The first condition for the equilibrium of a rigid body. Statics

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

To find the equilibrium conditions of an absolutely rigid body, it is necessary to mentally break it down into a large number of fairly 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 on 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 a 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 a body to be in equilibrium, it is necessary and sufficient that the geometric sum of all forces acting on any element of this body be equal to zero.

From this 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 up the equation for 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 magnitude and opposite in direction.

Hence,

.

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

This condition is necessary, but not sufficient. This is easy to verify by remembering the rotating action of a pair of forces, the geometric sum of which is also 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 of a rigid body in the case of an arbitrary number of external forces look like this:

.

Statics is a branch of mechanics that studies the equilibrium of bodies. Statics makes it possible to determine the conditions of equilibrium of bodies and answers some questions that relate to the movement of bodies, for example, it gives an answer in which direction the movement occurs if the balance is disturbed. It is worth looking around and you will notice that most bodies are in equilibrium - they are either moving at a constant speed or at rest. This conclusion can be drawn from Newton's laws.

An example is the person himself, a picture hanging on the wall, cranes, various buildings: bridges, arches, towers, buildings. The bodies around us are exposed to some forces. Different amounts of forces act on bodies, but if we find the resultant force, for a body in equilibrium it will be equal to zero.
There are:

• static equilibrium - the body is at rest;
• dynamic equilibrium - a body moves at a constant speed.

Static balance. If forces F1, F2, F3, and so on act on a body, then the main requirement for the existence of a state of equilibrium is (equilibrium). This is a vector equation in three-dimensional space, and represents three separate equations, one for each direction of space. .

The projections of all forces applied to the body in any direction must be compensated, that is, the algebraic sum of the projections of all forces in any direction must be equal to 0.

When finding the resultant force, you can transfer all the forces and place the point of their application at the center of mass. The center of mass is a point that is introduced to characterize the movement of a body or a system of particles as a whole, characterizes the distribution of masses in the body.

In practice, we very often encounter cases of both translational and rotational motion at the same time: a barrel rolling down an inclined plane, a dancing couple. With such a movement, the condition of equilibrium alone is not enough.

The necessary equilibrium condition in this case will be:

In practice and in life, it plays a big role stability of bodies, characterizing balance.

There are different types of balance:

• Stable balance;
• Unstable equilibrium;
• Indifferent balance.

Stable balance- this is equilibrium when, with a small deviation from the equilibrium position, a force arises that returns it to a state of equilibrium (a pendulum of a stopped clock, a tennis ball rolled into a hole, a Vanka-Vstanka or a tumbler, laundry on a line are in a state of stable equilibrium).

Unstable equilibrium– this is a state when a body, after being removed from an equilibrium position, deviates due to the resulting force even more from the equilibrium position (a tennis ball on a convex surface).

Indifferent Equilibrium- being left to its own devices, the body does not change its position after being removed from a state of equilibrium (a tennis ball lying on the table, a picture on the wall, scissors, a ruler hanging on a nail are in a state of indifferent equilibrium). The axis of rotation and the center of gravity coincide.

For two bodies, the body will be more stable, which has larger support area.

A body is at rest (or moves uniformly and rectilinearly) if the vector sum of all forces acting on it is equal to zero. They say that forces balance each other. When we are dealing with a body of a certain geometric shape, when calculating the resultant force, all forces can be applied to the center of mass of the body.

Condition for equilibrium of bodies

For a body that does not rotate to be in equilibrium, it is necessary that the resultant of all forces acting on it be equal to zero.

F → = F 1 → + F 2 → + . . + F n → = 0 .

The figure above shows the equilibrium of a rigid body. The block is in a state of equilibrium under the influence of three forces acting on it. The lines of action of the forces F 1 → and F 2 → intersect at point O. The point of application of gravity is the center of mass of the body C. These points lie on the same straight line, and when calculating the resultant force F 1 →, F 2 → and m g → are brought to point C.

The condition that the resultant of all forces be equal to zero is not enough if the body can rotate around a certain axis.

The arm of force d is the length of the perpendicular drawn from the line of action of the force to the point of its application. The moment of force M is the product of the force arm and its modulus.

The moment of force tends to rotate the body around its axis. Those moments that turn the body counterclockwise are considered positive. The unit of measurement of moment of force in the international SI system is 1 Newtonmeter.

Definition. Rule of Moments

If the algebraic sum of all moments applied to a body relative to a fixed axis of rotation is equal to zero, then the body is in a state of equilibrium.

M 1 + M 2 + . . +Mn=0

Important!

In the general case, for bodies to be in equilibrium, two conditions must be met: the resultant force must be equal to zero and the rule of moments must be observed.

In mechanics there are different types of equilibrium. Thus, a distinction is made between stable and unstable, as well as indifferent equilibrium.

A typical example of indifferent equilibrium is a rolling wheel (or ball), which, if stopped at any point, will be in a state of equilibrium.

Stable equilibrium is such an equilibrium of a body when, with its small deviations, forces or moments of force arise that tend to return the body to an equilibrium state.

Unstable equilibrium is a state of equilibrium, with a small deviation from which forces and moments of forces tend to throw the body out of balance even more.

In the figure above, the position of the ball is (1) - indifferent equilibrium, (2) - unstable equilibrium, (3) - stable equilibrium.

A body with a fixed axis of rotation can be in any of the described equilibrium positions. If the axis of rotation passes through the center of mass, indifference equilibrium occurs. In stable and unstable equilibrium, the center of mass is located on a vertical straight line that passes through the axis of rotation. When the center of mass is below the axis of rotation, the equilibrium is stable. Otherwise, it's the other way around.

A special case of balance is the balance of a body on a support. In this case, the elastic force is distributed over the entire base of the body, rather than passing through one point. A body is at rest in equilibrium when a vertical line drawn through the center of mass intersects the area of ​​support. Otherwise, if the line from the center of mass does not fall into the contour formed by the lines connecting the support points, the body tips over.

An example of body balance on a support is the famous Leaning Tower of Pisa. According to legend, Galileo Galilei dropped balls from it when he conducted his experiments on studying the free fall of bodies.

A line drawn from the center of mass of the tower intersects the base approximately 2.3 m from its center.

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The static calculation of engineering structures in many cases comes down to the consideration of the equilibrium conditions of a structure consisting of a system of bodies connected by some kind of connections. The connections connecting the parts of this structure will be called internal Unlike external connections connecting the structure to bodies not included in it (for example, to supports).

If, after discarding external connections (supports), the structure remains rigid, then statics problems are solved for it as for an absolutely rigid body. However, there may be engineering structures that do not remain rigid after discarding external connections. An example of such a design is a three-hinged arch. If we discard supports A and B, then the arch will not be rigid: its parts can rotate around hinge C.

Based on the principle of solidification, the system of forces acting on such a structure must, in equilibrium, satisfy the equilibrium conditions of a solid body. But these conditions, as indicated, while necessary, will not be sufficient; therefore, it is impossible to determine all unknown quantities from them. To solve the problem, it is necessary to additionally consider the equilibrium of one or more parts of the structure.

For example, by composing equilibrium conditions for the forces acting on a three-hinged arch, we obtain three equations with four unknowns X A, Y A, X B, Y B . Having additionally considered the equilibrium conditions of the left (or right) half of it, we obtain three more equations containing two new unknowns X C, Y C, in Fig. 61 not shown. By solving the resulting system of six equations, we find all six unknowns.

14. Special cases of reduction of a spatial system of forces

If, when bringing a system of forces to a dynamic screw, the main moment of the dynamo turns out to be equal to zero, and the main vector is different from zero, then this means that the system of forces is reduced to a resultant, and the central axis is the line of action of this resultant. Let us find out under what conditions related to the main vector Fp and the main moment M 0 this can happen. Since the main moment of the dynamism M* is equal to the component of the main moment M 0 directed along the main vector, the considered case M* = O means that the main moment M 0 is perpendicular to the main vector, i.e. / 2 = Fo*M 0 = 0. It immediately follows that if the main vector F 0 is not equal to zero, and the second invariant is equal to zero, Fo≠O, / 2 = F 0 *M 0 =0, (7.9) then the considered the system is reduced to the resultant.

In particular, if for any reduction center F 0 ≠0, and M 0 = 0, then this means that the system of forces is reduced to a resultant passing through this reduction center; in this case, condition (7.9) will also be satisfied. Let us generalize the theorem on the moment of the resultant (Varignon’s theorem) given in Chapter V to the case of a spatial system of forces. If the spatial system. forces are reduced to a resultant, then the moment of the resultant relative to an arbitrary point is equal to the geometric sum of the moments of all forces relative to the same point. P
Let the system of forces have a resultant R and a point ABOUT lies on the line of action of this resultant. If we bring a given system of forces to this point, we obtain that the main moment is equal to zero.
Let's take some other reduction center O1; (7.10)C
on the other hand, based on formula (4.14) we haveMo1=Mo+Mo1(Fo), (7.11) since M 0 = 0. Comparing expressions (7.10) and (7.11) and taking into account that in this case F 0 = R, we obtain (7.12).

Thus, the theorem is proven.

Let, for any choice of the reduction center, Fo=O, M ≠0. Since the main vector does not depend on the reduction center, it is equal to zero for any other choice of the reduction center. Therefore, the main moment also does not change when the center of reduction changes, and, therefore, in this case the system of forces is reduced to a pair of forces with a moment equal to M0.

Let us now compile a table of all possible cases of reduction of the spatial system of forces:

If all the forces are in the same plane, for example, in the plane Ooh, then their projections onto the axis G and moments about the axes X And at will be equal to zero. Therefore, Fz=0; Mox=0, Moy=0. Introducing these values ​​into formula (7.5), we find that the second invariant of a plane system of forces is equal to zero. We obtain the same result for a spatial system of parallel forces. Indeed, let all forces be parallel to the axis z. Then their projections on the axis X And at and the moments about the z axis will be equal to 0. Fx=0, Fy=0, Moz=0

Based on what has been proven, it can be argued that a plane system of forces and a system of parallel forces are not reduced to a dynamic screw.

11. Equilibrium of a body in the presence of sliding friction If two bodies / and // (Fig. 6.1) interact with each other, touching at a point A, then the reaction R A, acting, for example, from the side of the body // and applied to the body /, can always be decomposed into two components: N.4, directed along the common normal to the surface of the contacting bodies at point A, and T 4, lying in the tangent plane . Component N.4 is called normal reaction force T l is called sliding friction force - it prevents the body from sliding / along the body //. In accordance with the axiom 4 (Newton's 3rd z-on) a reaction force of equal magnitude and opposite direction acts on the body // from the side of the body /. Its component perpendicular to the tangent plane is called force of normal pressure. As mentioned above, the friction force T A = Oh, if the contacting surfaces are perfectly smooth. In real conditions, surfaces are rough and in many cases the friction force cannot be neglected. To clarify the basic properties of friction forces, we will carry out an experiment according to the scheme presented in Fig. 6.2, A. To body 5, located on a stationary plate D, is attached a thread thrown over block C, the free end of which is equipped with a support platform A. If the pad A gradually load, then with an increase in its total weight the thread tension will increase S, which tends to move the body to the right. However, as long as the total load is not too great, the frictional force T will hold the body IN at rest. In Fig. 6.2, b acts on the body are depicted IN forces, and P denotes the force of gravity, and N denotes the normal reaction of the plate D. If the load is insufficient to break the rest, the following equilibrium equations are valid: N- P = 0, (6.1) S-T = 0. (6.2). It follows from this that N = PAnd T = S. Thus, while the body is at rest, the friction force remains equal to the tension force of the thread S. Let us denote by Tmax friction force at the critical moment of the loading process, when the body IN loses balance and begins to slide on the slab D. Therefore, if the body is in equilibrium, then T≤Tmax.Maximum friction force T tah depends on the properties of the materials from which the bodies are made, their condition (for example, on the nature of surface treatment), as well as on the value of normal pressure N. As experience shows, the maximum friction force is approximately proportional to normal pressure, i.e. e. there is equality Tmax= fN. (6.4). This relation is called Amonton-Coulomb law. The dimensionless coefficient / is called sliding friction coefficient. As follows from experience, it the value does not depend within wide limits on the area of ​​contacting surfaces, but depends on the material and the degree of roughness of the contacting surfaces. The friction coefficient values ​​are determined empirically and can be found in reference tables. Inequality" (6.3) can now be written as T≤fN (6.5).The case of strict equality in (6.5) corresponds to the maximum value of the friction force. This means that the friction force can be calculated using the formula T = fN only in cases where it is known in advance that a critical incident is occurring. In all other cases, the friction force should be determined from the equilibrium equations. Consider a body located on a rough surface. We will assume that as a result of the action of active forces and reaction forces, the body is in limiting equilibrium. In Fig. 6.6, a the limiting reaction R and its components N and Tmax are shown (in the position shown in this figure, active forces tend to move the body to the right, the maximum friction force Tmax is directed to the left). Corner f between limit reaction R and the normal to the surface is called the friction angle. Let's find this angle. From Fig. 6.6, and we have tgφ=Tmax/N or, using expression (6.4), tgφ= f (6-7) From this formula it is clear that instead of the friction coefficient, you can set the friction angle (in the reference tables p

both quantities are given).