Stable and unstable equilibrium in physics. Statics

The equilibrium of a mechanical system is its state in which all points of the system under consideration are at rest with respect to the chosen reference frame.

The moment of force about any axis is the product of the magnitude of this force F and the arm d.

The easiest way to find out the equilibrium conditions is by the example of the simplest mechanical system - a material point. According to the first law of dynamics (see Mechanics), the condition for rest (or uniform rectilinear motion) of a material point in an inertial coordinate system is the equality to zero of the vector sum of all forces applied to it.

In the transition to more complex mechanical systems, this condition alone for their equilibrium is not enough. In addition to translational motion, which is caused by uncompensated external forces, a complex mechanical system can perform rotational motion or deform. Let us find out the equilibrium conditions for an absolutely rigid body - a mechanical system consisting of a collection of particles, the mutual distances between which do not change.

The possibility of translational motion (with acceleration) of a mechanical system can be eliminated in the same way as in the case of a material point, requiring that the sum of forces applied to all points of the system be equal to zero. This is the first condition for the equilibrium of a mechanical system.

In our case, a rigid body cannot be deformed, since we agreed that the mutual distances between its points do not change. But unlike a material point, a pair of equal and oppositely directed forces can be applied to an absolutely rigid body at its different points. Moreover, since the sum of these two forces is equal to zero, the considered mechanical system of translational motion will not perform. However, it is obvious that under the action of such a pair of forces, the body will begin to rotate about some axis with an ever-increasing angular velocity.

The occurrence of rotational motion in the system under consideration is due to the presence of uncompensated moments of forces. The moment of force about any axis is the product of the magnitude of this force $F$ by the arm $d,$ i.e. by the length of the perpendicular dropped from the point $O$ (see figure), through which the axis passes, by the direction of the force . Note that the moment of force with this definition is an algebraic quantity: it is considered positive if the force leads to counterclockwise rotation, and negative otherwise. Thus, the second condition for the equilibrium of a rigid body is the requirement that the sum of the moments of all forces about any axis of rotation be equal to zero.

In the case when both found equilibrium conditions are met, the rigid body will be at rest if, at the moment the forces began to act, the velocities of all its points were equal to zero. Otherwise, it will make uniform motion by inertia.

The considered definition of the equilibrium of a mechanical system does not say anything about what will happen if the system slightly leaves the equilibrium position. In this case, there are three possibilities: the system will return to its previous state of equilibrium; the system, despite the deviation, will not change its state of equilibrium; the system will be out of equilibrium. The first case is called a stable state of equilibrium, the second - indifferent, the third - unstable. The nature of the equilibrium position is determined by the dependence of the potential energy of the system on the coordinates. The figure shows all three types of balance on the example of a heavy ball located in a recess (stable balance), on a smooth horizontal table (indifferent), on top of a tubercle (unstable).

The above approach to the problem of equilibrium of a mechanical system was considered by scientists in the ancient world. So, the law of equilibrium of a lever (that is, a rigid body with a fixed axis of rotation) was found by Archimedes in the 3rd century. BC e.

In 1717, Johann Bernoulli developed a completely different approach to finding the equilibrium conditions for a mechanical system - the method of virtual displacements. It is based on the property of bond reaction forces arising from the energy conservation law: with a small deviation of the system from the equilibrium position, the total work of the bond reaction forces is zero.

When solving problems of statics (see Mechanics), on the basis of the equilibrium conditions described above, the connections existing in the system (supports, threads, rods) are characterized by the reaction forces arising in them. The need to take these forces into account when determining the equilibrium conditions in the case of systems consisting of several bodies leads to cumbersome calculations. However, due to the fact that the work of the bond reaction forces is equal to zero for small deviations from the equilibrium position, it is possible to avoid considering these forces in general.

In addition to reaction forces, external forces also act on the points of a mechanical system. What is their work with a small deviation from the equilibrium position? Since the system is initially at rest, for any movement of it, some positive work must be done. In principle, this work can be done by both external forces and reaction forces of bonds. But, as we already know, the total work of the reaction forces is zero. Therefore, in order for the system to leave the state of equilibrium, the total work of external forces for any possible displacement must be positive. Consequently, the condition of the impossibility of motion, i.e., the equilibrium condition, can be formulated as the requirement that the total work of external forces be nonpositive for any possible displacement: $ΔA≤0.$

Let us assume that when the points of the system $Δ\overrightarrow(γ)_1…\ Δ\overrightarrow(γ)_n$ move, the sum of the work of external forces turned out to be equal to $ΔA1.$ And what happens if the system moves $−Δ\overrightarrow(γ )_1,−Δ\overrightarrow(γ)_2,\ …,−Δ\overrightarrow(γ)_n?$ These displacements are possible in the same way as the first ones; however, the work of external forces will now change sign: $ΔA2 =−ΔA1.$ Arguing similarly to the previous case, we will come to the conclusion that now the equilibrium condition for the system has the form: $ΔA1≥0,$ i.e., the work of external forces must be non-negative. The only way to “reconcile” these two almost contradictory conditions is to require the exact equality to zero of the total work of external forces for any possible (virtual) displacement of the system from the equilibrium position: $ΔA=0.$ Possible (virtual) displacement here means an infinitesimal mental displacement of the system , which does not contradict the connections imposed on it.

So, the equilibrium condition of a mechanical system in the form of the principle of virtual displacements is formulated as follows:

"For the equilibrium of any mechanical system with ideal connections, it is necessary and sufficient that the sum of the elementary works acting on the system of forces for any possible displacement be equal to zero."

Using the principle of virtual displacements, the problems of not only statics, but also hydrostatics and electrostatics are solved.

This lecture covers the following questions:

1. Conditions for the equilibrium of mechanical systems.

2. Stability of equilibrium.

3. An example of determining equilibrium positions and studying 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", to solve problems in the disciplines "Theory of Machines and Mechanisms" and "Strength of Materials".

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

A mechanical system can oscillate for a sufficiently long period of time only near a position of stable equilibrium. Therefore, before compiling 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, confining and holonomic constraints are imposed, to be in equilibrium, it is necessary and sufficient that all generalized forces in this system be equal to zero:

where is the generalized force corresponding to j- oh generalized coordinate;

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

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

Balance stability

The definition of the concept of stability of an equilibrium position was given at the end of the 19th century in the works of the Russian scientist A. M. Lyapunov. Let's look at 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 another number , that in the case when the initial values ​​of the generalized coordinates and velocities will not exceed:

values ​​of generalized coordinates and velocities during further motion 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 it is always possible to find such sufficiently small initial values, at 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 visualized in the phase plane (Fig. 1).For a stable equilibrium position, the movement of the representative point, starting in the region [ ] , will not go beyond the area in the future.

Fig.1

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

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

Sufficient conditions for the stability of equilibrium positions for such systems are defined by 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 up to a constant. We 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), is 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 when all their arguments are zero are called sign-definite. Therefore, in order for the equilibrium position of a mechanical system to be stable, it is necessary and sufficient that, in the vicinity of this position, the potential energy be a positively defined function of 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 as 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 into 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 , in order to satisfy sufficient conditions for the stability of the equilibrium position, the potential energy must be a positive definite quadratic form of its generalized coordinates.

In mathematics there is Sylvester's criterion , which gives 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 principal 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 potential energy, one introduces into consideration the potential energy of the reduced system.

P 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 tubes with a spring (Fig. 2). Let us determine the equilibrium positions of the system and evaluate their stability for the following parameters: tube length l 2 = 1 m , rod length l 1 = 0,5 m . undeformed spring length l 0 = 0.6 m, spring rate c= 100 N/m. Tube weight m 2 = 2 kg, rod - m 1 = 1 kg and ball - m 3 = 0.5 kg. Distance OA equals l 3 = 0.4 m.

Let us write an 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 the field of gravity is equal to the product of the weight of the body and 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 coordinate values ​​in the equilibrium positions are the roots of the following system of equations.

A similar system of equations can be compiled for any mechanical system with two degrees of freedom. In some cases, it is possible to obtain an exact solution of the system. For system (5), such a solution does not exist, so 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 determine the generalized stiffness coefficients from them.

Equilibrium of a mechanical system is a state in which all points of a mechanical system are at rest with respect to the reference frame under consideration. If the frame of reference is inertial, the equilibrium is called absolute, if non-inertial — relative.

To find the equilibrium conditions for an absolutely rigid body, it is necessary to mentally divide it into a large number of sufficiently small elements, each of which can be represented by a material point. All these elements interact with each other - these interaction forces are called internal. In addition, external forces can act on a number of points of the body.

According to Newton's second law, for the acceleration of a point to be zero (and the acceleration of a point at rest to be zero), the geometric sum of the forces acting on that point must be zero. If the body is at rest, then all its points (elements) are also at rest. Therefore, for any point of the body, we can write:

where is the geometric sum of all external and internal forces acting on i th element of the body.

The equation means that for the equilibrium of a body it is necessary and sufficient that the geometric sum of all forces acting on any element of this body is equal to zero.

From it is easy to obtain the first condition for the equilibrium of a body (system of bodies). To do this, it is enough to sum the equation over all elements of the body:

.

The second sum is equal to zero according to Newton's third law: the vector sum of all internal forces of the system is equal to zero, since any internal force corresponds to a force equal in absolute value and opposite in direction.

Hence,

.

The first condition for the equilibrium of a rigid body(body systems) is the equality to zero of the geometric sum of all external forces applied to the body.

This condition is necessary but not sufficient. It is easy to verify this by remembering the rotating action of a pair of forces, the geometric sum of which is also equal to zero.

The second condition for the equilibrium of a rigid body is the equality to zero of the sum of the moments of all external forces acting on the body, relative to any axis.

Thus, the equilibrium conditions for a rigid body in the case of an arbitrary number of external forces look like this:

.

Class: 10

Presentation for the lesson

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Lesson Objectives: To study the state of equilibrium of bodies, to get acquainted with various types of equilibrium; find out the conditions under which the body is in equilibrium.

Lesson objectives:

• Training: To study two conditions of equilibrium, types of equilibrium (stable, unstable, indifferent). Find out under what conditions bodies are more stable.
• Developing: To promote the development of cognitive interest in physics. Development of skills to compare, generalize, highlight the main thing, draw conclusions.
• Educational: To cultivate attention, the ability to express one's point of view and defend it, to develop the communication skills of students.

Lesson type: lesson learning new material with computer support.

Equipment:

1. Disk "Work and power" from "Electronic lessons and tests.
2. Table "Equilibrium conditions".
3. Prism inclined with a plumb line.
4. Geometric bodies: cylinder, cube, cone, etc.
5. Computer, multimedia projector, interactive whiteboard or screen.
6. Presentation.

During the classes

Today in the lesson we will learn why the crane does not fall, why the Roly-Vstanka toy always returns to its original state, why the Leaning Tower of Pisa does not fall?

I. Repetition and updating of knowledge.

1. Formulate Newton's first law. What is the status of the law?
2. What question does Newton's second law answer? Formula and wording.
3. What question does Newton's third law answer? Formula and wording.
4. What is the resultant force? How is she?
5. From the disk "Movement and interaction of bodies" complete task No. 9 "The resultant of forces with different directions" (the rule of vector addition (2, 3 exercises)).

II. Learning new material.

1. What is called equilibrium?

Equilibrium is a state of rest.

2. Equilibrium conditions.(slide 2)

a) When is the body at rest? What law does this come from?

The first equilibrium condition: A body is in equilibrium if the geometric sum of the external forces applied to the body is zero. ∑ F = 0

b) Let two equal forces act on the board, as shown in the figure.

Will she be in balance? (No, she will turn)

Only the central point is at rest, while the others move. This means that for the body to be in equilibrium, it is necessary that the sum of all forces acting on each element be equal to 0.

The second equilibrium condition: The sum of the moments of forces acting clockwise must be equal to the sum of the moments of forces acting counterclockwise.

∑ M clockwise = ∑ M counterclockwise

Moment of force: M = F L

L - shoulder of force - the shortest distance from the fulcrum to the line of action of the force.

3. The center of gravity of the body and its location.(slide 4)

Center of gravity of the body- this is the point through which the resultant of all parallel gravity forces acting on individual elements of the body passes (at any position of the body in space).

Find the center of gravity of the following figures:

4. Types of balance.

a) (slides 5-8)

Conclusion: Equilibrium is stable if, with a small deviation from the equilibrium position, there is a force tending to return it to this position.

The position in which its potential energy is minimal is stable. (slide 9)

b) The stability of bodies located on the fulcrum or on the fulcrum.(slides 10-17)

Conclusion: For the stability of a body located on one point or line of support, it is necessary that the center of gravity be below the point (line) of support.

c) The stability of bodies on a flat surface.

(slide 18)

1) Support surface- this is not always a surface that is in contact with the body (but one that is limited by lines connecting the legs of the table, tripod)

2) Analysis of a slide from "Electronic lessons and tests", disk "Work and power", lesson "Types of balance".

Picture 1.

1. How are the stools different? (Square footing)
2. Which one is more stable? (with larger area)
3. How are the stools different? (Location of the center of gravity)
4. Which one is the most stable? (which center of gravity is lower)
5. Why? (Because it can be deflected to a larger angle without tipping over)

3) Experience with a deviating prism

1. Let's put a prism with a plumb line on the board and begin to gradually lift it over one edge. What do we see?
2. As long as the plumb line crosses the surface bounded by the support, the balance is maintained. But as soon as the vertical passing through the center of gravity begins to go beyond the boundaries of the support surface, the bookcase overturns.

Parsing slides 19–22.

Findings:

1. The body with the largest area of ​​support is stable.
2. Of two bodies of the same area, the body whose center of gravity is lower is stable, because it can be deflected without overturning at a large angle.

Parsing slides 23–25.

Which ships are the most stable? Why? (For which the cargo is located in the holds, and not on the deck)

What cars are the most stable? Why? (To increase the stability of cars on turns, the roadbed is tilted in the direction of the turn.)

Findings: Equilibrium can be stable, unstable, indifferent. The stability of the bodies is greater, the larger the area of ​​support and the lower the center of gravity.

III. Application of knowledge about the stability of bodies.

1. What specialties most need knowledge about the balance of bodies?
2. Designers and constructors of various structures (high-rise buildings, bridges, television towers, etc.)
3. Circus artists.
4. Drivers and other professionals.

(slides 28–30)

1. Why does Roly-Vstanka return to the equilibrium position at any inclination of the toy?
2. Why is the Leaning Tower of Pisa tilted and not falling?
3. How do cyclists and motorcyclists keep their balance?

Lesson takeaways:

1. There are three types of equilibrium: stable, unstable, indifferent.
2. The position of the body is stable, in which its potential energy is minimal.
3. The stability of bodies on a flat surface is greater, the larger the area of ​​support and the lower the center of gravity.

Homework: § 54 – 56 (G.Ya. Myakishev, B.B. Bukhovtsev, N.N. Sotsky)

Used sources and literature:

1. G.Ya. Myakishev, B.B. Bukhovtsev, N.N. Sotsky. Physics. Grade 10.
2. Filmstrip "Stability" 1976 (scanned by me on a film scanner).
3. Disk "Movement and interaction of bodies" from "Electronic lessons and tests".
4. Disk "Work and power" from "Electronic lessons and tests".

DEFINITION

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

This occurs 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 non-zero and is directed towards the equilibrium position. For example, a ball lying at the bottom of a spherical cavity (Fig. 1a).

DEFINITION

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

In this case, with a small 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 an equilibrium in which the body, taken out of equilibrium and left to itself, does not change its position (state).

In this case, with small displacements of the body from its original 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. Different types of body balance on a support: a) stable balance; b) unstable equilibrium; c) indifferent equilibrium.

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 equilibrium.

DEFINITION

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

dynamic balance- this is such an equilibrium when, under the action of forces, the body does not change its motion.

In a state of static equilibrium 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 friction forces.