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Formula for measuring body weight. Formula for measuring body weight What is the weight of a body length

Definition 1

Weight represents the force of influence of the body on the support (suspension, or other type of fastening), preventing the fall, and arising in the field of gravity. The SI unit of weight is the newton.

The concept of body weight

The concept of "weight" as such in physics is not considered necessary. So, more is said about the mass or the strength of the body. A more meaningful value is the force of impact on the support, the knowledge of which can help, for example, in assessing the ability of a structure to hold the body under study under given conditions.

Weight can be measured using spring scales, which also serve to indirectly measure mass with their appropriate graduation. At the same time, balances do not need this, since in such a situation the masses to be compared are those affected by equal acceleration of free fall or the sum of accelerations in non-inertial frames of reference.

When weighing with technical spring balances, variations in gravitational acceleration are usually not taken into account, since their influence is often less than what is required in practice in terms of weighing accuracy. To some extent, the measurement results may reflect the Archimedes force, provided that bodies of different densities are weighed on a balance scale and their comparative indicators.

Weight and mass in physics represent different concepts. Thus, weight is considered a vector quantity with which the body will directly act on a horizontal support or vertical suspension. Mass at the same time represents a scalar quantity, a measure of the body's inertia (inertial mass) or the charge of the gravitational field (gravitational mass). Such quantities will also have different units of measurement (in SI, mass is indicated in kilograms, and weight in newtons).

Situations with zero weight and also non-zero mass are also possible (when we are talking about the same body, for example, under weightlessness, the weight of each body will be equal to zero, but the mass will be different for everyone).

Important formulas for calculating body weight

The weight of a body ($P$), which is at rest in an inertial frame of reference, is equivalent to the force of gravity acting on it, and is proportional to the mass $m$, as well as the free fall acceleration $g$ at a given point.

Remark 1

The gravitational acceleration will depend on the height above the earth's surface, as well as on the geographical coordinates of the measurement point.

The result of the daily rotation of the Earth is a latitudinal decrease in weight. So, at the equator, the weight will be less in comparison with the poles.

Another factor influencing the value of $g$ can be considered gravitational anomalies, which are due to the peculiarities of the structure of the earth's surface. When the body is located near another planet (not the Earth), the acceleration of free fall is often determined by the mass and size of this planet.

The state of weightlessness (weightlessness) will occur when the body is far from the attracting object or is in free fall, that is, in a situation where

$(g - w) = 0$.

A body of mass $m$, whose weight is analyzed, may be subject to the application of certain additional forces, indirectly due to the presence of a gravitational field, in particular, the Archimedes force and the friction force.

Difference between body weight and gravity

Remark 2

Gravity and weight are two different concepts involved directly in the gravitational field theory of physics. These two completely different concepts are often misunderstood and used in the wrong context.

This situation is aggravated by the fact that in the standard understanding of the concept of mass (meaning the property of matter) and weight will also be perceived as identical. It is for this reason that the correct understanding of gravity and weight is considered very important for the scientific community.

Often, these two almost similar concepts are used interchangeably. The force that is directed at an object from the Earth or another planet in our Universe (in a broader sense - any astronomical body) will represent the force of gravity:

The force with which the body has a direct effect on the support or vertical suspension and will be considered the weight of the body, denoted as $W$ and representing a vector directed quantity.

Atoms (molecules) of the body will be repelled from the base particles. The result of this process is:

implementation of partial deformation not only of the support, but also of the object;
the emergence of elastic forces;
change in certain situations (to a small extent) of the shape of the body and support, which will occur at the macro level;
the emergence of a reaction force of the support with the appearance of an elastic force parallel on the surface of the body, which becomes a response to the support (this will represent the weight).

Definition 1

The concept of body weight

Important formulas for calculating body weight

Remark 1

The gravitational acceleration will depend on the height above the earth's surface, as well as on the geographical coordinates of the measurement point.

The result of the daily rotation of the Earth is a latitudinal decrease in weight. So, at the equator, the weight will be less in comparison with the poles.

The state of weightlessness (weightlessness) will occur when the body is far from the attracting object or is in free fall, that is, in a situation where

$(g - w) = 0$.

Difference between body weight and gravity

Remark 2

The force with which the body has a direct effect on the support or vertical suspension and will be considered the weight of the body, denoted as $W$ and representing a vector directed quantity.

Atoms (molecules) of the body will be repelled from the base particles. The result of this process is:

implementation of partial deformation not only of the support, but also of the object;
the emergence of elastic forces;
change in certain situations (to a small extent) of the shape of the body and support, which will occur at the macro level;
the emergence of a reaction force of the support with the appearance of an elastic force parallel on the surface of the body, which becomes a response to the support (this will represent the weight).

Quite a lot of mistakes and non-random reservations of students are connected with the strength of the weight. The phrase “power of weight” itself is not very familiar, because. we (teachers, authors of textbooks and problem books, teaching aids and reference literature) are more accustomed to speaking and writing “body weight”. Thus, the phrase itself leads us away from the concept that weight is strength, and leads to the fact that body weight is confused with body weight (we often hear in a store when they are asked to weigh a few kilograms of a product). The second common mistake students make is that they confuse the force of weight with the force of gravity. Let's try to deal with the force of weight at the level of a school textbook.

To begin with, let's look at the reference literature and try to understand the point of view of the authors on this issue. Yavorsky B.M., Detlaf A.A. (1) in a handbook for engineers and students, the weight of a body is the force with which this body acts due to gravity towards the Earth on a support (or suspension) that keeps the body from free fall. If the body and the support are stationary relative to the Earth, then the weight of the body is equal to its gravity. Let's ask some naive questions to the definition:

1. What reporting system are we talking about?

2. Is there one support (or suspension) or several (supports and suspensions)?

3. If the body gravitates not to the Earth, but, for example, to the Sun, will it have weight?

4. If a body in a spaceship moving with acceleration "almost" does not gravitate to anything in observable space, will it have weight?

5. How is the support located relative to the horizon, is the suspension vertical for the case of equality of body weight and gravity?

6. If the body moves uniformly and rectilinearly together with the support relative to the Earth, then the weight of the body is equal to its gravity?

In the reference guide to physics for applicants to universities and self-education, Yavorsky B.M. and Selezneva Yu.A. (2) provide an explanation on the last naive question, leaving the former unaddressed.

Koshkin N.I. and Shirkevich M.G. (3) it is proposed to consider the weight of the body as a vector physical quantity, which can be found by the formula:

The examples below will show that this formula works in cases where no other forces act on the body.

Kuchling H. (4) does not introduce the concept of weight as such at all, identifying it practically with the force of gravity; in the drawings, the force of weight is applied to the body, and not to the support.

In the popular "Physics Tutor" Kasatkina I.L. (5) body weight is defined as the force with which a body acts on a support or suspension due to attraction to the planet. In the following explanations and examples given by the author, answers are given only to the 3rd and 6th of the naive questions.

In most textbooks on physics, definitions of weight are given to some extent similar to the definitions of the authors (1), (2), (5). When studying physics in the 7th and 9th educational grades, perhaps this is justified. In the 10th profile classes with such a definition, when solving a whole class of problems, one cannot avoid various kinds of naive questions (in general, one should not at all strive to avoid any questions).

Authors Kamenetsky S.E., Orekhov V.P. in (6), delimiting and explaining the concepts of gravity and body weight, they write that body weight is a force that acts on a support or suspension. And that's it. You don't have to read between the lines. True, I still want to ask, how many supports and suspensions, and can the body have both support and suspension at once?

And, finally, let's look at the definition of body weight, which is given by Kasyanov V.A. (7) in a 10th grade physics textbook: “body weight is the total body elasticity force acting in the presence of gravity on all connections (supports, suspensions)”. If at the same time we remember that the force of gravity is equal to the resultant of two forces: the force of gravitational attraction to the planet and the centrifugal force of inertia, provided that this planet rotates around its axis, or some other force of inertia associated with the accelerated movement of this planet, then one could agree with this definition. Since, in this case, no one bothers us to imagine a situation where one of the components of gravity is negligible, for example, the case of a spaceship in deep space. And even with these reservations, it is tempting to remove the mandatory presence of gravity from the definition, because situations are possible when there are other forces of inertia that are not related to the movement of the planet or Coulomb forces of interaction with other bodies, for example. Or agree with the introduction of some "equivalent" gravity in non-inertial reference systems and define the force of weight for the case when there is no interaction of the body with other bodies, except for the body that creates gravitational attraction, supports and suspensions.

And yet, let's decide when the weight of the body is equal to the force of gravity in inertial frames of reference?

Suppose we have one support or one suspension. Is the condition sufficient that the support or suspension is motionless relative to the Earth (we consider the Earth to be an inertial frame of reference), or move uniformly and rectilinearly? Take a fixed support, located at an angle to the horizon. If the support is smooth, then the body slides along the inclined plane, i.e. is not resting on a support and is not in free fall. And if the support is so rough that the body is at rest, then either the inclined plane is not a support, or the weight of the body is not equal to the force of gravity (you can, of course, go further and question that the weight of the body is not equal in absolute value and is not opposite in direction support reaction force, and then there will be nothing to talk about at all). If we still consider the inclined plane as a support, and the sentence in brackets as irony, then, solving the equation for Newton's second law, which for this case will also be the equilibrium condition for the body on the inclined plane, written in projections onto the Y axis, we will obtain an expression for weight other than gravity:

So, in this case, it is not enough to say that the weight of the body is equal to the force of gravity, when the body and the support are motionless relative to the Earth.

Let us give an example with a suspension fixed relative to the Earth and a body on it. A positively charged metal ball on a thread is placed in a uniform electric field so that the thread makes some angle with the vertical. Let's find the weight of the ball from the condition that the vector sum of all forces is equal to zero for a body at rest.

As you can see, in the above cases, the weight of the body is not equal to the force of gravity when the condition of immobility of the support, suspension and body relative to the Earth is met. The features of the above cases are the existence of the friction force and the Coulomb force, respectively, the presence of which actually leads to the fact that the bodies are kept from moving. For vertical suspension and horizontal support, additional forces are not needed to keep the body from moving. Thus, to the condition of immobility of the support, suspension and body relative to the Earth, we could add that the support is horizontal and the suspension is vertical.

But would this addition solve our question? Indeed, in systems with a vertical suspension and a horizontal support, forces can act that reduce or increase the weight of the body. These can be the force of Archimedes, for example, or the force of Coulomb, directed vertically. To summarize for one support or one suspension: the weight of the body is equal to the force of gravity, when the body and the support (or suspension) are at rest (or move uniformly and rectilinearly) relative to the Earth, and only the reaction force of the support (or the elastic force of the suspension) and the force act on the body gravity. The absence of other forces, in turn, implies that the support is horizontal, the suspension is vertical.

Let us consider the cases when a body with several supports and/or suspensions is at rest (or moves uniformly and rectilinearly with them relative to the Earth) and no other forces act on it, except for the reaction forces of the support, the elastic forces of the suspensions, and attraction to the Earth. Using the definition of the weight force Kasyanov V.A. (7), we find the total force of elasticity of the body bonds in the first and second cases presented in the figures. The geometric sum of the forces of elastic bonds F, equal in modulus to the weight of the body, based on the equilibrium condition, is indeed equal to gravity and opposite to it in direction, and the angles of inclination of the planes to the horizon and the angles of deviation of the suspensions from the vertical do not affect the final result.

Let us consider an example (figure below), when in a system that is motionless relative to the Earth, a body has a support and a suspension, and no other forces act in the system, except for the forces of elastic bonds. The result is similar to the above. The weight of the body is equal to the force of gravity.

So, if the body is on several supports and (or) suspensions, and rests together with them (or moves uniformly and rectilinearly) relative to the Earth, in the absence of other forces, except for the force of gravity and the forces of elastic bonds, its weight is equal to the force of gravity. At the same time, the location of supports and suspensions in space and their number do not affect the final result.

Consider examples of finding body weight in non-inertial frames of reference.

Example 1 Find the weight of a body of mass m moving in a spaceship with acceleration a in "empty" space (so far from other massive bodies that their gravity can be neglected).

In this case, two forces act on the body: the force of inertia and the reaction force of the support. If the modulus of acceleration is equal to the acceleration of free fall on the Earth, then the weight of the body will be equal to the force of gravity on the Earth, and the astronauts will perceive the nose of the ship as the ceiling, and the tail as the floor.

The artificial gravity created in this way for the astronauts inside the ship will not differ in any way from the “real” earth.

In this example, due to its smallness, we neglect the gravitational component of gravity. Then the force of inertia on the spacecraft will be equal to the force of gravity. In view of this, we can agree that the cause of the body weight in this case is gravity.

Let's go back to Earth.

Example 2

With respect to the ground with acceleration a a trolley is moving, on which a body is fixed on a thread of mass m, deviated by an angle from the vertical. Find the weight of the body, neglect the air resistance.

A task with one suspension, therefore, the weight is equal in modulus to the elastic force of the thread.

Thus, you can use any formula to calculate the elastic force, and, therefore, the weight of the body (if the air resistance force is large enough, then it will need to be taken into account as a summand to the inertial force).

Let's work with the formula

Therefore, by introducing the "equivalent" force of gravity, we can assert that in this case the weight of the body is equal to the "equivalent" force of gravity. And finally, we can give three formulas for its calculation:

Example 3

Find the weight of a race car driver with mass m in a moving with acceleration a car.

At high accelerations, the reaction force of the seat back support becomes significant, and we will take it into account in this example. The total elastic force of the bonds will be equal to the geometric sum of both reaction forces of the support, which in turn is equal in absolute value and opposite in direction to the vector sum of the forces of inertia and gravity. For this problem, we find the module of the weight force by the formulas:

The effective free fall acceleration is found as in the previous problem.

Example 4

A ball on a thread of mass m is fixed on a platform rotating at a constant angular velocity ω at a distance r from its center. Find the weight of the ball.

Finding the body weight in non-inertial frames of reference in the given examples shows how well the formula for the body weight proposed by the authors in (3) works. Let's complicate the situation a bit in example 4. Let's assume that the ball is electrically charged, and the platform, together with its contents, is in a uniform vertical electric field. What is the weight of the ball? Depending on the direction of the Coulomb force, the weight of the body will decrease or increase:

It so happened that the question of weight naturally reduced to the question of gravity. If we define gravity as the resultant of the forces of gravitational attraction to a planet (or to any other massive object) and inertia, keeping in mind the principle of equivalence, leaving in the fog the origin of the force of inertia itself, then both components of gravity, or one of them, at least cause body weight. If there are other interactions in the system along with the force of gravitational attraction, the force of inertia and the forces of elastic bonds, then they can increase or decrease the weight of the body, lead to a state when the weight of the body becomes equal to zero. And these other interactions can cause weight gain in some cases. Let's charge a ball on a thin non-conducting thread in a spaceship moving uniformly and rectilinearly in a distant "empty" space (we will neglect the forces of gravity because of their smallness). Let's put the ball in the electric field, the thread will be stretched, the weight will appear.

Summarizing the above, we conclude that the weight of the body is equal to the force of gravity (or the equivalent force of gravity) in any system where no other forces act on the body, except for the forces of gravity, inertia and elasticity of bonds. Gravity, or "equivalent" gravity, is most often the cause of weight force. The force of weight and the force of gravity have a different nature and are applied to different bodies.

Bibliography.

1. Yavorsky B.M., Detlaf A.A. Handbook of physics for engineers and university students, M., Nauka, 1974, 944p.

2. Yavorsky B.M., Selezneva Yu.A. Physics Reference Guide for

entering universities and self-education., M., Nauka, 1984, 383p.

3. Koshkin N.I., Shirkevich M.G. Handbook of elementary physics., M., Nauka, 1980, 208s.

4. Kuhling H. Handbook of Physics., M., Mir, 1983, 520p.

5. Kasatkina I.L. Physics tutor. Theory. Mechanics. Molecular physics. Thermodynamics. Electromagnetism. Rostov-on-Don, Phoenix, 2003, 608s.

6. Kamenetsky S.E., Orekhov V.P. Methods for solving problems in physics in high school., M., Education, 1987, 336s.

7. Kasyanov V.A. Physics. Grade 10., M., Bustard, 2002, 416s.

In this paragraph, we will remind you about gravity, centripetal acceleration and body weight.

Every body on the planet is affected by the Earth's gravity. The force with which the Earth attracts each body is determined by the formula

The point of application is at the center of gravity of the body. The force of gravity always pointing vertically down.

The force with which a body is attracted to the Earth under the influence of the Earth's gravitational field is called gravity. According to the law of universal gravitation, on the surface of the Earth (or near this surface), a body of mass m is affected by the force of gravity

F t \u003d GMm / R 2

where M is the mass of the Earth; R is the radius of the Earth.
If only gravity acts on the body, and all other forces are mutually balanced, the body is in free fall. According to Newton's second law and the formula F t \u003d GMm / R 2 free fall acceleration modulus g is found by the formula

g=F t /m=GM/R 2 .

From formula (2.29) it follows that the free fall acceleration does not depend on the mass m of the falling body, i.e. for all bodies in a given place on the Earth it is the same. From formula (2.29) it follows that Fт = mg. In vector form

F t \u003d mg

In § 5 it was noted that since the Earth is not a sphere, but an ellipsoid of revolution, its polar radius is less than the equatorial one. From the formula F t \u003d GMm / R 2 it can be seen that for this reason the force of gravity and the acceleration of free fall caused by it is greater at the pole than at the equator.

The force of gravity acts on all bodies in the Earth's gravitational field, but not all bodies fall to the Earth. This is due to the fact that the movement of many bodies is hindered by other bodies, such as supports, suspension threads, etc. Bodies that restrict the movement of other bodies are called connections. Under the action of gravity, the bonds are deformed and the reaction force of the deformed bond, according to Newton's third law, balances the force of gravity.

The acceleration of free fall is affected by the rotation of the Earth. This influence is explained as follows. The frames of reference associated with the surface of the Earth (except for the two associated with the poles of the Earth) are not, strictly speaking, inertial frames of reference - the Earth rotates around its axis, and with it move along circles with centripetal acceleration and such frames of reference. This non-inertiality of reference systems is manifested, in particular, in the fact that the value of the acceleration of free fall turns out to be different in different places on the Earth and depends on the geographical latitude of the place where the reference frame associated with the Earth is located, relative to which the acceleration of gravity is determined.

Measurements carried out at different latitudes showed that the numerical values of the gravitational acceleration differ little from each other. Therefore, with not very accurate calculations, one can neglect the non-inertiality of reference systems associated with the Earth's surface, as well as the difference in the shape of the Earth from spherical, and assume that the acceleration of free fall in any place on the Earth is the same and equal to 9.8 m / s 2.

From the law of universal gravitation it follows that the force of gravity and the acceleration of free fall caused by it decrease with increasing distance from the Earth. At a height h from the Earth's surface, the gravitational acceleration module is determined by the formula

g=GM/(R+h) 2.

It has been established that at a height of 300 km above the Earth's surface, the free fall acceleration is less than at the Earth's surface by 1 m/s2.
Consequently, near the Earth (up to heights of several kilometers), the force of gravity practically does not change, and therefore the free fall of bodies near the Earth is a uniformly accelerated motion.

Body weight. Weightlessness and overload

The force in which, due to attraction to the Earth, the body acts on its support or suspension, is called body weight. Unlike gravity, which is a gravitational force applied to a body, weight is an elastic force applied to a support or suspension (i.e., to a connection).

Observations show that the weight of the body P, determined on a spring balance, is equal to the force of gravity F t acting on the body only if the balance with the body relative to the Earth is at rest or moving uniformly and rectilinearly; In this case

P \u003d F t \u003d mg.

If the body is moving with acceleration, then its weight depends on the value of this acceleration and on its direction relative to the direction of free fall acceleration.

When a body is suspended on a spring balance, two forces act on it: the force of gravity F t =mg and the elastic force F yp of the spring. If at the same time the body moves vertically up or down relative to the direction of free fall acceleration, then the vector sum of the forces F t and F yn gives the resultant, causing the acceleration of the body, i.e.

F t + F pack \u003d ma.

According to the above definition of the concept of "weight", we can write that P=-F yp. From the formula: F t + F pack \u003d ma. taking into account the fact that F t =mg, it follows that mg-ma=-F yp . Therefore, P \u003d m (g-a).

The forces F t and F yn are directed along one vertical straight line. Therefore, if the acceleration of the body a is directed downward (i.e., it coincides in direction with the acceleration of free fall g), then modulo

P=m(g-a)

If the acceleration of the body is directed upwards (i.e., opposite to the direction of free fall acceleration), then

P \u003d m \u003d m (g + a).

Consequently, the weight of a body whose acceleration coincides in direction with the acceleration of free fall is less than the weight of a body at rest, and the weight of a body whose acceleration is opposite to the direction of acceleration of free fall is greater than the weight of a body at rest. The increase in body weight caused by its accelerated movement is called overload.

In free fall a=g. From the formula: P=m(g-a)

it follows that in this case P=0, i.e., there is no weight. Therefore, if bodies move only under the influence of gravity (i.e., fall freely), they are in a state weightlessness. A characteristic feature of this state is the absence of deformations and internal stresses in freely falling bodies, which are caused in resting bodies by gravity. The reason for the weightlessness of bodies is that the force of gravity imparts the same accelerations to a freely falling body and its support (or suspension).

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Attention! The slide preview is for informational purposes only and may not represent the full extent of the presentation. If you are interested in this work, please download the full version.

This presentation is intended to help students in grades 9-10 in preparing the topic "Body Weight".

Presentation goals:

Repeat and deepen the concepts: "gravity"; "body weight"; "weightlessness".
Emphasize to students that gravity and body weight are different forces.
To teach students to determine the weight of a body moving vertically.

In everyday life, body weight is determined by weighing. From the 7th grade physics course, it is known that the force of gravity is directly proportional to the mass of the body. Therefore, the weight of a body is often identified with its mass or gravity. From the point of view of physics, this is a gross mistake. The weight of a body is a force, but gravity and the weight of a body are different forces.

The force of gravity is a special case of the manifestation of the forces of universal gravitation. Therefore, it is appropriate to recall the law of universal gravitation, as well as the fact that the forces of gravitational attraction appear when the bodies or one of the bodies have huge masses (slide 2).

When applying the law of universal gravitation for terrestrial conditions (slide 3), the planet can be considered as a homogeneous ball, and small bodies near its surface as point masses. The radius of the earth is 6400 km. The mass of the Earth is 6∙10 24 kg.

= ,
where g is the free fall acceleration.

Near the Earth's surface g = 9.8 m/s 2 ≈ 10 m/s 2.

Body weight - the force with which this body acts on a horizontal support or stretches the suspension.

Fig.1

On fig. 1 shows a body on a support. The reaction force of the support N (F control) is applied not to the support, but to the body located on it. The modulus of the reaction force of the support is equal to the modulus of the weight according to Newton's third law. The weight of the body is a special case of the manifestation of the force of elasticity. The most important feature of the weight is that its value depends on the acceleration with which the support or suspension is moving. Weight is equal to gravity only for a body at rest (or a body moving at a constant speed). If the body moves with acceleration, then the weight can be greater or less than the force of gravity, and even equal to zero.

In the presentation, using the example of solving problem 1, various cases of determining the weight of a load with a mass of 500 g suspended from a dynamometer spring are considered, depending on the nature of the movement:

a) the load is lifted up with an acceleration of 2 m / s 2;
b) the load is lowered down with an acceleration of 2 m / s 2;
c) the load is evenly lifted up;
d) the load falls freely.

Tasks for calculating body weight are included in the "Dynamics" section. The solution of problems on dynamics is based on the use of Newton's laws, followed by projection onto the selected coordinate axes. This determines the sequence of actions.

A drawing is made showing the forces acting on the body(s) and the direction of acceleration. If the direction of acceleration is unknown, it is chosen arbitrarily, and the solution of the problem gives an answer about the correctness of the choice.
Write down Newton's second law in vector form.
Select axes. It is usually convenient to direct one of the axes along the direction of the body's acceleration, the other one - perpendicular to the acceleration. The choice of axes is determined by considerations of convenience: so that the expressions for the projections of Newton's laws would have the simplest form.
The vector equations obtained in projections on the axis are supplemented with relations arising from the text of the problem conditions. For example, the equations of kinematic connection, definitions of physical quantities, Newton's third law.
Using the resulting system of equations, they try to answer the question of the problem.

Setting up animation in a presentation allows you to focus on the sequence of actions when solving problems. This is important, because the skills acquired while solving problems for calculating body weight will be useful to students when studying other topics and sections of physics.

Solution of problem 1.

1a. The body moves with an acceleration of 2 m / s 2 up (slide 7).

Fig.2

1b. The body moves with acceleration downwards (slide 8). We direct the OY axis down, then the projections of gravity and elasticity in equation (2) change signs, and it looks like:

(2) mg – F control = ma.

Therefore, P \u003d m (g-a) \u003d 0.5 kg ∙ (10 m / s 2 - 2 m / s 2) \u003d 4 N.

1c. With uniform motion (slide 9), equation (2) has the form:

(2) mg - F control = 0, since there is no acceleration.

Therefore, P \u003d mg \u003d 5 N.

1g In free fall = (slide 10). We use the result of solving Problem 1b:

P \u003d m (g - a) \u003d 0.5 kg (10 m / s 2 - 10 m / s 2) \u003d 0 H.

The state in which the weight of the body is zero is called the state of weightlessness.

Only the force of gravity acts on the body.

Speaking of weightlessness, it should be noted that astronauts experience a prolonged state of weightlessness during flight with the spacecraft engines turned off.

ship, and to experience a short-term state of weightlessness, just jump up. A running person at the moment when his feet do not touch the ground is also in a state of weightlessness.

The presentation can be used in the lesson when explaining the topic "Body Weight". Depending on the level of preparation of the class, students may not be offered all the slides with solutions to problem 1. For example, in classes with increased motivation to study physics, it is enough to explain how to calculate the weight of a body moving with acceleration upwards (task 1a), and the rest of the tasks (b , c, d) provide for an independent solution with subsequent verification. The conclusions obtained as a result of solving problem 1, students should try to draw on their own.

Conclusions (slide 11).

Body weight and gravity are different forces. They have a different nature. These forces are applied to different bodies: gravity - to the body; body weight - to the support (suspension).
The weight of the body coincides with the force of gravity only when the body is motionless or moves uniformly and rectilinearly, and other forces, except for the force of gravity and the support reaction (suspension tension), do not act on it.
The weight of the body is greater than the force of gravity (P> mg), if the acceleration of the body is directed in the direction opposite to the direction of gravity.
Body weight is less than gravity (P< mg), если ускорение тела совпадает по направлению с силой тяжести.
The state in which the weight of the body is zero is called the state of weightlessness. A body is in a state of weightlessness when it moves with free fall acceleration, that is, when only gravity acts on it.

Tasks 2 and 3 (slide 12) can be offered to students as homework.

The Body Weight presentation can be used for distance learning. In this case, it is recommended:

when viewing the presentation, write down the solution to problem 1 in a notebook;
independently solve problems 2, 3, using the sequence of actions proposed in the presentation.

The presentation on the topic “Body weight” allows you to show the theory of solving problems on dynamics in an interesting, accessible interpretation. The presentation activates the cognitive activity of students and allows you to form the right approach to solving physical problems.

Literature:

Grinchenko B.I. Physics 10-11. Theory of problem solving. For high school students and college students. - Velikiye Luki: Velikie Luki City Printing House, 2005.
Gendenstein L.E. Physics. Grade 10. At 2 p.m. H 1./L.E. Gendenstein, Yu.I. Dick. – M.: Mnemosyne, 2009.
Gendenstein L.E. Physics. Grade 10. At 2 o'clock. H 2. Task book./L.E. Gendenstein, L.A. Kirik, I.M. Gelgafgat, I.Yu. Nenashev.- M.: Mnemosyne, 2009.

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