Introduction

In kinematics, the description of the simplest types of mechanical movements is considered. At the same time, the reasons causing changes in the position of the body relative to other bodies were not touched upon, and the frame of reference is chosen for reasons of convenience when solving a particular problem. In dynamics, first of all, of interest are the reasons due to which some bodies begin to move relative to other bodies, as well as the factors that cause the appearance of acceleration. However, the laws in mechanics, strictly speaking, have different forms in different frames of reference. It has been established that there are such frames of reference in which the laws and regularities do not depend on the choice of the frame of reference. Such reference systems are called inertial systems(ISO). In these frames of reference, the value of acceleration depends only on the acting forces and does not depend on the choice of frame of reference. The inertial frame of reference is heliocentric frame of reference, whose origin is at the center of the Sun. Frames of reference moving uniformly rectilinearly relative to the inertial frame are also inertial, and frames of reference moving with acceleration relative to the inertial frame are non-inertial. For these reasons, the earth's surface is, strictly speaking, a non-inertial frame of reference. In many problems, the frame of reference associated with the Earth can be considered inertial with a good degree of accuracy.

Basic laws of dynamics in inertial and non-inertial

reference systems

The ability of a body to maintain a state of uniform rectilinear motion or to rest in ISO is called body inertia. The measure of body inertia is weight. Mass is a scalar quantity, in the SI system it is measured in kilograms (kg). The measure of interaction is a quantity called force. Force is a vector quantity, in the SI system it is measured in Newtons (N).

Newton's first law. In inertial frames of reference, a point moves uniformly in a straight line or is at rest if the sum of all forces acting on it is zero, i.e.:

where are the forces acting on a given point.

Newton's second law. In inertial systems, a body moves with acceleration if the sum of all forces acting on it is not equal to zero, and the product of the body's mass and its acceleration is equal to the sum of these forces, i.e.:

Newton's third law. The forces with which the bodies act on each other are equal in magnitude and opposite in direction, i.e.: .

Forces, as measures of interaction, are always born in pairs.

To successfully solve most problems using Newton's laws, it is necessary to adhere to a certain sequence of actions (a kind of algorithm).

The main points of the algorithm.

1. Analyze the condition of the problem and find out with which bodies the considered body interacts. Based on this, determine the number of forces acting on the body in question. Suppose the number of forces acting on the body is . Then perform a schematically correct drawing, on which to build all the forces acting on the body.

2. Using the condition of the problem, determine the direction of acceleration of the body in question, and depict the acceleration vector in the figure.

3. Write in vector form Newton's second law, i.e.:

where forces acting on the body.

4. Choose an inertial frame of reference. Draw a rectangular Cartesian coordinate system in the figure, the OX axis of which is directed along the acceleration vector, the OY and OZ axes are directed perpendicular to the OX axis.

5. Using the main property of vector equalities, write Newton's second law for the projections of vectors on the coordinate axes, i.e.:

6. If in the problem, in addition to forces and accelerations, it is required to determine the coordinates and speed, then, in addition to Newton's second law, it is necessary to use kinematic equations of motion. Having written the system of equations, it is necessary to pay attention to the fact that the number of equations is equal to the number of unknowns in this problem.

Consider a non-inertial frame of reference rotating with a constant angular velocity around an axis moving translationally at a speed relative to the inertial frame. In this case, the acceleration of a point in the inertial frame () is related to the acceleration in the non-inertial frame () by the relation:

where is the acceleration of the non-inertial frame relative to the inertial frame , the linear velocity of the point in the non-inertial frame. From the last relation, instead of acceleration, we substitute into equality (1), we obtain the expression:

This ratio is called Newton's second law in a non-inertial frame of reference.

Forces of inertia. Let us introduce the notation:

1. – translational inertia force;

2. – Coriolis force;

3 – centrifugal force of inertia.

In tasks, the translational force of inertia is depicted against the vector by the acceleration of the translational motion of a non-inertial frame of reference (), the centrifugal force of inertia - from the center of rotation along the radius (); the direction of the Coriolis force is determined by the rule gimlet for the cross product of vectors .

Strictly speaking, the forces of inertia are not forces in the full sense, because Newton's third law does not hold for them, i.e. they are not paired.

Forces

The force of gravity. The force of universal gravitation arises in the process of interaction between bodies with masses, and is calculated from the ratio:

. (4)

The coefficient of proportionality is called gravitational constant. Its value in the SI system is .

Reaction force. Reaction forces arise when a body interacts with various structures that limit its position in space. For example, a body suspended by a thread is subjected to a reaction force, usually called the force tension. The force of the thread tension is always directed along the thread. There is no formula for calculating its value. Usually its value is found either from the first or from the second law of Newton. Reaction forces also include forces acting on a particle on a smooth surface. They call her normal reaction force, denote . The reaction force is always directed perpendicular to the considered surface. A force acting on a smooth surface from a body is called force of normal pressure(). According to Newton's third law, the reaction force is equal in magnitude to the force of normal pressure, but the vectors of these forces are opposite in direction.

Elastic force. Elastic forces arise in bodies if the bodies are deformed, i.e. if the shape of the body or its volume is changed. When the deformation stops, the elastic forces disappear. It should be noted that, although elastic forces arise during deformations of bodies, deformation does not always lead to the emergence of elastic forces. Elastic forces arise in bodies capable of restoring their shape after the termination of external influence. Such bodies, and their corresponding deformations, are called elastic. With plastic deformation, the changes do not completely disappear after the termination of the external influence. A striking example of the manifestation of elastic forces can be the forces arising in springs subject to deformation. For elastic deformations that occur in deformed bodies, the elastic force is always proportional to the magnitude of the deformation, i.e.:

, (5)

where is the coefficient of elasticity (or stiffness) of the spring, the strain vector of the spring.

This statement is called Hooke's law.

Friction force. When one body moves along the surface of another, forces arise that prevent this movement. Such forces are called sliding friction forces. The magnitude of the static friction force can vary depending on the applied external force. At a certain value of the external force, the static friction force reaches its maximum value. After that, the sliding of the body begins. It has been experimentally established that the force of sliding friction is directly proportional to the force of normal pressure of the body on the surface. According to Newton's third law, the force of normal pressure of a body on a surface is always equal to the reaction force with which the surface itself acts on a moving body. With this in mind, the formula for calculating the magnitude of the sliding friction force has the form:

, (6)

where is the magnitude of the reaction force; coefficient of sliding friction. The sliding friction force acting on a moving body is always directed against its speed, along the contacting surfaces.

The power of resistance. When bodies move in liquids and gases, friction forces also arise, but they differ significantly from the forces of dry friction. These forces are called viscous friction forces, or resistance forces. The forces of viscous friction arise only with the relative motion of bodies. The resistance forces depend on many factors, namely: on the size and shape of bodies, on the properties of the medium (density, viscosity), on the speed of relative motion. At low speeds, the resistance force is directly proportional to the speed of the body relative to the medium, i.e.:

. (7)

At high speeds, the resistance force is proportional to the square of the speed of the body relative to the medium, i.e.:

, (8)

where some coefficients of proportionality, called drag coefficients.

Basic equation of dynamics

The basic equation of the dynamics of a material point is nothing more than a mathematical expression of Newton's second law:

. (9)

In an inertial frame of reference, the sum of all forces includes only forces that are measures of interactions; in non-inertial frames, the sum of forces includes the forces of inertia.

From a mathematical point of view, relation (9) is a differential equation of point motion in vector form. Its solution is the main problem of the dynamics of a material point.

Examples of problem solving

Task number 1. A glass is placed on a sheet of paper. With what acceleration must the sheet be set in motion in order to pull it out from under the glass, if the coefficient of friction between the glass and the sheet of paper is 0.3?

Let us assume that for some force acting on a sheet of paper, the glass moves together with the sheet. Let us depict separately the forces acting on a glass with mass . The following bodies act on the glass: Earth with gravity, a sheet of paper with a reaction force, a sheet of paper with a friction force directed along the speed of the glass. The movement of the glass is uniformly accelerated, therefore, the acceleration vector is directed along the speed of the glass.

Let's depict the glass acceleration vector in the figure. We write Newton's second law in vector form for the forces acting on the glass:

.

Let's direct the OX axis along the glass acceleration vector, and the OY axis ¾ vertically upwards. We write Newton's second law in projections onto these coordinate axes, we obtain the following equations:

(1.1)

With an increase in the force acting on a sheet of paper, the magnitude of the friction force with which a sheet of paper acts on a glass increases. At a certain value of the force, the magnitude of the friction force reaches its maximum value, which is equal in magnitude to the sliding friction force. From this moment, the glass begins to slide relative to the surface of the paper. The limiting value of the friction force is related to the reaction force acting on the glass by the following relationship:

From equality (1.2) we express the magnitude of the reaction force, and then we substitute it into the last relation, we have . From the obtained relation, we find the value of the friction force and put it into equation (1.1), we obtain an expression for determining the maximum acceleration of the glass:

Substituting the numerical values ​​of the quantities in the last equality, we find the value of the maximum acceleration of the glass:

.

The obtained value of the glass acceleration is equal to the minimum acceleration of a sheet of paper, at which it can be “pulled out” from under the glass.

Answer: .

Let's depict all the forces acting on the body. In addition to the external force, the Earth acts on the body with the force of gravity, a horizontal surface with the reaction force and the force of friction, directed against the speed of the body. The body moves uniformly accelerated, and, therefore, the vector of its acceleration is directed along the speed of movement. Let's draw a vector in the figure. Choose a coordinate system as shown in the figure. We write Newton's second law in vector form:

.

Using the main property of vector equalities, we write down the equations for the projections of the vectors included in the last vector equality:

We write the ratio for the force of sliding friction

From equality (2.2) we find the magnitude of the reaction force

From the resulting expression, we substitute into equality (2.3) instead of the magnitude of the reaction force , we obtain the expression

Substituting the resulting expression for the friction force into equation (2.1), we will have a formula for calculating the acceleration of the body:

In the last formula we substitute numerical data in the SI system, we find the value of the acceleration of the movement of the load:

Answer: .

For the minimum value of the force, we determine the direction of the friction force that acts on the resting bar. Imagine that the force is less than the minimum force sufficient to keep the body at rest. In this case, the body will move down, and the friction force applied to it will be directed vertically upwards. In order to stop the body, you need to increase the magnitude of the applied force. In addition, this body is affected by the Earth with a force of gravity directed vertically downwards, as well as a wall with a reaction force directed horizontally to the left. Let's depict in the figure all the forces acting on the body. We take a rectangular Cartesian coordinate system, the axes of which we direct as shown in the figure. For a body at rest, we write Newton's first law in vector form:

.

For the found vector equality, we write the equalities for the projections of vectors on the coordinate axes, we obtain the following equations:

At the minimum value of the external force, the magnitude of the static friction force reaches a maximum value equal to the magnitude of the sliding friction force:

From equality (3.1) we find the value of the reaction force , and substitute it into equation (3.3), we obtain the following expression for the friction force:

.

Let us substitute the right side of this relation instead of the friction force into equation (3.2), we obtain a formula for calculating the magnitude of the applied force:

From the last formula we find the magnitude of the force:

.

Answer: .

Let's depict all the forces acting on a ball moving vertically downwards in the air. It is acted upon by the Earth with the force of gravity and the air with the drag force. We depict the considered forces in the figure. At the initial moment of time, the resultant of all forces has a maximum value, since the speed of the ball is zero and the resistance force is also zero. At this moment, the ball has a maximum acceleration equal to . As the ball moves, the speed of its movement increases, and, consequently, the force of air resistance increases. At some point in time, the drag force reaches a value equal to the value of gravity. From this point in time, the ball moves uniformly. Let's write Newton's first law in vector form for the uniform motion of the ball:

.

Let's direct the OY axis vertically down. For a given vector equality, we write an equality for the projections of vectors onto the OY axis:

. (4.1)

The resistance force depends on the cross-sectional area of ​​the ball and the magnitude of its speed as follows:

, (4.2)

where is the coefficient of proportionality, called the drag coefficient.

Equations (4.1) and (4.2) imply the following relationship:

. (4.3)

We express the mass of the ball in terms of its density and volume, and the volume, in turn, in terms of the radius of the ball:

. (4.4)

From this expression we find the mass and substitute it into equality (4.3), we obtain the following equality:

. (4.5)

We express the cross-sectional area of ​​the ball in terms of its radius:

Taking into account relation (4.6), equality (4.5) takes the following form:

.

Denote as the radius of the first ball; as the radius of the second ball. Let us write the formulas for the velocities of the steady motion of the first and second balls:

From the obtained equalities we find the ratio of speeds:

.

From the condition of the problem, the ratio of the radii of the balls is equal to two. Using this condition, we find the ratio of speeds:

.

Answer: .

On a body moving up along an inclined plane, external bodies act: a) Earth with gravity directed vertically downward; b) an inclined plane with a reaction force directed perpendicular to the inclined plane; c) an inclined plane with friction force directed against the movement of the body; d) an external body with a force directed upwards along an inclined plane. Under the action of these forces, the body moves uniformly accelerated up the inclined plane, and, therefore, the acceleration vector is directed along the movement of the body. Let's depict the acceleration vector in the figure. Let's write Newton's second law in vector form:

.

We choose a rectangular Cartesian coordinate system, the OX axis of which is directed along the acceleration of the body, and the OY axis is perpendicular to the inclined plane. We write Newton's second law in projections onto these coordinate axes, we obtain the following equations:

The sliding friction force is related to the reaction force by the following relationship:

From equality (5.2) we find the magnitude of the reaction force and substitute it into equation (5.3), we have the following expression for the friction force:

. (5.4)

We substitute the right side of equation (5.4) instead of the friction force into equation (5.1), we obtain the following equation for calculating the magnitude of the desired force:

Let's calculate the magnitude of the force:

Answer: .

Let's depict all the forces acting on the bodies and on the block. Consider the process of movement of bodies connected by a thread thrown over a block. The thread is weightless and inextensible, therefore, the magnitude of the tension force in any section of the thread will be the same, i.e. And .

The movements of bodies for any time intervals will be the same, and, therefore, at any moment of time, the velocities and accelerations of these bodies will be the same. From the fact that the block rotates without friction and is weightless, it follows that the tension force of the thread on both sides of the block will be the same, i.e.: .

This implies the equality of the tension forces of the thread acting on the first and second bodies, i.e. . Let us depict the acceleration vectors of the first and second bodies in the figure. Let's draw two x-axes. Let's direct the first axis along the acceleration vector of the first body, the second - along the acceleration vector of the second body. We write Newton's second law for each body in projection onto these coordinate axes:

Taking into account that , and expressing from the first equation , we substitute into the second equation, we get

From the last equality we find the value of acceleration:

.

From equality (1) we find the magnitude of the tension force:

Answer: , .

Two forces act on a small ring as it rotates around a circle: gravity, directed vertically downward, and reaction force, directed towards the center of the ring. We depict these forces in the figure, and also show on it the trajectory of the ringlet. The centripetal acceleration vector of the ring lies in the plane of the trajectory and is directed towards the axis of rotation. Let's show in the picture. Let's write Newton's second law in vector form for a rotating ringlet:

.

We choose a rectangular coordinate system, the OX axis of which will be directed along the centripetal acceleration, and the OY axis - vertically upwards along the axis of rotation. We write Newton's second law in projections onto these coordinate axes:

From equality (7.2) we find the magnitude of the reaction force and substitute it into equation (7.1), we obtain the expression:

. (7.3)

Centripetal acceleration is related to the rotational speed by the ratio: , where is the radius of rotation of the small ring. Let us substitute the right side of the last equality in formula (7.3), we obtain the following relation:

. (7.4)

From the figure we find the value of the tangent of the angle alpha . Taking this expression into account, equality (7.4) takes the form:

From the last equation we find the required height:

Answer: .

Three forces act on a body rotating with the disk: gravity, reaction force and friction force, directed towards the axis of rotation. Let's depict all the forces in the figure. Let's show in this figure the direction of the centripetal acceleration vector. We write Newton's second law in vector form:

.

We choose a rectangular Cartesian coordinate system as shown in the figure. Let's write Newton's second law in projections on the coordinate axes:

; (8.1)

. (8.2)

We write the relation for centripetal acceleration:

. (8.3)

We substitute the right side of equality (8.3) instead of centripetal acceleration into equality (8.1), we get:

. (8.4)

From equation (8.4) it can be seen that the value of the friction force is directly proportional to the radius of rotation, therefore, with an increase in the radius of rotation, the static friction force increases, and at a certain value, the static friction force reaches a maximum value equal to the sliding friction force ().

Taking into account equality (8.2), we obtain expressions for the maximum static friction force:

.

We substitute the right side of the obtained equality instead of the friction force by equality (4), we obtain the following relation:

From this equation we find the limit value of the radius of rotation:

Answer: .

During the flight of a drop, two forces act on it: gravity and drag. Let's depict all the forces in the figure. We choose a vertically directed axis OY, the origin of which is located on the surface of the Earth. Let us write down the basic equation of dynamics:

.

Projecting equality onto the OY axis, we will have the relation:

We divide both parts of the last equality by and simultaneously multiply both parts by , taking into account that , we get the expression:

We divide both parts of this expression into , we get the ratio:

.

We integrate the last relation, we obtain the dependence of the speed on time: .

We find the constant from the initial conditions ( ), we obtain the desired dependence of the speed on time:

.

Determine the maximum speed from the condition :

.

Answer: ; .

Let us depict in the figure the forces acting on the washer. We write Newton's second law in projections on the axes OX, OY and OZ

Because , then for the entire trajectory of the washer for the friction force, the formula is valid, which, taking into account the equality for OZ, is converted to the form:

Taking this relation into account, the equality for the OX axis takes the form

Projecting Newton's second law on the tangent to the trajectory of the puck at the point under consideration, we obtain the relation:

where is the value of tangential acceleration. Comparing the right parts of the last equalities, we conclude that .

Since and , taking into account the previous relation, we have the equality , whose integration leads to the expression , where is the integration constant. Substitute in the last expression , we obtain the dependence of the speed on the angle :

We determine the constant from the initial conditions (when . ) . With this in mind, we write the final dependence

.

The minimum velocity value is reached when , and the velocity vector is directed parallel to the OX axis and its value is equal to .

The general equation of dynamics for a system with any constraints (joined d'Alembert-Lagrange principle or general equation of mechanics):

where is the active force applied to the -th point of the system; is the strength of the bond reaction; - point inertia force; - possible movement.

In the case of equilibrium of the system, when all the forces of inertia of the points of the system go to zero, it passes into the principle of possible displacements. It is usually used for systems with ideal constraints for which the condition

In this case, (229) takes one of the forms:

,

,

. (230)

In this way, according to the general equation of dynamics, at any moment of motion of a system with ideal constraints, the sum of the elementary works of all active forces and inertia forces of the points of the system is equal to zero at any possible displacement of the system allowed by the constraints.

The general equation of dynamics can be given other equivalent forms. Expanding the scalar product of vectors, it can be expressed as

where are the coordinates of the -th point of the system. Taking into account that the projections of inertial forces on the coordinate axes through the projections of accelerations on these axes are expressed by the relations

,

the general equation of dynamics can be given the form

In this form it is called the general equation of dynamics in analytical form.

When using the general equation of dynamics, it is necessary to be able to calculate the elementary work of the system inertia forces on possible displacements. For this, the corresponding formulas for elementary work obtained for ordinary forces are used. Let us consider their application to the forces of inertia of a rigid body in particular cases of its motion.

With forward movement. In this case, the body has three degrees of freedom and, due to the imposed constraints, can only perform translational motion. Possible movements of the body, which allow connections, are also translational.

The forces of inertia in translational motion are reduced to the resultant . For the sum of elementary work of inertia forces on the translational possible displacement of the body, we obtain

where is the possible displacement of the center of mass and any point of the body, since the translational possible displacement is the same for all points of the body: the accelerations are the same, i.e. .

When a rigid body rotates around a fixed axis. The body in this case has one degree of freedom. It can rotate around a fixed axis. Possible displacement, which is allowed by superimposed constraints, is also a rotation of the body through an elementary angle around a fixed axis.

The forces of inertia, reduced to a point on the axis of rotation, are reduced to the main vector and the main moment. The main vector of inertial forces is applied to a fixed point, and its elementary work on a possible displacement is zero. For the main moment of inertia forces, elementary work not equal to zero will be performed only by its projection onto the axis of rotation. Thus, for the sum of the work of the inertial forces on the considered possible displacement, we have

,

if the angle is reported in the direction of the arc arrow of the angular acceleration .

in flat motion. The constraints imposed on a rigid body in this case allow only a possible plane displacement. In the general case, it consists of a translational possible movement along with the pole, for which we choose the center of mass, and rotation by an elementary angle around the axis passing through the center of mass and perpendicular to the plane, parallel to which the body can perform a plane motion.

Since the forces of inertia in a plane motion of a rigid body can be reduced to the main vector and the main moment (if the center of mass is chosen as the center of reference), then the sum of the elementary work of the forces of inertia on a possible plane displacement will be reduced to the elementary work of the inertial force vector on the possible displacement of the center of mass and the elementary work of the main moment of inertia forces on the elementary rotational movement around the axis passing through the center of mass. In this case, elementary work not equal to zero can only be performed by the projection of the main moment of inertia forces on the axis , i.e. . Thus, in the case under consideration, we have

if the rotation by an elementary angle is directed along the arc arrow for .

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The external forces of a mechanical system are the forces with which they act on the points of the body system and points that are not included in the system under consideration. The internal forces of the mechanical

Theorem on the motion of the center of mass
The center of mass of the system moves in the same way as a material point, whose mass is equal to the mass of the entire system, if all external forces applied to the mechanical system act on the point:

Number of movement point and system
The amount of motion of a material point is a vector equal to the product of the mass of the point

Theorem on the change in momentum of a point
The theorem on the change in the momentum of a point in differential form: the first time derivative of the momentum of a point is equal to the force acting on the point:

Theorem on the change in the momentum of the system
The theorem on the change in the momentum of the system in differential form: the time derivative of the momentum of the system is equal to the vector sum of all external forces acting on the system

Laws of conservation of momentum
The laws of conservation of the momentum of the system are obtained as special cases of the theorem on the change in the momentum for the system depending on the features of the system of external forces applied to the race.

Theorem on the change of the kinetic moment
For a material point with a mass moving at a speed

Theorem on the change in the angular momentum of a point
The first time derivative of the angular momentum of a point with respect to any center is equal to the moment of force with respect to the same center:

Theorem on the change in the angular momentum of the system
The first time derivative of the angular momentum of the system with respect to any point is equal to the vector sum of the moments of external forces acting on the system with respect to the same point.

Laws of conservation of momentum
1. If the main moment of the external forces of the system relative to the point is equal to zero, i.e.

Differential equation of rotation of a rigid body around a fixed axis
From the theorem on the change of the kinetic moment (172") follows the differential equation for the rotation of a rigid body around a fixed axis

Theorem on the change in the angular momentum of the system in relative motion with respect to the center of mass
Let the mechanical system move relative to the main coordinate system. Let's take a mobile system

Differential equations of plane motion of a rigid body
For a rigid body that moves in a plane and, therefore, has three degrees of freedom, we respectively obtain the following three differential equations:

Force work
The work of a force on any displacement is one of the main characteristics that evaluate the action of a force on this displacement.

Kinetic energy
Kinetic energy of a point and system. The kinetic energy of a material point is called half the product of the mass of the point and the square of its speed, i.e.

Theorem on the change in the kinetic energy of a point
The theorem on the change in the kinetic energy of a point in differential form: the differential of the kinetic energy of a point is equal to the elementary work of the force acting on the point.

Theorem on the change in the kinetic energy of the system
The theorem on the change in the kinetic energy of the system in differential form: the differential of the kinetic energy of the system is equal to the sum of the elementary works of all external and internal systems

d'Alembert's principle for a material point
The d'Alembert principle for a free material point is equivalent to the basic law of dynamics. For a non-free point, it is equivalent to the basic law together with the axiom of connections. The equation is movable

d'Alembert's principle for a system of material points
Consider a system of material points. To each point of the system, in the general case, the resultant active is applied

Forces of inertia of a rigid body in particular cases of its motion
With forward movement. If a rigid body moves forward, then the accelerations of its points are the same. The inertial forces of these points constitute a system of parallel forces directed in one

Possible movements
For one point, a possible (virtual) movement is such an infinitely cute (elementary) mental movement, which is allowed at the considered moment of time superimposed on t

The elementary work of a force on a possible displacement. Perfect Connections
The elementary work of a force on the possible displacement of its point of application is calculated using the usual formulas for elementary work, i.e.

The principle of possible movements
The principle of possible displacements, or the Lagrange principle, contains the necessary and sufficient conditions for the equilibrium of some mechanical systems. It is formulated as follows: for

Generalized system coordinates
Let the system consist of points and, consequently, its position in space at each moment of time is determined by

Generalized forces
Let us write down the sum of the elementary work of the forces acting on the points of the system on the possible displacement of the system:

Generalized Force Calculation
1. The generalized force can be calculated by formula (227), which determines it, i.e. . 2. Generalized

Lagrange equations of the second kind
The Lagrange equations can be considered as an algorithm for obtaining differential equations of system motion, i.e. differential equations with respect to generalized coordinates. Lagr equations

Solving Dynamics Problems
Example 7. On a vertical section

Bibliographic list
1. Nikitin N.N. Theoretical mechanics course: textbook for mechanical engineering. and instrument maker. specialist. universities / N.N. Nikitin. - M .: Higher. school, 1990. 607p. 2. Butenin N.V. Theoretical mechanics course

An example of solving a problem using the general equation of dynamics (the d'Alembert-Lagrange principle) for a system with rigid bodies, weights, pulleys and a block connected by threads.

Content

The task

The mechanical system consists of homogeneous stepped pulleys 1 and 2 wrapped with threads, weights 3-6 attached to these threads, and a weightless block. The system moves in a vertical plane under the action of gravity and a pair of forces with a moment M = 10 Nm applied to pulley 1. The radii of the steps of pulley 1 are: R 1 \u003d 0.2 m, r 1 \u003d 0.1 m, and pulley 2 - R 2 \u003d 0.3 m, r 2 \u003d 0.15 m; their radii of gyration relative to the axes of rotation are, respectively, ρ 1 = 0.1 m and ρ 2 = 0.2 m.

Neglecting friction, determine the acceleration of the load 5. The weights of the pulleys and loads are given: P 1 = 40 N, P 2 = 0 , P 3 = 0 , P 4 = 20 N, P 5 = 30 N, P 6 = 10 N. Cargoes whose weights are equal to zero should not be shown on the drawing.

indication. When solving a problem, use general equation of dynamics (d'Alembert-Lagrange principle).

The solution of the problem

Given: R 1 \u003d 0.2 m, r 1 \u003d 0.1 m, R 2 \u003d 0.3 m, r 2 \u003d 0.15 m, ρ 1 = 0.1 m, ρ 2 = 0.2 m. P 1 = 40 N, P 2 = 0 , P 3 = 0 , P 4 = 20 N, P 5 = 30 N, P 6 = 10 N, M = 10 Nm.

To find: a 5 .

Establishment of kinematic relationships

Let us establish kinematic relations. Let V 4 , V 5 , V 6 , a 4 , a 5 , a 6 , δS 4 , δS 5 , δS 6 - speeds, accelerations and small displacements of loads 4,5 and 6. Let ω 1 , ω 2 , ε 1 , ε 2 , δφ 1 , δφ 2 - angular velocities, angular accelerations and small angles of rotation of pulleys 1 and 2.

Thread speed between bodies 2, 4 and 5:
. From here.
Thread speed between pulleys 1 and 2:
. From here
.
Thread speed between bodies 1 and 6:
.

So, we have found a connection between the speeds of bodies.
;
;
.

Since accelerations are derivatives of velocities with respect to time, ,
then differentiating the previous formulas with respect to time, we find the relationship between the accelerations:
;
;
.

Since velocities are derivatives of time displacements, there is a similar relationship between infinitesimal displacements.
;
;
.

Active external forces

Consider external forces acting on the system.
These are the gravity forces of bodies P 1 = 40 N, P 4 = 20 N, P 5 = 30 N and P 6 = 10 N pointing down;
given pair of forces with moment M = 10 Nm;
axes pressure force N 1 , N 2 and N pulleys 1, 2 and weightless block;
reaction force N 4 and N 6 acting on loads from surfaces perpendicular to these surfaces.

Forces of inertia

We will solve this problem using the general equation of dynamics, applying the d'Alembert-Lagrange principle. It lies in the fact that first we introduce the forces of inertia. After the introduction of inertial forces, the problem of dynamics turns into a problem of statics. That is, we need to find the unknown forces of inertia so that the system is in equilibrium. We solve this problem of statics by applying the d'Alembert principle. That is, we consider that the system has made a small displacement. Then in equilibrium, the sum of the work of all forces, with such a displacement, is equal to zero.

So, in the first step, we introduce forces of inertia. To do this, we assume that the system moves with some, not yet determined, acceleration. That is, pulleys 1 and 2 rotate with angular accelerations ε 1 and ε 2 , respectively; loads 4,5 and 6 perform translational motion with accelerations a 4 , a 5 and a 6 , respectively. There are connections between these accelerations that we found earlier. That is, all these accelerations can be expressed in terms of one acceleration a 5 . The forces of inertia are defined so that they are equal in absolute value and opposite in direction to those forces (and moments of forces) that, according to the laws of dynamics, would create the expected accelerations (in the absence of other forces).

We determine the modules (absolute values) of forces and moments of inertia and express them through a 5 .
Let - masses of bodies;
- the moment of inertia of the pulley 1.
Moment of inertia acting on pulley 1:
.
Inertia forces acting on weights 4, 5 and 6:
;
;
.

We depict the forces of inertia in the drawing, given that their directions are opposite to accelerations.

Application of the general equation of dynamics

We give the system an infinitesimal displacement. Let the load 5 move a small distance δS 5 . Then the angle of rotation δφ 1 pulley 1 and displacement δS 4 and δS 6 loads 4 and 6 are determined using the previously established kinematic relationships. Since the threads are inextensible, they do not do work during such a movement. This means that the system has ideal connections. Therefore, we can apply the general equation of dynamics:
,
according to which the sum of the work of all active forces and inertia forces, with such a movement, is equal to zero.

Determination of the sum of work of external active forces and inertia forces

The work that the force does when moving the point of its application to a small displacement is equal to the scalar product of the vectors, that is, the product of the modules of the vectors F and ds and the cosine of the angle between them.

The work done by the moment of forces is calculated in the same way:
.

We determine the work of all active forces and inertial forces. Since the centers of the axes of pulleys 1, 2 and the weightless block do not move, the forces P 1 , N 1 , N 2 and N do no work. Since the forces N 4 and N 6 are perpendicular to the movements of loads 4 and 6, then these forces also do not do work.

We find the sum of the work of the remaining active forces and the forces of inertia.

.
We substitute expressions for the forces of inertia and apply kinematic relations.

.
We reduce by δS 5 and transform.

.
We substitute numerical values.

;
;

The principle of possible movements: for the equilibrium of a mechanical system with ideal connections, it is necessary and sufficient that the sum of the elementary works of all active forces acting on it for any possible displacement be equal to zero. or in projections: .

The principle of possible displacements gives in a general form the equilibrium conditions for any mechanical system, gives a general method for solving problems of statics.

If the system has several degrees of freedom, then the equation of the principle of possible displacements is made up for each of the independent displacements separately, i.e. there will be as many equations as the system has degrees of freedom.

The principle of possible displacements is convenient in that when considering a system with ideal connections, their reactions are not taken into account and it is necessary to operate only with active forces.

The principle of possible movements is formulated as follows:

To the mother. the system, subject to ideal constraints, was at rest, it is necessary and sufficient that the sum of elementary works performed by active forces on possible displacements of the points of the system be positive

General dynamics equation - when a system moves with ideal connections at any given moment of time, the sum of elementary works of all applied active forces and all inertia forces on any possible movement of the system will be equal to zero. The equation uses the principle of possible displacements and the d'Alembert principle and allows one to compose differential equations of motion for any mechanical system. Gives a general method for solving problems of dynamics.

Compilation sequence:

a) the specified forces acting on it are applied to each body, and also the forces and moments of pairs of inertia forces are conditionally applied;

b) inform the system of possible movements;

c) compose the equations of the principle of possible displacements, considering the system to be in equilibrium.

It should be noted that the general equation of dynamics can also be applied to systems with non-ideal bonds, only in this case the reactions of non-ideal bonds, such as, for example, the friction force or the rolling friction moment, must be classified as active forces.

The work on the possible displacement of both active and inertia forces is sought in the same way as the elementary work on the actual displacement:

Possible work of force: .

Possible work of the moment (pair of forces): .

The generalized coordinates of a mechanical system are independent parameters q 1 , q 2 , …, q S of any dimension, which uniquely determine the position of the system at any time.

The number of generalized coordinates is S - the number of degrees of freedom of the mechanical system. The position of each νth point of the system, that is, its radius vector, in the general case, can always be expressed as a function of generalized coordinates:

The general equation of dynamics in generalized coordinates looks like a system of S equations as follows:

;

;

……..………. ;

(25)

………..……. ;

,

here is the generalized force corresponding to the generalized coordinate :

(26)

a is the generalized inertia force corresponding to the generalized coordinate :

The number of independent possible displacements of the system is called the number of degrees of freedom of this system. For example. the ball on the plane can move in any direction, but any possible movement can be obtained as the geometric sum of two movements along two mutually perpendicular axes. A free rigid body has 6 degrees of freedom.

Generalized forces. For each generalized coordinate, one can calculate the corresponding generalized force Qk.

The calculation is made according to this rule.

To determine the generalized force Qk corresponding to the generalized coordinate q k, you need to give this coordinate an increment (increase the coordinate by this amount), leaving all other coordinates unchanged, calculate the sum of the work of all forces applied to the system on the corresponding displacements of the points and divide it by the increment of the coordinate:

(7)

where is displacement i-that point of the system obtained by changing k-th generalized coordinate.

The generalized force is determined using elementary work. Therefore, this force can be calculated differently:

And since there is an increment of the radius vector due to the increment of the coordinates with the remaining coordinates and time unchanged t, the ratio can be defined as a partial derivative of . Then

where the coordinates of the points are functions of the generalized coordinates (5).

If the system is conservative, that is, the movement occurs under the action of potential field forces whose projections are , where , and the coordinates of the points are functions of generalized coordinates, then

The generalized force of a conservative system is a partial derivative of the potential energy with respect to the corresponding generalized coordinate with a minus sign.

Of course, when calculating this generalized force, the potential energy should be defined as a function of the generalized coordinates

P = P( q 1 , q 2 , q 3 ,…,qs).

Remarks.

First. When calculating the generalized reaction forces, ideal bonds are not taken into account.

Second. The dimension of the generalized force depends on the dimension of the generalized coordinate.

Lagrange equations of the 2nd kind are derived from the general equation of dynamics in generalized coordinates. The number of equations corresponds to the number of degrees of freedom:

(28)

To compose the Lagrange equation of the 2nd kind, generalized coordinates are chosen and generalized velocities are found . The kinetic energy of the system is found, which is a function of the generalized velocities , and, in some cases, generalized coordinates. The operations of differentiation of the kinetic energy are performed, provided for by the left-hand sides of the Lagrange equations. The resulting expressions are equated to generalized forces, for which, in addition to formulas (26), the following are often used when solving problems:

(29)

In the numerator of the right side of the formula - the sum of the elementary work of all active forces on the possible displacement of the system, corresponding to the variation of the i-th generalized coordinate - . With this possible displacement, all other generalized coordinates do not change. The resulting equations are differential equations of motion of a mechanical system with S degrees of freedom.

The principle of possible displacements, which provides a general method for solving problems of statics, can be applied to solving problems of dynamics. As is known, according to the d'Alembert principle, the totality of all forces acting on a mechanical system and the forces of inertia forms a balanced system of forces at each moment of time. Then, applying the principle of possible displacements to these forces, we obtain the equation for the mechanical system

This equation expresses the following d'Alembert-Lagrange principle: when a mechanical system moves at each moment of time, the sum of elementary works of all forces acting on the system and all inertia forces on any possible displacement of the system is equal to zero. Equation (24.1) is called the general equation of dynamics.

The first term of equation (24.1) includes the work of active forces and the work of the reactions of bonds. If ideal constraints are imposed on the system, then for their reactions

and the general equation of dynamics for a system with ideal constraints takes the form

Since equations (24.1), (24.2) include the work of inertial forces, the magnitude of which is expressed in terms of accelerations of points, these equations make it possible to draw up differential equations of motion of a mechanical system. If the system is a collection of some solid bodies, then it is advisable to replace the set of inertia forces of all points of each body with their force equivalents: a force applied in any center equal to the main vector of the body's inertia forces, and a pair of inertia forces with a moment equal to the main moment inertia about this center.

For a system with s degrees of freedom, work equation

(24.2) can be written in terms of generalized forces and generalized coordinates in the form

where Qj- generalized active force; Q 1*- generalized inertia force corresponding to the generalized coordinate q f .

Since the possible movements 8q, are independent of each other and each of them is generally not equal to zero, then condition (24.3) will be satisfied if

where s- the number of generalized coordinates or the number of degrees of freedom of the system.

Equations (24.4) express general equation of dynamics in generalized forces.

Problem 24.1. The mechanical system (Fig. 24.1) consists of a two-stage pulley I(weight R ] - 20 N, step radii R- 0.4 m G - 0.2 m, radius of inertia about the axis of rotation p = 0.3 m), wrapped with threads, at the ends of which a load is attached BUT(weighing R 2= 10 N) and a roller (a solid homogeneous cylinder weighing P 3 = 80 N). The roller rolls without slipping on a rough inclined surface with an inclination angle a = 30°. The system moves in a vertical plane under the action of gravity and torque M - 6 N m applied to the pulley I. Determine the angular acceleration of the pulley, assuming the bodies are absolutely rigid and the threads are inextensible.

Solution. 1. Consider the motion of a mechanical system consisting of bodies 1, 2, 3, connected by threads. The connections imposed on the system are ideal. The system has one degree of freedom. Let us choose as a generalized coordinate the angle cp, - the angle of rotation of the pulley 1.

To determine the angular acceleration e of the pulley, we apply the general equation of dynamics (24.2)

where 28/4 ^ is the sum of elementary works of active forces; 28/4” - the sum of the elementary work of the forces of inertia.

2. We depict active forces in the drawing P x, P 2, P 3 and torque M. The reactions of ideal bonds (at points O and L) are not shown in the drawing.

We set the direction of the angular acceleration s of the pulley counterclockwise. In accordance with this, we depict the acceleration in the drawing a 2 load and acceleration and in center of gravity IN cylindrical roller. Now we add inertia forces to the active forces acting on the system, directing them opposite to the corresponding acceleration. The numerical values ​​of these quantities are determined by the formulas

The values ​​of the moments of inertia are substituted into these formulas J0 pulley and J in solid homogeneous cylinder 3.

3. We inform the system of the possible displacement 5φj >0; while the cargo L will get move 5 s2, dot IN roller - moving 5 s B , and the skating rink 3 will turn through an angle 5f 3 directed counterclockwise.

Compiling equation (a), we get

To solve this equation and determine the angular acceleration e, it is necessary to perform two preparatory operations: to express all displacements in terms of an increment of the generalized coordinate and to express the magnitude of all accelerations in terms of the desired acceleration.

All movements involved in equation (c) are expressed in terms of 5cpj:

When compiling the last equality, it was taken into account that the point TO cylinder 3 is the instantaneous center of velocities.

Acceleration values a 2, a b, s 3 involved in formulas (b), we express through the desired angular acceleration s:

Substituting quantities (b), taking into account equalities (e) and relations (d), into equation (c), after simplifications, we bring it to the form

Since bf, 0, then we equate to zero the expression in curly brackets. From the equation obtained as a result of this equation, we find the desired value

Calculations give the following answer: s = 2.4 s; the sign indicates that the angular acceleration of the pulley is directed as it was assumed at the beginning of the calculation, i.e., as shown in fig. 24.1.

For example, if in the same problem the torque was equal to M = 2 N m, then as a result of calculations according to the formula (g) we would get e \u003d -2.4 s -1; this would mean that in the case under consideration, the angular acceleration of the pulley would be directed opposite to that shown in Fig. 24.1.

Solutions of problems of dynamics contain, as a special case, the solution of the corresponding problem of statics. If for the considered mechanical system (see Fig. 24.1) the equilibrium condition was determined according to the principle of possible displacements, then we would get the calculation equation

As you can see, on the left side of the equality is the expression of the numerator of the formula (g), i.e., the condition is defined under which 8 = 0 (which corresponds to the rest of the system or movement with uniform rotation of the pulley). The meaning of this equality lies in the fact that the generalized active force of the system at a possible displacement 8f is equal to zero, i.e. Q" = 0.