Equilibrium mechanical system they call its state in which all points of the system under consideration are at rest with respect to the chosen reference system.

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

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

When moving to more complex mechanical systems, this condition alone is not enough for their equilibrium. Except forward movement, which is caused by uncompensated external forces, a complex mechanical system can rotate or become deformed. 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, by 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, the solid body cannot deform, since we have 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 different points. Moreover, since the sum of these two forces is zero, the mechanical system under consideration will not perform translational motion. However, it is obvious that under the influence of such a pair of forces the body will begin to rotate relative to a certain axis with an ever-increasing angular velocity.

Occurrence in the system under consideration rotational movement due to the presence of uncompensated moments of force. The moment of a 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 lowered 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 if otherwise. Thus, the second condition for the equilibrium of a rigid body is the requirement that the sum of the moments of all forces relative to any axis of rotation be equal to zero.

In the case when both found equilibrium conditions are met, the solid 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 commit uniform motion by inertia.

The considered definition of equilibrium of a mechanical system does not say anything about what will happen if the system moves slightly out of its 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 equilibrium state; the system will go out of equilibrium. The first case is called steady state 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 equilibrium using the example of a heavy ball located in a depression (stable equilibrium), on a smooth horizontal table (indifferent), on the top of a tubercle (unstable).

The above approach to the problem of equilibrium of a mechanical system was considered by scientists back in ancient world. Thus, the law of equilibrium of a lever (i.e., 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 of a mechanical system - the method of virtual displacements. It is based on the property of bond reaction forces arising from the law of conservation of energy: 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) based on 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 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 altogether.

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

Let us assume that when moving the points of the system $Δ\overrightarrow(γ)_1…\ Δ\overrightarrow(γ)_n$ the sum of the work of external forces turned out to be equal to $ΔA1.$ And what happens if the system makes movements $−Δ\overrightarrow(γ )_1,−Δ\overrightarrow(γ)_2,\ …,−Δ\overrightarrow(γ)_n?$ These movements are possible in the same way as the first ones; however, the work of external forces will now change sign: $ΔA2 =−ΔA1.$ Reasoning similarly to the previous case, we will come to the conclusion that now the equilibrium condition of 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 demand the exact equality to zero of the total work of external forces for any possible (virtual) movement of the system from the equilibrium position: $ΔA=0.$ By possible (virtual) movement here we mean an infinitesimal mental movement 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 amount basic work the forces acting on the system for any possible movement were equal to zero.”

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

In order to judge the behavior of the body in real conditions, it is not enough to know that it is in equilibrium. We still need to evaluate this balance. There are 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 nearby positions. Position 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, with the slightest deviation from it, the resultant of the forces acting on the body causes a further deviation of the body from the equilibrium position (Fig. 1, position 1). In an unstable equilibrium position, the height of the center of gravity is maximum and the potential energy is maximum in relation to other close positions of the body.

Equilibrium, in which the displacement of a 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 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 with an axis of rotation (for example, a uniform ruler that can rotate around an axis passing through point O, shown in Figure 2) is in equilibrium if a vertical straight 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 equilibrium is stable. If the center of gravity and the axis of rotation coincide (Fig. 2,3), then the equilibrium position is indifferent.

A body having a support area is in equilibrium if the vertical line passing through the center of gravity of the body does not go beyond the support area of ​​this body, i.e. beyond 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 shaped like a cylinder. If you tilt it at a small angle, it will return to initial position 1 or 2. If it is tilted at an angle (position 3), the body will tip over. For a given mass and support area, the stability of a body is higher, the lower its center of gravity is located, i.e. the smaller the angle between the straight line connecting the center of gravity of the body and extreme point contact of the support area with the horizontal plane.

It follows that if geometric sum of all external forces applied to the body is equal to zero, then the body is at rest or performs a uniform rectilinear movement. In this case, it is customary to say that the forces applied to the body balance each other. When calculating the resultant, all forces acting on the body can be applied to the center of mass.

For a non-rotating body to be in equilibrium, it is necessary that the resultant of all forces applied to the body be equal to zero.

$(\overrightarrow(F))=(\overrightarrow(F_1))+(\overrightarrow(F_2))+...= 0$

If a body can rotate about a certain axis, then for its equilibrium it is not enough for the resultant of all forces to be zero.

The rotating effect of a force depends not only on its magnitude, but also on the distance between the line of action of the force and the axis of rotation.

The length of the perpendicular drawn from the axis of rotation to the line of action of the force is called the arm of the force.

The product of the force modulus $F$ and the arm d is called the moment of force M. The moments of those forces that tend to rotate the body counterclockwise are considered positive.

Rule of moments: a body having a fixed axis of rotation is in equilibrium if algebraic sum the moments of all forces applied to the body relative to this axis are equal to zero:

IN general case, when a body can move translationally and rotate, for equilibrium it is necessary to fulfill both conditions: the resultant force being equal to zero and the sum of all moments of forces being equal to zero. Both of these conditions are not sufficient for peace.

Figure 1. Indifferent equilibrium. Wheel rolling along horizontal surface. The resultant force and moment of forces are equal to zero

A wheel rolling on a horizontal surface is an example of indifferent equilibrium (Fig. 1). If the wheel is stopped at any point, it will be in equilibrium. Along with indifferent equilibrium, mechanics distinguishes between states of stable and unstable equilibrium.

A state of equilibrium is called stable if, with small deviations of the body from this state, forces or moments of force arise that tend to return the body to an equilibrium state.

With a small deviation of the body from a state of unstable equilibrium, forces or moments of force arise that tend to remove the body from the equilibrium position. A ball lying on a flat horizontal surface is in a state of indifferent equilibrium.

Figure 2. Different kinds equilibrium of the ball on the support. (1) -- indifferent equilibrium, (2) -- unstable equilibrium, (3) -- stable equilibrium

A ball located at the top point of a spherical protrusion is an example of unstable equilibrium. Finally, the ball at the bottom of the spherical recess is in a state of stable equilibrium (Fig. 2).

For a body with a fixed axis of rotation, all three types of equilibrium are possible. Indifference 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 straight line passing through the axis of rotation. Moreover, if the center of mass is below the axis of rotation, the state of equilibrium turns out to be stable. If the center of mass is located above the axis, the equilibrium state is unstable (Fig. 3).

Figure 3. Stable (1) and unstable (2) equilibrium of a homogeneous circular disk fixed on the O axis; point C is the center of mass of the disk; $(\overrightarrow(F))_t\ $-- gravity; $(\overrightarrow(F))_(y\ )$-- elastic force of the axis; d -- shoulder

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

Problem 1

The inclined plane is inclined at an angle of 30o to the horizontal (Fig. 4). There is a body P on it, the mass of which is m = 2 kg. Friction can be neglected. A thread thrown through a block makes an angle of 45o with inclined plane. At what weight of the load Q will the body P be in equilibrium?

Figure 4

The body is under the influence of three forces: the force of gravity P, the tension of the thread with the load Q and the elastic force F from the side of the plane pressing on it in the direction perpendicular to the plane. Let's break down the force P into its components: $\overrightarrow(P)=(\overrightarrow(P))_1+(\overrightarrow(P))_2$. Condition $(\overrightarrow(P))_2=$ For equilibrium, taking into account the doubling of the force by the moving block, it is necessary that $\overrightarrow(Q)=-(2\overrightarrow(P))_1$. Hence the equilibrium condition: $m_Q=2m(sin \widehat((\overrightarrow(P))_1(\overrightarrow(P))_2)\ )$. Substituting the values ​​we get: $m_Q=2\cdot 2(sin \left(90()^\circ -30()^\circ -45()^\circ \right)\ )=1.035\ kg$.

When there is wind, the tethered balloon does not hang above the point on the Earth to which the cable is attached (Fig. 5). The cable tension is 200 kg, the angle with the vertical is a=30$()^\circ$. What is the force of wind pressure?

\[(\overrightarrow(F))_в=-(\overrightarrow(Т))_1;\ \ \ \ \left|(\overrightarrow(F))_в\right|=\left|(\overrightarrow(Т)) _1\right|=Тg(sin (\mathbf \alpha )\ )\] \[\left|(\overrightarrow(F))_в\right|=\ 200\cdot 9.81\cdot (sin 30()^\circ \ )=981\ N\]

« Physics - 10th grade"

Remember what a moment of force is.
Under what conditions is the body at rest?

If a body is at rest relative to the chosen frame of reference, then this body is said to be in equilibrium. Buildings, bridges, beams with supports, machine parts, a book on a table and many other bodies are at rest, despite the fact that forces are applied to them from other bodies. The task of studying the conditions of equilibrium of bodies is of great importance practical significance for mechanical engineering, construction, instrument making and other fields of technology. All real bodies, under the influence of forces applied to them, change their shape and size, or, as they say, are deformed.

In many cases encountered in practice, the deformations of bodies when they are in equilibrium are insignificant. In these cases, deformations can be neglected and calculations can be carried out, considering the body absolutely hard.

For brevity, we will call an absolutely rigid body solid body or simply body. Having studied the equilibrium conditions solid, we will find the equilibrium conditions real bodies in cases where their deformations can be ignored.

Remember the definition of an absolutely rigid body.

The branch of mechanics in which the conditions of equilibrium of absolutely rigid bodies are studied is called static.

In statics, the size and shape of bodies are taken into account; in this case, not only the value of the forces is significant, but also the position of the points of their application.

Let us first find out, using Newton's laws, under what condition any body will be in equilibrium. To this end, let us mentally break down the entire body into big number small elements, each of which can be considered as a material point. As usual, we will call the forces acting on the body from other bodies external, and the forces with which the elements of the body itself interact internal (Fig. 7.1). So, a force of 1.2 is a force acting on element 1 from element 2. A force of 2.1 acts on element 2 from element 1. These are internal forces; these also include forces 1.3 and 3.1, 2.3 and 3.2. It is obvious that the geometric sum of internal forces is equal to zero, since according to Newton’s third law

12 = - 21, 23 = - 32, 31 = - 13, etc.

Statics - special case dynamics, since the rest of bodies when forces act on them is a special case of motion ( = 0).

In general, several external forces can act on each element. By 1, 2, 3, etc. we will understand all external forces applied respectively to elements 1, 2, 3, .... In the same way, through "1, "2, "3, etc. we denote the geometric sum of internal forces applied to elements 2, 2, 3, ... respectively (these forces are not shown in the figure), i.e.

" 1 = 12 + 13 + ... , " 2 = 21 + 22 + ... , " 3 = 31 + 32 + ... etc.

If the body is at rest, then the acceleration of each element is zero. Therefore, according to Newton’s second law, the geometric sum of all forces acting on any element will also be equal to zero. Therefore, we can write:

1 + "1 = 0, 2 + "2 = 0, 3 + "3 = 0. (7.1)

Each of these three equations expresses the equilibrium condition of a rigid body element.

The first condition for the equilibrium of a rigid body.

Let us find out what conditions external forces applied to a solid body must satisfy in order for it to be in equilibrium. To do this, we add equations (7.1):

(1 + 2 + 3) + ("1 + "2 + "3) = 0.

In the first brackets of this equality we write vector sum all external forces applied to the body, and secondly, the vector sum of all internal forces acting on the elements of this body. But, as is known, the vector sum of all internal forces of the system is equal to zero, since, according to Newton’s third law, any inner strength corresponds to a force equal to it in magnitude and opposite in direction. Therefore, on the left side of the last equality only the geometric sum of external forces applied to the body will remain:

1 + 2 + 3 + ... = 0 . (7.2)

In the case of an absolutely rigid body, condition (7.2) is called the first condition for its equilibrium.

It is necessary, but not sufficient.

So, if a rigid body is in equilibrium, then the geometric sum of external forces applied to it is equal to zero.

If the sum of external forces is zero, then the sum of the projections of these forces on the coordinate axes is also zero. In particular, for the projections of external forces on the OX axis, we can write:

F 1x + F 2x + F 3x + ... = 0. (7.3)

The same equations can be written for the projections of forces on the OY and OZ axes.

The second condition for the equilibrium of a rigid body.

Let us make sure that condition (7.2) is necessary, but not sufficient for the equilibrium of a rigid body. Let us apply it to the board lying on the table in various points two equal in magnitude and oppositely directed forces as shown in Figure 7.2. The sum of these forces is zero:

+ (-) = 0. But the board will still rotate. In the same way, two forces of equal magnitude and opposite directions turn the steering wheel of a bicycle or car (Fig. 7.3).

What other condition for external forces, besides their sum being equal to zero, must be satisfied for a rigid body to be in equilibrium? Let's use the theorem about the change in kinetic energy.

Let us find, for example, the equilibrium condition for a rod hinged on a horizontal axis at point O (Fig. 7.4). This simple device, as you know from the basic school physics course, is a lever of the first kind.

Let forces 1 and 2 be applied to the lever perpendicular to the rod.

In addition to forces 1 and 2, a vertically upward force acts on the lever normal reaction 3 from the side of the lever axis. At lever equilibrium, the sum of all three forces is equal to zero: 1 + 2 + 3 = 0.

Let's calculate the work done by external forces when turning the lever through a very small angle α. The application points of forces 1 and 2 will travel along the paths s 1 = BB 1 and s 2 = CC 1 (arcs BB 1 and CC 1 at small angles α can be considered straight segments). The work A 1 = F 1 s 1 of force 1 is positive, because point B moves in the direction of the force, and the work A 2 = -F 2 s 2 of force 2 is negative, because point C moves to the side, opposite direction forces 2. Force 3 does not do any work, since the point of its application does not move.

The traveled paths s 1 and s 2 can be expressed in terms of the angle of rotation of the lever a, measured in radians: s 1 = α|VO| and s 2 = α|СО|. Taking this into account, let us rewrite the expressions for work as follows:

A 1 = F 1 α|BO|, (7.4)
A 2 = -F 2 α|CO|.

The radii BO and СО of the circular arcs described by the points of application of forces 1 and 2 are perpendiculars lowered from the axis of rotation on the line of action of these forces

As you already know, the arm of a force is the shortest distance from the axis of rotation to the line of action of the force. We will denote the force arm by the letter d. Then |VO| = d 1 - arm of force 1, and |СО| = d 2 - arm of force 2. In this case, expressions (7.4) will take the form

A 1 = F 1 αd 1, A 2 = -F 2 αd 2. (7.5)

From formulas (7.5) it is clear that the work of each force is equal to the product of the moment of force and the angle of rotation of the lever. Consequently, expressions (7.5) for work can be rewritten in the form

A 1 = M 1 α, A 2 = M 2 α, (7.6)

A full time job external forces can be expressed by the formula

A = A 1 + A 2 = (M 1 + M 2)α. α, (7.7)

Since the moment of force 1 is positive and equal to M 1 = F 1 d 1 (see Fig. 7.4), and the moment of force 2 is negative and equal to M 2 = -F 2 d 2, then for work A we can write the expression

A = (M 1 - |M 2 |)α.

When the body begins to move, it kinetic energy increases. To increase kinetic energy, external forces must do work, i.e. in this case A ≠ 0 and, accordingly, M 1 + M 2 ≠ 0.

If the work done by external forces is zero, then the kinetic energy of the body does not change (remains equal to zero) and the body remains motionless. Then

M 1 + M 2 = 0. (7.8)

Equation (7 8) is second condition for the equilibrium of a rigid body.

When a rigid body is in equilibrium, the sum of the moments of all external forces acting on it relative to any axis is equal to zero.

So, in case any number external forces, the equilibrium conditions for an absolutely rigid body are as follows:

1 + 2 + 3 + ... = 0, (7.9)
M 1 + M 2 + M 3 + ... = 0.

The second equilibrium condition can be derived from the basic equation of the dynamics of the rotational motion of a rigid body. According to this equation where M is the total moment of forces acting on the body, M = M 1 + M 2 + M 3 + ..., ε - angular acceleration. If the rigid body is motionless, then ε = 0, and, therefore, M = 0. Thus, the second equilibrium condition has the form M = M 1 + M 2 + M 3 + ... = 0.

If the body is not absolutely solid, then under the action of external forces applied to it it may not remain in equilibrium, although the sum of external forces and the sum of their moments relative to any axis are equal to zero.

Let us apply, for example, to the ends of a rubber cord two forces equal in magnitude and directed along the cord in opposite sides. Under the influence of these forces, the cord will not be in equilibrium (the cord is stretched), although the sum of external forces is equal to zero and the sum of their moments relative to the axis passing through any point of the cord is equal to zero.

TYPES OF EQUILIBRIUM

In the statics of an absolutely rigid body, three types of equilibrium are distinguished.

1. Consider a ball that is on a concave surface. In the position shown in Fig. 88, the ball is in equilibrium: the reaction force of the support balances the force of gravity .

If the ball is deflected from the equilibrium position, then the vector sum of the forces of gravity and the reaction of the support is no longer equal to zero: a force arises , which tends to return the ball to its original equilibrium position (to the point ABOUT).

This is an example of stable equilibrium.

S u t i a t i o n This type of equilibrium is called, upon exiting which forces or moments of forces arise that tend to return the body to an equilibrium position.

The potential energy of the ball at any point on the concave surface is greater than the potential energy at the equilibrium position (at the point ABOUT). For example, at the point A(Fig. 88) potential energy is greater than the potential energy at a point ABOUT by the amount E P ( A) - E n(0) = mgh.

In a position of stable equilibrium, the potential energy of the body has a minimum value compared to neighboring positions.

2. A ball on a convex surface is in an equilibrium position at the top point (Fig. 89), where the force of gravity is balanced by the support reaction force. If you deflect the ball from the point ABOUT, then a force appears directed away from the equilibrium position.

Under the influence of force, the ball will move away from the point ABOUT. This is an example of an unstable equilibrium.

Unstable This type of equilibrium is called, upon exiting which forces or moments of forces arise that tend to take the body even further from the equilibrium position.

The potential energy of a ball on a convex surface is highest value(maximum) at point ABOUT. At any other point the potential energy of the ball is less. For example, at the point A(Fig. 89) potential energy is less than at a point ABOUT, by the amount E P ( 0 ) - E p ( A) = mgh.

In a position of unstable equilibrium, the potential energy of the body has maximum value compared to neighboring positions.

3. On a horizontal surface, the forces acting on the ball are balanced at any point: (Fig. 90). If, for example, you move the ball from the point ABOUT exactly A, then the resultant force
gravity and ground reaction are still zero, i.e. at point A the ball is also in an equilibrium position.

This is an example of indifferent equilibrium.

Indifferent This type of equilibrium is called, upon exiting which the body remains in a new position in equilibrium.

The potential energy of the ball at all points of the horizontal surface (Fig. 90) is the same.

In positions of indifferent equilibrium, the potential energy is the same.

Sometimes in practice it is necessary to determine the type of equilibrium of bodies various shapes in the field of gravity. To do this you need to remember following rules:

1. The body can be in a position of stable equilibrium if the point of application of the ground reaction force is above the center of gravity of the body. Moreover, these points lie on the same vertical (Fig. 91).

In Fig. 91, b The role of the support reaction force is played by the tension force of the thread.

2. When the point of application of the ground reaction force is below the center of gravity, two cases are possible:

If the support is point-like (the surface area of ​​the support is small), then the balance is unstable (Fig. 92). With a small deviation from the equilibrium position, the moment of force tends to increase the deviation from initial position;

If the support is non-point (the surface area of ​​the support is large), then the equilibrium position is stable in the case when the line of action of gravity AA" intersects the surface of the body support
(Fig. 93). In this case, with a slight deviation of the body from the equilibrium position, a moment of force and occurs, which returns the body to its original position.

??? ANSWER THE QUESTIONS:

1. How does the position of the center of gravity of a body change if the body is removed from the position of: a) stable equilibrium? b) unstable equilibrium?

2. How does the potential energy of a body change if its position is changed in indifferent equilibrium?