Bringing a flat system of forces to a given center. Bringing a flat system of forces to a given point

31.05.2020

Theorem . StrengthF , without changing its effect on the body, it can be transferred from the point of its application A to any center of reduction O, while attaching to the body a pair of forces with a momentM , geometrically equal to the momentM O (F ) of this force relative to the center of reference.

Let the force be given F, lying in the horizontal plane OXY parallel to the axis OX (Fig. 1.41).

According to the Poinsot method instead of force F applied at point A, the force is obtained F 1 , equal in magnitude to the force F, but applied at the point O and attached pair of forces , whose vector moment M= M O ( F).

According to the equivalence theorem for pairs of forces, an attached pair of forces can be replaced by any other pair of forces with the same vector moment.

1.15. Bringing an arbitrary system of forces to a given center

Theorem . Any arbitrary system of forces acting on a body can be reduced in the general case to a force and a pair of forces.

This process of replacing a system of forces with one force and a pair of forces is called bringing the system of forces to a given center .

Let an arbitrary system of forces be given ( F 1 , …, F n) (Fig. 1.42).

By successively applying the Poinsot method to each of the given system of forces, we bring it to an arbitrary center O. As a result, we obtain a system of forces ( F 1 , …, F n) applied at the center O, and an attached pair of forces with a moment M= Σ M O ( F i). Adding forces F 1 , …, F n by the parallelogram rule, we obtain their resultant R* , equal to the geometric sum of the given forces and applied in the center of reduction.

The geometric sum of all the forces of the system is called the main vector of the system of forces and, in contrast to the resultant R, denote R * .

Vector M= Σ M O ( F i) call the main moment of the system of forces relative to the center of reduction.

This result can be formulated as follows: forces arbitrarily located in space can be reduced to one force equal to their main vector and applied at the center of reduction and to a pair of forces with a moment equal to the main moment of all forces relative to the center of reduction.

The choice of the center of reduction does not affect the modulus and the direction of the main vector R* , but affects the modulus and direction of the main moment M. Main vector R* is a free vector and can be applied anywhere on the body.

1.16. Analytical equilibrium conditions for a plane arbitrary system of forces

Planar arbitrary system of forces – a system of forces whose lines of action are arbitrarily located in the same plane.

The lines of action of a plane arbitrary system of forces intersect at different points.

and fig. 1.43 shows a given flat arbitrary system of forces ( F 1 , …, F n) whose action lines lie in the OYZ plane.

Consistently applying the Poinsot method for each of the forces F i , we will carry out a parallel transfer of forces from points A i to the beginning O of the OXYZ reference frame. According to this method, the force F i will be equivalent to the force F i applied at the point O, and the attached pair of forces with the moment M i = M O ( F i ) . In this case, M i = ± F i h i , where h i is the arm of the force F i with respect to the reduction center O. Upon completion of this work, we obtain a converging system of forces ( F i ,…, F n) and a convergent system of vector moments M i = M O ( F i) coupled pairs of forces applied at the center of reduction. Adding the vectors of forces, we get the heads

vector R* = Σ F i and the main moment of the equivalent pair of forces M = Σ M O ( F i).

In this way, plane arbitrary system of forces (F i ,…, F n ) is equivalent to one force R* = Σ F i and a pair of forces with moment M = Σ M O (F i ).

When solving static problems, projections of forces on coordinate axes and algebraic moments of forces relative to a point are used.

On fig. 1.44 shows a flat arbitrary system of forces, reduced to the main vector of forces, the module of which is R * =
and equivalent pair forces with algebraic moment M = Σ M О ( F i).

these formulas Σ F iО X , Σ F iОY are the sums of projections of forces on the coordinate axes; Σ M O ( F i) - the sum of the algebraic moments of forces about the point O.

Geometric equilibrium condition of any system of forces is expressed by vector equalities: R* = Σ F i = 0; M= Σ M O ( F i) = 0.

When solving problems, it is required to determine the reactions R i E external constraints imposed on the mechanical system. At the same time, the active forces F i E applied to this system are known. Since the active forces F i E and bond reactions R i E belong to the category of external forces, then it is expedient to express the geometric equilibrium condition of the system of external forces by vector equalities:

Σ F i E + Σ R i E = 0;

Σ M A( F i E) + Σ M A( R i E) = 0.

For the equilibrium of the system of external forces, it is necessary and sufficient that the geometric sum of the active forces F i E and reactions R i E external bonds and the geometric sum of the moments of active forces M A ( F i E ) and reactions of external relations M A ( R i E ) with respect to an arbitrary point A were equal to zero.

Projecting these vector equalities onto the coordinate axes of the reference system, we obtain analytical conditions for the equilibrium of a system of external forces . For a plane arbitrary system of forces, these equations take the following form:

Σ
+ Σ
= 0;

Σ M A ( F i E) + Σ M A ( R i E) = 0,

where Σ
, Σ
- respectively, the sum of the projections of active forces on the coordinate axes OX, OY; Σ
, Σ
are the sums of projections of external bond reactions on the coordinate axes OX, OY; Σ M A ( F i E) is the sum of algebraic moments of active forces F i E relative to point A; Σ M A ( R i E) is the sum of algebraic moments of reactions R i E external links relative to point A.

The combination of these formulas is the first (basic) form of the equilibrium equations for a plane arbitrary system of external forces .

In this way , for the equilibrium of a flat arbitrary system of external forces applied to a mechanical system, it is necessary and sufficient that the sum of the projections of active forces and reactions of external links on two coordinate axes and the sum of algebraic moments of active forces and reactions of external links relative to an arbitrary point A are equal to zero.

There are other forms of equilibrium equations for a plane arbitrary system of forces.

Second form expressed by a set of formulas:

Σ
+ Σ
= 0;

Σ M A ( F i E) + Σ M A ( R i E) = 0;

Σ M B ( F i E) + Σ M В ( R i E) = 0.

For the equilibrium of a flat arbitrary system of external forces applied to the body, it is necessary and sufficient that the sum of the projections of forces on the coordinate axis and the sum of the algebraic moments of forces relative to arbitrary points A and B equal zero.

Third form equilibrium equations are expressed by a set of formulas:

Σ M A ( F i E) + Σ M A ( R i E) = 0;

Σ M B ( F i E) + Σ M В ( R i E) = 0;

Σ M C ( F i E) + Σ M С ( R i E) = 0.

For the equilibrium of a flat arbitrary system of external forces applied to the body, it is necessary and sufficient that the sums of the algebraic moments of these forces with respect to arbitrary points A, B and C are equal to zero.

When using the third form of equilibrium equations, points A, B and C should not lie on the same line.

The method of bringing one force to a given point can be applied to any number of forces. Let us assume that at some points of the body (Fig. 1.24) forces are applied F 1 F 2 , F 3 and F4. It is required to bring these forces to the point O planes. Let us first give the force applied at the point A. Let us apply (see § 16) at the point O two forces equal separately in value to a given force parallel to it and directed in opposite directions. As a result of bringing the force, we get the force , applied at point O, and a pair of forces with a shoulder . Doing the same with force , applied at the point V, get power , applied at the point O, and a pair of forces with a shoulder, etc. A flat system of forces applied at points A, B, C and D, we have replaced by converging forces , attached at a point O, and pairs of forces with moments equal to the moments of given forces about the point O:

fig.1.24

Forces converging at a point can be replaced by one force equal to the geometric sum of the components,

This force, equal to the geometric sum of the given forces, is called the main vector of the system of forces and denote .

By the magnitude of the projections of the main vector on the coordinate axes, we find the module of the main vector:

Based on the rule for adding pairs of forces, they can be replaced by the resulting pair, the moment of which is equal to the algebraic sum of the moments of the given forces about the point O and called highlight relative to the reference point

Thus, an arbitrary flat system of forces can be reduced to one force(the main vector of the system of forces) and one moment(principal moment of the system of forces).

It is necessary to learn that the hundred main vector is not the resultant of this system of forces, since this system is not equivalent to one force. Since the main vector is equal to the geometric sum of the forces of a given system, neither the module nor its direction depends on the choice of the center of reduction. The value and sign of the main moment depends on the position of the center of reduction, since the shoulders of the constituent pairs depend on the mutual position of the forces and the point (center) relative to which the moments are taken.

Special cases of bringing the system of forces:

one) ; the system is in equilibrium, i.e. for the equilibrium of a flat system of forces, it is necessary and sufficient that its main vector and main moment are simultaneously equal to zero.

The plane system of forces is also reduced to a force equal to and applied in an arbitrarily chosen center O, and a pair with a moment

in this case, the vector can be determined either geometrically by constructing a force polygon (see Section 4), or analytically. Thus, for a flat system of forces

R x = F kx , R y = F ky ,

where all the moments in the last equality are algebraic and the sum is also algebraic.

Let us find the simplest form to which a given flat system of forces that is not in equilibrium can be reduced. The result depends on the values of R and M O .

1. If for a given system of forces R = 0, a M O ?0, then it is reduced to one pair with a moment M O, the value of which does not depend on the choice of the center O.
2. If for a given system of forces R? 0, then it is reduced to one force, i.e., to the resultant. In this case, two cases are possible:
- a) R?0, M O =0. In this case, the system, which is immediately evident, is reduced to the resultant R passing through the center O;
- b) R?0, M O?0. In this case, a pair with a moment M O can be represented by two forces R" and R", taking R"=R, and R"= - R. In this case, if d=OC is the arm of the pair, then it should be Rd=|MO | .

Now discarding the forces R and R "as balanced, we find that the entire system of forces is replaced by the resultant R" = R passing through the point C. The position of the point C is determined by two conditions: 1) the distance OC = d () must satisfy the equality Rd = | MO |; 2) the sign of the moment relative to the center O of the force R "applied at point C, i.e., the sign of m O (R") must coincide with the sign of M O.

The theorem on reduction of the system of forces:

Any system of forces acting on an absolutely solid, can be replaced by one force R, equal to the main vector of this system of forces and applied to an arbitrarily chosen center O, and one pair of forces with a moment L O , equal to the main moment of the system of forces about the center O.

Such an equivalent replacement of a given system of forces by a force R and a couple of forces with a moment L O call bringing the system of forces to the center O.

Consider here special case bringing a flat system of forces to the center O, which lies in the same plane. In this case, the system of forces is replaced by one force and one pair of forces lying in the plane of action of the forces of the system. The moment of this pair of forces can be considered as an algebraic quantity LO and depicted in the figures by an arc arrow (the algebraic main moment of a flat system of forces).

As a result of bringing a flat system of forces to the center, the following cases are possible:

if R = 0, L O = 0, then the given system is equilibrium;
if at least one of the values R or L O is not equal to zero, then the system of forces not in balance.
Wherein:

16 question. Equilibrium equation

For the equilibrium of a rigid body under the action of a plane system of forces, it is necessary and sufficient that the main vector of this system of forces and its algebraic principal moment be equal to zero, that is R\u003d 0, L O \u003d 0, where O is any center located in the plane of action of the forces of the system.

The resulting analytical equilibrium conditions (equilibrium equations) for a flat system of forces can be formulated in the following three forms:

The main form of the equilibrium equations:

for the equilibrium of an arbitrary planar system of forces, it is necessary and sufficient that the sum of the projections of all forces on each of the coordinate axes and the sum of their algebraic moments with respect to any center lying in the plane of action of the forces are equal to zero:

Fix = 0; F iy = 0; M O ( F i) = 0. (I)

The second form of equilibrium equations:

for the equilibrium of an arbitrary planar system of forces, it is necessary and sufficient that the sum of the algebraic moments of all forces about two centers A and B and the sum of their projections on the Ox axis, which is not perpendicular to the Ox axis, be equal to zero:

Fix = 0; M A ( F i) = 0; M B ( F i) = 0. (II)

The third form of equilibrium equations (equations of three moments):

for the equilibrium of an arbitrary planar system of forces, it is necessary and sufficient that the sums of the algebraic moments of all forces with respect to any three centers A, B and C, not lying on the same straight line, were equal to zero:

M A ( F i) = 0; M B ( F i) = 0; M C ( F i) = 0. (III)

Equilibrium equations in the form (I) are considered basic, since when using them there are no restrictions on the choice of coordinate axes and the center of moments.

Question

Varignon's theorem. If the plane system of forces under consideration is reduced to a resultant, then the moment of this resultant relative to any point is equal to the algebraic sum of the moments of all forces of the given system relative to that point itself. Suppose that the system of forces is reduced to the resultant R passing through the point O. Let us now take another point O 1 as the center of reduction. The main moment (5.5) about this point is equal to the sum moments of all forces: M O1Z =åM o1z (F k) (5.11). On the other hand, we have M O1Z =M Olz (R), (5.12) since the principal moment for the reduction center O is equal to zero (M Oz =0). Comparing relations (5.11) and (5.12), we obtain M O1z (R)=åM OlZ (F k); (5.13) h.e.d. Using the Varignon theorem, you can find the equation for the line of action of the resultant. Let the resultant R 1 be applied at some point O 1 with coordinates x and y (Fig. 5.5) and the main vector F o and the main moment M Oya at the center of reduction at the origin are known. Since R 1 \u003d F o, then the components of the resultant along the x and y axes are R lx \u003d F Ox \u003d F Ox i and R ly \u003d F Oy \u003d F oy j. According to the Varignon theorem, the moment of the resultant relative to the origin is equal to the main moment at the center of reduction at the origin, i.e. M oz \u003d M Oz (R 1) \u003d xF Oy -yF Ox. (5.14). The values M Oz , F Ox and F oy do not change when the point of application of the resultant is moved along its line of action, therefore, the coordinates x and y in equation (5.14) can be viewed as the current coordinates of the line of action of the resultant. Thus, equation (5.14) is the equation of the line of action of the resultant. For F ox ≠0, it can be rewritten as y=(F oy /F ox)x–(M oz /F ox).

Question

termination one body into another (for example, a rod into a fixed wall) does not allow this body to move and rotate relative to another. In the case of termination, the force reaction R A is not the only factor in the interaction between the body and the support. In addition to this force, the embedding reaction is also determined by a pair of forces with a moment M A unknown in advance. If force R A present its constituents X A , Y A , then to find the termination reaction it is necessary to determine three unknown scalar quantities: X A , Y A , M A .

Let us give examples of the replacement of flat systems of parallel distributed forces by their resultant ones.

For such a system of forces, the intensity has a constant value: q = const.

When solving problems of statics, this system of forces can be replaced by a concentrated resultant force Q, equal in absolute value to the product of the intensity q and the length of the segment AB = a (Q = q · a) and applied in the middle of the segment AB.

For such a system of forces, the intensity q is a variable that varies from zero to the maximum value q max according to a linear law.

Resultant Q of this system of forces is equal in absolute value to Q =0.5 a q max and is applied at the point K dividing the segment AB in relation to AK: KB = 2: 1.

19. Calculation of composite structures
1.1. Calculation with the division of a system of bodies into separate bodies
1.1.1. The system of bodies is divided by internal connection C into separate bodies and their equilibrium is considered.
1.1.2. All connections are discarded from each of the bodies, replacing their action with reactions. In the given mechanisms, the following types of connections are applied: a fixed axial hinge (the reaction is decomposed into components parallel to the coordinate axes X, Y); movable axial hinge (reaction N is perpendicular to the supporting surface, directed away from it); rigid termination (the reaction is a combination of the reaction of the fixed hinge X, Y and a pair of forces with a reactive
moment m). The components of the reaction of the internal hinge C, applied to different bodies of the system, according to the principle of action and reaction, are equal in magnitude and directed oppositely. The distributed load is replaced by a concentrated force applied in the middle of the interval and equal to the modulus of the product of the load intensity q and the length of the interval.
1.1.3. Compose equilibrium equations, including equations of projections onto standard axes and equations of moments (calculated and verification). The center of the calculated equation of moments is chosen at the intersection of the lines of action of the maximum number of unknown reactions, the verification equation - at the intersection of the lines of action of known forces, through which none of the unverified unknown reactions passes. It is recommended to draw up equilibrium equations, considering the forces in turn as follows: determine the acute angle α between the line of force and the line of one of the axes; the force projection on this axis will contain cos α, on the second axis – sin α; the projection is positive if the angle of alignment of the force vector with the axis is acute, and negative if it is obtuse; determine the shoulder of the force, lowering the perpendicular from the center to the line of action of the force, and the sign of the moment in the direction of rotation of the shoulder by the force around the center (when the shoulder is rotated clockwise, the moment is negative, counterclockwise is positive). At an arbitrary position of the force, to determine the moment, it is decomposed into components parallel to the coordinate axes (their magnitudes are equal to the corresponding projections of the force) and the sum of the moments of these components is found using the Varignon theorem.
Thus, for each of the bodies, 3 calculated and 1 verification equation are compiled.
1.1.4. Solve a system of 6 calculation equations for unknown reactions.
The found reactions are substituted into the verification equations, the module of the resulting sum should not exceed 0.02 Rav, where Rav is the average value of the modules of the tested reactions.
1.2. Calculation using the solidification principle
1.2.1. Replace the internal hinge with a rigid joint and consider the balance of the resulting body. The second is considered one of the bodies of the system (clause 1.1.1).
1.2.2. Draw up a drawing for each of the considered bodies in the same way as in clause 1.1.2.
1.2.3. For the first body, 3 calculated and 1 verification equations are made similarly to clause 1.1.3. For the second body, one calculation equation of the moments of forces relative to the center C is compiled.
1.2.4. Solve a system of 4 calculation equations and make a check similar to paragraph 1.1.4. 2.

2. Calculation using the principle of possible movements. Reactions of bonds are determined by considering them in turn.

20. Lever balance condition. Stability of bodies during capsizing. The distance from the fulcrum to the straight line along which the force acts is called the shoulder of this force. Let F1 and F2 denote the forces acting on the lever from the side of the loads (see diagrams on the right side of Fig. 25.2). Let us denote the shoulders of these forces as l1 and l2, respectively. Our experiments have shown that the lever is in equilibrium if the forces F1 and F2 applied to the lever tend to rotate it in opposite directions, and the modules of the forces are inversely proportional to the shoulders of these forces: F1 / F2 \u003d l2 / l1. Stability of bodies during overturning. These are the tasks that arise in the design of various lifting mechanisms and in the calculation of the safe conditions for their operation, stipulated in the rules for working with these mechanisms. A feature of solving these not very difficult problems for a flat system of forces is that when solving them, equilibrium equations are not compiled. Separately, the following are determined: a) overturning moment (Mopr) - the sum of the moments of forces that tend to overturn the mechanism in question relative to some axis projected on the drawing (support points); c) holding moment (Mud) - the sum of the moments of forces that prevent overturning. For the stable operation of the mechanism, it is necessary that the holding moment with a certain margin be greater than the overturning one. The ratio Mud, / Mopr =k is commonly called the stability coefficient. The value of k should, of course, be more than one. For various lifting mechanisms and for different conditions of their operation, the value of the stability coefficient is determined from SNiP, TU and other sources. Taking this coefficient into account, calculations are made of the value of the counterweight load or its position on the mechanism, options are calculated - at what reach of the boom and with what loads it is safe to work. An example of solving one of the stability problems is given below. It is especially important to be able to perform elementary stability calculations in production conditions, when you have to work with the maximum loads for the available crane.

21 Sliding friction. The laws of friction. Friction coefficient. A sliding friction force arises between moving bodies in the plane of their contact. This is primarily due to the roughness of the contacting surfaces and the presence of adhesion of the pressed bodies. In engineering calculations, empirically established regularities are usually used, which reflect the action of the friction force with a certain degree of accuracy. These patterns are called the laws of sliding friction (Coulomb). They can be formulated as follows.
1. When trying to move one body relative to another in the plane of their contact, a friction force arises F, the modulus of which can take any value from zero to Fmax, i.e. 0<=F<=Fmax . Сила трения приложена к телу и направлена в сторону, противоположную возможному направлению скорости точки приложения силы.
2. The maximum friction force is equal to the product of the friction coefficient f and the normal pressure force N: Fmax=fN.
The coefficient of friction f is a dimensionless quantity that depends on the materials and the state of the surfaces of the contacting bodies (roughness, temperature, humidity, etc.). Determine it empirically.
There are coefficients of static friction and sliding friction, and the latter, as a rule, also depends on the sliding speed. The coefficient of static friction corresponds to such a maximum friction force Fmax, at which there is a limit state of equilibrium. The slightest increase in external forces can cause movement. The coefficient of static friction is usually slightly greater than the coefficient of sliding friction. With an increase in the sliding speed, the value of the coefficient of sliding friction first slightly decreases, and then remains practically unchanged. The values of friction coefficients for some friction pairs are as follows: wood on wood 0.4-0.7; metal to metal 0.15-0.25; steel on ice 0.027.
3. The maximum friction force in a fairly wide range does not depend on the area of the contact surfaces.
The sliding friction force is sometimes called the dry friction force.

Angle and friction cone

The reaction of a real (rough) connection will be composed of two components: from the normal reaction and the friction force perpendicular to it. Consequently, the total reaction will deviate from the normal to the surface by some angle. When the friction force changes from zero to F pr, the force R will change from N to R pr, and its angle with the normal will increase from zero to a certain limit value (Fig. 26).

Fig.26

The largest angle that the total rough bond reaction makes with the normal to the surface is called friction angle. It can be seen from the drawing that

Since , from here we find the following relationship between the angle of friction and the coefficient of friction:

At equilibrium, the total reaction R, depending on the shear forces, can take place anywhere within the angle of friction. When the equilibrium becomes limiting, the reaction will deviate from the normal by an angle.

Friction cone called the cone described by the limiting reaction force of a rough bond around the direction of the normal reaction.

If a force P is applied to a body lying on a rough surface, forming an angle with the normal (Fig. 27), then the body will move only when the shear force Psin is greater (we consider N=Pcos, neglecting the weight of the body). But the inequality in which is satisfied only for , i.e. at . Therefore, no force that forms an angle with the normal less than the angle of friction can move the body along a given surface. This explains the well-known phenomena of jamming or self-braking of bodies.

Fig.27

For the equilibrium of a solid body on a rough surface, it is necessary and sufficient that the line of action of the resultant of the active forces acting on the solid body passes inside the friction cone or along its generatrix through its top.

The body cannot be unbalanced by any modulo active force if its line of action passes inside the cone of friction.

rolling friction

The origin of rolling friction can be visualized as follows. When a ball or cylinder rolls on the surface of another body, it is slightly pressed into the surface of this body, while itself is compressed slightly. Thus, the rolling body all the time, as it were, rolls up the hill.

Fig.33

At the same time, there is a detachment of sections of one surface from another, and the adhesion forces acting between these surfaces prevent this. Both of these phenomena cause rolling friction forces. The harder the surfaces, the less indentation and the less rolling friction.

rolling friction called the resistance that occurs when one body rolls on the surface of another.

While , the skating rink is at rest; when rolling begins.

The linear quantity k included in the formula is called rolling friction coefficient. The value of k is usually measured in centimeters. The value of the coefficient k depends on the material of the bodies and is determined empirically.

The coefficient of rolling friction during rolling in the first approximation can be considered independent of the angular velocity of the roller and its sliding speed along the plane.

For a wagon wheel along a rail, k=0.5 mm.

Consider the movement of the driven wheel.

Wheel rolling will start when the condition QR>M or Q>M max /R=kN/R is met

Wheel slip will start when the condition Q>F max =fN is met.

Usually attitude and rolling starts before sliding.

If , then the wheel will slide on the surface, without rolling.

The ratio for most materials is much less than the static coefficient of friction. This explains why in technology, whenever possible, they try to replace sliding by rolling (wheels, rollers, ball bearings, etc.).

Lecture 3

Summary: Bringing an arbitrary and flat system of forces to the center. The theorem on parallel transfer of force, the main theorem of statics Bringing the system of forces to a given center The main vector and the main moment of the system of forces. Dependence of the main moment on the choice of the center. Analytical definition of the main vector and the main moment of the system of forces. Force system invariants. Bringing the system of forces to the simplest form. Special cases of reduction of an arbitrary system of forces, dynamic screw. Varignon's theorem on the moment of the resultant.

Bringing a force to a given center (Poinsot's Lemma)

The resultant system of converging forces is directly found using the addition of forces according to the parallelogram rule. Obviously, a similar problem can be solved for an arbitrary system of forces, if we find a method for them that allows us to transfer all the forces to one point.

Poinsot Lemma on Parallel Transfer of Force. .Without changing the action of a force on a rigid body, it can be transferred parallel to itself to any point of the body, while adding a pair, the moment of which is equal to the moment of this force relative to the new point of application.

Let a force be applied at point A. The action of this force does not change if two balanced forces are applied at point B. The resulting system of three forces is a force equal to but applied at point B and a pair with a moment. The process of replacing a force with a force and a pair of forces is called bringing the force to a given center B. ■

Bringing the system of forces to a given center.

The main vector of the force system is called a vector equal to the vector sum of these forces.

The main moment of the system of forces relative to the point O of the body, is called a vector equal to the vector sum of the moments of all the forces of the system relative to this point.

Poinsot theorem (Basic theorem of statics)

An arbitrary system of forces acting on a rigid body can be replaced by an equivalent system consisting of a force and a pair of forces. The force is equal to the main vector of the system of forces and is applied at an arbitrarily chosen point (center of reduction), the moment of the pair is equal to the main

moment of the system of forces about this point.

PROOF.

Dot O- referral center. By Poinsot's lemma, we transfer the force F1 exactly O. In this case, instead of F1, we have at the point O the same force F1 'and additionally a pair of forces with a moment m1.

Similarly, we transfer all other forces. As a result, we obtain a system of converging forces and a system of pairs of forces. By the theorem on the existence of a resultant system of convergent

forces can be replaced by one force R, equal to the main vector. According to the pair addition theorem, the system of pairs can be replaced by one pair, the moment of which is equal to the main moment Mo. ■

Static invariants

Invariants of statics are characteristics of the system of forces that do not depend on the choice of the center of reduction.

First invariant statics - the main vector of the system of forces (by definition).

Second invariant statics is the scalar product of the main vector and the main moment.

Indeed, the main point obviously depends on the choice of the center of reduction. Consider an arbitrary system of forces . We bring it first to the center O, and then to the center O 1 .

It can be seen from the figure that .Therefore, the formula for takes the form

Or .

We multiply both parts of this equality by respectively, given that the main vector of the system of forces is first invariant statics: . By

property mixed product vectors , hence:

If we use the definition of the scalar product, then for the second invariant we can get one more form:

Since , then the previous expression will take the form:

Thus, the projection of the main moment on the direction of the main vector is a constant value for a given system of forces and does not depend on the choice of the center of reduction.

Particular cases of reducing an arbitrary system of forces to the simplest form

1) If when bringing the system of forces to the center O then on the basis of (6.4) we can write

resultant, applied in the center of reduction and coinciding in magnitude and direction with the main vector.

2) If, when bringing the system of forces to the center O

then presenting in the form of a pair of forces with a shoulder ,

we get: .

In this case, the system of forces is reduced to resultant, coinciding in magnitude and direction with the main vector, and the line of action of the resultant is separated from the line of action of the main vector at a distance.

3) If, when bringing the system of forces to the center O then you can write

, that is, the system of forces is reduced to couple of forces with a moment equal to the main moment of the system of forces.

4) If, when bringing the system of forces to the center O then you can write

Those. force system is in equilibrium.

Definition: A system consisting of a force and a pair of forces whose moment is collinear to the force (the plane of the pair is perpendicular to the line of action of the force) is called dynamo or dynamic screw.

If, when the system of forces is reduced to the center O, the second invariant is not equal to zero, then this system of forces is reduced to dynamo.

Having decomposed into two components - along the main vector and - perpendicular to the main vector, for and we will have case 2), and the vector, as a free one, can be transferred parallel to itself to the point O 1:

The vectors represent a dynamo, where , .

In the considered case of reduction of the system of forces, the main moment has a minimum value. This value of the moment is preserved when the given system of forces is brought to any point lying on the line of action of the main vector and the main moment . The equation of this line (the central helical axis of the system of forces) is determined from the condition of the collinarity of the vectors and : .