DEFINITION

Stable balance- this is a balance in which the body, taken out of a position of balance and left to itself, returns to its previous position.

This happens if, with a slight displacement of the body in any direction from the initial position, the resultant of the forces acting on the body becomes nonzero and is directed towards the equilibrium position. For example, a ball lying at the bottom of a spherical depression (Fig. 1 a).

DEFINITION

Unstable equilibrium- this is an equilibrium in which the body, taken out of the equilibrium position and left to itself, will deviate even more from the equilibrium position.

In this case, with a slight displacement of the body from the equilibrium position, the resultant of the forces applied to it is nonzero and is directed from the equilibrium position. An example is a ball located at the top of a convex spherical surface (Fig. 1 b).

DEFINITION

Indifferent balance- this is a balance in which the body, taken out of a position of balance and left to itself, does not change its position (state).

In this case, at small displacements of the body from the initial position, the resultant of the forces applied to the body remains equal to zero. For example, a ball lying on a flat surface (Fig. 1, c).

Fig. 1. Various types of body balance on a support: a) stable balance; b) unstable balance; c) indifferent balance.

Static and dynamic balance of bodies

If, as a result of the action of forces, the body does not receive acceleration, it can be at rest or move uniformly in a straight line. Therefore, we can talk about static and dynamic balance.

DEFINITION

Static balance- this is such a balance when, under the action of the applied forces, the body is at rest.

Dynamic balance- this is such a balance when, under the action of forces, the body does not change its motion.

In a state of static equilibrium there is a lantern suspended on cables, any building structure. As an example of dynamic equilibrium, we can consider a wheel that rolls on a flat surface in the absence of frictional forces.

The main types of balance points

Let a second-order linear homogeneous system with constant coefficients be given: \ [\ left \ (\ begin (array) (l) \ frac ((dx)) ((dt)) = (a_ (11)) x + (a_ (12 )) y \\ \ frac ((dy)) ((dt)) = (a_ (21)) x + (a_ (22)) y \ end (array) \ right .. \] This system of equations is autonomous, since the right-hand sides of the equations do not explicitly contain the independent variable \ (t. \)

In matrix form, the system of equations is written as \ [(\ mathbf (X ") = A \ mathbf (X), \; \; \ text (where) \; \; \ mathbf (X) = \ left ((\ begin ( array) (* (20) (c)) x \\ y \ end (array)) \ right),) \; \; (A = \ left ((\ begin (array) (* (20) (c) ) ((a_ (11))) & ((a_ (12))) \\ ((a_ (21))) & ((a_ (22))) \ end (array)) \ right).) \] Equilibrium positions are found from the solution of the stationary equation \ This equation has a unique solution \ (\ mathbf (X) = \ mathbf (0), \) if the matrix \ (A \) is non-degenerate , i.e. under the condition \ (\ det A \ ne 0. \) In the case degenerate matrix the system has an infinite set of equilibrium points.

The classification of equilibrium positions is determined eigenvalues \ ((\ lambda _1), (\ lambda _2) \) matrices \ (A. \) The numbers \ ((\ lambda _1), (\ lambda _2) \) are found from the solution characteristic equation \ [(\ lambda ^ 2) - \ left (((a_ (11)) + (a_ (22))) \ right) \ lambda + (a_ (11)) (a_ (22)) - (a_ (12 )) (a_ (21)) = 0. \] In the general case, when the matrix \ (A \) is nondegenerate, there are \ (4 \) different types of equilibrium points:

The stability of equilibrium positions is determined general stability theorems... So, if the real eigenvalues ​​(or real parts of complex eigenvalues) are negative, then the equilibrium point is asymptotically stable ... Examples of such equilibrium positions are and steady focus .

If the real part of at least one eigenvalue is positive, then the corresponding equilibrium position is unstable ... For example, it might be.

Finally, in the case of purely imaginary roots (the equilibrium point is center) we are dealing with a classical stability in the sense of Lyapunov .

Our next goal is to study the behavior of solutions near equilibrium positions. For systems of \ (2 \) th order, it is convenient to do this graphically using phase portrait , which is a collection phase trajectories on the coordinate plane. Arrows on the phase trajectories show the direction of movement of a point (i.e., some specific state of the system) over time.

Let us consider in more detail each type of equilibrium point and the corresponding phase portraits.

Stable and unstable knot

The eigenvalues ​​\ (((\ lambda _1), (\ lambda _2)) \) points of the "node" type satisfy the following conditions: \ [(\ lambda _1), (\ lambda _2) \ in \ Re, \; \; ( \ lambda _1) \ cdot (\ lambda _2)> 0. \] The following special cases may occur.

The roots \ (((\ lambda _1), (\ lambda _2)) \) are different \ (\ left (((\ lambda _1) \ ne (\ lambda _2)) \ right) \) and negative \ (\ left ( ((\ lambda _1)
Let's construct a schematic phase portrait of such an equilibrium point. Let, for definiteness, \ (\ left | ((\ lambda _1)) \ right |
Since both eigenvalues ​​are negative, the solution \ (\ mathbf (X) = \ mathbf (0) \) is asymptotically stable ... This equilibrium position is called stable knot ... As \ (t \ to \ infty \), the phase curves tend to the origin \ (\ mathbf (X) = \ mathbf (0). \)

Let us specify the direction of the phase trajectories. Since \ [(x \ left (t \ right) = (C_1) (V_ (11)) (e ^ ((\ lambda _1) t)) + (C_2) (V_ (12)) (e ^ ((\ lambda _2) t)),) \; \; (y \ left (t \ right) = (C_1) (V_ (21)) (e ^ ((\ lambda _1) t)) + (C_2) (V_ (22)) (e ^ ((\ lambda _2) t)),) \] then the derivative \ (\ large \ frac ((dy)) ((dx)) \ normalsize \) is \ [\ frac ((dy)) ((dx)) = \ frac ((( C_1) (V_ (21)) (\ lambda _1) (e ^ ((\ lambda _1) t)) + (C_2) (V_ (22)) (\ lambda _2) (e ^ ((\ lambda _2) t )))) (((C_1) (V_ (11)) (\ lambda _1) (e ^ ((\ lambda _1) t)) + (C_2) (V_ (12)) (\ lambda _2) (e ^ ((\ lambda _2) t)))). \] Divide the numerator and denominator by \ (((e ^ ((\ lambda _1) t))): \) \ [\ frac ((dy)) ((dx )) = \ frac (((C_1) (V_ (21)) (\ lambda _1) + (C_2) (V_ (22)) (\ lambda _2) (e ^ (\ left (((\ lambda _2) - (\ lambda _1)) \ right) t)))) (((C_1) (V_ (11)) (\ lambda _1) + (C_2) (V_ (12)) (\ lambda _2) (e ^ (\ left (((\ lambda _2) - (\ lambda _1)) \ right) t)))). \] In this case \ ((\ lambda _2) - (\ lambda _1)
In the case \ ((C_1) = 0 \) the derivative for any \ (t \) is \ [\ frac ((dy)) ((dx)) = \ frac (((V_ (22)))) ((( V_ (12)))), \] ie the phase trajectory lies on a straight line directed along the eigenvector \ ((\ mathbf (V) _2). \)

Now consider the behavior of the phase trajectories for \ (t \ to - \ infty. \) Obviously, the coordinates \ (x \ left (t \ right), y \ left (t \ right) \) tend to infinity, and the derivative \ ( \ large \ frac ((dy)) ((dx)) \ normalsize \) with \ ((C_2) \ ne 0 \) takes the following form: \ [\ frac ((dy)) ((dx)) = \ frac (((C_1) (V_ (21)) (\ lambda _1) (e ^ (\ left (((\ lambda _1) - (\ lambda _2)) \ right) t)) + (C_2) (V_ (22 )) (\ lambda _2))) (((C_1) (V_ (11)) (\ lambda _1) (e ^ (\ left (((\ lambda _1) - (\ lambda _2)) \ right) t) ) + (C_2) (V_ (12)) (\ lambda _2))) = \ frac (((V_ (22)))) (((V_ (12)))), \] i.e. the phase curves at infinity become parallel to the vector \ ((\ mathbf (V) _2). \)

Accordingly, for \ ((C_2) = 0 \) the derivative is \ [\ frac ((dy)) ((dx)) = \ frac (((V_ (21)))) (((V_ (11))) ). \] In this case, the phase trajectory is determined by the direction of the eigenvector \ ((\ mathbf (V) _1). \)

Taking into account the considered properties of phase trajectories, the phase portrait stable node has the form shown schematically in the figure \ (1. \)

Similarly, one can study the behavior of phase trajectories for other types of equilibrium positions. Further, omitting the detailed analysis, let us carry out the main qualitative characteristics of other equilibrium points.

The roots \ (((\ lambda _1), (\ lambda _2)) \) are distinct \ (\ left (((\ lambda _1) \ ne (\ lambda _2)) \ right) \) and are positive \ (\ left ( ((\ lambda _1)> 0, (\ lambda _2))> 0 \ right). \)
In this case, the point \ (\ mathbf (X) = \ mathbf (0) \) is called unstable knot ... Her phase portrait is shown in the figure \ (2. \)

Note that in the case of both a stable and an unstable node, the phase trajectories are tangent to a straight line, which is directed along the eigenvector corresponding to the smaller in absolute value eigenvalue \ (\ lambda. \)

Dystopian node

Let the characteristic equation have one zero root of multiplicity \ (2, \) i.e. consider the case \ ((\ lambda _1) = (\ lambda _2) = (\ lambda) \ ne 0. \) In this case, the system has a basis of two eigenvectors, i.e. the geometric multiplicity of the eigenvalue \ (\ lambda \) is \ (2. \) In terms of linear algebra, this means that the dimension of the eigenspace of the matrix \ (A \) is \ (2: \) \ (\ dim \ ker A = 2 . \) This situation is realized in systems like \ [(\ frac ((dx)) ((dt)) = \ lambda x,) \; \; (\ frac ((dy)) ((dt)) = \ lambda y.) \] The direction of the phase trajectories depends on the sign \ (\ lambda. \) The following two cases are possible here:

Case \ ((\ lambda _1) = (\ lambda _2) = (\ lambda) This equilibrium is called stable dicritic knot (picture \ (3 \)).

Case \ ((\ lambda _1) = (\ lambda _2) = (\ lambda)> 0. \) This combination of eigenvalues ​​corresponds to unstable dicritic node (figure \ (4 \)).

Degenerate knot

Let the eigenvalues ​​of the matrix \ (A \) coincide again: \ ((\ lambda _1) = (\ lambda _2) = (\ lambda) \ ne 0. \) In contrast to the previous case of a dicritical knot, we assume that the geometric multiplicity of the eigenvalue values ​​(or in other words, the dimension of the eigen subspace) is now \ (1. \) This means that the matrix \ (A \) has only one eigenvector \ ((\ mathbf (V) _1). \) The second linearly independent vector, needed to compose the basis, is defined as the vector \ ((\ mathbf (W) _1), \) appended to \ ((\ mathbf (V) _1). \)

In the case \ ((\ lambda _1) = (\ lambda _2) = (\ lambda) the equilibrium point is called stable degenerate knot (figure \ (5 \)).

For \ ((\ lambda _1) = (\ lambda _2) = (\ lambda)> 0 \) the equilibrium position is called unstable degenerate knot (drawing \ (6 \)).

The equilibrium position is under the conditions \ [(\ lambda _1), (\ lambda _2) \ in \ Re, \; \; (\ lambda _1) \ cdot (\ lambda _2) 0. \) Eigenvalues ​​\ ((\ lambda _1) \) and \ ((\ lambda _2) \) are associated with the corresponding eigenvectors \ ((\ mathbf (V) _1) \) and \ ((\ mathbf (V) _2). \) Lines directed along eigenvectors vectors \ ((\ mathbf (V) _1), \) \ ((\ mathbf (V) _2), \) are called separatrices ... They are asymptotes for the remaining phase trajectories, which have the form of hyperbolas. Each of the separatrices can be associated with a certain direction of movement. If the separatrix is ​​associated with a negative eigenvalue \ ((\ lambda _1) 0, \) i.e. for the separatrix associated with the vector \ ((\ mathbf (V) _2), \) the motion is directed from the origin. The phase portrait of the saddle is shown schematically in the figure \ (7. \)

Steady and unstable focus

Now let the eigenvalues ​​\ ((\ lambda _1), (\ lambda _2) \) be complex numbers , the real parts of which are not equal to zero. If the matrix \ (A \) consists of real numbers, then the complex roots will be represented as complex conjugate numbers: \ [(\ lambda _ (1,2)) = \ alpha \ pm i \ beta. \] Let us find out what form the phase trajectories have in the vicinity of the origin. Let's construct a complex solution \ ((\ mathbf (X) _1) \ left (t \ right) \) corresponding to the eigenvalue \ ((\ lambda _1) = \ alpha + i \ beta: \) \ [((\ mathbf (X ) _1) \ left (t \ right) = (e ^ ((\ lambda _1) t)) (\ mathbf (V) _1)) = ((e ^ (\ left ((\ alpha + i \ beta) \ right) t)) \ left ((\ mathbf (U) + i \ mathbf (W)) \ right),) \] where \ ((\ mathbf (V) _1) = \ mathbf (U) + i \ mathbf (W) \) is a complex-valued eigenvector associated with the number \ ((\ lambda _1), \) \ (\ mathbf (U) \) and \ (\ mathbf (W) \) are real vector functions. As a result of the transformations we get \ [((\ mathbf (X) _1) \ left (t \ right) = (e ^ (\ alpha t)) (e ^ (i \ beta t)) \ left ((\ mathbf (U ) + i \ mathbf (W)) \ right)) = ((e ^ (\ alpha t)) \ left ((\ cos \ beta t + i \ sin \ beta t) \ right) \ left ((\ mathbf (U) + i \ mathbf (W)) \ right)) = ((e ^ (\ alpha t)) \ left ((\ mathbf (U) \ cos \ beta t + i \ mathbf (U) \ sin \ beta t + i \ mathbf (W) \ cos \ beta t - \ mathbf (W) \ sin \ beta t) \ right)) = ((e ^ (\ alpha t)) \ left ((\ mathbf (U) \ cos \ beta t + - \ mathbf (W) \ sin \ beta t) \ right)) + (i (e ^ (\ alpha t)) \ left ((\ mathbf (U) \ sin \ beta t + \ mathbf (W) \ cos \ beta t) \ right).) \] The real and imaginary parts in the last expression form the general solution of the system, which looks like: \ [(\ mathbf (X) \ left (t \ right) = ( C_1) \ text (Re) \ left [((\ mathbf (X) _1) \ left (t \ right)) \ right] + (C_2) \ text (Im) \ left [((\ mathbf (X) _1 ) \ left (t \ right)) \ right]) = ((e ^ (\ alpha t)) \ left [((C_1) \ left ((\ mathbf (U) \ cos \ beta t - \ mathbf (W ) \ sin \ beta t) \ right)) \ right.) + (\ left. ((C_2) \ left ((\ mathbf (U) \ sin \ beta t + \ mathbf (W) \ cos \ beta t) \ right) ) \ right]) = ((e ^ (\ alpha t)) \ left [(\ mathbf (U) \ left (((C_1) \ cos \ beta t + (C_2) \ sin \ beta t) \ right) ) \ right. ) + (\ left. (\ mathbf (W) \ left (((C_2) \ cos \ beta t - (C_1) \ sin \ beta t) \ right)) \ right].) \] Represent the constants \ (( C_1), (C_2) \) in the form \ [(C_1) = C \ sin \ delta, \; \; (C_2) = C \ cos \ delta, \] where \ (\ delta \) is some auxiliary angle. Then the solution is written as \ [(\ mathbf (X) \ left (t \ right) = C (e ^ (\ alpha t)) \ left [(\ mathbf (U) \ left ((\ sin \ delta \ cos \ beta t + \ cos \ delta \ sin \ beta t) \ right)) \ right.) + (\ left. (\ mathbf (W) \ left ((\ cos \ delta \ cos \ beta t - \ sin \ delta \ sin \ beta t) \ right)) \ right]) = (C (e ^ (\ alpha t)) \ left [(\ mathbf (U) \ sin \ left ((\ beta t + \ delta) \ right )) \ right. + \ left. (\ mathbf (W) \ cos \ left ((\ beta t + \ delta) \ right)) \ right].) \] So the solution is \ (\ mathbf (X) \ left (t \ right) \) is expanded in the basis given by the vectors \ (\ mathbf (U) \) and \ (\ mathbf (W): \) \ [\ mathbf (X) \ left (t \ right) = \ mu \ left (t \ right) \ mathbf (U) + \ eta \ left (t \ right) \ mathbf (W), \] where the expansion coefficients \ (\ mu \ left (t \ right), \) \ (\ eta \ left (t \ right) \) are defined by the formulas: \ [(\ mu \ left (t \ right) = C (e ^ (\ alpha t)) \ sin \ left ((\ beta t + \ delta ) \ right),) \; \; (\ eta \ left (t \ right) = C (e ^ (\ alpha t)) \ cos \ left ((\ beta t + \ delta) \ right). ) \] This shows that the phase trajectories are spirals. For \ (\ alpha steady focus... Accordingly, for \ (\ alpha> 0 \) we have erratic focus .

The direction of twisting of the spirals can be determined by the sign of the coefficient \ ((a_ (21)) \) in the original matrix \ (A. \) Indeed, consider the derivative \ (\ large \ frac ((dy)) ((dt)) \ normalsize, \) for example, at point \ (\ left ((1,0) \ right): \) \ [\ frac ((dy)) ((dt)) \ left ((1,0) \ right) = (a_ (21)) \ cdot 1 + (a_ (22)) \ cdot 0 = (a_ (21)). \] A positive factor \ ((a_ (21))> 0 \) corresponds to a counterclockwise spiral as shown in the figure \ (8. \) For \ ((a_ (21))
Thus, taking into account the direction of the twisting of the spirals, there are \ (4 \) different types of focus. They are shown schematically in Figures \ (8-11. \)

If the eigenvalues ​​of the matrix \ (A \) are imaginary numbers, then this equilibrium is called center... For a matrix with real elements, the imaginary eigenvalues ​​will be complex conjugate. In the case of the center, the phase trajectories are formally obtained from the spiral equation for \ (\ alpha = 0 \) and are ellipses, i.e. describe the periodic motion of a point on the phase plane. Equilibrium positions of the "center" type are Lyapunov stable.

There are two types of center, differing in the direction of movement of the points (Figures \ (12, 13 \)). As in the case of spirals, the direction of movement can be determined, for example, by the sign of the derivative \ (\ large \ frac ((dy)) ((dt)) \ normalsize \) at some point. If we take the point \ (\ left ((1,0) \ right), \) then \ [\ frac ((dy)) ((dt)) \ left ((1,0) \ right) = (a_ (21 )). \] ie the direction of rotation is determined by the sign of the coefficient \ ((a_ (21)). \)

So, we have considered various types of equilibrium points in the case non-degenerate matrix \ (A \) \ (\ left ((\ det A \ ne 0) \ right). \) Taking into account the direction of the phase trajectories, there are \ (13 \) different phase portraits shown, respectively, in the figures \ (1- 13.\)

Now let's turn to the case degenerate matrix \ (A. \)

Degenerate matrix

If the matrix is ​​degenerate, then one or both of its eigenvalues ​​are equal to zero. In this case, the following special cases are possible:

Case \ ((\ lambda _1) \ ne 0, (\ lambda _2) = 0 \).
Here the general solution is written as \ [\ mathbf (X) \ left (t \ right) = (C_1) (e ^ ((\ lambda _1) t)) (\ mathbf (V) _1) + (C_2) (\ mathbf (V) _2), \] where \ ((\ mathbf (V) _1) = (\ left (((V_ (11)), (V_ (21))) \ right) ^ T), \) \ ((\ mathbf (V) _2) = (\ left (((V_ (12)), (V_ (22))) \ right) ^ T), \) are eigenvectors corresponding to numbers \ ((\ lambda _1 ) \) and \ ((\ lambda _2). \) It turns out that in this case the whole line passing through the origin and directed along the vector \ ((\ mathbf (V) _2), \) consists of equilibrium points (these points do not have a special name). Phase trajectories are rays parallel to another eigenvector \ ((\ mathbf (V) _1). \) Depending on the sign \ ((\ lambda _1) \), the motion at \ (t \ to \ infty \) occurs either in direction of the straight line \ ((\ mathbf (V) _2) \) (fig. \ (14 \)), or from it (fig. \ (15 \)). Case \ ((\ lambda _1) = (\ lambda _2) = 0, \ dim \ ker A = 2. \)
In this case, the dimension of the eigenspace of the matrix is ​​\ (2 \) and, therefore, there are two eigenvectors \ ((\ mathbf (V) _1) \) and \ ((\ mathbf (V) _2). \) Such a situation is possible at zero matrix \ (A. \) The general solution is expressed by the formula \ [\ mathbf (X) \ left (t \ right) = (C_1) (\ mathbf (V) _1) + (C_2) (\ mathbf (V) _2). \ ] It follows that any point of the plane is the equilibrium position of the system.

Case \ ((\ lambda _1) = (\ lambda _2) = 0, \ dim \ ker A = 1. \)
This case of a degenerate matrix differs from the previous one in that there is only \ (1 \) eigenvector (Matrix \ (A \) will be non-zero). To construct a basis, as the second linearly independent vector, we can take the vector \ ((\ mathbf (W) _1), \) attached to \ ((\ mathbf (V) _1). \) The general solution of the system is written as \ [\ mathbf (X) \ left (t \ right) = \ left (((C_1) + (C_2) t) \ right) (\ mathbf (V) _1) + (C_2) (\ mathbf (W) _1). \] Here all points of the straight line passing through the origin and directed along the eigenvector \ ((\ mathbf (V) _1), \) are unstable equilibrium positions. The phase trajectories are straight lines parallel to \ ((\ mathbf (V) _1). \) The direction of motion along these straight lines for \ (t \ to \ infty \) depends on the constant \ ((C_2): \) for \ (( C_2) 0 \) - in the opposite direction (fig. \ (16 \)).

Recall that followed by a matrix called a number equal to the sum of diagonal elements: \ [(A = \ left ((\ begin (array) (* (20) (c)) ((a_ (11))) & ((a_ (12))) \\ ((a_ (21))) & ((a_ (22))) \ end (array)) \ right),) \; \; (\ text (tr) \, A = (a_ (11)) + (a_ (22)),) \; \; (\ det A = (a_ (11)) (a_ (22)) - (a_ (12)) (a_ (21)).) \] Indeed, the characteristic equation of the matrix has the following form: \ [(\ lambda ^ 2 ) - \ left (((a_ (11)) + (a_ (22))) \ right) \ lambda + (a_ (11)) (a_ (22)) - (a_ (12)) (a_ (21) ) = 0. \] It can be written in terms of the determinant and trace of the matrix: \ [(\ lambda ^ 2) - \ text (tr) \, A \ cdot \ lambda + \ det A = 0. \] The discriminant of this quadratic equation is determined by the relation \ Thus, bifurcation curve , delimiting various stability modes, is a parabola on the plane \ (\ left ((\ text (tr) \, A, \ det A) \ right) \) (Fig. \ (17 \)): \ [\ det A = (\ left ((\ frac (\ text (tr) \, A) (2)) \ right) ^ 2). \] Above the parabola are the focus and center equilibrium points. Points of the "center" type are located on the positive semiaxis \ (Oy, \) i.e. under the condition \ (\ text (tr) \, A = 0. \) Below the parabola there are points of the "knot" or "saddle" type. The parabola itself contains dicritical or degenerate nodes.

Stable modes of motion exist in the upper left quadrant of the bifurcation diagram. The remaining three quadrants correspond to unstable equilibrium positions.

Phase portrait construction algorithm

To schematically construct a phase portrait of a linear autonomous system of \ (2 \) th order with constant coefficients \ [(\ mathbf (X ") = A \ mathbf (X),) \; \; (A = \ left ((\ begin (array) (* (20) (c)) ((a_ (11))) & ((a_ (12))) \\ ((a_ (21))) & ((a_ (22))) \ end (array)) \ right),) \; \; (\ mathbf (X) = \ left ((\ begin (array) (* (20) (c)) x \\ y \ end (array)) \ right )) \] you need to do the following:

Find the eigenvalues ​​of the matrix by solving the characteristic equation \ [(\ lambda ^ 2) - \ left (((a_ (11)) + (a_ (22))) \ right) \ lambda + (a_ (11)) (a_ ( 22)) - (a_ (12)) (a_ (21)) = 0. \]

Determine the type of equilibrium position and the nature of stability.

Note: The type of equilibrium position can also be determined based on the bifurcation diagram (Fig. \ (17 \)), knowing the trace and determinant of the matrix: \ [(\ text (tr) \, A = (a_ (11)) + (a_ ( 22)),) \; \; (\ det A = \ left | (\ begin (array) (* (20) (c)) ((a_ (11))) & ((a_ (12))) \\ ((a_ (21))) & ((a_ (22))) \ end (array)) \ right |) = ((a_ (11)) (a_ (22)) - (a_ (12)) (a_ (21)).) \]

Find Equation isocline: \ [(\ frac ((dx)) ((dt)) = (a_ (11)) x + (a_ (12)) y) \; \; (\ left (\ text (vertical isocline) \ right),) \] \ [(\ frac ((dy)) ((dt)) = (a_ (21)) x + (a_ (22)) y) \ ; \; (\ left (\ text (horizontal isocline) \ right).) \]

If the equilibrium position is knot or, then it is necessary to calculate the eigenvectors and draw parallel asymptotes to them passing through the origin.

Sketch the phase portrait.

Show the direction of motion along the phase trajectories (it depends on the stability or instability of the equilibrium point). When focus the direction of twisting of the trajectories should be determined. This can be done by calculating the velocity vector \ (\ left ((\ large \ frac ((dx)) ((dt)) \ normalsize, \ large \ frac ((dy)) ((dt)) \ normalsize) \ right) \) at an arbitrary point, for example, at the point \ (\ left ((1,0) \ right). \) The direction of movement is determined in a similar way if the equilibrium position is center .

All forces applied to the body relative to any arbitrary axis of rotation are also equal to zero.

In a state of equilibrium, the body is at rest (the velocity vector is zero) in the selected frame of reference either moves uniformly in a straight line or rotates without tangential acceleration.

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Subtitles

Definition through the energy of the system

Since energy and forces are linked by fundamental relationships, this definition is equivalent to the first. However, the definition in terms of energy can be extended in order to obtain information about the stability of the equilibrium position.

Balance types

Let's give an example for a system with one degree of freedom. In this case, a sufficient condition for the equilibrium position will be the presence of a local extremum at the point under study. As is known, the condition for a local extremum of a differentiable function is the equality to zero of its first derivative. To determine when this point is the minimum or maximum, it is necessary to analyze its second derivative. The stability of the equilibrium position is characterized by the following options:

• unstable equilibrium;
• stable balance;
• indifferent balance.

In the case when the second derivative is negative, the potential energy of the system is in the state of a local maximum. This means that the equilibrium position unstable... If the system is displaced a short distance, then it will continue to move due to the forces acting on the system. That is, when the body is unbalanced, it does not return to its original position.

Stable balance

Second derivative> 0: potential energy at a local minimum, equilibrium position steadily(see Lagrange's theorem on the stability of an equilibrium). If the system is displaced a short distance, it will return back to a state of equilibrium. Equilibrium is stable if the center of gravity of the body is in the lowest position in comparison with all possible neighboring positions. With this balance, the unbalanced body returns to its original place.

Indifferent balance

Second derivative = 0: in this region, the energy does not vary, and the equilibrium position is indifferent... If the system is moved a short distance, it will remain in the new position. If you deflect or move the body, it will remain in balance.

• Types of sustainability

The section of mechanics in which the conditions of equilibrium of bodies are studied is called statics. It follows from Newton's second law that if the vector sum of all forces applied to the body is equal to zero, then the body retains its speed unchanged. In particular, if the initial velocity is zero, the body remains at rest. The condition for the invariability of the speed of the body can be written in the form:

or in projections on the coordinate axes:

.

Obviously, the body can be at rest only in relation to one specific coordinate system. In statics, the conditions of equilibrium of bodies are studied in just such a system. The necessary condition for equilibrium can also be obtained by considering the motion of the center of mass of a system of material points. Internal forces do not affect the movement of the center of mass. The acceleration of the center of mass is determined by the vector sum of external forces. But if this sum is equal to zero, then the acceleration of the center of mass, and, consequently, the speed of the center of mass. If at the initial moment, then the center of mass of the body remains at rest.

Thus, the first condition for the balance of bodies is formulated as follows: the speed of a body does not change if the sum of external forces applied at each point is equal to zero. The obtained condition of rest of the center of mass is a necessary (but insufficient) condition for the equilibrium of a rigid body.

Example

It may be that all the forces acting on the body are balanced; nevertheless, the body will accelerate. For example, if you apply two equal and oppositely directed forces (they are called a pair of forces) to the center of mass of a wheel, then the wheel will be at rest if its initial speed was zero. If these forces are applied to different points, then the wheel will begin to rotate (Fig. 4.5). This is because the body is in equilibrium when the sum of all forces is zero at every point of the body. But if the sum of external forces is equal to zero, and the sum of all forces applied to each element of the body is not equal to zero, then the body will not be in equilibrium, perhaps (as in the considered example) rotational motion. Thus, if the body can rotate about some axis, then for its equilibrium it is not enough for the resultant of all forces to be zero.

To obtain the second equilibrium condition, we use the equation of rotational motion, where is the sum of the moments of external forces relative to the axis of rotation. When, then b = 0, which means that the angular velocity of the body does not change. If at the initial moment w = 0, then the body will not rotate in the future. Consequently, the second condition for mechanical equilibrium is the requirement that the algebraic sum of the moments of all external forces relative to the axis of rotation be equal to zero:

In the general case of an arbitrary number of external forces, the equilibrium conditions can be represented as follows:

,

.

These conditions are necessary and sufficient.

Example

Equilibrium is stable, unstable, and indifferent. Equilibrium is stable if, at small displacements of the body from the equilibrium position, the forces and moments of forces acting on it tend to return the body to the equilibrium position (Fig. 4.6a). Equilibrium is unstable if the acting forces move the body further from the equilibrium position (Fig. 4.6b). If at small displacements of the body the acting forces are still balanced, then the equilibrium is indifferent (Fig. 4.6c). A ball lying on a flat horizontal surface is in a state of indifferent equilibrium. A ball at the top of a spherical protrusion is an example of an unstable balance. Finally, the ball at the bottom of the spherical depression is in a state of stable equilibrium.

An interesting example of the balance of a body on a support is the leaning tower in the Italian city of Pisa, which, according to legend, was used by Galileo when studying the laws of free fall of bodies. The tower has the shape of a cylinder with a radius of 7 m. The top of the tower is deviated from the vertical by 4.5 m.

The Leaning Tower of Pisa is famous for the fact that it is strongly tilted. The tower "falls". The height of the tower is 55.86 meters from the ground on the lowest side and 56.70 meters on the highest side. Its weight is estimated at 14,700 tons. The current tilt is about 5.5 °. A vertical line drawn through the tower's center of gravity intersects the base at approximately 2.3 m from its center. Thus, the tower is in a state of equilibrium. The balance will be disturbed and the tower will fall when the deviation of its top from the vertical reaches 14 m. Apparently, this will happen very soon.

It was believed that the curvature of the tower was originally conceived by the architects - for the sake of demonstrating their extraordinary skill. But something else is much more likely: the architects knew that they were building on an extremely unreliable foundation, and therefore incorporated the possibility of easy deviation into the structure.

When the real threat of the tower collapse arose, modern engineers took over it. She was pulled into a steel corset of 18 cables, the foundation was weighted with lead blocks and, in parallel, the soil was strengthened by pumping concrete underground. With the help of all these measures, it was possible to reduce the angle of inclination of the falling tower by half a degree. Experts say that now it will be able to stand for at least another 300 years. From the point of view of physics, the measures taken mean that the equilibrium conditions of the tower have become more reliable.

For a body with a fixed axis of rotation, all three types of balance are possible. Indifferent equilibrium occurs when the axis of rotation passes through the center of mass. In stable and unstable equilibrium, the center of mass is on a vertical line passing through the axis of rotation. In this case, if the center of mass is below the axis of rotation, the state of equilibrium is stable (Fig. 4.7a). If the center of mass is located above the axis, the state of equilibrium is unstable (Fig. 4.7b).

A special case of balance is the balance of the body on the support. In this case, the elastic support force is applied not to one point, but is distributed over the base of the body. The body is in equilibrium if the vertical line drawn through the center of mass of the body passes through the area of ​​support, that is, inside the contour formed by the lines connecting the points of support. If this line does not intersect the support area, then the body overturns.

Balance types

In order to judge the behavior of a body in real conditions, it is not enough to know that it is in equilibrium. We must also assess this balance. Distinguish between stable, unstable and indifferent equilibrium.

The balance of the body is called sustainable if, when deviating from it, forces arise that return the body to the equilibrium position (Fig. 1, position 2). In stable equilibrium, the center of gravity of the body occupies the lowest of all close positions. The position of stable equilibrium is associated with a minimum of potential energy in relation to all close neighboring positions of the body.

The balance of the body is called unstable if at the slightest deviation from it, the resultant forces acting on the body cause further deviation of the body from the equilibrium position (Fig. 1, position 1). In a position of unstable equilibrium, the height of the center of gravity is maximum and potential energy is maximum in relation to other close body positions.

Equilibrium in which the displacement of the body in any direction does not cause a change in the forces acting on it and the balance of the body is maintained is called indifferent(fig. 1 position 3).

Indifferent equilibrium is associated with the constant potential energy of all close states, and the height of the center of gravity is the same in all sufficiently close positions.

A body that has an axis of rotation (for example, a uniform ruler that can rotate around an axis passing through the point O, shown in Figure 2) is in equilibrium if the vertical line passing through the center of gravity of the body passes through the axis of rotation. Moreover, if the center of gravity C is higher than the axis of rotation (Fig. 2.1), then for any deviation from the equilibrium position, the potential energy decreases and the moment of gravity relative to the O axis deflects the body further from the equilibrium position. This is an unstable equilibrium position. If the center of gravity is below the axis of rotation (Fig. 2.2), then the balance is stable. If the center of gravity and the axis of rotation coincide (Fig. 2, 3), then the equilibrium position is indifferent.

balance physics displacement

A body with an area of ​​support is in equilibrium if the vertical line passing through the center of gravity of the body does not go beyond the area of ​​support of this body, i.e. beyond the boundaries of the contour formed by the points of contact of the body with the support, Equilibrium in this case depends not only on the distance between the center of gravity and the support (i.e., on its potential energy in the gravitational field of the Earth), but also on the location and size of the support area of ​​this body.

Figure 2 shows a body in the form of a cylinder. If you tilt it at a small angle, then it will return to its original position 1 or 2. If you tilt it at an angle (position 3), then the body will overturn. For a given mass and support area, the stability of the body is the higher, the lower its center of gravity is, i.e. the smaller the angle between the straight line connecting the center of gravity of the body and the extreme point of contact of the support area with the horizontal plane.