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Gravity and the force of universal gravitation. Gravity: formula, definition

We are all accustomed in life to using the word strength in comparative characteristics speaking men stronger than women, a tractor is stronger than a car, a lion is stronger than an antelope.

Force in physics is defined as a measure of the change in the speed of a body that occurs when bodies interact. If force is a measure and we can compare the application of different forces, then it is a physical quantity that can be measured. In what units is force measured?

Force units

In honor of the English physicist Isaac Newton, who did enormous research into the nature of existence and use various types force, the unit of force in physics is 1 newton (1 N). What is a force of 1 N? In physics, units of measurement are not chosen just like that, but special coordination is made with those units that are already accepted.

We know from experience and experiments that if a body is at rest and a force acts on it, then the body, under the influence of this force, changes its speed. Accordingly, to measure force, a unit was chosen that would characterize the change in body speed. And don’t forget that there is also body mass, since it is known that with the same force the impact on various items will be different. We can throw a ball far, but a cobblestone will fly a much shorter distance. That is, taking into account all the factors, we come to the determination that a force of 1 N will be applied to a body if a body weighing 1 kg under the influence of this force changes its speed by 1 m/s in 1 second.

Unit of gravity

We are also interested in the unit of gravity. Since we know that the Earth attracts all bodies on its surface, it means that there is an attractive force and it can be measured. And again, we know that the force of gravity depends on the mass of the body. The greater the mass of a body, the more strongly the Earth attracts it. It has been experimentally established that The force of gravity acting on a body weighing 102 grams is 1 N. And 102 grams is approximately one tenth of a kilogram. To be more precise, if 1 kg is divided into 9.8 parts, then we will get approximately 102 grams.

If a force of 1 N acts on a body weighing 102 grams, then a force of 9.8 N acts on a body weighing 1 kg. Acceleration free fall denoted by the letter g. And g is equal to 9.8 N/kg. This is the force that acts on a body weighing 1 kg, accelerating it by 1 m/s every second. It turns out that a body falling from high altitude, during the flight it gains very high speed. Why then do snowflakes and raindrops fall quite calmly? They have very little mass, and the earth pulls them towards itself very weakly. And the air resistance for them is quite high, so they fly towards the Earth at a not very high, rather uniform speed. But meteorites, for example, when approaching the Earth, gain a very high speed and upon landing, a decent explosion is formed, which depends on the size and mass of the meteorite, respectively.

« Physics - 10th grade"

Why does the Moon move around the Earth?
What happens if the moon stops?
Why do the planets revolve around the Sun?

Chapter 1 discussed in detail that the globe imparts to all bodies near the surface of the Earth the same acceleration - the acceleration of gravity. But if the globe imparts acceleration to the body, then, according to Newton’s second law, it acts on the body with some force. The force with which the Earth acts on a body is called gravity. First, we will find this force, and then we will consider the force of universal gravity.

Acceleration in absolute value is determined from Newton's second law:

In general, it depends on the force acting on the body and its mass. Since the acceleration of gravity does not depend on mass, it is clear that the force of gravity must be proportional to mass:

The physical quantity is the acceleration of gravity, it is constant for all bodies.

Based on the formula F = mg, you can specify a simple and practically convenient method for measuring the mass of bodies by comparing the mass of a given body with a standard unit of mass. The ratio of the masses of two bodies is equal to the ratio of the forces of gravity acting on the bodies:

This means that the masses of bodies are the same if the forces of gravity acting on them are the same.

This is the basis for determining masses by weighing on spring or lever scales. By ensuring that the force of pressure of a body on a pan of scales, equal to the force of gravity applied to the body, is balanced by the force of pressure of weights on another pan of scales, equal to the force of gravity applied to the weights, we thereby determine the mass of the body.

The force of gravity acting on given body near the Earth, can be considered constant only at a certain latitude near the Earth’s surface. If the body is lifted or moved to a place with a different latitude, then the acceleration of gravity, and therefore the force of gravity, will change.

The force of universal gravity.

Newton was the first to strictly prove that the cause of a stone falling to the Earth, the movement of the Moon around the Earth and the planets around the Sun are the same. This force of universal gravity, acting between any bodies in the Universe.

Newton came to the conclusion that if not for air resistance, then the trajectory of a stone thrown from high mountain(Fig. 3.1) with a certain speed, could become such that it would never reach the surface of the Earth at all, but would move around it in the same way as planets describe their orbits in celestial space.

Newton found this reason and was able to accurately express it in the form of one formula - the law of universal gravitation.

Since the force of universal gravity imparts the same acceleration to all bodies regardless of their mass, it must be proportional to the mass of the body on which it acts:

“Gravity exists for all bodies in general and is proportional to the mass of each of them... all planets gravitate towards each other...” I. Newton

But since, for example, the Earth acts on the Moon with a force proportional to the mass of the Moon, then the Moon, according to Newton’s third law, must act on the Earth with the same force. Moreover, this force must be proportional to the mass of the Earth. If the force of gravity is truly universal, then from the side of a given body a force must act on any other body proportional to the mass of this other body. Consequently, the force of universal gravitation must be proportional to the product of the masses of interacting bodies. From this follows the formulation of the law of universal gravitation.

Law of universal gravitation:

The force of mutual attraction between two bodies is directly proportional to the product of the masses of these bodies and inversely proportional to the square of the distance between them:

The proportionality factor G is called gravitational constant.

The gravitational constant is numerically equal to the force of attraction between two material points weighing 1 kg each, if the distance between them is 1 m. Indeed, with masses m 1 = m 2 = 1 kg and distance r = 1 m, we obtain G = F (numerically).

It must be borne in mind that the law of universal gravitation (3.4) as a universal law is valid for material points. In this case, the forces of gravitational interaction are directed along the line connecting these points (Fig. 3.2, a).

It can be shown that homogeneous bodies shaped like a ball (even if they cannot be considered material points, Fig. 3.2, b) also interact with the force determined by formula (3.4). In this case, r is the distance between the centers of the balls. The forces of mutual attraction lie on a straight line passing through the centers of the balls. Such forces are called central. The bodies that we usually consider falling to Earth have dimensions much smaller than the Earth's radius (R ≈ 6400 km).

Such bodies can, regardless of their shape, be considered as material points and determine the force of their attraction to the Earth using the law (3.4), keeping in mind that r is the distance from a given body to the center of the Earth.

A stone thrown to the Earth will deviate under the influence of gravity from a straight path and, having described a curved trajectory, will finally fall to the Earth. If you throw it at a higher speed, it will fall further." I. Newton

Determination of the gravitational constant.

Now let's find out how to find the gravitational constant. First of all, note that G has a specific name. This is due to the fact that the units (and, accordingly, the names) of all quantities included in the law of universal gravitation have already been established earlier. The law of gravity gives new connection between known quantities with certain names of units. That is why the coefficient turns out to be a named quantity. Using the formula of the law of universal gravitation, it is easy to find the name of the unit of gravitational constant in SI: N m 2 / kg 2 = m 3 / (kg s 2).

For quantification G it is necessary to independently determine all the quantities included in the law of universal gravitation: both masses, force and distance between bodies.

The difficulty is that the gravitational forces between bodies of small masses are extremely small. It is for this reason that we do not notice the attraction of our body to surrounding objects and the mutual attraction of objects to each other, although gravitational forces are the most universal of all forces in nature. Two people with masses of 60 kg at a distance of 1 m from each other are attracted with a force of only about 10 -9 N. Therefore, to measure the gravitational constant, fairly subtle experiments are needed.

The gravitational constant was first measured by the English physicist G. Cavendish in 1798 using an instrument called a torsion balance. The diagram of the torsion balance is shown in Figure 3.3. A light rocker with two identical weights at the ends is suspended from a thin elastic thread. Two heavy balls are fixed nearby. Gravitational forces act between the weights and the stationary balls. Under the influence of these forces, the rocker turns and twists the thread until the resulting elastic force becomes equal to the gravitational force. By the angle of twist you can determine the force of attraction. To do this, you only need to know the elastic properties of the thread. The masses of the bodies are known, and the distance between the centers of interacting bodies can be directly measured.

From these experiments the following value for the gravitational constant was obtained:

G = 6.67 10 -11 N m 2 / kg 2.

Only in the case when bodies of enormous mass interact (or at least the mass of one of the bodies is very large) does the gravitational force reach of great importance. For example, the Earth and the Moon are attracted to each other with a force F ≈ 2 10 20 N.

Dependence of the acceleration of free fall of bodies on geographic latitude.

One of the reasons for the increase in the acceleration of gravity when the point where the body is located moves from the equator to the poles is that the globe is somewhat flattened at the poles and the distance from the center of the Earth to its surface at the poles is less than at the equator. Another reason is the rotation of the Earth.

Equality of inertial and gravitational masses.

The most striking property of gravitational forces is that they impart the same acceleration to all bodies, regardless of their masses. What would you say about a football player whose kick would be equally accelerated by an ordinary leather ball and a two-pound weight? Everyone will say that this is impossible. But the Earth is just such an “extraordinary football player” with the only difference that its effect on bodies is not of the nature of a short-term blow, but continues continuously for billions of years.

In Newton's theory, mass is the source of the gravitational field. We are in the Earth's gravitational field. At the same time, we are also sources of the gravitational field, but due to the fact that our mass is significantly less than the mass of the Earth, our field is much weaker and surrounding objects do not react to it.

The extraordinary property of gravitational forces, as we have already said, is explained by the fact that these forces are proportional to the masses of both interacting bodies. The mass of a body, which is included in Newton’s second law, determines the inertial properties of the body, i.e. its ability to acquire a certain acceleration under the influence of a given force. This inert mass m and.

It would seem, what relation can it have to the ability of bodies to attract each other? The mass that determines the ability of bodies to attract each other is the gravitational mass m r.

It does not at all follow from Newtonian mechanics that the inertial and gravitational masses are the same, i.e. that

m and = m r . (3.5)

Equality (3.5) is a direct consequence of experiment. It means that we can simply talk about the mass of a body as a quantitative measure of both its inertial and gravitational properties.

A particular, but extremely important for us, type of universal gravitational force is force of attraction of bodies to the Earth. This force is called gravity . According to the law of universal gravitation, it is expressed by the formula

$~F_T = G \frac(mM)((R+h)^2)$ , (1)

Where m– body weight, M– mass of the Earth, R– radius of the Earth, h– height of the body above the Earth’s surface. The force of gravity is directed vertically downward, towards the center of the Earth.

More precisely, in addition to this force, in the reference frame associated with the Earth, the body is affected by the centrifugal force of inertia $~\vec F_c$, which arises due to the daily rotation of the Earth, and is equal to $~F_c = m \cdot \ omega^2 \cdot r$ , where m– body weight; r– the distance between the body and the earth’s axis. If the height of a body above the Earth’s surface is small compared to its radius, then $~r = R \cos \varphi$, where R– radius of the Earth, φ – geographical latitude at which the body is located (Fig. 1). Taking this into account, $~F_c = m \cdot \omega^2 \cdot R \cos \varphi$ .

Gravity is the force acting on anything nearby earth's surface body.

It is defined as the geometric sum of the force of gravitational attraction to the Earth $~\vec F_g$ acting on the body and the centrifugal force of inertia $~\vec F_c$, taking into account the effect of the daily rotation of the Earth around its own axis, i.e. $~\vec F_T = \vec F_g + \vec F_c$ . The direction of gravity is the direction of the vertical at a given point on the earth's surface.

BUT the magnitude of the centrifugal force of inertia is very small compared to the force of gravity of the Earth (their ratio is approximately 3∙10 -3), then the force $~\vec F_c$ is usually neglected. Then $~\vec F_T \approx \vec F_g$ .

Acceleration of gravity

The force of gravity imparts an acceleration to the body, called the acceleration of gravity. According to Newton's second law

$~\vec g = \frac(\vec F_T)(m)$ .

Taking into account expression (1) for the gravitational acceleration module we will have

$~g_h = G \frac(M)((R+h)^2)$ . (2)

On the Earth's surface (h = 0) the gravitational acceleration modulus is equal to

$~g = G \frac(M)(R^2)$ ,

and the force of gravity is

$~\vec F_T = m \vec g$ .

The module of gravity acceleration included in the formulas is approximately 9.8 m/s 2 .

From formula (2) it is clear that the acceleration of gravity does not depend on the mass of the body. It decreases as the body rises above the Earth's surface: the acceleration of gravity is inversely proportional to the square of the distance of the body from the center of the Earth.

However, if the height h body above the Earth's surface does not exceed 100 km, then in calculations allowing an error of ≈ 1.5%, this height can be neglected in comparison with the radius of the Earth (R = 6370 km). The acceleration of free fall at altitudes up to 100 km can be considered constant and equal to 9.8 m/s 2 .

And yet At the surface of the Earth, the acceleration of gravity is not the same everywhere. It depends on latitude: more at the Earth's poles than at the equator. The fact is that the globe is somewhat flattened at the poles. The equatorial radius of the Earth is 21 km greater than the polar radius.

Another, more significant reason for the dependence of the acceleration of gravity on geographic latitude is the rotation of the Earth. Newton's second law is valid in an inertial frame of reference. Such a system is, for example, the heliocentric system. The reference system associated with the Earth, strictly speaking, cannot be considered inertial. The Earth rotates around its axis and moves in a closed orbit around the Sun.

The rotation of the Earth and its oblateness at the poles leads to the fact that the acceleration of gravity relative to the geocentric reference system at different latitudes is different: at the poles g floor ≈ 9.83 m/s 2, at the equator g eq ≈ 9.78 m/s 2 , at latitude 45° g≈ 9.81 m/s 2. However, in our calculations we will assume the acceleration of gravity to be approximately equal to 9.8 m/s 2 .

Due to the rotation of the Earth around its axis, the acceleration of gravity in all places except the equator and poles is not directed exactly towards the center of the Earth.

In addition, the acceleration of gravity depends on the density of rocks located in the bowels of the Earth. In areas where rocks occur whose density is greater than the average density of the Earth (for example, iron ore), g more. And where there are oil deposits, g less. Geologists use this when searching for minerals.

Body weight

Body weight- this is the force with which a body, due to its attraction to the Earth, acts on a support or suspension.

Consider, for example, a body suspended from a spring, the other end of which is fixed (Fig. 2). The body is affected by the force of gravity $~\vec F_T = m \vec g$ directed downwards. It therefore begins to fall, dragging the lower end of the spring with it. Because of this, the spring will be deformed, and the elastic force $~\vec F_(ynp)$ of the spring will appear. It is attached to the upper edge of the body and directed upward. The upper edge of the body will therefore “lag behind” in its fall from its other parts, to which the elastic force of the spring is not applied. As a result, the body is deformed. Another elastic force arises - the elastic force of the deformed body. It is attached to the spring and directed downwards. This force is the weight of the body.

According to Newton's third law, both of these elastic forces are equal in magnitude and directed in opposite directions. After several oscillations, the body on the spring is at rest. This means that the force of gravity $~m \vec g$ is equal in modulus to the force of elasticity F spring control But this same force is also equal to the weight of the body.

Thus, in our example, the weight of the body, which we denote by the letter $~\vec P$, is equal in absolute value to the force of gravity:

$~P = m g$ .

Second example. Let the body A is on a horizontal support IN(Fig. 3). On the body A the force of gravity $~m \vec g$ and the ground reaction force $~\vec N$ act. But if the support acts on the body with a force $~\vec N$ then the body acts on the support with a force $~\vec P$, which, in accordance with Newton’s third law, is equal in magnitude and opposite in direction $~ \vec N$ \[~\vec P = -\vec N\] . The force $~\vec P$ is the weight of the body.

If the body and the support are stationary or moving uniformly and rectilinearly, i.e. without acceleration, then, according to Newton’s second law,

$~\vec N + m \vec g = 0$ .

$~\vec N = -\vec P$ , then $~-\vec P + m \vec g = 0$ .

Hence,

$~\vec P = m \vec g$ .

Means, if acceleration a = 0, then the weight of the body is equal to the force of gravity.

But this does not mean that the weight of a body and the force of gravity applied to it are the same thing. Gravity is applied to the body, and weight is applied to the support or suspension. The nature of gravity and weight is also different. If gravity is the result of the interaction of the body and the Earth (gravitational force), then weight appears as a result of a completely different interaction: the interaction of the body A and supports IN. Support IN and body A in this case, they are deformed, which leads to the appearance of elastic forces. Thus, body weight(same as ground reaction force) is a special type of elastic force.

Weight has characteristics that significantly distinguish it from gravity.

Firstly, weight is determined by the totality of forces acting on a body, and not just by gravity (for example, the weight of a body in liquid or air is less than in a vacuum due to the appearance of buoyant (Archimedean) force). Secondly, the weight of the body depends significantly on the acceleration with which the support (suspension) moves.

Body weight when the support or suspension moves with acceleration

Is it possible to increase or decrease body weight without changing the body itself? It turns out yes. Let the body be in an elevator car moving with acceleration $~\vec a$ (Fig. 4 a, b).

Rice. 4

According to Newton's second law

$~\vec N + m \vec g = m \vec a$ , (3)

Where N– support reaction force (elevator floor), m– body weight.

According to Newton's third law, the weight of the body is $~\vec P = -\vec N$ . Therefore, taking into account (3), we obtain

$~\vec P = m (\vec g - \vec a)$ .

Let's direct the coordinate axis Y reference system associated with the Earth, vertically downwards. Then the projection of the body weight on this axis will be equal to

$~P_y = m (g_y - a_y)$ .

Since the vectors $~\vec P$ and $~\vec g$ are aligned with the coordinate axis Y, That R y= R And g y= g. If the acceleration $~\vec a$ is directed downward (see Fig. 4, a), then a y= A, and the equality takes the following form:

$~P = m (g - a)$ .

From the formula it follows that only when A= 0 body weight is equal to gravity. At A≠ 0 body weight differs from gravity. When the elevator moves with acceleration directed downward (for example, at the beginning of the elevator's descent or during its stop when moving upward) and in magnitude less than the acceleration of free fall, the weight of the body is less than the force of gravity. Consequently, in this case, the weight of the body is less than the weight of the same body if it is on a stationary or uniformly moving support (suspension). For the same reason, the weight of a body at the equator is less than at the Earth’s poles, since due to the daily rotation of the Earth, a body at the equator moves with centripetal acceleration.

Let us now consider what happens if the body moves with acceleration $~\vec a$ directed vertically upward (see Fig. 4, b). IN in this case we get

$~P = m (g + a)$ .

The weight of a body in an elevator moving with acceleration directed vertically upward is more weight body at rest. The increase in body weight caused by the accelerated movement of the support (or suspension) is called overload. The overload can be estimated by finding the ratio of the weight of an accelerating body to the weight of a body at rest:

$~k = \frac(m (g + a))(m g) = 1 + \frac(a)(g)$ .

A trained person is able to briefly withstand approximately six times the overload. This means that the acceleration of the spacecraft, according to the formula obtained, should not exceed five times the acceleration of gravity.

Weightlessness

Let's take in our hands a spring with a load suspended from it, or better yet, a spring scale. The spring scale can be used to measure body weight. If the hand holding the scale is at rest relative to the Earth, the scale will show that the absolute weight of the body is equal to the force of gravity mg. Let's release the scales from our hands, they, along with the load, will begin to fall freely. In this case, the scale arrow is set to zero, indicating that the body weight has become zero. And this is understandable. In free fall, both the scale and the load move with the same acceleration equal to g. The lower end of the spring is not carried away by the load, but follows it itself, and the spring is not deformed. Therefore, there is no elastic force that would act on the load. This means that the load is not deformed and does not act on the spring. The weight is gone! The load, as they say, has become weightless.

Weightlessness is explained by the fact that the force of universal gravitation, and therefore the force of gravity, imparts the same acceleration to all bodies (in our case, the load and the spring) g. Therefore, every body on which acts only gravity or, in general, the force of universal gravity, is in a state of weightlessness. Freely falling bodies, for example bodies in a spacecraft, are found in such conditions. After all, spacecraft, and the bodies in it are also in a state of prolonged free fall. However, each of you is in a state of weightlessness, although not for long, jumping from a chair to the floor or jumping up.

This can also be proven mathematically. When the body is in free fall $~\vec a = \vec g$ and $~P = m (g - g) = 0$ .

Literature

Kikoin I.K., Kikoin A.K. Physics: Textbook. for 9th grade. avg. school – M.: Prosveshchenie, 1992. – 191 p.
Lutsevich A.A., Yakovenko S.V. Physics: Textbook. allowance. – Mn.: Higher. school, 2000. – 495 p.
Physics: Mechanics. 10th grade: Textbook. for in-depth study of physics / M.M. Balashov, A.I. Gomonova, A.B. Dolitsky and others; Ed. G.Ya. Myakisheva. – M.: Bustard, 2002. – 496 p.

If a body accelerates, then something acts on it. How to find this “something”? For example, what kind of forces act on a body near the surface of the earth? This is the force of gravity directed vertically downward, proportional to the mass of the body and for heights much smaller than the radius of the earth $(\large R)$, almost independent of the height; it is equal

$(\large F = \dfrac (G \cdot m \cdot M)(R^2) = m \cdot g )$

$(\large g = \dfrac (G \cdot M)(R^2) )$

so-called acceleration due to gravity. In the horizontal direction the body will move with constant speed, however, movement in the vertical direction according to Newton’s second law:

$(\large m \cdot g = m \cdot \left (\dfrac (d^2 \cdot x)(d \cdot t^2) \right) )$

after contracting $(\large m)$, we find that the acceleration in the direction $(\large x)$ is constant and equal to $(\large g)$. This is good famous movement freely falling body, which is described by the equations

$(\large v_x = v_0 + g \cdot t)$

$(\large x = x_0 + x_0 \cdot t + \dfrac (1)(2) \cdot g \cdot t^2)$

How is strength measured?

In all textbooks and smart books, it is customary to express force in Newtons, but except in the models that physicists operate, Newtons are not used anywhere. This is extremely inconvenient.

Newton newton (N) is a derived unit of force in the International System of Units (SI).
Based on Newton's second law, the unit newton is defined as the force that changes the speed of a body weighing one kilogram by 1 meter per second in one second in the direction of the force.

Thus, 1 N = 1 kg m/s².

Kilogram-force (kgf or kg) is a gravitational metric unit of force equal to the force that acts on a body weighing one kilogram in the gravitational field of the earth. Therefore, by definition, a kilogram-force is equal to 9.80665 N. A kilogram-force is convenient because its value is equal to the weight of a body weighing 1 kg.
1 kgf = 9.80665 newtons (approximately ≈ 10 N)
1 N ≈ 0.10197162 kgf ≈ 0.1 kgf

1 N = 1 kg x 1 m/s2.

Law of gravitation

Every object in the Universe is attracted to every other object with a force proportional to their masses and inversely proportional to the square of the distance between them.

$(\large F = G \cdot \dfrac (m \cdot M)(R^2))$

We can add that any body reacts to a force applied to it with acceleration in the direction of this force, in magnitude inversely proportional to the mass of the body.

$(\large G)$ — gravitational constant

$(\large M)$ — mass of the earth

$(\large R)$ — radius of the earth

$(\large G = 6.67 \cdot (10^(-11)) \left (\dfrac (m^3)(kg \cdot (sec)^2) \right) )$

$(\large M = 5.97 \cdot (10^(24)) \left (kg \right) )$

$(\large R = 6.37 \cdot (10^(6)) \left (m \right) )$

Within the framework of classical mechanics, gravitational interaction is described by Newton's law of universal gravitation, according to which the force of gravitational attraction between two bodies of mass $(\large m_1)$ and $(\large m_2)$ separated by a distance $(\large R)$ is

$(\large F = -G \cdot \dfrac (m_1 \cdot m_2)(R^2))$

Here $(\large G)$ is the gravitational constant equal to $(\large 6.673 \cdot (10^(-11)) m^3 / \left (kg \cdot (sec)^2 \right) )$. The minus sign means that the force acting on the test body is always directed along the radius vector from the test body to the source of the gravitational field, i.e. gravitational interaction always leads to the attraction of bodies.
The gravity field is potential. This means that you can introduce the potential energy of gravitational attraction of a pair of bodies, and this energy will not change after the bodies move along closed loop. The potentiality of the gravitational field entails the law of conservation of the sum of kinetic and potential energy, which, when studying the motion of bodies in a gravitational field, often significantly simplifies the solution.
Within the framework of Newtonian mechanics, gravitational interaction is long-range. This means that no matter how massive a body moves, at any point in space the gravitational potential and force depend only on the position of the body in at the moment time.

Heavier - Lighter

The weight of a body $(\large P)$ is expressed by the product of its mass $(\large m)$ and the acceleration due to gravity $(\large g)$.

$(\large P = m \cdot g)$

When on earth the body becomes lighter (presses less on the scales), this is due to a decrease masses. On the moon, everything is different; the decrease in weight is caused by a change in another factor - $(\large g)$, since the acceleration of gravity on the surface of the moon is six times less than on the earth.

mass of the earth = $(\large 5.9736 \cdot (10^(24))\ kg )$

moon mass = $(\large 7.3477 \cdot (10^(22))\ kg )$

acceleration of gravity on Earth = $(\large 9.81\ m / c^2 )$

gravitational acceleration on the Moon = $(\large 1.62 \ m / c^2 )$

As a result, the product $(\large m \cdot g )$, and therefore the weight, decreases by 6 times.

But it is impossible to describe both of these phenomena with the same expression “make it easier.” On the moon, bodies do not become lighter, but only fall less rapidly; they are “less epileptic”))).

Vector and scalar quantities

A vector quantity (for example, a force applied to a body), in addition to its value (modulus), is also characterized by direction. A scalar quantity (for example, length) is characterized only by its value. All classical laws of mechanics are formulated for vector quantities.

Figure 1.

In Fig. 1 shown various options location of the vector $( \large \overrightarrow(F))$ and its projection $( \large F_x)$ and $( \large F_y)$ on the axis $( \large X)$ and $( \large Y)$, respectively:

A. the quantities $( \large F_x)$ and $( \large F_y)$ are non-zero and positive
B. the quantities $( \large F_x)$ and $( \large F_y)$ are non-zero, while $(\large F_y)$ is a positive quantity, and $(\large F_x)$ is negative, because the vector $(\large \overrightarrow(F))$ is directed in the direction opposite to the direction of the $(\large X)$ axis
C.$(\large F_y)$ is a positive non-zero quantity, $(\large F_x)$ is equal to zero, because the vector $(\large \overrightarrow(F))$ is directed perpendicular to the axis $(\large X)$

moment of force

A moment of power is called the vector product of the radius vector drawn from the axis of rotation to the point of application of the force and the vector of this force. Those. According to the classical definition, the moment of force is a vector quantity. Within the framework of our problem, this definition can be simplified to the following: the moment of force $(\large \overrightarrow(F))$ applied to a point with coordinate $(\large x_F)$, relative to the axis located at point $(\large x_0 )$ is a scalar quantity equal to the product of the force modulus $(\large \overrightarrow(F))$ and the force arm - $(\large \left | x_F - x_0 \right |)$. And the sign of this scalar quantity depends on the direction of the force: if it rotates the object clockwise, then the sign is plus, if counterclockwise, then the sign is minus.

It is important to understand that we can choose the axis arbitrarily - if the body does not rotate, then the sum of the moments of forces about any axis is zero. The second important note is that if a force is applied to a point through which an axis passes, then the moment of this force about this axis is equal to zero (since the arm of the force will be equal to zero).

Let's illustrate the above example, in Fig. 2. Let us assume that the system shown in Fig. 2 is in equilibrium. Consider the support on which the loads stand. It is acted upon by 3 forces: $(\large \overrightarrow(N_1),\ \overrightarrow(N_2),\ \overrightarrow(N),)$ points of application of these forces A, IN And WITH respectively. The figure also contains forces $(\large \overrightarrow(N_(1)^(gr)),\ \overrightarrow(N_2^(gr)))$. These forces are applied to the loads, and according to Newton's 3rd law

$(\large \overrightarrow(N_(1)) = - \overrightarrow(N_(1)^(gr)))$

$(\large \overrightarrow(N_(2)) = - \overrightarrow(N_(2)^(gr)))$

Now consider the condition for the equality of the moments of forces acting on the support relative to the axis passing through the point A(and, as we agreed earlier, perpendicular to the drawing plane):

$(\large N \cdot l_1 - N_2 \cdot \left (l_1 +l_2 \right) = 0)$

Please note that the moment of force $(\large \overrightarrow(N_1))$ was not included in the equation, since the arm of this force relative to the axis in question is equal to $(\large 0)$. If for some reason we want to select an axis passing through the point WITH, then the condition for equality of moments of forces will look like this:

$(\large N_1 \cdot l_1 - N_2 \cdot l_2 = 0)$

It can be shown that, from a mathematical point of view, the last two equations are equivalent.

Center of gravity

Center of gravity in a mechanical system is the point relative to which the total moment of gravity acting on the system is equal to zero.

Center of mass

The point of the center of mass is remarkable in that if a great many forces act on the particles forming a body (no matter whether it is solid or liquid, a cluster of stars or something else) (meaning only external forces, since all internal forces compensate each other), then the resulting the force leads to such an acceleration of this point as if the entire mass of the body $(\large m)$ were in it.

The position of the center of mass is determined by the equation:

$(\large R_(c.m.) = \frac(\sum m_i\, r_i)(\sum m_i))$

This is a vector equation, i.e. in fact, three equations - one for each of the three directions. But consider only the $(\large x)$ direction. What does the following equality mean?

$(\large X_(c.m.) = \frac(\sum m_i\, x_i)(\sum m_i))$

Suppose the body is divided into small pieces with the same mass $(\large m)$, and the total mass of the body will be equal to the number of such pieces $(\large N)$ multiplied by the mass of one piece, for example 1 gram. Then this equation means that you need to take the $(\large x)$ coordinates of all the pieces, add them and divide the result by the number of pieces. In other words, if the masses of the pieces are equal, then $(\large X_(c.m.))$ will simply be the arithmetic mean of the $(\large x)$ coordinates of all the pieces.

Mass and density

Mass is a fundamental physical quantity. Mass characterizes several properties of a body at once and in itself has a number of important properties.

Mass serves as a measure of the substance contained in a body.
Mass is a measure of the inertia of a body. Inertia is the property of a body to maintain its speed unchanged (in the inertial frame of reference) when external influences are absent or compensate each other. Subject to availability external influences the inertia of a body is manifested in the fact that its speed does not change instantly, but gradually, and the more slowly, the greater the inertia (i.e. mass) of the body. For example, if a billiard ball and a bus are moving at the same speed and are braked by the same force, then it takes much less time to stop the ball than to stop the bus.
The masses of bodies are the reason for their gravitational attraction to each other (see the section “Gravity”).
The mass of a body is equal to the sum of the masses of its parts. This is the so-called additivity of mass. Additivity allows you to use a standard of 1 kg to measure mass.
The mass of an isolated system of bodies does not change with time (law of conservation of mass).
The mass of a body does not depend on the speed of its movement. Mass does not change when moving from one frame of reference to another.
Density of a homogeneous body is the ratio of the mass of the body to its volume:

$(\large p = \dfrac (m)(V) )$

Density does not depend on the geometric properties of the body (shape, volume) and is a characteristic of the substance of the body. Densities various substances presented in reference tables. It is advisable to remember the density of water: 1000 kg/m3.

Newton's second and third laws

The interaction of bodies can be described using the concept of force. Force is a vector quantity, which is a measure of the influence of one body on another.
Being a vector, force is characterized by its modulus (absolute value) and direction in space. In addition, the point of application of the force is important: the same magnitude and direction of the force applied in different points body, may have different effects. So, if you grab the rim of a bicycle wheel and pull tangentially to the rim, the wheel will begin to rotate. If you pull along the radius, there will be no rotation.

Newton's second law

The product of the body mass and the acceleration vector is the resultant of all forces applied to the body:

$(\large m \cdot \overrightarrow(a) = \overrightarrow(F) )$

Newton's second law relates acceleration and force vectors. This means that the following statements are true.

$(\large m \cdot a = F)$, where $(\large a)$ is the acceleration modulus, $(\large F)$ is the resulting force modulus.
The acceleration vector has the same direction as the resultant force vector, since the mass of the body is positive.

Newton's third law

Two bodies act on each other with forces equal in magnitude and opposite in direction. These forces have the same physical nature and are directed along the straight line connecting their points of application.

Superposition principle

Experience shows that if several other bodies act on a given body, then the corresponding forces add up as vectors. More precisely, the principle of superposition is valid.
The principle of superposition of forces. Let the forces act on the body$(\large \overrightarrow(F_1), \overrightarrow(F_2),\ \ldots \overrightarrow(F_n))$ If you replace them with one force$(\large \overrightarrow(F) = \overrightarrow(F_1) + \overrightarrow(F_2) \ldots + \overrightarrow(F_n))$ , then the result of the impact will not change.
The force $(\large \overrightarrow(F))$ is called resultant forces $(\large \overrightarrow(F_1), \overrightarrow(F_2),\ \ldots \overrightarrow(F_n))$ or resulting by force.

Forwarder or carrier? Three secrets and international cargo transportation

Forwarder or carrier: who to choose? If the carrier is good and the forwarder is bad, then the first. If the carrier is bad and the forwarder is good, then the latter. This choice is simple. But how can you decide when both candidates are good? How to choose from two seemingly equivalent options? The fact is that these options are not equivalent.

Horror stories of international transport

BETWEEN A HAMMER AND A HILL.

It is not easy to live between the customer of transportation and the very cunning and economical owner of the cargo. One day we received an order. Freight for three kopecks, additional conditions on two sheets, the collection is called.... Loading on Wednesday. The car is in place on Tuesday, and by lunchtime next day the warehouse begins to slowly throw into the trailer everything that your forwarder has collected for its recipient customers.

AN ENCHANTED PLACE - PTO KOZLOVICHY.

According to legends and experience, everyone who transported goods from Europe by road knows how scary place is PTO Kozlovichi, Brest Customs. What chaos the Belarusian customs officers create, they find fault in every possible way and charge exorbitant prices. And it's true. But not all...

ON THE NEW YEAR'S TIME WE WERE BRINGING POWDERED MILK.

Loading with groupage cargo at a consolidation warehouse in Germany. One of the goods - powdered milk from Italy, the delivery of which was ordered by the Forwarder.... A classic example of the work of a forwarder-“transmitter” (he doesn’t delve into anything, he just transmits along the chain).

Documents for international transport

International road transport of goods is very organized and bureaucratic; as a result, a bunch of unified documents are used to carry out international road transport of goods. It doesn’t matter if it’s a customs carrier or an ordinary one - he won’t travel without documents. Although this is not very exciting, we tried to explain in a simpler way the purpose of these documents and the meaning they have. They gave an example of filling out TIR, CMR, T1, EX1, Invoice, Packing List...

Axle load calculation for road freight transport

The goal is to study the possibility of redistributing loads on the axles of the tractor and semi-trailer when the location of the cargo in the semi-trailer changes. And applying this knowledge in practice.

In the system we are considering there are 3 objects: a tractor $(T)$, a semi-trailer $(\large ((p.p.)))$ and a load $(\large (gr))$. All variables related to each of these objects will be marked with the superscript $T$, $(\large (p.p.))$ and $(\large (gr))$ respectively. For example, the tare weight of a tractor will be denoted as $m^(T)$.

Why don't you eat fly agarics? The customs officer exhaled a sigh of sadness.

What is happening in the international road transport market? The Federal Customs Service of the Russian Federation has banned the issuance of TIR Carnets without additional guarantees for several years already federal districts. And she notified that from December 1 of this year she will completely terminate the agreement with the IRU as not meeting the requirements of the Customs Union and is putting forward financial claims that are not childish.
IRU in response: “The explanations of the Federal Customs Service of Russia regarding the alleged debt of ASMAP in the amount of 20 billion rubles are a complete fiction, since all the old TIR claims have been fully settled..... What do we, common carriers, think?

Stowage Factor Weight and volume of cargo when calculating the cost of transportation

The calculation of the cost of transportation depends on the weight and volume of the cargo. For sea transport most often crucial has volume, for air - weight. For road transport of goods, a complex indicator is important. Which parameter for calculations will be chosen in a particular case depends on specific gravity of the cargo (Stowage Factor) .

Gravity is the force with which the Earth attracts a body located near its surface. .

The phenomena of gravity can be observed everywhere in the world around us. A ball thrown up falls down, a stone thrown horizontally will end up on the ground after some time. An artificial satellite launched from the Earth, due to the effects of gravity, does not fly in a straight line, but moves around the Earth.

Gravity always directed vertically downward, towards the center of the Earth. It is designated Latin letter F t (T- heaviness). The force of gravity is applied to the center of gravity of the body.

To find the center of gravity of an arbitrary shape, you need to hang a body on a thread at its different points. The point of intersection of all directions marked by the thread will be the center of gravity of the body. The center of gravity of bodies of regular shape is at the center of symmetry of the body, and it is not necessary that it belong to the body (for example, the center of symmetry of a ring).

For a body located near the surface of the Earth, the force of gravity is:

where is the mass of the Earth, m- body weight, R- radius of the Earth.

If only this force acts on the body (and all others are balanced), then it undergoes free fall. The acceleration of this free fall can be found by applying Newton's second law:

(2)

From this formula we can conclude that the acceleration of gravity does not depend on the mass of the body m, therefore, it is the same for all bodies. According to Newton's second law, gravity can be defined as the product of the mass of a body and its acceleration (in this case, the acceleration due to gravity g);

Gravity, acting on the body, is equal to the product of the mass of the body and the acceleration of gravity.

Like Newton's second law, formula (2) is valid only in inertial frames of reference. On the Earth's surface, inertial reference systems can only be systems associated with the Earth's poles, which do not take part in its daily rotation. All other points on the earth's surface move in circles with centripetal accelerations and the reference systems associated with these points are non-inertial.

Due to the rotation of the Earth, the acceleration of gravity at different latitudes is different. However, the acceleration of free fall in different regions of the globe varies very little and differs very little from the value calculated by the formula

Therefore, in rough calculations, the non-inertiality of the reference system associated with the Earth’s surface is neglected, and the acceleration of gravity is considered to be the same everywhere.