Bringing an arbitrary system of forces to the simplest form. Particular cases of reduction of systems of forces to the simplest form. Cases of reduction to the simplest form
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TwitterConsider some special cases of the previous theorem.
1. If for a given system of forces R = 0, M 0 = 0, then it is in equilibrium.
2. If for a given system of forces R = 0, M 0 0, then it is reduced to one pair with the moment M 0 = m 0 (F i). In this case, the value of M 0 does not depend on the choice of the center O.
3. If for a given system of forces R 0, then it is reduced to one resultant, and if R 0 and M 0 = 0, then the system is replaced by one force, i.e. resultant R passing through the center O; if R 0 and M 0 0, then the system is replaced by one force passing through some point C, and OS = d(OCR) and d = |M 0 |/R.
Thus, a flat system of forces, if it is not in equilibrium, is reduced either to one resultant (when R 0) or to one pair (when R = 0).
Example 2 Forces applied to the disk:
(Fig. 3.16) bring this system of forces to the simplest form.
Solution: choose the Oxy coordinate system. For the center of reduction we choose the point O. The main vector R:
R x \u003d F ix \u003d -F 1 cos30 0 - F 2 cos30 0 + F 4 cos45 0 \u003d 0; Rice. 3.16
R y \u003d F iy \u003d -F 1 cos60 0 + F 2 cos60 0 - F 3 + F 4 cos45 0 \u003d 0. Therefore, R \u003d 0.
The main moment of the system M 0:
M 0: \u003d m 0 (F i) \u003d F 3 *a - F 4 *a * sin45 0 \u003d 0, where a is the radius of the disk.
Answer: R = 0; M 0 = 0; the body is in balance.
Bring to the simplest form the system of forces F 1, F 2, F 3 shown in the figure (Fig. 3.17). The forces F 1 and F 2 are directed on opposite sides, and the force F 3 is directed along the diagonal of the rectangle ABCD, the side AD of which is equal to a. |F 1 | = |F 2 | = |F 3 |/2 = F.
Solution: direct the coordinate axes as shown in the figure. We define the projections of all forces on the coordinate axes:
The module of the main vector R is: 
;
.
The direction cosines will be: 
;
.
Hence: (x, R) = 150 0 ; (y, R) = 60 0 .
ABOUT 
let us limit the main moment of the system of forces relative to the center of reduction A. Then
m A \u003d m A (F 1) + m A (F 2) + m A (F 3).
Considering that m A (F 1) \u003d m A (F 3) \u003d 0, since the direction of forces passes through point A, then
m A \u003d m A (F 2) \u003d F * a.
Thus, the system of forces is reduced to a force R and a pair of forces with a moment m A directed counterclockwise (Fig. 3.18).
Answer: R = 2F; (x, ^ R) \u003d 150 0; (y,^ R) = 60 0 ; m A = F*a.
Questions for self-control
What is the moment of force about the center?
What is a couple of forces?
Bringing an arbitrary flat system of forces to a given center?
Addition of parallel forces?
Literature:,,.
Lecture 4
Basic form of equilibrium conditions. 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 two coordinate axes and the sum of their moments relative to any center lying in the plane of action of the forces are equal to zero:
Fix = 0; F iy = 0; m 0 (F i) = 0.
The second form of equilibrium conditions: For the equilibrium of an arbitrary flat system of forces, it is necessary and sufficient that the sum of the moments of all these forces relative to any two centers A and B and the sum of their projections on the Ox axis not perpendicular to the straight line AB are equal to zero:
m A (F i) = 0; m B (F i) = 0; Fix = 0.
The third form of equilibrium conditions (the equation of three moments): For the equilibrium of an arbitrary flat system of forces, it is necessary and sufficient that the sum of all these forces with respect to any three centers A, B, C, not lying on one straight line, be equal to zero:
m A (F i) = 0; m B (F i) = 0; m С (F i) = 0.
P 
Example 1. Determine the reaction of the embedment of a cantilever beam under the action of a uniformly distributed load, one concentrated force and two pairs of forces (Fig. 4.1); load intensity q \u003d 3 * 10 4 H / m; F \u003d 4 * 10 4 H; m 1 \u003d 2 * 10 4 H * m; m 2 \u003d 3 * 10 4 H * m. BN = 3m; NC = 3m; CA = 4m.
R 
Solution:
According to the principle of release from bonds, we will replace bonds with the corresponding reactions. With a rigid seal in the wall, a reaction force R A of an unknown direction and an unknown moment m A arises (Fig. 4.2). We replace the distributed load with an equivalent concentrated force Q applied at point K (VK = 1.5 m). We choose the coordinate system VHU and compose the equilibrium conditions for the beam in the main form:
force projections on the X axis: - Fcos45 0 – R Ax = 0 (1)
force projections on the Y axis: -Q - Fsin45 0 + R Ax = 0 (2)
sum of moments: m A (F) \u003d m 1 - m 2 + m A + Q * KA + F ”* CA \u003d 0 (3)
We decompose the force F at point C into two mutually perpendicular components F ”and F ’; force F’ does not create a moment relative to point A, since the line of action of the force passes through point A. The modulus of force F” = Fcos45 0 = F(2) 1/2 /2.
Substituting numerical values into equations (1), (2) and (3), we obtain:
There are three unknowns in this system of three equations, so the system has a solution and, moreover, only one.
4*10 4 *0.7 = R Ax R Ax = 2.8*10 4 H
3*10 4 *3 – 4*10 4 *0.7 + R Ay = 0 R Ay = 11.8*10 4 H
m A – 10 4 + 3*10 4 *3*8.5 + 4*10 4 *2.8 = 0 m A = - 86.8*10 4 H*m
Answer: R Ax \u003d 2.8 * 10 4 H; R Ay \u003d 11.8 * 10 4 H; m A \u003d - 86.8 * 10 4 H * m.
Example 2. Determine the reactions of the supports A, B, C and the hinge D of the composite beam (Fig. 4.3).
q 
\u003d 1.75 * 10 4 H / m; F \u003d 6 * 10 4 H; P \u003d 5 * 10 4 H.
Solution: According to the principle of release from bonds, we will replace the bonds with the corresponding reactions.
We replace the distributed load q with the equivalent concentrated force Q = q * KA applied at the point M (AM = 2m). Number of unknown reaction forces: R Ax , R Ay , R B , R C and two pairs of component reaction forces in hinge D.
R 
Let us consider separately the reactions in the hinge D. To do this, consider separately the beams AD and DE (Fig. 4.5a, 4.5b).
According to Newton's third law in the hinge D, the system of forces R Dx and R Dy acts on the beam KD, and the opposite system of forces acts on the beam DE: R' Dx and R' Dy , and the modules of forces are pairwise equal, i.e. R Dx = R Dx and R Dy = R Dy . These are the internal forces of the composite beam, so the number of unknown reaction forces is six. To determine them, it is necessary to compose six independent equations of equilibrium states. The following options for compiling the equations of state are possible.
We compose the equilibrium conditions for the entire structure (3 equations) and for a separate element of this structure: beams KD or beams DE. When compiling the equilibrium equations for the entire structure, internal forces are not taken into account, since when summed they cancel each other out.
Equilibrium condition equations for the entire structure:
R Ax – Fcos60 0 = 0
Q - R Ay - Fsin60 0 + R B + R C - P = 0
m A (F) = Q*m A - Fsin60 0 *AN + R B *AB + R C *AC - P*AE = 0
Equilibrium condition equations for element DE:
R’ Dy , + R C – P*DE = 0
M D (F) = R C *DC - P*DE = 0
Thus, six independent equations with six unknowns are composed, so the system of equations has a solution, and moreover, only one. Solving the system of equations, we determine the unknown reaction forces.
Cases of reduction to the simplest form
Bringing to a couple
Let, as a result of bringing the forces to the center O, it turned out that the main vector is equal to zero, and the main moment is different from zero: . Then, by virtue of the fundamental theorem of statics, we can write
This means that the original system of forces in this case is equivalent to a pair of forces with moment .
The moment of a couple does not depend on which point is chosen as the center of moments when calculating the moment of a couple. Therefore, in this case, the main moment should not depend on the choice of the center of reduction. But it is to this conclusion that the relation
linking the main points with respect to two different centers. For , the additional term is also equal to zero, and we get
Reduction to the resultant
Let now the main vector is not equal to zero, and the main moment is equal to zero: . By virtue of the fundamental theorem of statics, we have
that is, the system of forces turns out to be equivalent to one force - the main vector. Therefore, in this case, the original system of forces is reduced to the resultant, and this resultant coincides with the main vector applied at the reduction center: .
The system of forces is reduced to the resultant even in the case when the main vector and the main moment are both non-zero, but mutually perpendicular: . The proof is carried out using the following sequence of actions.
Through the reduction center O we draw a plane perpendicular to the main moment (Fig. 50, a). In the figure, this plane is aligned with the plane of the drawing, and the main vector is located in it. In this plane, we build a pair with a moment, and we choose the forces of the pair equal in absolute value to the main vector; then the leverage of the pair will be equal to . Next, we move the pair in its plane in such a way that one of the forces of the pair is applied at the reduction center O opposite to the main one; the second force of the pair will be applied at point C, which is away from the center O in the desired direction, determined by the direction, at a distance OS equal to the shoulder of the pair h (Fig. 50, b). Discarding now the balanced forces R and - applied at point O, we arrive at one force applied at point C (Fig. 50, c). It will serve as the resultant of this system of forces.
It can be seen that the resultant is still equal to the main vector , but differs from the main vector in its point of application. If the main vector is applied at the reduction center O, then the resultant is at point C, the position of which requires a special definition. The geometric way of finding the point C is visible from the above construction.
For the moment of the resultant relative to the center of reduction O, one can write (see Fig. 50):
or, omitting intermediate values:
If we project this vector equality onto any axis passing through the point O, we obtain the corresponding equality in projections:
Recalling that the projection of the moment of force about a point on the axis passing through this point is the moment of force about the axis, we rewrite this equality as follows:
The resulting equalities express Varignon's theorem in its general form (in Lecture 2 the theorem was formulated only for converging forces): if the system of forces has a resultant, then the moment of this resultant (with respect to a point, with respect to an axis) is equal to the sum of the moments of all given forces - components (with respect to that same point, same axis). It is clear that in the case of a point the summation of the moments is vector, in the case of an axis it is algebraic.
Bringing to the dynamo
A dynamo or a dynamic screw is a combination of a pair of forces and a force directed perpendicular to the plane of action of the pair. It can be shown that in the general case of reduction, when and is not perpendicular to , the original system of forces is equivalent to some dynamo.
Case I.
If the main vector of the system of forces is equal to zero and its main moment relative to the center of reduction is equal to zero, then the forces are mutually balanced.
Case II.
If the main vector of the system of forces is equal to zero, and its main moment relative to the center of reduction is not equal to zero, then the forces are reduced to a pair of forces. The moment of this pair of forces is equal to the main moment of the system of forces relative to the center of reduction.
In this case, the main moments of the system of forces with respect to all points in space are geometrically equal.
Case III.
If the main vector of the system of forces is not equal to zero, and its main moment relative to the center of reduction is equal to zero, then the forces are reduced to the resultant, the line of action of which passes through the center of the ghost.
Case IV. And .
If the main moment of the system of forces relative to the center of reduction is perpendicular to the main vector, then the forces are reduced to the resultant, the line of action of which does not pass through the center of reduction (Fig. 145).
Case V. and 
.
If the main moment of the system of forces relative to the center of reduction is not perpendicular to the main vector, then the forces are reduced to two crossing forces or to a power screw (dynamo), i.e. to a combination of a force and a pair of forces whose plane is perpendicular to the force.
Reduction to two crossing forces (Fig. 147):
Equilibrium equations for various systems of forces
For forces arbitrarily located in space, two equilibrium conditions correspond:
The modules of the main moment and the main vector for the considered system of forces are determined by the formulas:
The conditions are satisfied only with the corresponding six basic equations of the balance of forces, located arbitrarily in space:
The first three equations are called the equations of the moments of forces relative to the coordinate axes, and the last three are the equations of the projections of forces on the axis.
Forms of equilibrium equations for a plane system of forces
For forces arbitrarily located on a plane, there are two equilibrium conditions:
Two conditions for the equilibrium of forces arbitrarily located on a plane can be expressed as a system of three equations:
These equations are called the basic equations for the equilibrium of a plane system of forces. The center of moments and the direction of the coordinate axes for this system of equations can be chosen arbitrarily.
There are two other systems of three equations of the system of forces.
At the same time, the axis in the system u must not be perpendicular to the line passing through points A and B.
Since the main moments of the system of forces with respect to two centers are equal to zero, the considered system of forces is not reduced to a pair of forces. The projection of the resultant on any axis is equal to the sum of the projections of the component forces, i.e. therefore, the supposed resultant. Thus, the system of forces is not reduced to either a pair of forces or to a resultant, and, therefore, is balanced.
where points A, B, C do not lie on one straight line. In this case, the forces are not reduced to a pair of forces, since the main moments of the forces about the three centers are equal to zero. Forces are not reduced to a resultant either, since if it exists, then the line of action of it cannot pass through three points that do not lie on one straight line. Thus, the system of forces is not reduced either to a pair of forces or to a resultant, and, therefore, is balanced.
Center of Parallel Forces
When two parallel forces are added, two parallel forces are reduced to one force - the resultant, the line of action of which is directed parallel to the lines of action of the forces. The resultant is applied at a point dividing the straight line, at distances inversely proportional to the magnitude of the forces.
Since the force can be transferred along the line of its action, the point of application of the resultant is not defined. If the forces are rotated through the same angle and the forces are added again, then we get a different direction of the line of action of the resultant. The point of intersection of these two resultant lines can be considered as the point of application of the resultant, which does not change its position when all forces are rotated simultaneously through the same angle. Such a point is called the center of parallel forces.
As it was proved above, an arbitrary system of forces, arbitrarily located in space, can be reduced to one force equal to the main vector of the system and applied in an arbitrary center of reduction ABOUT, and one pair with a moment equal to the principal moment of the system about the same center. By
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Let us consider some cases of reduction of the system of forces. 
1. Flat system of forces. Let, for definiteness, all forces be in a plane OXY. Then in the most general case
Principal vector is non-zero, principal moment is non-zero, their dot product is zero, indeed
consequently, the main vector is perpendicular to the main moment: a flat system of forces is reduced to a resultant.
2. System of parallel forces. Let, for definiteness, all forces be parallel to the axis oz. Then in the most general case
Here also the principal vector is non-zero, the principal moment is non-zero, and their scalar product is zero, indeed
therefore, the main vector is perpendicular to the main moment: the system of parallel forces is reduced to the resultant. In a particular case, if it is equal to zero, then the main vector of forces is equal to zero, and the system of forces is reduced to a pair of forces, the moment vector of which is in the plane OXY. We now systematize the considered cases. Recall that an arbitrary spatial system of forces applied to a rigid body is statically equivalent to a force equal to the main vector applied at an arbitrary point of the body (reduction center) and a pair of forces with a moment equal to the main moment of the system of forces relative to the specified reduction center.
1) Let =0,≠0. This is the case when the system of forces is reduced to one force, which we will call the resultant system of forces. An example of such a system of forces can be considered a convergent system of forces, for which the lines of action of all forces intersect at one point.
2) ≠0,=0 . A system of forces is equivalent to a pair of forces.
3) ≠0,≠0, but. The main vector is not equal to zero, the main moment is not equal to zero, their scalar product is equal to zero, i.e. principal vector and principal moment are orthogonal. Any system of vectors in which the main vector and the main moment are not equal to zero and they are perpendicular is equivalent to the resultant, the line of action of which passes through the point ABOUT*(Figure 8). An example of such a system of forces can be considered a flat system of forces or a system of parallel forces.
4) ≠0,≠0, and the principal vector and principal moment are non-orthogonal. In this case, the system of forces is reduced to a dynamo or to two non-intersecting forces.
As shown in § 12, any one is generally reduced to a force equal to the main vector R and applied at an arbitrary center O, and to a pair with a moment equal to the main moment (see Fig. 40, b). Let us find the simplest form to which a spatial system of forces that is not in equilibrium can be reduced. The result depends on the values that R and R have for this system.
1. If for a given system of forces , and then it is reduced to a pair of forces, the moment of which is equal to and can be calculated using formulas (50). In this case, as was shown in § 12, the value does not depend on the choice of the center O.
2. If for a given system of forces, then it is reduced to a resultant equal to R, the line of action of which passes through the center O. The value of R can be found by formulas (49).
3. If for a given system of forces but then this system is also reduced to a resultant equal to R, but not passing through the center O.
Indeed, at , the pair represented by the vector and the force R lie in the same plane (Fig. 91).
Then, choosing the forces of the pair equal in modulo R and arranging them as shown in Fig. 91, we find that the forces are mutually balanced, and the system is replaced by one resultant line of action of which passes through the point O (see, § 15, p. 2, b). The distance ) is determined in this case by formula (28), where
It is easy to verify that the considered case will, in particular, always take place for any system of parallel forces or forces lying in the same plane, if the main vector of this system If for a given system of forces and at the same time the vector is parallel to R (Fig. 92, a) , then this means that the system of forces is reduced to the totality of the force R and the pair P, P, lying in a plane perpendicular to the force (Fig. 92, b). Such a combination of force and a pair is called a dynamic screw, and the straight line along which the vector R is directed is called the axis of the screw. Further simplification of this system of forces is impossible. Indeed, if we take any other point C as the center of reduction (Fig. 92, a), then the vector can be transferred to the point C as free, and when the force R is transferred to the point C (see § 11), another pair with moment perpendicular to the vector R, and hence, and . As a result, the moment of the resulting pair will be numerically larger, thus, the moment of the resulting pair in this case, when reduced to the center O, has the smallest value. This system of forces cannot be reduced to one force (resultant) or to one pair.
If one of the forces of a pair, for example P, is added to the force R, then the system of forces under consideration can still be replaced by two intersecting, i.e., forces Q and not lying in the same plane (Fig. 93). Since the resulting system of forces is equivalent to a dynamic screw, it also does not have a resultant.
5. If for a given system of forces and at the same time, the vectors and R are not perpendicular to each other and not parallel, then such a system of forces is also reduced to a dynamic screw, but the axis of the screw will not pass through the center O.
To prove this, we decompose the vector into components: directed along R, and perpendicular to R (Fig. 94). In this case, where are the vectors and R. The pair represented by the vector and the force R can, as in the case shown in Fig. 91, be replaced by one force R applied at point O, Then this system of forces will be replaced by a force and a pair of moments parallel, and the vector as a free one can also be applied at point O. The result will indeed be a dynamic screw, but with an axis passing through the point
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