons of water (0.00003 m3). Let's calculate the mass of displaced water: 1000 kg/m3 × 0.00003 m3 = 0.03 kg.

   5. Compare the mass of displaced water with the mass of the submerged body. If the mass of the submerged body is greater than the mass of the displaced water, then the body will sink. If the mass of displaced water is greater than the mass of the body, then it floats. Therefore, in order for a body to float, it must displace an amount of water with a mass greater than the mass of the body itself.

      • Thus, bodies with a small mass but a large volume have the best buoyancy. These two parameters are typical for hollow bodies. Think of a boat - it has excellent buoyancy because it is hollow and displaces a lot of water with a small mass of the boat itself. If the boat were not hollow, it would not float at all (but would sink).
      • In our example, the mass of the car (0.05 kg) is greater than the mass of the displaced water (0.03 kg). That's why the car sank.
   • Use a scale that can be reset to 0 before each new weighing. In this case, you will get accurate results.

Lesson objectives: to verify the existence of a buoyant force, to understand the reasons for its occurrence and to derive rules for its calculation, to contribute to the formation of a worldview idea of ​​the knowability of phenomena and properties of the surrounding world.

Lesson objectives: Work on developing the skills to analyze properties and phenomena based on knowledge, highlight the main reason influencing the result. Develop communication skills. At the stage of putting forward hypotheses, develop oral speech. To check the level of independent thinking of the student in terms of the students’ application of knowledge in various situations.

Archimedes is an outstanding scientist of Ancient Greece, born in 287 BC. in the port and shipbuilding city of Syracuse on the island of Sicily. Archimedes received an excellent education from his father, the astronomer and mathematician Phidias, a relative of the Syracusan tyrant Hiero, who patronized Archimedes. In his youth, he spent several years in the largest cultural center in Alexandria, where he developed friendly relations with the astronomer Conon and the geographer-mathematician Eratosthenes. This was the impetus for the development of his outstanding abilities. He returned to Sicily as a mature scientist. He became famous for his numerous scientific works, mainly in the fields of physics and geometry.

The last years of his life, Archimedes was in Syracuse, besieged by the Roman fleet and army. I was in 2nd place Punic War. And the great scientist, sparing no effort, organizes the engineering defense of his hometown. He built many amazing combat vehicles that sank enemy ships, smashed them to pieces, and destroyed soldiers. However, the army of the city’s defenders was too small compared to the huge Roman army. And in 212 BC. Syracuse was taken.

The genius of Archimedes aroused admiration among the Romans and the Roman commander Marcellus ordered his life to be spared. But the soldier, who did not know Archimedes by sight, killed him.

One of his most important discoveries was the law, later called Archimedes' law. There is a legend that the idea of ​​this law came to Archimedes while he was taking a bath, with the exclamation “Eureka!” he jumped out of the bath and ran naked to write down the scientific truth that had come to him. The essence of this truth remains to be clarified; we need to verify the existence of a buoyant force, understand the reasons for its occurrence and derive rules for calculating it.

The pressure in a liquid or gas depends on the depth of the body's immersion and leads to the appearance of a buoyant force acting on the body and directed vertically upward.

If a body is lowered into a liquid or gas, then under the action of a buoyant force it will float up from deeper layers to shallower ones. Let us derive a formula for determining the Archimedes force for a rectangular parallelepiped.

The fluid pressure on the upper face is equal to

where: h1 is the height of the liquid column above the top edge.

Pressure force on the top the edge is equal

F1= p1*S = w*g*h1*S,

Where: S – area of ​​the upper face.

The fluid pressure on the lower face is equal to

where: h2 is the height of the liquid column above the bottom edge.

The pressure force on the lower edge is equal to

F2= p2*S = w*g*h2*S,

Where: S is the area of ​​the bottom face of the cube.

Since h2 > h1, then р2 > р1 and F2 > F1.

The difference between the forces F2 and F1 is equal to:

F2 – F1 = w*g*h2*S – w*g*h1*S = w*g*S* (h2 – h1).

Since h2 – h1 = V is the volume of a body or part of a body immersed in a liquid or gas, then F2 – F1 = w*g*S*H = g* w*V

The product of density and volume is the mass of the liquid or gas. Therefore, the difference in forces is equal to the weight of the fluid displaced by the body:

F2 – F1= mf*g = Pzh = Fout.

The buoyancy force is the Archimedes force, which defines Archimedes' law

The resultant of the forces acting on the side faces is zero, therefore it is not involved in the calculations.

Thus, a body immersed in a liquid or gas experiences a buoyant force equal to the weight of the liquid or gas displaced by it.

Archimedes' Law was first mentioned by Archimedes in his treatise On Floating Bodies. Archimedes wrote: “bodies heavier than the liquid, immersed in this liquid, will sink until they reach the very bottom, and in the liquid they will become lighter by the weight of the liquid in a volume equal to the volume of the immersed body.”

Let's consider how the Archimedes force depends and whether it depends on the weight of the body, the volume of the body, the density of the body and the density of the liquid.

Based on the Archimedes force formula, it 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 directions (the force of gravity is downward, and the Archimedean force is upward), then the weight of the body in the liquid will be less than the weight of the body in vacuum by the Archimedean force:

P A = m t g – m f g = g (m t – m f)

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.

Hence:

The Archimedes force depends on the density of the liquid and the volume of the body or its immersed part and does not depend on the density of the body, its weight and the volume of the liquid.

Determination of Archimedes' force by laboratory method.

Equipment: a glass of clean water, a glass of salt water, a cylinder, a dynamometer.

Work progress:

• determine the weight of the body in the air;
• determine the weight of the body in the liquid;
• find the difference between the weight of a body in air and the weight of a body in liquid.

4. Measurement results:

Conclude how the Archimedes force depends on the density of the liquid.

The buoyancy force acts on bodies of any geometric shape. In technology, the most common bodies are cylindrical and spherical shapes, bodies with a developed surface, hollow bodies in the shape of a ball, a rectangular parallelepiped, or a cylinder.

The gravitational force is applied to the center of mass of a body immersed in a liquid and is directed perpendicular to the surface of the liquid.

The lifting force acts on the body from the side of the liquid, is directed vertically upward, and is applied to the center of gravity of the displaced volume of liquid. The body moves in a direction perpendicular to the surface of the liquid.

Let's find out the conditions for floating bodies, which are based on Archimedes' law.

The behavior of a body located in a liquid or gas depends on the relationship between the modules of gravity F t and the Archimedes force F A , which act on this body. The following three cases are possible:

• F t > F A - the body drowns;
• F t = F A - the body floats in a liquid or gas;
• F t< F A - тело всплывает до тех пор, пока не начнет плавать.

Another formulation (where P t is the density of the body, P s is the density of the medium in which it is immersed):

• P t > P s - the body sinks;
• P t = P s - the body floats in a liquid or gas;
• Pt< P s - тело всплывает до тех пор, пока не начнет плавать.

The density of organisms living in water is almost the same as the density of water, so they don’t need strong skeletons! Fish regulate their diving depth by changing the average density of their body. To do this, they only need to change the volume of the swim bladder by contracting or relaxing the muscles.

If a body lies at the bottom in a liquid or gas, then the Archimedes force is zero.

Archimedes' principle is used in shipbuilding and aeronautics.

Floating body diagram:

The line of action of the force of gravity of the body G passes through the center of gravity K (center of displacement) of the displaced volume of fluid. In the normal position of a floating body, the center of gravity of the body T and the center of displacement K are located along the same vertical, called the axis of swimming.

When rolling, the center of displacement K moves to point K1, and the force of gravity of the body and the Archimedean force FA form a pair of forces that tends to either return the body to its original position or increase the roll.

In the first case, the floating body has static stability, in the second case there is no stability. The stability of the body depends on relative position the center of gravity of the body T and the metacenter M (the point of intersection of the line of action of the Archimedean force during a roll with the axis of navigation).

In 1783, the MONTGOLFIER brothers made a huge paper ball, under which they placed a cup of burning alcohol. The balloon filled with hot air and began to rise, reaching a height of 2000 meters.

The buoyant force acting on a body immersed in a liquid is equal to the weight of the liquid displaced by it.

"Eureka!" (“Found!”) - this is the exclamation, according to legend, made by the ancient Greek scientist and philosopher Archimedes, who discovered the principle of repression. Legend has it that the Syracusan king Heron II asked the thinker to determine whether his crown was made of pure gold without harming the royal crown itself. It was not difficult to weigh the crown of Archimedes, but this was not enough - it was necessary to determine the volume of the crown in order to calculate the density of the metal from which it was cast and determine whether it was pure gold.

Then, according to legend, Archimedes, preoccupied with thoughts about how to determine the volume of the crown, plunged into the bath - and suddenly noticed that the water level in the bath had risen. And then the scientist realized that the volume of his body displaced an equal volume of water, therefore, the crown, if lowered into a basin filled to the brim, would displace a volume of water equal to its volume. A solution to the problem was found and, according to the most common version of the legend, the scientist ran to report his victory in royal palace without even bothering to get dressed.

However, what is true is true: it was Archimedes who discovered buoyancy principle. If a solid body is immersed in a liquid, it will displace a volume of liquid equal to the volume of the part of the body immersed in the liquid. The pressure that previously acted on the displaced liquid will now act on the solid body that displaced it. And, if the buoyant force acting vertically upward turns out to be greater than the force of gravity pulling the body vertically downward, the body will float; otherwise it will sink (drown). Speaking modern language, the body floats if it average density less than the density of the liquid in which it is immersed.

Archimedes' principle can be interpreted in terms of molecular kinetic theory. In a fluid at rest, pressure is produced by the impacts of moving molecules. When a certain volume of liquid is displaced solid body, the upward impulse of the collisions of the molecules will fall not on the liquid molecules displaced by the body, but on the body itself, which explains the pressure exerted on it from below and pushing it towards the surface of the liquid. If the body is completely immersed in the liquid, the buoyant force will continue to act on it, since the pressure increases with increasing depth, and the lower part of the body is subjected to more pressure than the upper, which is where the buoyant force arises. This is the explanation of buoyant force at the molecular level.

This pushing pattern explains why a ship made of steel, which is much denser than water, remains afloat. The fact is that the volume of water displaced by a ship is equal to the volume of steel submerged in water plus the volume of air contained inside the ship's hull below the waterline. If we average the density of the hull shell and the air inside it, it turns out that the density of the ship (as a physical body) is less than the density of water, therefore the buoyancy force acting on it as a result of upward impulses of impact of water molecules turns out to be higher gravitational force the gravity of the Earth pulling the ship to the bottom - and the ship floats.

Archimedes' law is formulated as follows: a body immersed in a liquid (or gas) is acted upon by a buoyant force equal to the weight of the liquid (or gas) displaced by this body. The force is called by the power of Archimedes:

where is the density of the liquid (gas), is the acceleration of free fall, and is the volume of the submerged body (or the part of the volume of the body located below the surface). If a body floats on the surface or moves uniformly up or down, then the buoyant force (also called the Archimedean force) is equal in magnitude (and opposite in direction) to the force of gravity acting on the volume of liquid (gas) displaced by the body, and is applied to the center of gravity of this volume.

A body floats if the Archimedes force balances the force of gravity of the body.

It should be noted that the body must be completely surrounded by liquid (or intersect with the surface of the liquid). So, for example, Archimedes' law cannot be applied to a cube that lies at the bottom of a tank, hermetically touching the bottom.

As for a body that is in a gas, for example in air, to find the lifting force it is necessary to replace the density of the liquid with the density of the gas. For example, a helium balloon flies upward due to the fact that the density of helium is less than the density of air.

Archimedes' law can be explained using the difference in hydrostatic pressure using the example of a rectangular body.

Where P A , P B- pressure at points A And B, ρ - fluid density, h- level difference between points A And B, S- horizontal cross-sectional area of ​​the body, V- volume of the immersed part of the body.

18. Equilibrium of a body in a fluid at rest

A body immersed (fully or partially) in a liquid experiences a total pressure from the liquid, directed from bottom to top and equal to the weight of the liquid in the volume of the immersed part of the body. P vyt = ρ and gV Pogr

For a homogeneous body floating on the surface, the relation is true

Where: V- volume of the floating body; ρ m- body density.

The existing theory of a floating body is quite extensive, so we will limit ourselves to considering only the hydraulic essence of this theory.

The ability of a floating body, removed from a state of equilibrium, to return to this state again is called stability. The weight of liquid taken in the volume of the immersed part of the vessel is called displacement, and the point of application of the resultant pressure (i.e., the center of pressure) is displacement center. In the normal position of the ship, the center of gravity WITH and center of displacement d lie on the same vertical line O"-O", representing the axis of symmetry of the vessel and called the axis of navigation (Fig. 2.5).

Let under the influence external forces the ship tilted at a certain angle α, part of the ship KLM came out of the liquid, and part K"L"M", on the contrary, plunged into it. At the same time, we received a new position for the center of displacement d". Let's apply it to the point d" lift R and continue the line of its action until it intersects with the axis of symmetry O"-O". Received point m called metacenter, and the segment mC = h called metacentric height. Let's count h positive if point m lies above the point C, and negative - otherwise.

Rice. 2.5. Cross profile of the vessel

Now consider the equilibrium conditions of the ship:

1) if h> 0, then the ship returns to its original position; 2) if h= 0, then this is a case of indifferent equilibrium; 3) if h<0, то это случай неостойчивого равновесия, при котором продолжается дальнейшее опрокидывание судна.

Consequently, the lower the center of gravity and the greater the metacentric height, the greater will be the stability of the vessel.

ARCHIMEDES' LAW– the law of statics of liquids and gases, according to which a body immersed in a liquid (or gas) is acted upon by a buoyant force equal to the weight of the liquid in the volume of the body.

The fact that a certain force acts on a body immersed in water is well known to everyone: heavy bodies seem to become lighter - for example, our own body when immersed in a bath. When swimming in a river or in the sea, you can easily lift and move very heavy stones along the bottom - ones that we cannot lift on land; the same phenomenon is observed when, for some reason, a whale is washed up on the shore - the animal cannot move outside the aquatic environment - its weight exceeds the capabilities of its muscular system. At the same time, lightweight bodies resist immersion in water: sinking a ball the size of a small watermelon requires both strength and dexterity; It will most likely not be possible to immerse a ball with a diameter of half a meter. It is intuitively clear that the answer to the question - why a body floats (and another sinks) is closely related to the effect of the liquid on the body immersed in it; one cannot be satisfied with the answer that light bodies float and heavy ones sink: a steel plate, of course, will sink in water, but if you make a box out of it, then it can float; however, her weight did not change. To understand the nature of the force acting on a submerged body from the side of a liquid, it is enough to consider a simple example (Fig. 1).

Cube with an edge a immersed in water, and both the water and the cube are motionless. It is known that the pressure in a heavy liquid increases in proportion to depth - it is obvious that a higher column of liquid presses more strongly on the base. It is much less obvious (or not at all obvious) that this pressure acts not only downwards, but also sideways and upwards with the same intensity - this is Pascal's law.

If we consider the forces acting on the cube (Fig. 1), then due to the obvious symmetry, the forces acting on the opposite side faces are equal and oppositely directed - they try to compress the cube, but cannot affect its balance or movement. There remain forces acting on the upper and lower faces. Let h– depth of immersion of the upper face, r– fluid density, g– acceleration of gravity; then the pressure on the upper face is equal to

r· g · h = p 1

and on the bottom

r· g(h+a)= p 2

The pressure force is equal to the pressure multiplied by the area, i.e.

F 1 = p 1 · a\up122, F 2 = p 2 · a\up122 , where a- cube edge,

and strength F 1 is directed downwards and the force F 2 – up. Thus, the action of the liquid on the cube is reduced to two forces - F 1 and F 2 and is determined by their difference, which is the buoyancy force:

F 2 – F 1 =r· g· ( h+a)a\up122 – r gha· a 2 = pga 2

The force is buoyant, since the lower edge is naturally located below the upper one and the force acting upward is greater than the force acting downward. Magnitude F 2 – F 1 = pga 3 is equal to the volume of the body (cube) a 3 multiplied by the weight of one cubic centimeter of liquid (if we take 1 cm as a unit of length). In other words, the buoyant force, which is often called the Archimedean force, is equal to the weight of the liquid in the volume of the body and is directed upward. This law was established by the ancient Greek scientist Archimedes, one of the greatest scientists on Earth.

If a body of arbitrary shape (Fig. 2) occupies a volume inside the liquid V, then the effect of a liquid on a body is completely determined by the pressure distributed over the surface of the body, and we note that this pressure is completely independent of the material of the body - (“the liquid doesn’t care what to press on”).

To determine the resulting pressure force on the surface of the body, you need to mentally remove from the volume V given body and fill (mentally) this volume with the same liquid. On the one hand, there is a vessel with a liquid at rest, on the other hand, inside the volume V- a body consisting of a given liquid, and this body is in equilibrium under the influence of its own weight (the liquid is heavy) and the pressure of the liquid on the surface of the volume V. Since the weight of liquid in the volume of a body is equal to pgV and is balanced by the resultant pressure forces, then its value is equal to the weight of the liquid in the volume V, i.e. pgV.

Having mentally made the reverse replacement - placing it in volume V given body and noting that this replacement will not affect the distribution of pressure forces on the surface of the volume V, we can conclude: a body immersed in a heavy liquid at rest is acted upon by an upward force (Archimedean force), equal to the weight of the liquid in the volume of the given body.

Similarly, it can be shown that if a body is partially immersed in a liquid, then the Archimedean force is equal to the weight of the liquid in the volume of the immersed part of the body. If in this case the Archimedean force is equal to the weight, then the body floats on the surface of the liquid. Obviously, if, during complete immersion, the Archimedean force is less than the weight of the body, then it will drown. Archimedes introduced the concept of "specific gravity" g, i.e. weight per unit volume of a substance: g = pg; if we assume that for water g= 1, then a solid body of matter for which g> 1 will drown, and when g < 1 будет плавать на поверхности; при g= 1 a body can float (hover) inside a liquid. In conclusion, we note that Archimedes' law describes the behavior of balloons in the air (at rest at low speeds).

Vladimir Kuznetsov