Pulling force. Law of Archimedes

academic year

Lesson topic: Archimedean strength.

Law of Archimedes

Goalslesson:

educational: o detect the presence of a force that pushes the body out of the liquid;

developing: teach to apply the law of Archimedes;

educational: to form intellectual skills to analyze, compare, systematize knowledge. Instill in students an interest in science.

Type of lesson: a lesson in the assimilation of new knowledge.

Equipment (for teacher): tripod, a glass vessel with a hole for water to flow out, a dynamometer, a set of weights, a glass

for students: dynamometer, thread, set of weights, vessels with water, plasticine, ball.

Demonstration: experience according to Fig. 139 of the textbook, a wooden block, a ball, a vessel with water.

movelesson

1. Organizational moment.

Message about the objectives of the lesson.

2.Updating knowledge.

Answer the questions:

1. How is Pascal's law formulated?

2. How is the liquid pressure on the bottom and walls of the vessel calculated?

3. Preparation for the assimilation of new material.

Statement of educational problems:

a/ Does the liquid act on the body immersed in it?

b/ Does a liquid always act on a submerged body?

c/ how to theoretically explain this action of a liquid on a body immersed in it?

Let's turn to experience. We lower a wooden block into the water. The block floats on the surface of the water. Why does a wooden block float on water?

We lower the ball into the water and remove our hand. The ball bounces onto the surface of the water. Why does the ball jump out of the water?

A buoyant force acts on submerged bodies in water.

Does a liquid always act on a submerged body? A metal cylinder immersed in water sinks. Is the effect of water on this body noticeable?

4. Explanationnewmaterial:

Let's do an experiment. Hang the cylinder on a dynamometer and observe the stretching of the spring in air and then in water.

1. Buoyancy detection experience:

1. Determine the weight of the load in the air P1.

2. Determine the weight of the cargo in water P2.

3.Compare the measurement results and draw a conclusion.

Conclusion: body weight in water is less than body weight in air: P1 > P2.

Why is body weight in water less than body weight in air?

Answer: the liquid acts on any body immersed in it. This force is directed vertically upwards.

- And how can you find the magnitude of the buoyant force?

Answer: the weight of the body in the air must be subtracted from the weight of the body in the water.

We have come to the following conclusion. Two forces act on a body immersed in a liquid: one force is gravity, directed downwards, the other is pushing, directed upwards.

https://pandia.ru/text/78/176/images/image003_168.gif" width="12" height="75"> 2

Today we will study the buoyant force acting on bodies immersed in a liquid. Let us find out on what factors this force depends. Let's learn how to calculate this force. It is called pushing, or Archimedean power in honor of the ancient Greek scientist Archimedes, who first pointed out its existence and calculated its value.

Archimedes (287-212 BC) -

Ancient Greek scientist, physicist and mathematician. Established the rule of the lever, discovered the law of hydrostatics. Material about Archimedes is attached at the end of the development of the lesson.

5. Work in groups.

What does Archimedean force depend on?

To answer this question, we will work in groups. Each group receives a task and answers the question.

Assignment to the first group

Determine the dependence of the Archimedean force on the density of the body.

Equipment: a vessel with water, a dynamometer, bodies of the same volume and different densities (aluminum and copper cylinders), a thread.

1.Determine the weight of the aluminum cylinder in the air. P1= …….. H

2. Determine the weight of the aluminum cylinder in water. P2= …...... N

3. Find the Archimedean force acting on the aluminum cylinder. P1 - P2=………. H

4. Determine the weight of the copper cylinder in the air. P3=………. H

5.Determine the weight of the copper cylinder in water. P4= ………H

6. Find the Archimedean force acting on the copper cylinder. P3 - P4 = ……..N

7. Make a conclusion about dependencies (independence) Archimedean force on the density of the body.

Answer: Archimedean force………………………………………………………………………………………………………………………………………………………………………………………………………

Assignment to the second group

Determine the dependence of the Archimedean force on the volume of the body.

Equipment: a vessel with water, bodies of different volumes (aluminum cylinders), a dynamometer, a thread.

1. Determine the weight of the large cylinder in the air. P1= H

2. Determine the weight of the large cylinder in water. P2= H

3.Find the Archimedean force acting on the large cylinder. Р1 –Р2= Н

4. Determine the weight of the small cylinder in the air. P3= H

5. Determine the weight of the small cylinder in water. P4= H

6.Find the Archimedean force acting on a small cylinder. Р3 –Р4= H

7. Make a conclusion about dependencies (independence) of the Archimedean force on the volume of the body.

Answer: Archimedean force ………………………………… on the volume of the body.

Assignment to the third group

Determine the dependence of the Archimedean force on the density of the liquid.

Equipment: dynamometer, thread, vessels with fresh water and salt water, ball.

1. Determine the weight of the ball in the air. P1= H

2. Determine the weight of the ball in fresh water. P2= H

3.Find the Archimedean force acting on the ball in fresh water. P1 - P2 = H

4. Determine the weight of the ball in the air. P1= H

5. Determine the weight of the ball in salt water. P3= H

6.Find the Archimedean force acting on the ball in salt water. P1-P2 = H

7. Make a conclusion about dependencies (independence) of the Archimedean force on the density of the liquid.

Answer: Archimedean force ………………………………… on the density of the liquid.

Task for the fourth group

Determine the dependence of the Archimedean force on the depth of immersion.

Equipment: dynamometer, thread, beaker with water, aluminum cylinder.

1.Determine the weight of the aluminum cylinder in the air. P1= H

2. Determine the weight of the aluminum cylinder in water at a depth of 5 cm. P2= H

3.Find the Archimedean force acting on an aluminum cylinder in water.

P1 - P2 = H

4. Determine the weight of the aluminum cylinder in the air. P1= H

5. Determine the weight of the aluminum cylinder in water at a depth of 10 cm. P3 = H

6. Find the Archimedean force acting on the aluminum cylinder in the second case.

P1 - P3 \u003d H

7. Make a conclusion about dependencies (independence) of the Archimedean force on the depth of immersion of the body.

Answer: Archimedean force ……………………………………………………………………………………………………………………………………………….

Assignment to the fifth group

Determine the dependence of the Archimedean force on the shape of the body.

Equipment: dynamometer, thread, a vessel with water, a piece of plasticine.

1. Give a piece of plasticine the shape of a cube.

2. Determine the weight of plasticine in the air. P1= H

3. Determine the weight of plasticine in water. P2 = H

4. Find the Archimedean force acting on a piece of plasticine. P1 - P2 = H

5. Give a piece of plasticine the shape of a ball.

6. Determine the weight of plasticine in the air. P3= H

7. Determine the weight of plasticine in water. P4= H

8. Find the Archimedean force acting on a piece of plasticine. P3-P4= H

9. Compare these forces and draw a conclusion about dependencies (independence) of the Archimedean force on the shape of the body.

Answer: Archimedean force …………………………………………………… from the shape of the body.

After receiving the results, each group verbally reports on their work and reports their findings. The conclusions are written by students in notebooks, and by the teacher - on the board in the form of a table:

We learned that the Archimedean force depends on the volume of the body and the density of the liquid. How to theoretically explain the effect of a liquid on a body immersed in it. Experiments show that the action of the liquid is directed upwards.

The value of the buoyancy force can be determined using the instrument in front of you.

The device is called "Archimedes' bucket". This is a spring with a pointer, a scale, a bucket, a cylinder of the same volume, a pouring vessel, a glass.

Here the spring acts as a dynamometer.

1. Show that the volume of the bucket is equal to the volume of the cylinder.

2. Pour water into the drain vessel just above the level of the drain pipe. Excess water will pour into the glass. We drain the water.

3. Let's hang the bucket to the spring, and to it - the cylinder. We note the stretching of the spring with the help of a pointer. The arrow shows the weight of the body in the air.

4. Having lifted the body, we substitute a pouring vessel under it. After immersion in the pouring vessel, part of the water will pour into the glass. The spring pointer will move up and the spring will contract, indicating a decrease in the weight of the body in the fluid.

Why does the spring shrink?

In this case, in addition to gravity, the body is also affected by the force pushing it out of the liquid.

In which direction is the buoyant force directed?

The buoyant force is directed upward.

5. Pour water from a glass into a bucket.

Pay attention to the spring indicator. Where did the spring indicator stop after we poured water from a glass into a bucket?

The pointer returned to its original position.

Why did the spring pointer return to its previous position?

In addition to gravity and buoyancy, the spring is affected by the weight of the water in the bucket.

The weight of the water is equal to the buoyant force.

Notice how much water came out.

Full bucket.

Compare the volume of water poured into the bucket and the volume of the cylinder.

They are the same.

Based on this experience, we conclude: the buoyant force is equal to the weight of the fluid displaced by the body.

6. The law of Archimedes is formulated: a buoyant force acts on a body immersed in a liquid, equal in magnitude to the weight of the liquid displaced by the body.

Based on this experience, it can be concluded that the force pushing out a body completely immersed in a liquid is equal to the weight of the liquid in the volume of this body.

If a similar experiment were done with a body immersed in a gas, it would show that strength, pushing the body out of the gas is also equal to the weight of the gas taken in the volume of the body.

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. FA=РЖ= g m zh.

The mass of liquid m f, displaced by the body, can be expressed in terms of its density (ρf) and the volume of the body (Vt) immersed in the liquid (since Vl - the volume of the liquid displaced by the body is equal to Vt - the volume of the body immersed in the liquid, Vl = Vt), t i.e. mzh = ρzhVt.

Then we get FA =gρzhVt.

As it was found, 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 P1 will be less than the weight of the body in vacuum P = g m (m is the mass of the body) by the Archimedean force FA \u003d g m f (m f is the mass of the fluid displaced by the body) i.e. P1 \u003d P - FA, or P1 \u003d g m - g m f.

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

It should be remembered that when calculating the Archimedes force, V is understood only as that part of the volume of the body that is completely in the liquid.

This can be part of the volume of the body (if it floats on the surface, not completely submerged), and the entire volume (if the body drowned).

In figure 2, this volume is shaded.

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Archimedes' law can be obtained mathematically.

To explain, we use the idea of ​​the pressure of a fluid on a body. Pressure inside the liquid: p=gρzhh. Consider Figure 3. There is a parallelepiped in the liquid. If the top face is at depth h1 and the bottom face is at depth h2, then р2 >р1. The pressure on the side faces is compensated, since, according to Pascal's law, (on the side faces) the pressure at the same level is the same in all directions.

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Conclusion: the expulsion of the body occurs as a result of the action of different pressures on the lower and upper faces:

Lower > Upper.

We find the forces with which the fluid acts on the upper and lower faces of the parallelepiped.

F1=p1S= gρzh h1.

F2=p2S= gρzh h2.

F2 - F1=gρЖ h2- gρЖh1=gρЖ (h2 –h1).

Since (h2 –h1)= h is the height of the parallelepiped, then Sh=V is the volume of the parallelepiped. As a result, F2 - F1 =gρzhV.

Finally: FA \u003d gρzhV.

What is gρzhV? According to the formula, this is the weight of the liquid displaced by these bodies.

5. An example of solving a problem

Determine the buoyant force acting in sea water on a stone with a volume of 1.6 m3.

Given: Solution:

https://pandia.ru/text/78/176/images/image010_85.gif" width="2 height=86" height="86">V= 1.6 m3 FA =gρzhV. FA=9.8 m /kg 1030 kg/m3 1.6 m3 = N ≈ 16.5 kN.

ρzh =1030 kg/m3

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18. Two steel cylinders of the same mass are suspended from the balance beam. Will the balance of the scales be disturbed if one cylinder is immersed in water, and the second cylinder is immersed in kerosene. The density of water is 1000 kg/m3, and the density of kerosene is 800 kg/m3.

7. Work on the book.

Solving problems from exercise 32 (3.4) of the textbook.

8. Checking the assimilation by students of the material covered.

Students receive cards with tasks of different difficulty levels:

The first task is to determine the buoyancy force, the second is to determine the volume, the third is a combined one.

Card 1.

2. What is the volume of a steel cylinder if the difference between the weight of the cylinder in air and in water is 4 N? The density of water is 1000 kg/m3.

3. A granite slab measuring 1.2 x 0.6 x 0.3 m is immersed in water for half of its volume. How much lighter is the board? The density of water is 1000 kg/m3.

Card 2.

1. The volume of the ball is 0.002 m3. What is the buoyant force acting on the ball when it is immersed in water? The density of water is 1000 kg/m3.

3. A lead cylinder weighing 200 g is suspended from a spring balance. Then the cylinder is immersed in water. What are the readings of the scales in the first and second cases? The density of water is 1000 kg/m3. lead density 11300 kg/m3.

Card 3.

1. With what force is a cork bar 4 x 5 x 10 cm in size pushed out of kerosene? Density 800 kg/m3.

2. The Archimedean force acting on the part in water is 1000 N. Find the volume of the part. The density of water is 1000 kg/m3.

Card 4.

1. What is the buoyant force acting on a metal bar with a volume of 0.8 dm3 when it is completely immersed in water? The density of water is 1000 kg/m3.

2. The Archimedean force acting on the beam in water is 1000 N. Find the volume of the part. The density of water is 1000 kg/m3.

3. What force must be applied to keep a granite slab in water, on which a gravity force of 27,000 N acts? The plate volume is 1 m3. water density - 1000 kg/m3.

Card 5.

1. The volume of the steel bar is 6 dm3. What is the buoyant force acting on the block? The density of water is 1000 kg/m3.

2. The steel plate weighed 1960 N in the air, after immersion in water the plate began to weigh 1708.7 N. What is the volume of the steel plate? The density of water is 1000 kg/m3.

3. A wooden ball with a density of 500 kg/m3 floats in water. What part of the volume of the sphere is immersed in water if the density of water is 1000 kg/m3.

9. Summing up the lesson.

In this lesson we studied the principle of Archimedes. What have we learned? Have we achieved the lesson objectives?

Those who stand out are judged. Thank you very much for the lesson!

10. Homework: § 49, exercise 32 (1,2)

§8. Legend of Archimedes. Page 163.

For capable students, complete task 29.

Additional material for the lesson

On page 106 of the book “Entertaining Physics” there are articles “Eternal” water engine”, “How was Sadko raised? I recommend to read.

Archimedes and his inventions.

Undoubtedly, Archimedes (circa 287-212 BC) is the most brilliant scientist of Ancient Greece. He ranks with Newton, Gauss, Euler, Lobachevsky and other great mathematicians of all time. His works are devoted not only to mathematics. He made remarkable discoveries in mechanics, knew astronomy, optics, hydraulics well and was truly a legendary person.

The son of the astronomer Phidias, who wrote an essay on the diameters of the Sun and Moon, Archimedes was born and lived in the Greek city of Syracuse in Sicily. He was close to the court of King Hieron II and his son-heir.

The story of the sacrificial crown of Hiero is well known. Archimedes was instructed to check the honesty of the jeweler and determine whether the crown was made of pure gold or with impurities of other metals and whether there were voids inside it. One day, thinking about this, Archimedes plunged into the bath, and noticed that the water displaced by his body spilled over the edge. The brilliant scientist immediately had a bright idea, and with a cry of “Eureka, eureka!” he, as he was naked, rushed to conduct the experiment.

Archimedes' idea is very simple. A body immersed in water displaces as much fluid as the volume of the body itself. By placing the crown in a cylindrical vessel with water, you can determine how much liquid it will displace, that is, find out its volume. And, knowing the volume and weighing the crown, it is easy to calculate the specific gravity. This will make it possible to establish the truth: after all, gold is a very heavy metal, and lighter impurities, and especially voids, reduce the specific gravity of the product.

But Archimedes did not stop there. In his work “On Floating Bodies”, he formulated a law that says: “A body immersed in a liquid loses as much in its weight as the weight of the displaced liquid”. Archimedes' law is (along with other, later discovered facts) the basis of hydraulics - a science that studies the laws of motion and equilibrium of fluids. It is this law that explains why a steel ball (without voids) sinks in water, while a wooden body floats. In the first case, the weight of the displaced water is less than the weight of the ball itself, i.e. the Archimedean “buoyant” force is insufficient to keep it on the surface. And a heavily laden ship, whose hull is made of metal, does not sink, sinking only to the so-called waterline. Since there is a lot of space filled with air inside the ship's hull, the average specific gravity of the ship is less than the density of water and the buoyancy force keeps it afloat. Archimedes' principle also explains why a balloon filled with warm air or a gas that is lighter than air (hydrogen, helium) flies up.

Knowledge of hydraulics allowed Archimedes to invent a screw pump for pumping water. Until recently, such a pump (kohl) was used in Spanish and Mexican silver mines.

From the course of physics, everyone is familiar with the Archimedean rule of the lever. According to legend, the scientist uttered the catchphrase: “Give me a foothold, and I will lift the Earth!” . Of course, Archimedes had in mind the use of a lever, but he was somewhat self-confident: in addition to a fulcrum, he would also need an absolutely fantastic lever - an incredibly long and at the same time unbending rod.

Reliable facts and numerous legends indicate that Archimedes invented many interesting machines and devices.

List of used literature:

Independent work in physics.

Entertaining experiments in physics.

6th class of physicsdan problems dәreslәr.

Book for reading in physics.

Collection of problems in physics grade 7-8.

Thematic and lesson planning.

Entertaining physics. Book 2. (p. 106).

Pourochnye development in physics.

A. V Postnikov. Testing students' knowledge of physics.

Qualitative problems in physics.

Independent work of students in physics.

Didactic material in physics.

Additional tasks on the topic

Tasks:

Tasks of the first level of complexity.

To determine the buoyant force.

1. The volume of the steel bar is 0.2 m3. What is the buoyant force acting on the block when it is immersed in water? The density of water is 1000 kg/m3.

2. The volume of the ball is 0.002 m3. What is the buoyant force acting on the ball when it is immersed in water? The density of water is 1000 kg/m3.

3. With what force is a cork bar 4 x 5 x 10 cm in size pushed out of kerosene? Density 800 kg/m3.

4. What is the buoyancy force acting on a metal bar with a volume of 0.8 dm3 when it is completely immersed in water? The density of water is 1000 kg/m3.

5. The volume of the steel bar is 6 dm3. What is the buoyant force acting on the block? The density of water is 1000 kg/m3.

6. A cylinder with a volume of 0.02 m3 is lowered into the water. Find the Archimedean force. The density of water is 1000 kg/m3.

7. Calculate the buoyancy force acting on a granite block, which, when completely immersed in water, displaces some of it. The volume of displaced water is 0.8 m3. The density of water is 1000 kg/m3.

8. Reinforced concrete slab measuring 3.5 x 1.5 x 0.2 m is completely submerged in water. Calculate the Archimedean force acting on the plate. The density of water is 1000 kg/m3.

Tasks of the second level of complexity.

To determine the volume:

1. What is the volume of a steel cylinder if the difference in the weight of the cylinder in air and in water is

4 N? The density of water is 1000 kg/m3.

2. Determine the volume of a body completely submerged in water if the buoyant force acting on it is 29.4 N. The density of water is 1000 kg/m3.

3. The Archimedean force acting on the part in water is 1000 N. Find the volume of the part. The density of water is 1000 kg/m3.

4. The Archimedean force acting on the beam in water is 1000 N. Find the volume of the part. The density of water is 1000 kg/m3.

5. The steel plate weighed 1960 N in air, after immersion in water the plate began to weigh 1708.7 N. What is the volume of the steel plate? The density of water is 1000 kg/m3.

Tasks of the third level.

1. A granite slab measuring 1.2 x 0.6 x 0.3 m is immersed in water for half of its volume. How much lighter is the board? The density of water is 1000 kg/m3.

2. A lead cylinder weighing 200 g is suspended from a spring balance. Then the cylinder is immersed in water. What are the readings of the scales in the first and second cases? The density of water is 1000 kg/m3. lead density 11300 kg/m3.

3. What force must be applied to a ball with a volume of 5 dm3 and a mass of 0.5 kg to keep it under water? The density of water is 1000 kg/m3. Where is this force directed?

4. What force must be applied to keep a granite slab in water, on which a gravity force of 27,000 N acts? The plate volume is 1 m3. water density - 1000 kg/m3.

5. A wooden ball with a density of 500 kg/m3 floats in water. What part of the volume of the sphere is immersed in water if the density of water is 1000 kg/m3.

Tasks:

practical tasks.

card work:

1. Aluminum and iron bars are suspended at the ends of the balance beam (see Fig.). Their masses are chosen so that the scales in the water are in equilibrium. Which bar will outweigh if the water from their vessel is poured out?

2. Two identical steel balls are suspended at the ends of the balance beam. Will the equilibrium be preserved if the balls are lowered into different liquids (see the figure)?

Kerosene Water

3. The figure shows two spherical bodies floating in water. Which body has the highest density?

4. A body floats on the surface of the water. Graphically depict the forces acting on this body (see fig.).

5. A glass ball without air and a lead ball are balanced on a balance scale (see Fig.) Will the balance of the scales be disturbed if the scales together with the balls are moved to the top of the mountain?

6. Balls of equal mass but different volumes are suspended from identical springs. From below, a vessel with water is brought to the balls and raised to such a level that the balls are completely immersed in water (see Fig.). Which spring will contract more?

7. Bodies of equal mass and equal volume are suspended from springs of equal elasticity (see Fig.). Which spring will be the shortest when immersed in a liquid?

8. Which of the steel balls lowered into the water is affected by the greatest buoyant force? Why?

9. Identical balls suspended from the balance beam were immersed in liquid as shown in the figure. a and then, as shown in the figure b. In what case will the equilibrium of the scales be disturbed? Why?

Density of some substances needed in solving problems.

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A body immersed in a liquid or gas is subjected to a buoyant force equal to the weight of the liquid or gas displaced by this body.

In integral form

Archimedean force always directed opposite to gravity, so the weight of a body in a liquid or gas is always less than the weight of this body in a vacuum.

If a body floats on a surface or moves up or down uniformly, then the buoyant force (also called Archimedean force) is equal in absolute value (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.

As for bodies that are in a gas, for example, in air, to find the lifting force (Archimedes Force), you need to replace the density of the liquid with the density of the gas. For example, a balloon with helium flies upwards due to the fact that the density of helium is less than the density of air.

In the absence of a gravitational field (Gravity), that is, in a state of weightlessness, law of Archimedes does not work. Astronauts are familiar with this phenomenon quite well. In particular, in zero gravity there is no convection phenomenon (the natural movement of air in space), therefore, for example, air cooling and ventilation of the living compartments of spacecraft are forced by fans

In the formula we used

Observations and experiments show that bodies placed in liquid and gas are under pressure. The pressure of a liquid and a gas at the same height is the same in all directions. With a change in altitude, there is a change in pressure. For this reason, a buoyant force arises, which is called the Archimedean force. What is the Archimedean force in liquid and gas.

What is the pressure in gases and liquids

Recall the definition of pressure. By pressure p called a physical quantity equal to the ratio of the force F, directed perpendicular to the surface with area S:

$p=(F\over S)$ (1)

The French researcher Blaise Pascal discovered the law, later named after him, which sounds like this: liquids and gases transmit the pressure produced on them equally in all directions.

Based on Pascal's law and formula (1), the pressure of the liquid column can be calculated:

$p=(F\over S)=(m*g\over S)$ (2)

where: m is the mass of liquid, g= 9.8 N/kg is the free fall acceleration.

Then, if we express the mass of the liquid in terms of the density ρ and volume V, we get:

$p=(ρ*V*g\over S)$ (3)

Expressing volume V across the square S and height h, we get the final formula for pressure:

$p=(ρ*g*h)$ (4)

In physics, it is always necessary to know how a physical quantity is measured. In honor of Pascal, not only the law is named, but also the unit of pressure. Since force is measured in newtons, and area in square meters, then:

$$=( \over )$$

Multiple units of pressure are often used: kilopascal (kPa) and megapascal (MPa).

Law of Archimedes

A heavy object, which we tear off the ground with great difficulty, can be lifted quite easily when it is in the water. If you take an empty plastic bottle with a closed cork, immerse it completely in water and release it, the bottle will float. Why is this happening?

To explain these phenomena, it suffices to look at the last formula (4). Pressure dependence p in liquid or gas from depth h(height), leads to the appearance of a buoyant force acting on any body immersed in a liquid or gas. This force is called the Archimedean force.

Rice. 1. Portrait, image of Archimedes

The ancient Greek mathematician, engineer and physicist Archimedes (287-212 BC) not only discovered this phenomenon, but was able to find an explanation for it and derived a formula for calculating the buoyancy force. In addition to the law of Archimedes, he discovered the famous rule of the lever, was the first to derive mathematical formulas for calculating the areas and volumes of complex geometric surfaces, opened the first planetarium, and invented many useful devices.

Rice. 2. The action of the buoyant force on a body immersed in water

A drawing showing a rectangular parallelepiped (height h and base area S) placed in a liquid will help answer the question: how to find the Archimedean force. The pressure forces on the side faces balance each other, and the forces F 2 and F 1 differ, because according to formula (4), the pressure on the upper and lower faces will be different due to the fact that h 2 > h 1 :

We get the formula for the resulting force F A, equal to the difference F 2 and F 1 :

$F_A=F_2−F_1=p_2*S−p_1*S=ρ*g*h_2*S−ρ*g*h_1*S=$
$ρ*g*S*((h_2− h_1))=ρ*g*S*h$ (5)

where: $S*h=V$ is the volume, and $ρ*V=m$ is the mass of the fluid displaced by the body. Then, since m* g is the weight of the displaced fluid, then we obtain the final formula for the Archimedean force F A:

$F_A =m*g=ρ*V*g$ (6)

The resulting formula allows us to formulate the law of Archimedes:

The force that pushes a body immersed in a liquid (or gas) is equal to the weight of the liquid (or gas) displaced by the body.

Immersion, balance, ascent

Now it becomes clear why we easily lift heavy stones in water: the Archimedean force “helps” us, because. it is directed opposite to gravity. For the same reason, the weight of a body when weighed in liquid will always be less than the weight measured in air.

From formula (6) it follows that the magnitude of the Archimedean force depends in direct proportion to the density of the liquid ρ and on the volume of the immersed body V. The density of the substance from which the body is made can be any - it does not affect the magnitude of the buoyant force. Depending on the ratio of the Archimedean force F A and gravity Fg There are three possible positions of the body in the liquid:

• If FA > Fg, then the body will be pushed up - “float”;
• If FA
• If FA = Fg, then the body can be in the liquid at any depth in a state of equilibrium.

Archimedes' law is the basis of the hydrometer, a device for measuring the density of a liquid. The hydrometer is a glass, sealed flask, weighted from the lower end with a weight. The upper part is made in the form of a long process, on which a measuring scale is applied. When placed in a liquid, the hydrometer sinks to a greater or lesser depth, depending on the density of the liquid. The greater the density of the liquid, the less the hydrometer sinks. The reading on the scale indicates the density of a given liquid when the hydrometer is at equilibrium.

Rice. 3. Hydrometer

What have we learned?

So, we have learned why the Archimedean force arises in gases and liquids, and on what quantities its value depends. A body immersed in a liquid (or gas) is subjected to a buoyant force. The force that pushes a body immersed in a liquid (or gas) is equal to the weight of the liquid (or gas) displaced by the body. For a more detailed report on the Archimedean force, interesting examples can be prepared with various liquids other than water, such as kerosene or mercury. The topic of this article is closely related to the features of swimming and aeronautics of bodies, which we will consider in the next chapters of the 7th grade physics course.

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Law of Archimedes, or How to recognize a lie?

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Discovery history

In ancient Syracuse, there lived an engineer, mathematician, and physicist named Archimedes. He received an excellent education, his inventions were valued and he did not need funds. And from time to time the powers that be turned to him to solve all sorts of complex problems. And one of these tasks was to determine the authenticity of the crown of King Hieron.

It would seem that this is difficult?

Use the formula

ρ t = m t / V t (1).

Divide m t the mass of the ingot that was given to the jeweler by the volume of the crown V t, you will get the density of the crown ρ t. Compare the result with the known density of gold, and the trick is in the bag. And the jeweler will receive either a payment for the work, or a close acquaintance with the court executioner.

However, this formula works well with objects of a simple shape: a sphere, a cube, a box. And we remember that we are examining the crown, which has many teeth, bulges and openwork weaves.

How can you determine the volume of an object of such a complex shape? Do not know? Archimedes didn't know either.

The textbook complies with the Federal State Educational Standard for Basic General Education. A large number of colorful illustrations, a variety of questions and tasks, as well as additional information and interesting facts contribute to the effective assimilation of educational material.

For a long time, the scientist thought about the problem, and one day, in thought, sinking into a bath filled with water, he noticed that part of the water splashed over the edge. Contemporaries say that it was at this moment that Archimedes shouted: “Eureka!”, which in Greek means “Found!” and, without even getting dressed, ran to the royal palace.

It took the researcher a couple more days to invent a device with which he could measure the volume of water that poured out when the crown sank. This device, later called the bucket of Archimedes, can be seen on

Then, using experiments with gold and silver ingots, prove that the volume of the liquid is equal to the volume of the ingot, and therefore will be equal to the volume of the crown. And the last step is to determine the density of the crown.

They say that the king was right in his suspicions, and the jeweler was dishonest. And Archimedes received all the payment that was due for the crown to the master.

The action of liquid and gas on a body immersed in them

What did Archimedes discover through his experiments?

The scientist has identified a certain force that acts in the opposite direction to the force of gravity and allows objects to float in water and air. This force is rightly called the force of Archimedes or the buoyant force.

Definition Archimedes' law: a body immersed into a liquid, loses as much in its weight as the liquid displaced by it weighs.

Formulas

On planet Earth, all objects are affected by the force of gravity. For objects on the earth's surface, the force of attraction can be calculated by the formula:

F t = m t g, (2)

where m m - body weight, and g- free fall acceleration equal to 9.8 m/s 2 .

When an object is immersed in a liquid or gas, a buoyant force or Archimedes force begins to act on it, which is calculated by the formula:

F A = m and g, (3)

where m W is the mass of the liquid displaced by the whole object or its part, which is in the liquid.

The mass of the displaced liquid, in turn, can be determined using the formula:

m f = ρ f V f, (4)

and convert accordingly formula of Archimedes' law:

F A = ρ w V and g. (5)

What is the relationship between gravity and buoyancy? It's simple:

• if the force of attraction is greater than the force of expulsion, the object will sink;
• if the forces are approximately equal, the object will float in the thickness of the liquid or gas;
• and if the force of expulsion is greater than the force of attraction, the object will float.

Despite the fact that Archimedes first discovered the buoyancy force in water, the Archimedes force is also characteristic of gases, and it was thanks to it that the first balloon was able to rise into the air, and the writer Jules Verne, inspired and delighted by this event, wrote his novel.

And now let's help the king solve his problem with the crown.

Suppose that the crown of King Hiero weighs 22 N in air and 19.75 N in water, calculate the density of the corona.

As we learned at the beginning of the article, the density of a substance is found by the formula:

ρ t = m t / V t. (1)

Looking at the formula, we understand that to solve the problem, we do not know either the mass of the corona or its volume.

From the previous physics course (§ 27 of the textbook), we remember that for a motionless body, the weight P equal to gravity F t and is calculated by the formula:

P = F t = m t g, (2)

where g- free fall acceleration and its value is equal to g= 9.8 N/kg. However, if greater accuracy in calculations is not required, the value can be rounded up to 10 N/kg

• the force with which Archimedes moved the tub;
• force that lifts up a body in a liquid or gas; (+)
• muscle strength of Archimedes;
• force with which a rigid body acts on a surface.

2. The force of Archimedes acts:

• on bodies immersed only in gas;
• on bodies immersed only in liquid;
• on bodies immersed in gas or liquid; (+)
• on bodies in weightlessness.

3. What is the free fall acceleration g?

• 9.8 m / s 3;
• 9.8 N/kg; +
• 9.8 km/h;
• 8.9 m / s 2.

4. A body is suspended from a spring. If a body is immersed in a container of liquid, what will happen to the spring?

• stretch more;
• will shrink; (+)
• Will not change;
• depends on body weight.

5. Two friends went swimming in the river. One of them, when immersed, displaces a volume of 60 dm 3, the second 40 dm 3. Which of the guys will be affected by the great force of Archimedes?

• on the one who knows how to swim better;
• on the one who displaced more water; (+)
• on the one who cannot swim;
• on the one who displaced less water.

6. The formula for pushing force is:

• F A = ρ w V and g; (+)
• F A = ρ t V and g;
• F A = ρ w V t g;
• F A = m t g.

7. If gravity is greater than the force of Archimedes, the body:

• take off;
• pop up;
• drown; (+)
• will float.

8. 4 identical steel balls were immersed in 4 different liquids: pure water, dead sea water, gasoline, olive oil. In which liquid will the buoyancy force be the least?

The density of oil is 915 kg/m 3 , the density of gasoline is 750 kg/m 3 .

• petrol; (+)
• Dead Sea water;
• olive oil;
• pure water.

9. Gravity depends:

• on the density of the liquid;
• from the displaced volume of liquid;
• from body weight; +
• from the time the body is in the liquid.

10. Two balls of equal volume float in two containers. Is the pull force the same?

• is the same, because the volume of the balls is the same;
• the pushing force is greater in a container of kerosene, because the density is less than water;
• the pushing force is greater in a container of water because its density is greater than that of kerosene. (+)

Liquids and gases, according to which, on any body immersed in a liquid (or gas), a buoyant force acts from this liquid (or gas), equal to the weight of the liquid (gas) displaced by the body and directed vertically upwards.

This law was discovered by the ancient Greek scientist Archimedes in the III century. BC e. Archimedes described his research in the treatise On Floating Bodies, which is considered one of his last scientific works.

The following are the findings from Archimedes' law.

The action of liquid and gas on a body immersed in them.

If you submerge an air-filled ball in water and release it, it will float. The same will happen with wood chips, cork and many other bodies. What force makes them float?

A body immersed in water is subjected to water pressure from all sides (Fig. a). At each point of the body, these forces are directed perpendicular to its surface. If all these forces were the same, the body would experience only all-round compression. But at different depths, the hydrostatic pressure is different: it increases with increasing depth. Therefore, the pressure forces applied to the lower parts of the body turn out to be greater than the pressure forces acting on the body from above.

If we replace all pressure forces applied to a body immersed in water with one (resulting or resultant) force that has the same effect on the body as all these individual forces together, then the resulting force will be directed upwards. This is what makes the body float. This force is called the buoyant force, or Archimedean force (after Archimedes, who first pointed out its existence and established what it depends on). On the image b it is labeled as F A.

The Archimedean (buoyant) force acts on the body not only in water, but also in any other liquid, since in any liquid there is hydrostatic pressure, which is different at different depths. This force also acts in gases, due to which balloons and airships fly.

Due to the buoyancy force, the weight of any body in water (or in any other liquid) is less than in air, and less in air than in airless space. It is easy to verify this by weighing the weight with the help of a training spring dynamometer, first in the air, and then lowering it into a vessel with water.

Weight reduction also occurs when a body is transferred from vacuum to air (or some other gas).

If the weight of a body in a vacuum (for example, in a vessel from which air is pumped out) is equal to P0, then its weight in air is:

,

where F' A is the Archimedean force acting on a given body in air. For most bodies, this force is negligible and can be neglected, i.e., we can assume that P air =P 0 =mg.

The weight of the body in liquid decreases much more than in air. If the weight of the body in the air P air =P 0, then the weight of the body in the fluid is P liquid \u003d P 0 - F A. Here F A is the Archimedean force acting in the fluid. Hence it follows that

Therefore, in order to find the Archimedean force acting on a body in any liquid, this body must be weighed in air and in the liquid. The difference between the obtained values ​​will be the Archimedean (buoyant) force.

In other words, taking into account formula (1.32), we can say:

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

The Archimedean force can also be determined theoretically. To do this, suppose that a body immersed in a fluid consists of the same fluid in which it is immersed. We have the right to assume this, since the pressure forces acting on a body immersed in a liquid do not depend on the substance from which it is made. Then the Archimedean force applied to such a body F A will be balanced by the downward force of gravity mandg(where m f is the mass of liquid in the volume of a given body):

But the force of gravity is equal to the weight of the displaced fluid R f. In this way.

Given that the mass of a liquid is equal to the product of its density ρ w on volume, formula (1.33) can be written as:

where Vand is the volume of the displaced fluid. This volume is equal to the volume of that part of the body that is immersed in the liquid. If the body is completely immersed in the liquid, then it coincides with the volume V of the whole body; if the body is partially immersed in the liquid, then the volume Vand volume of displaced fluid V bodies (Fig. 1.39).

Formula (1.33) is also valid for the Archimedean force acting in a gas. Only in this case, it is necessary to substitute the density of the gas and the volume of the displaced gas, and not the liquid, into it.

In view of the foregoing, Archimedes' law can be formulated as follows:

Any body immersed in a liquid (or gas) at rest is affected by a buoyant force from this liquid (or gas), equal to the product of the density of the liquid (or gas), the free fall acceleration and the volume of that part of the body that is immersed in the liquid ( or gas).