Which mechanical system is called unstable. Equilibrium of a mechanical system. Equilibrium conditions for mechanical systems

It is known that for the equilibrium of a system with ideal constraints, it is necessary and sufficient that or. (7)

Since the variations of the generalized coordinates are independent of each other and, in the general case, are not equal to zero, it is necessary that
,
,…,
.

For the equilibrium of a system with holonomic confining, stationary, ideal constraints, it is necessary and sufficient that all generalized forces corresponding to the selected generalized coordinates be equal to zero.

Potential Force Case:

If the system is in a potential force field, then

,
,…,

,
,…,

That is, the equilibrium positions of the system can be only for those values ​​of the generalized coordinates at which the force function U and potential energy NS have extreme values ​​( max or min).

The concept of equilibrium stability.

Having determined the positions in which the system can be in equilibrium, it is possible to determine which of these positions are realizable and which are unrealizable, that is, to determine which position is stable and which is unstable.

In general, the necessary equilibrium stability sign according to Lyapunov can be formulated as follows:

Let us take the system out of the equilibrium position by reporting small modulus values ​​of the generalized coordinates and their velocities. If, upon further consideration of the system, the generalized coordinates and their velocities remain small in magnitude, that is, the system will not deviate far from the equilibrium position, then such an equilibrium position is stable.

A Sufficient Condition for Equilibrium Stability system is determined the Lagrange-Dirichlet theorem :

If in the state of equilibrium of a mechanical system with ideal connections the potential energy has a minimum value, then such an equilibrium position is stable.

,
- stable.

This lecture addresses the following issues:

1. Equilibrium conditions for mechanical systems.

2. Stability of balance.

3. An example of the determination of equilibrium positions and the study of their stability.

The study of these issues is necessary to study the oscillatory movements of a mechanical system relative to the equilibrium position in the discipline "Machine parts", for solving problems in the disciplines "Theory of machines and mechanisms" and "Resistance of materials".

An important case of motion of mechanical systems is their oscillatory motion. Oscillations are repetitive movements of a mechanical system relative to some of its position, occurring more or less regularly in time. The course work examines the oscillatory motion of a mechanical system relative to the equilibrium position (relative or absolute).

A mechanical system can vibrate for a sufficiently long period of time only near a stable equilibrium position. Therefore, before composing the equations of oscillatory motion, it is necessary to find the equilibrium positions and investigate their stability.

Equilibrium conditions for mechanical systems.

According to the principle of possible displacements (the basic equation of statics), in order for a mechanical system, on which ideal, stationary, holding and holonomic constraints are imposed, to be in equilibrium, it is necessary and sufficient that all generalized forces are equal to zero in this system:

where - generalized force corresponding j - th generalized coordinate;

s- the number of generalized coordinates in the mechanical system.

If for the system under study, differential equations of motion were compiled in the form of Lagrange equations of the second kind, then to determine the possible equilibrium positions, it is enough to equate the generalized forces to zero and solve the resulting equations with respect to generalized coordinates.

If the mechanical system is in equilibrium in a potential force field, then from equations (1) we obtain the following equilibrium conditions:

Therefore, in the equilibrium position, the potential energy has an extreme value. Not every equilibrium defined by the above formulas can be realized in practice. Depending on the behavior of the system when deviating from the equilibrium position, one speaks of the stability or instability of this position.

Stable balance

The definition of the concept of stability of an equilibrium position was given in late XIX century in the works of the Russian scientist A.M. Lyapunov. Let's consider this definition.

To simplify the calculations, we will further agree on the generalized coordinates q 1 , q 2 ,...,q s count from the equilibrium position of the system:

where

An equilibrium position is called stable if for any arbitrarily small numberyou can find such a different number , that in the case when the initial values ​​of the generalized coordinates and velocities will not exceed:

the values ​​of the generalized coordinates and velocities during the further movement of the system will not exceed .

In other words, the equilibrium position of the system q 1 = q 2 = ...= q s = 0 is called sustainable if one can always find such sufficiently small initial valuesat which the motion of the systemwill not leave any given arbitrarily small neighborhood of the equilibrium position... For a system with one degree of freedom, the stable motion of the system can be graphically depicted in the phase plane (Fig. 1).For a stable equilibrium position, the movement of the representing point, starting in the area [ ] , will not go beyond the area in the future.

Fig. 1

The equilibrium position is called asymptotically stable , if over time the system approaches the equilibrium position, that is

Determining the stability conditions for an equilibrium position is a rather complicated problem, therefore, we restrict ourselves to the simplest case: the study of the equilibrium stability of conservative systems.

Sufficient conditions for the stability of equilibrium positions for such systems are determined the Lagrange - Dirichlet theorem : the equilibrium position of a conservative mechanical system is stable if, in the equilibrium position, the potential energy of the system has an isolated minimum .

The potential energy of a mechanical system is determined to within a constant. Let us choose this constant so that in the equilibrium position the potential energy is equal to zero:

P (0) = 0.

Then, for a system with one degree of freedom, a sufficient condition for the existence of an isolated minimum, along with the necessary condition (2), will be the condition

Since in the equilibrium position the potential energy has an isolated minimum and P (0) = 0 , then in some finite neighborhood of this position

П (q) = 0.

Functions that have a constant sign and are equal to zero only for zero values ​​of all their arguments are called definite... Consequently, in order for the equilibrium position of the mechanical system to be stable, it is necessary and sufficient that in the vicinity of this position the potential energy is a positive definite function of the generalized coordinates.

For linear systems and for systems that can be reduced to linear for small deviations from the equilibrium position (linearized), the potential energy can be represented in the form of a quadratic form of generalized coordinates

where - generalized stiffness coefficients.

Generalized coefficientsare constant numbers that can be determined directly from the expansion of the potential energy in a series or from the values ​​of the second derivatives of the potential energy with respect to the generalized coordinates in the equilibrium position:

It follows from formula (4) that the generalized stiffness coefficients are symmetric with respect to the indices

For so that sufficient conditions stability of the equilibrium position, the potential energy should be a positive definite quadratic form of its generalized coordinates.

In mathematics, there is Sylvester criterion giving necessary and sufficient conditions for the positive definiteness of quadratic forms: the quadratic form (3) will be positive definite if the determinant composed of its coefficients and all its main diagonal minors are positive, i.e. if the coefficients will satisfy the conditions

.....

In particular, for a linear system with two degrees of freedom, the potential energy and the conditions of the Sylvester criterion will have the form

In a similar way, one can study the positions of relative equilibrium if, instead of the potential energy, one introduces the potential energy of the reduced system.

NS An example of determining equilibrium positions and studying their stability

Fig. 2

Consider a mechanical system consisting of a tube AB, which is the pivot OO 1 connected to the horizontal axis of rotation, and a ball that moves through the tube without friction and is connected to a point A tube with a spring (Fig. 2). Let us determine the equilibrium positions of the system and estimate their stability for the following parameters: tube length l 2 = 1 m , rod length l 1 = 0,5 m . length of undeformed spring l 0 = 0.6 m, spring rate c= 100 N / m. Tube weight m 2 = 2 kg, rods - m 1 = 1 kg and the ball - m 3 = 0.5 kg. Distance OA equals l 3 = 0.4 m.

Let us write down the expression for the potential energy of the system under consideration. It consists of the potential energy of three bodies in a uniform gravity field and the potential energy of a deformed spring.

The potential energy of a body in a gravity field is equal to the product of the body's weight by the height of its center of gravity above the plane in which the potential energy is considered to be zero. Let the potential energy be zero in the plane passing through the axis of rotation of the rod OO 1, then for gravity

For the elastic force, the potential energy is determined by the amount of deformation

Let us find the possible equilibrium positions of the system. The values ​​of the coordinates at the equilibrium positions are the roots of the following system of equations.

A similar system of equations can be drawn up for any mechanical system with two degrees of freedom. In some cases, an exact solution to the system can be obtained. For system (5), such a solution does not exist; therefore, the roots must be sought using numerical methods.

Solving the system of transcendental equations (5), we obtain two possible equilibrium positions:

To assess the stability of the obtained equilibrium positions, we find all the second derivatives of the potential energy with respect to the generalized coordinates, and from them we determine the generalized stiffness coefficients.

Allows analysis general patterns motion, if the dependence of the potential energy on the coordinates is known. Consider, for example, the one-dimensional motion of a material point (particle) along the axis 0x in the potential field shown in Fig. 4.12.

Figure 4.12. Particle motion near stable and unstable equilibrium positions

Since in a uniform field of gravity forces the potential energy is proportional to the height of the body's rise, one can imagine an ice slide (neglecting friction) with a profile corresponding to the function N (x) on the image.

From the law of conservation of energy E = K + P and from the fact that kinetic energy K = E - P is always non-negative, it follows that the particle can be located only in regions where E> P... The figure shows a particle with full energy E can only move in areas

In the first area, its movement will be limited (finitely): with a given stock full energy the particle cannot overcome the "slides" on its way (they are called potential barriers) and is doomed to forever remain in the "valley" between them. Forever - from the point of view of classical mechanics, which we are studying now. At the end of the course, we will see how quantum mechanics helps a particle to get out of imprisonment in a potential hole - the region

In the second region, the particle motion is not limited (infinite), it can move infinitely far from the origin to the right, but on the left its motion is still limited by a potential barrier:

Video 4.6. Demonstration of finite and infinite movements.

At the points of the potential energy extremum x MIN and x MAX the force acting on the particle is zero, because the derivative of the potential energy is zero:

If a particle at rest was placed at these points, then it would remain there ... again, forever, if it were not for the fluctuations of its position. There is nothing strictly at rest in this world, a particle can experience small deviations (fluctuations) from the equilibrium position. In this case, naturally, forces arise. If they return the particle to the equilibrium position, then such equilibrium is called sustainable... If, when the particle is deflected, the arising forces move it even further away from the equilibrium position, then we are dealing with unstable equilibrium, and the particle usually does not stay in this position for a long time. By analogy with the ice slide, one can guess that the position will be stable at the minimum potential energy, and unstable - at the maximum.

Let us prove that this is indeed the case. For a particle at the extremum point x M (x MIN or x MAX) the force acting on it F x (x M) = 0... Let, due to fluctuations, the particle coordinate changes by a small amount x... With such a change in the coordinate, the force will begin to act on the particle

(the prime denotes the derivative with respect to the coordinate x). Considering that F x = -П ", we obtain the expression for the force

At the minimum point, the second derivative of the potential energy is positive: U "(x MIN)> 0... Then, for positive deviations from the equilibrium position x > 0 the resulting force is negative, and at x<0 the force is positive. In both cases, the force prevents the particle coordinate from changing, and the equilibrium position at the minimum potential energy is stable.

Conversely, at the maximum point, the second derivative is negative: U "(x MAX)<0 ... Then an increase in the coordinate of the particle Δx leads to the appearance of a positive force, which further increases the deviation from the equilibrium position. At x<0 the force is negative, that is, in this case, it also contributes to the further deflection of the particle. This equilibrium position is unstable.

Thus, a stable equilibrium position can be found by jointly solving the equation and inequality

Video 4.7. Potential holes, potential barriers and equilibrium: stable and unstable.

Example... The potential energy of a diatomic molecule (for example, H 2 or About 2) is described by an expression of the form

where r is the distance between atoms, and A, B- positive constants. Determine the equilibrium distance r M between the atoms of the molecule. Is a diatomic molecule stable?

Solution... The first term describes the repulsion of atoms at small distances (the molecule resists compression), the second describes the attraction at large distances (the molecule resists breaking). In accordance with the above, the equilibrium distance is found by solving the equation

Differentiating the potential energy, we obtain

We now find the second derivative of the potential energy

and substitute there the value of the equilibrium distance r M :

The equilibrium position is stable.

In fig. 4.13 presents an experiment on the study of potential curves and conditions of equilibrium of the ball. If the ball is placed on the potential curve model at a height greater than the height of the potential barrier (the energy of the ball is greater than the energy of the barrier), then the ball overcomes the potential barrier. If the initial height of the ball is less than the height of the barrier, then the ball remains within the potential well.

The ball, placed at the highest point of the potential barrier, is in unstable equilibrium, since any external influence leads to the transition of the ball to the lowest point of the potential well. At the bottom point of the potential well, the ball is in stable equilibrium, since any external influence leads to the return of the ball to the bottom point of the potential well.

Rice. 4.13. Experimental study of potential curves

Additional Information

http://vivovoco.rsl.ru/quantum/2001.01/KALEID.PDF - Supplement to the journal "Kvant" - reasoning about stable and unstable equilibrium (A. Leonovich);

http://mehanika.3dn.ru/load/24-1-0-3278 - Targ S.M. A short course in theoretical mechanics, Publishing House, Higher School, 1986 - pp. 11–15, §2 - the initial provisions of statics.

We represent equations (16) from § 107 and (35) or (38) in the form:

Let us show that from these equations, which are consequences of the laws presented in § 74, all the original results of statics are obtained.

1. If a mechanical system is at rest, then the velocities of all its points are equal to zero and, therefore, where O is any point. Then equations (40) give:

Thus, conditions (40) are necessary conditions for the equilibrium of any mechanical system. This result contains, in particular, the solidification principle formulated in § 2.

But for any system conditions (40), obviously, are not sufficient equilibrium conditions. For example, if shown in Fig. 274 points and are free, then under the action of forces they can move towards each other, although conditions (40) for these forces will be fulfilled.

The necessary and sufficient conditions for the equilibrium of a mechanical system will be set forth in Sections 139 and 144.

2. Let us prove that conditions (40) are not only necessary, but also sufficient equilibrium conditions for forces acting on an absolutely rigid body. Let a system of forces begin to act on a free rigid body at rest, satisfying conditions (40), where O is any point, i.e., in particular, point C. Then equations (40) give, and since the body initially was at rest, then At point C is motionless and the body can only have rotation with an angular velocity c around some instantaneous axis (see § 60). Then, according to formula (33), the body will have. But there is a projection of the vector onto the axis, and since then and whence it follows that and that is, when conditions (40) are satisfied, the body remains at rest.

3. From the previous results follow, in particular, the initial positions 1 and 2, formulated in § 2, since it is obvious that the two forces shown in Fig. 2, satisfy conditions (40) and are balanced, and that if we add (or subtract from) a balanced system of forces to the forces acting on the body, i.e., satisfying conditions (40), then neither these conditions nor equations (40), determining body movement will not change.

I will consider a material point, the movement of which is limited in such a way that it has only one degree of freedom.

This means that its position can be determined using a single quantity, such as the x coordinate. An example is a ball sliding without friction on a wire fixed motionlessly, bent in a vertical plane (Fig. 26.1, a).

Another example is a ball attached to the end of a spring, sliding without friction to a horizontal guide (Fig. 26.2, a).

A conservative force acts on the ball: in the first case, it is the force of gravity, in the second, the elastic force of the deformed spring. Potential energy plots are shown in Fig. 26.1, b and 26.2, b.

Since the balls move along the wire without friction, the force with which the wire acts on the ball is in both cases perpendicular to the speed of the ball and, therefore, does not work on the ball. Therefore, energy conservation takes place:

From (26.1) it follows that the kinetic energy can increase only due to a decrease in the amusing energy. Therefore, if the ball is in such a state that its speed is zero, and the potential energy has a minimum value, then without external influence it will not be able to move, that is, it will be in equilibrium.

The U minima correspond to the values ​​on the graphs equal (in Fig. 26.2 is the length of the undeformed squad) The condition for the minimum potential energy has the form

In accordance with t (22.4), condition (26.2) is equivalent to the fact that

(in the case when U is a function of only one variable). Thus, the position corresponding to the minimum of potential energy has the property that the force acting on the body is zero.

In the case shown in Fig. 26.1, conditions (26.2) and (26.3) are also satisfied for x equal to (i.e., for the maximum of U). The position of the ball determined by this value will also be in equilibrium. However, this equilibrium, in contrast to equilibrium at, will be unstable: it is enough to slightly remove the ball from this position, as a force arises that will remove the ball from the position. The forces arising from the displacement of the ball from a stable equilibrium position (for which) are directed in such a way that they tend to return the ball to the equilibrium position.

Knowing the form of the function t, which expresses the potential energy, it is possible to draw a number of conclusions about the nature of the movement of the part. Let us explain this using the graph shown in Fig. 26.1, b. If the total energy has the value indicated in the figure, then the particle can move either in the range from to or in the range from to infinity. In the region, the particle cannot penetrate, since the potential energy cannot, become greater than the total energy (if this happened, then the kinetic energy would become negative). Thus, the region is a potential barrier through which the particle cannot penetrate, having a given supply of total energy. The area is called a potential pit.

If the particle during its motion cannot move away to infinity, the motion is called finite. If the particle can go arbitrarily far, the motion is called infinite. A particle in a potential well makes a finite motion. The motion of a particle with negative total energy in the central field of attractive forces will also be finite (it is assumed that the potential energy vanishes at infinity).