Theorem on the reduction of a system of forces:
Any system of forces acting on an absolutely rigid body can be replaced by a single 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 relative to the center O.
Such an equivalent replacement of a given system of forces by force R and a couple of forces with a moment L O is called bringing the system of forces to the center O.
Let us consider here a special case of 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 (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:
Question 16. 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= 0, L O = 0, where O is any center located in the plane of action of the system forces.
The resulting analytical equilibrium conditions (equilibrium equations) for a flat system of forces can be formulated in the following three forms:
- Basic form of 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 relative 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)
- Second form of equilibrium equations:
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 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 V ( F i) = 0. (II)
- The third form of equilibrium equations (three moment equations):
for the equilibrium of an arbitrary plane system of forces, it is necessary and sufficient that the sums of the algebraic moments of all forces relative to any three centers A, B and C that do not lie on the same straight line are equal to zero:
M A ( F i) = 0; M V ( F i) = 0; M C ( F i) = 0. (III)
Equilibrium equations in form (I) are considered basic, since when using them there are no restrictions on the choice of coordinate axes and 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 same point. Let us assume that the system of forces is reduced to a resultant R passing through 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 of the 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 main 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 Varignon's theorem, one can find the equation of 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 let the main vector F o and the main moment M O be known at the center of reduction at the origin. Since R 1 =F o, the components of the resultant along the x and y axes are equal to R lx =F Ox =F Ox i and R ly =F Oy =F oy j. According to Varignon’s theorem, the moment of the resultant relative to the origin is equal to the principal moment at the center of reduction at the origin, i.e. Моz =M Oz (R 1)=xF Oy –yF Ox. (5.14). The quantities M Oz, F Ox and Foy do not change when the point of application of the resultant is moved along its line of action; therefore, the x and y coordinates of 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. When F ox ≠0 it can be rewritten as y=(F oy /F ox)x–(M oz /F ox).
Question
Sealing one body into another (for example, a rod into a stationary wall) does not allow this body to move and rotate relative to the other. In case of embedding, 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 previously unknown moment M A . If force R A imagine it by its components X A, Y A, then to find the embedding reaction it is necessary to determine three unknown scalar quantities: X A, Y A, M A.
Let us give examples of replacing plane systems of parallel distributed forces with their resultants.
For such a system of forces, the intensity has a constant value: q = const.
When solving statics problems, this system of forces can be replaced by a concentrated resultant force Q, equal in magnitude to the product of 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 value, changing from zero to the maximum value q max according to a linear law.
Resultant Q of this system of forces is equal in magnitude to Q =0.5 · a · q max and is applied at point K dividing the segment AB in the ratio AK: KB = 2: 1.
19.Calculation of composite structures
1.1. Calculation with division of a system of bodies into separate bodies
1.1.1. The system of bodies according to the internal connection C is divided into separate bodies and their equilibrium is considered.
1.1.2. All connections are discarded from each of the bodies, replacing their actions with reactions. The following types of connections are applied in the given mechanisms: 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 embedding (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. Equilibrium equations are compiled, including equations of projections onto standard axes and equations of moments (calculated and verified). The center of the calculated moment equation is chosen at the intersection of the lines of action of the maximum number of unknown reactions, the verification equation is chosen at the intersection of the lines of action of known forces, through which none of the untested unknown reactions passes. It is recommended to draw up equilibrium equations by considering the forces one by one as follows: determine the acute angle α between the line of force and the line of one of the axes; the projection of force 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 arm of the force by 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 arm by the force around the center (when the arm is rotated clockwise, the moment is negative, counter-clockwise, it 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 Varignon’s theorem.
Thus, for each of the bodies, 3 calculation and 1 verification equations are made.
1.1.4. Solve a system of 6 calculated equations for unknown reactions.
Substitute the found reactions into the verification equations; the modulus of the resulting sum should not exceed 0.02 Ravg, where Ravg is the average value of the moduli of the reactions being tested.
1.2. Calculation using the solidification principle
1.2.1. Replace the internal hinge with a rigid connection and consider the equilibrium of the resulting body. The second one is considered one of the bodies of the system (section 1.1.1).
1.2.2. Draw up a drawing for each of the bodies under consideration in the same way as in paragraph 1.1.2.
1.2.3. For the first body, 3 calculation and 1 verification equations are made similar to paragraph 1.1.3. For the second body, one calculated equation of the moments of forces relative to the center C is drawn up.
1.2.4. Solve a system of 4 design equations and do a check similar to paragraph 1.1.4. 2.
2. Calculation using the principle of possible displacements. The reactions of the connections are determined by considering them in turn.
20. Balance condition of the lever. Stability of bodies when tipping over. The distance from the fulcrum to the straight line along which the force acts is called the arm of this force. Let us denote by F1 and F2 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 arms of these forces: F1/F2 = l2/l1. Stability of bodies when overturning. These are tasks that arise when designing various lifting mechanisms and when calculating the safe conditions for their operation, specified in the rules for working with these mechanisms. The peculiarity of solving these not very complex problems on a plane system of forces is that when solving them, equilibrium equations are not drawn up. The following are determined separately: a) overturning moment (Mopr) - the sum of the moments of forces that tend to overturn the mechanism in question relative to some axis (support point) projected in the drawing; c) holding moment (Md) - the sum of the moments of forces that prevent overturning. For stable operation of the mechanism, it is necessary that the holding moment be greater than the overturning moment by some margin. The ratio Mud,/Mopr =k is usually called the stability coefficient. The value k must, naturally, be greater than unity. For various lifting mechanisms and for different operating conditions, the value of the stability coefficient is determined from SNiP, TU and other sources. Taking into account this coefficient, calculations are made of the size of the counterweight load or its position on the mechanism, and options are calculated - at what boom radius and what loads can be safely worked with. An example of solving one of the stability problems is given below. It is especially important to be able to perform basic stability calculations in production conditions, when you have to work with the maximum loads available to the crane.
21 Sliding friction. Laws of friction. Friction coefficient. Between moving bodies in the plane of their contact, a sliding friction force arises. This is primarily due to the roughness of the contacting surfaces and the presence of adhesion between the pressed bodies. In engineering calculations, they usually use experimentally established laws that reflect the effect of friction force with some degree of accuracy. These laws 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 F arises, 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 friction coefficient f is a dimensionless quantity that depends on the materials and state of the surfaces of contacting bodies (roughness, temperature, humidity, etc.). It is determined experimentally.
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 the maximum friction force Fmax at which there is a limiting state of equilibrium. The slightest increase in external forces can cause movement. The coefficient of static friction is, as a rule, slightly greater than the coefficient of sliding friction. With increasing sliding speed, the value of the sliding friction coefficient first decreases slightly and then remains practically unchanged. The friction coefficient values 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, within a fairly wide range, does not depend on the area of contacting surfaces.
The sliding friction force is sometimes called the dry friction force.
Angle and cone of friction
The reaction of a real (rough) connection will consist of two components: the normal reaction and the friction force perpendicular to it. Consequently, the total reaction will be deviated from the normal to the surface by some angle. When the friction force changes from zero to F, the force R will change from N to R, and its angle with the normal will increase from zero to a certain limiting value (Fig. 26).
Fig.26
The largest angle that the total reaction of a rough bond makes with the normal to the surface is called friction angle. From the drawing it is clear that
Since , from here we find the following relationship between the friction angle and the friction coefficient:
At equilibrium, the total reaction R, depending on the shear forces, can take place anywhere within the friction angle. When equilibrium becomes limiting, the reaction will be deviated from the normal by an angle .
Friction cone called a cone described by the maximum 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 when , i.e. at . Consequently, no force forming an angle with the normal that is smaller than the friction angle 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 active forces acting on the solid body pass inside the friction cone or along its generatrix through its apex.
A body cannot be thrown out of balance by any modulus active force if its line of action passes inside the friction cone.
Rolling friction
The origin of rolling friction can be visualized as follows. When a ball or cylinder rolls along the surface of another body, it is slightly pressed into the surface of this body, and itself is slightly compressed. Thus, a rolling body always seems to be rolling up a hill.
Fig.33
At the same time, sections of one surface are separated 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 is the resistance that occurs when one body rolls over the surface of another.
For now, the skating rink is at rest; when rolling begins.
The linear quantity k included in the formula is called rolling friction coefficient. The k value is usually measured in centimeters. The value of the coefficient k depends on the material of the bodies and is determined experimentally.
As a first approximation, the rolling friction coefficient during rolling can be considered independent of the angular velocity of the roller and its sliding speed along the plane.
For a carriage wheel on a rail k=0.5 mm.
Let's consider the movement of the driven wheel.
The wheel will start rolling when the condition QR>M or Q>M max /R=kN/R is met
The wheel will start sliding when the condition Q>F max =fN is met.
Typically, attitude and rolling begin before sliding.
If , then the wheel will slide on the surface without rolling.
The ratio for most materials is significantly less than the static coefficient of friction. This explains that in technology, whenever possible, they strive to replace sliding with rolling (wheels, rollers, ball bearings, etc.).
Let us assume that an arbitrary plane system of forces is reduced to one force equal to the main vector and applied to the center of reduction, and to one pair with a moment equal to the main moment 
(Figure 57, A). Let us prove that the arbitrary plane system of forces under consideration is reduced in this general case to the resultant force 
, the line of action of which passes through the point A, distanced from the selected reduction center ABOUT on distance 
. To do this, we transform the couple with moment 
so that strength 
And 
, making up this pair, turned out to be equal in magnitude to the main vector R". In this case, it is necessary to select the arm of the pair so that its moment t 
remained equal to M 0. For this, the leverage of the pair 
one must obviously find from the equality
.
(1)
Taking advantage of the fact that a pair can always be moved in its plane of action as desired, let us move the pair 
so that her strength 
found itself attached to the adduction center ABOUT and opposite to the main vector 
(Figure 57, b).
The arbitrary plane system of forces under consideration is therefore equivalent to the force 
and couple 
. Throwing back the forces 
And 
as balanced, we obtain that the entire system of forces under consideration is replaced by one force 
, which is therefore the resultant. In this case, the line of action of the resultant 
will pass through the point A, the position of which relative to the selected reduction center is determined by formula (1).
If, as a result of bringing an arbitrary plane system of forces, it turns out that 
, A 
, then in this particular case this system of forces is immediately replaced by one force, i.e., the resultant 
, the line of action of which passes through the selected center of reduction.
Problem 7. To the dots IN And WITH bodies are respectively applied equal in magnitude and mutually perpendicular forces 
And 
, distant from the point ABOUT bodies at equal distances 
. Bring this system of forces to a point ABOUT(Figure 58).
Solution. Let's transfer the forces 
And 
parallel to ourselves to a point ABOUT. As a result of this transfer we obtain (Figure 58) forces 
And 
, applied at the point ABOUT, and adjoint pairs 
And 
, lying in the same plane with the moments 
And 
(the forces that form these pairs are marked in the figure with 58 dashes). From geometric addition of forces 
And 
, applied at the point ABOUT, we obtain the main vector of this system of forces
whose modulus is obviously equal to
From the addition of the adjoint pairs we obtain a resultant pair, the moment of which is equal to the main moment 
given system of forces with respect to the point ABOUT:
Therefore, this system of two forces 
And 
has a resultant
,
applied at the point A, which is distant from the point ABOUT on distance
.
;
,
i.e., the resultant forms with both given forces 
And 
equal angles of 45 0.
Task 8. Vertical forces act on the bridge truss (Figure 59) 
t and 
t respectively at a distance of 10 m and 40 m from the left end of the truss and horizontal force 
t at the level of the upper chord of the truss, the height of the truss is 6 m. Bring a system of forces 
,
And 
the simplest form.
Solution. We draw the coordinate axes as shown in Figure 59, taking the origin of coordinates at the point A. Let us find the projections of the main vector of a given system of forces on the axis of the selected coordinate system:
where we find the module of the main vector 
:
T 
.
Let us now find the main moment of the given system of forces relative to the origin of coordinates A:
t m 
.
Consequently, this system of forces has a resultant 
, whose module 
T.
Now let's find the line of action of the resultant. Resultant moment 
relative to the origin A will be determined by the formula
,
Where X And y
- coordinates of a point lying on the line of action of the resultant. Because 
t and 
t, then
.
WITH on the other hand, by Varignon’s theorem on the moment of the resultant (5, § 11) we have
Hence,
.
This is the equation of the line of action of the resultant.
Assuming in this equation 
, we find that the point of intersection of the line of action of the resultant 
with the top chord of the truss located at a distance 
m from the left end of the farm. Assuming 
m, we find that the point of intersection of the line of action of the resultant 
with the lower chord of the truss is at a distance 
m from the left end of the farm. Connections defined in this way are the points of intersection of the lines of action of the resultant 
with the upper and lower chords of the truss in a straight line, we find the line of action of the resultant 
.
We will replace the flat system of forces applied at points A, B, C, D:
1) forces F 1 ’, F 2 ’, F 3 ’, F 4 ’, applied at point O;
2) in pairs of forces:
F 1 F 1 ': M 1 = M o (F 1) = F 1 a 1
F 2 F 2 ': M 2 = M o (F 2) = F 2 a 2
F 3 F 3 ': M 3 = M o (F 3) = F 3 a 3
F 4 F 4 ': M 4 = M o (F 4) = F 4 a 4
The forces F 1 ’, F 2 ’, F 3 ’, F 4 ’ converging at point O can be replaced by one force (resultant) F ch:
F gl = F 1 ' + F 2 ' + F 3 ' + F 4 ' = F 1 + F 2 + F 3 + F 4
F Ch– the main vector of the force system.
The resulting force pairs can be replaced by the resulting pair, the moment of which M ch:
M gl = M 1 + M 2 + M 3 + M 4 = Σ M i = Σ M o (F i)
M gl - main point regarding the reference point.
The plane system of forces at a given point O is replaced by an equivalent system consisting of one force (the main vector) and one pair (the main moment).
Theorem on the moment of the resultant (Varignon’s theorem)
The moment of the resultant plane system of forces relative to an arbitrary point is equal to the algebraic sum of the moments of the component forces relative to the same point.
M o (F Σ)= Σ M o (F i)
Equilibrium equations for a plane system of forces
F GL = 0;
M gl = ΣM o (F i) = 0.
The module of the main vector can be determined through projections onto the coordinate axes of all forces of the system.
F GL = (ΣF іх) 2 +(ΣF іу) 2 =0 from this the equilibrium equations follow:
Σ F іх =0
Σ F іу =0
Σ M o (F i)=0
Other forms of equilibrium equations:
Σ M A (F i)=0
Σ М В (F і)=0 (ABCs do not lie on the same
Σ M C (F i)=0 direct)
Σ M A (F i)=0 (x axis is not perpendicular
Σ M B (F і)=0 direct AB)
Σ F іх =0
For a system of parallel forces, choosing one of the projection axes parallel to these forces (y-axis), and the other perpendicular to them (x-axis), we obtain two equilibrium equations:
Σ F іу =0
Σ M o (F i)=0
Σ M A (F i)=0
Σ M V (F i)=0
Algorithm for solving problems
1. Select an object of equilibrium (body or point): we will consider equilibrium relative to...
We show in the figure all the acting forces, including the reactions of the connections.
3. Select a coordinate system - it is advisable to direct the coordinate axes parallel or perpendicular to the required forces.
We compose equilibrium equations for the object of study.
Σ F іх =0
Σ F іу =0
Σ M o (F i)=0
From the resulting equations we determine unknown quantities (we determine reactions).
We check the correctness of the solution of the equations.
Σ M p (F i)=0
Σ M e (F i)=0
5. Supporting devices for beam systems
Articulating support
Hinged-fixed form and rigid seal (pinching)
Subject:
"Center of gravity.
Geometric characteristics of flat sections"
Plan
1. Center of parallel forces and its coordinates.
2. Center of gravity of areas. Statistical moments of areas.
3. Solving problems to determine the coordinates of the center of gravity of a flat composite figure.
4. Polar and axial moments of inertia.
5. Axial moments of inertia about parallel axes.
6. Determination of moments of inertia of composite sections using tables of normal assortment.
1. Center of parallel forces and its coordinates
Let a system of parallel forces F 1, F 2, F 3, ..., Fn be given; the coordinates of points C 1, C2, C3, ..., Cn of application of these forces are known (Fig. 42, b). Let us denote the point of application by the resultant letter C, and its coordinates we denote x c, y c.
FΣ = F 1 + F 2 + F 3+…. + Fn = ΣF i . (1)
FΣ xс = F 1 x 1 + F 2 x 2 + F 3 x 3 +… + Fnxn = Σ F i x i ,
x c = F 1 x 1 + F 2 x 2 + F 3 x 3 +… + Fnxn / FΣ = Σ F i x i / FΣ
FΣ = F 1+ F 2+ F 3+…+ Fn= Σ F іх c =
= F 1 x 1 + F 2 x 2 + F 3 x 3 +… + Fnxn / F 1+ F 2+ F 3+…+ Fn= Σ F i x i / F i (2)
Lecture 3
Summary: Bringing an arbitrary and flat system of forces to the center. Theorem on parallel transfer of force, the main theorem of statics Bringing a 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 center. Analytical determination of the main vector and main moment of a system of forces. Invariants of the force system. Bringing the system of forces to its simplest form. Special cases of bringing 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 of a 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's lemma on parallel force transfer. .Without changing the action of the force on a rigid body, it can be transferred parallel to itself to any point of the body, adding a pair whose moment is equal to the moment of the given force relative to the new point of application.
Let a force be applied at point A. The effect 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 a 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 point of the system of forces relative to point O of the body, a vector is called equal to the vector sum of the moments of all forces of the system relative to this point.
Poinsot's theorem (Fundamental 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 force system and is applied at an arbitrarily selected point (center of reduction), the moment of the pair is equal to the main
the moment of the system of forces relative to this point.
PROOF.
Dot ABOUT- adduction center. By Poinsot's lemma we transfer the force F1 exactly ABOUT. In this case, instead of F1, at point O we have the same force F1’ and an additional pair of forces with a moment m1.
Let us similarly 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, a system of pairs can be replaced by one pair whose moment is equal to the main moment Mo. ■
Statics invariants
Statics invariants are characteristics of a force system that do not depend on the choice of the center of gravity.
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 
. Let's bring it first to the center O, and then to the center O 1.
From the figure it is clear that 
.Therefore, the formula for will take the form
Or 
.
Let us multiply both sides of this equality by respectively, taking into account that the main vector of the force system is first invariant statics: . By
property of mixed product of vectors 
, hence:
.
If we use the definition of the scalar product, then for the second invariant we can get another form:
Since , the previous expression will take the form:
Thus, the projection of the main moment onto 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.
Special 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 based on (6.4) we can write
.
resultant, applied at the reduction center and coinciding in magnitude and direction with the main vector.
2) If, when bringing the system of forces to the center O
then presenting it as 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 located at a distance from the line of action of the main vector.
3) If, when bringing the system of forces to the center O 
then we can write
, that is, the system of forces is reduced to a 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 we can write
Those. the system of forces is in equilibrium.
Definition: A system consisting of a force and a pair of forces, the moment of which 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 bringing a system of forces to the center O, the second invariant is not equal to zero, then this system of forces is reduced to dynamo.
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 free, can be transferred parallel to itself to point O 1:
Vectors represent dynamism, where 
, 
.
In the considered case of bringing a system of forces, the main moment has a minimum value. This value of the moment is preserved when a 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 force system) is determined from the condition of collinearity of the vectors and: .
Theorem . StrengthF , without changing its action on the body, can be transferred from the point of its application A to any center of adduction O, while attaching a couple of forces with a moment to the bodyM , geometrically equal to the momentM ABOUT (F ) of this force relative to the center of reduction.
Let the force be given F, lying in the horizontal plane OXY parallel to the OX axis (Fig. 1.41).
According to the Poinsot method, instead of force F, applied at point A, a force is obtained F 1, equal in magnitude to force F, but applied at point O and attached couple of forces , whose vector moment M= M ABOUT ( F).
According to the theorem on the equivalence of pairs of forces, the associated 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 .
P
Consistently applying the Poinsot method to each of the given system of forces, we will bring it to an arbitrary center O. As a result of this, we obtain a system of forces ( F 1 , …, F n), applied at the center O, and an associated pair of forces with a moment M= Σ M ABOUT ( F i). Adding up strength F 1 , …, F n according to the parallelogram rule, we obtain their resultant R* , equal to the geometric sum of the given forces and applied at the center of reduction.
The geometric sum of all the forces of the system is called the main vector of the force system and, in contrast to the resultant R, denote R * .
Vector M= Σ M ABOUT ( F i) are called 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 magnitude and 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 to any point on the body.
1.16. Analytical equilibrium conditions for a plane arbitrary system of forces
Flat arbitrary force system – 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 various points.
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Consistently applying the Poinsot method for each of the forces F i, let us carry out a parallel transfer of forces from points A i to the beginning O of the reference system OXYZ. According to this method, the force F i will be equivalent to the force F i applied at point O, and the attached pair of forces with a moment M i = M ABOUT ( F i ) . In this case, M i = ± F i h i , where h i is the force arm F i relative to the center of reduction O. At the end of this work we obtain a converging system of forces ( F i ,…, F n) and a convergent system of vector moments M i = M ABOUT ( F i) attached pairs of forces applied at the center of reduction. Adding the force vectors, we get chapters
new vector R* = Σ F i and the main moment of an equivalent pair of forces M = Σ M ABOUT ( F i).
Thus, flat 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 ABOUT (F i ).
When solving problems, statics use projections of force on coordinate axes and algebraic moments of forces relative to a point.
In Fig. Figure 1.44 shows a flat arbitrary system of forces reduced to the main vector of forces, the module of which R*= 
and an equivalent pair of forces with algebraic moment M = Σ M O ( F i).
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Geometric equilibrium condition of any system of forces is expressed by vector equalities: R* = Σ F i = 0; M= Σ M ABOUT ( F i) = 0.
When solving problems, it is necessary to determine reactions R i E external connections imposed on the mechanical system. At the same time, active forces F i E applied to this system are known. Since active forces F i E and bond reactions R i E belong to the category of external forces, then it is advisable to express the geometric condition for the equilibrium of a 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 system of external forces to be in equilibrium, it is necessary and sufficient that the geometric sum of the active forces F i E and reactions R i E external connections 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 ) relative 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 flat arbitrary system of forces, these equations take the following form:
Σ 
+ Σ 
= 0;
Σ 
+ Σ 
= 0;
Σ M A ( F i E) + Σ M A ( R i E) = 0,
where Σ 
,
Σ 
– respectively, the sum of projections of active forces onto the coordinate axes OX, OY; Σ 
,
Σ 
– the sum of projections of reactions of external connections on the coordinate axes OX, OY; Σ M A ( F i E) – sum of algebraic moments of active forces F i E relative to point A; Σ M A ( R i E) – sum of algebraic moments of reactions R i E external connections relative to point A.
The set of these formulas is the first (basic) form of the equilibrium equations for a plane arbitrary system of external forces .
Thus , 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 connections on two coordinate axes and the sum of the algebraic moments of active forces and reactions of external connections 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 is expressed by a set of formulas:
Σ 
+ Σ 
= 0;
Σ M A ( F i E) + Σ M A ( R i E) = 0;
Σ M V ( F i E) + Σ M V ( R i E) = 0.
For the equilibrium of a flat arbitrary system of external forces applied to a body, it is necessary and sufficient that the sum of the projections of the forces on the coordinate axis and the sum of the algebraic moments of forces relative to arbitrary points A and B are equal to zero.
Third form equilibrium equations is expressed by a set of formulas:
Σ M A ( F i E) + Σ M A ( R i E) = 0;
Σ M V ( F i E) + Σ M V ( R i E) = 0;
Σ M С ( F i E) + Σ M С ( R i E) = 0.
For the equilibrium of a flat arbitrary system of external forces applied to a body, it is necessary and sufficient that the sums of the algebraic moments of these forces relative 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 straight line.