Special cases of bringing an arbitrary spatial system of forces to the center. Cases of reduction to the simplest form Forms of equilibrium equations of a plane system of forces

Let several pairs of forces with moments acting in different planes be simultaneously applied to a rigid body. Is it possible to reduce this system of pairs to a simpler form? It turns out that it is possible, and the answer is suggested by the following theorem on the addition of two pairs.

Theorem. Two pairs of forces acting in different planes are equivalent to one pair of forces with a moment equal to the geometric sum of the moments of the given pairs.

Let the pairs be defined by their moments and (Fig. 36, a). Let's construct two planes perpendicular to these vectors (the plane of action of the pairs) and, choosing a certain segment AB on the line of intersection of the planes for the shoulder common to both pairs, we will construct the corresponding pairs: (Fig. 36, b).

In accordance with the definition of the moment of a couple, we can write

At points A and B we have converging forces. Applying the rule of parallelogram of forces (axiom 3), we will have:

The given pairs turn out to be equivalent to two forces, which also form a pair. Thus, the first part of the theorem is proven. The second part of the theorem is proven by direct calculation of the moment of the resulting pair:

If there are a number of pairs, then by adding them in pairs in accordance with this theorem, any number of pairs can be reduced to one pair. As a result, we come to the following conclusion: a set (system) of pairs of forces applied to an absolutely rigid body can be reduced to one pair with a moment equal to the geometric sum of the moments of all given pairs.

Mathematically, this can be written as follows:

In Fig. Figure 37 gives a geometric illustration of the resulting conclusion.

For equilibrium of force pairs, it is required that the moment of the resulting pair be equal to zero, which leads to the equality

This condition can be expressed in geometric and analytical form. Geometric condition for the equilibrium of pairs of forces: for a system of pairs of forces to be in equilibrium, it is necessary and sufficient that the vector polygon constructed from the moments of all pairs be closed.

Analytical condition for the equilibrium of force pairs: for a system of force pairs to be in equilibrium, it is necessary and sufficient that the algebraic sums of the projections of the moment vectors of all pairs onto arbitrarily chosen coordinate axes Oxyz are equal to zero:

If all the pairs lie in the same plane, that is, they form a flat system of pairs, only one analytical equilibrium condition is obtained—the sum of the algebraic moments of the pairs is equal to zero.

Self-test questions

1. What is the force polygon rule? What is the power polygon used for?

2. How to find the resultant of converging forces analytically?

3. What is the geometric condition for the equilibrium of converging forces? How is this same condition formulated analytically?

4. State the three force theorem.

5. Which statics problems are called statically defined and which are called statically indeterminate? Give an example of a statically indeterminate problem.

6. What is called a pair of forces?

7. What is called the moment (vector-moment) of a pair of forces? What are the direction, magnitude and point of application of the moment?

8. What is called the algebraic moment of a pair?

9. Formulate a rule for adding pairs arbitrarily located in space.

10. What are the vector, geometric and analytical conditions for the equilibrium of a system of pairs of forces?

The main theorem of statics about bringing an arbitrary system of forces to a given center: Any plane system of forces is equivalent to one force equal to the main vector of the system applied at some point (center of reduction) and a pair of forces, the moment of which is equal to the main moment of the forces of the system relative to the center of reduction.

The proof of the theorem is carried out in the following sequence: select a certain point (for example, a point ABOUT) as the center of reduction and transfer each force to this point, adding, according to the theorem on parallel force transfer, the corresponding pairs of forces. As a result, a system of converging forces applied at the point is obtained ABOUT, where , and a system of added force pairs whose moments are . Then the system of converging forces is replaced by a resultant equal to the main vector of the system, and the system of pairs of forces is replaced by one pair of forces with a moment equal to the main moment of the system relative to the center of reduction . As a result, we get that ~. Therefore, the theorem is proven.

Cases of reducing the spatial system of forces to the simplest form:

1, a – the system is reduced to one pair of forces with a moment equal to the main moment of the system, and the value of the main moment of the system does not depend on the choice of the center of reduction.

2, a – the system of forces is reduced to a resultant equal to the main vector of the system, the line of action of which passes through the center O of the reduction.

3, and – such a system of forces is reduced to one resultant, equal to the main vector of the system, the line of action of which is shifted from the previous center of reduction by a distance.

4 If the main vector and the main moment are , then the system of forces will be balanced, i.e. ~0.

2.1.5 Equilibrium conditions for a plane system of forces

Necessary and sufficient conditions for equilibrium of any plane system of forces are determined by the equations:

The magnitude of the main vector of a plane system of forces is determined by the dependencies: , and the main moment by the dependency .

The main vector will be equal to zero only when simultaneously . Consequently, the equilibrium conditions are satisfied when the following analytical equations are satisfied:

These equations are the basic ( first ) the form of analytical conditions for the equilibrium of an arbitrary plane system of forces, which are formulated as follows: for the equilibrium of an arbitrary plane 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 algebraic sum of the moments of these forces relative to any point on the plane of action of the forces are equal to zero.

Note that the number of equilibrium equations for an arbitrary plane system of forces in the general case is three. They can be presented in different forms.

There are two more forms of equilibrium equations for an arbitrary plane system of forces, the fulfillment of which expresses the equilibrium conditions ().

Second the form of analytical equilibrium conditions provides: for the equilibrium of an arbitrary plane system of forces, it is necessary and sufficient that the sum of the moments of all forces relative to two points and the sum of the projections of these forces on an axis not perpendicular to the straight line drawn through these points are equal to zero:

(line AB not perpendicular to the axis Oh)

Let's formulate third the form of analytical conditions for the equilibrium of the system of forces under consideration: for the equilibrium of an arbitrary plane system of forces, it is necessary and sufficient that the sum of the moments of the forces of the system relative to any three points not lying on the same straight line is equal to zero:

In the case of a plane system of parallel forces, you can direct the axis OU parallel to the forces of the system. Then the projections of each of the forces of the system onto the axis Oh will be equal to zero. As a result, for a plane system of parallel forces there will remain two forms of equilibrium conditions.

For the equilibrium of a plane system of parallel forces, it is necessary and sufficient that the sum of the projections of all forces onto the axis parallel to them and the sum of the moments of all forces relative to any point are equal to zero:

This first form of analytical equilibrium conditions for a plane system of parallel forces follows from equations ().

We obtain the second form of equilibrium conditions for a plane system of parallel forces from equations ().

For the equilibrium of a plane system of parallel forces, it is necessary and sufficient that the sum of the moments of all forces of the system relative to two points that do not lie on a straight line parallel to the forces is equal to zero:

As shown in § 12, any one is reduced in the general case 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 to what simplest form a spatial system of forces that is not in equilibrium can be reduced. The result depends on the values ​​that this system has for the quantities R and

1. If for a given system of forces , then it is reduced to a pair of forces whose moment is equal and can be calculated using formulas (50). In this case, as was shown in § 12, the value does not depend on the choice of 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 using 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, when the couple, represented by the vector, and the force R lie in the same plane (Fig. 91).

Then, choosing the forces of the pair to be equal in modulus R and arranging them as shown in Fig. 91, we find that the forces will be mutually balanced, and the system will be replaced by one resultant action line of which passes through point O (see, § 15, paragraph 2, b). Distance ) is determined by formula (28), where

It is easy to verify that the case considered will, in particular, always occur 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 the vector is parallel to R (Fig. 92, a) , this means that the system of forces is reduced to a combination of force R and a pair P, P lying in a plane perpendicular to the force (Fig. 92, b). Such a combination of force and couple is called a dynamic screw, and the straight line along which the vector R is directed is the axis of the screw. Further simplification of this system of forces is impossible. In fact, if we take any other point C as the center of reduction (Fig. 92, a), then the vector can be transferred to point C as free, and when the force R is transferred to point C (see § 11), another pair with moment perpendicular to the vector R, and therefore . As a result, the moment of the resulting pair will be numerically greater, so the moment of the resulting pair has the smallest value in this case when brought to the center O. This system of forces cannot be reduced to one force (resultant) or to one pair.

If one of the forces of the pair, for example P, is added to the force R, then the system of forces under consideration can also be replaced by two crossing forces, that is, 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, let us decompose the vector into components: directed along R, and perpendicular to R (Fig. 94). In this case, where are vectors and R. The pair represented by the vector and the force R can be, 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 parallel torques, and the vector as a free one can also be applied at point O. The result will actually be a dynamic screw, but with an axis passing through the point

If, after bringing the spatial system of forces to the selected center O, the main vector and the main moment are equal to zero, i.e.

The system of forces is balanced. Under the influence of such a system of forces, the solid body will be in equilibrium. It is obvious that in the general case, two vector equations (4.1) correspond to six scalar equations, reflecting the equality to zero of the projections of these vectors on the axes of the chosen coordinate system (for example, Cartesian).

If, after bringing the spatial system of forces to the selected center O, the main vector is equal to zero, and the main moment is not equal to zero, i.e.

A resulting pair of forces acts on the body, tending to rotate it. Note that in this case the choice of the reduction center does not affect the result.

If, after bringing the spatial system of forces to the selected center O, the main vector is not equal to zero, and the main moment is equal to zero, i.e.

The body is acted upon by the resultant system of forces passing through the center of reduction and tending to move the body along the line of its action. It is obvious that relations (4.3.) are valid for all points of the line of action of the resultant.

Note that the action of a system of converging forces is reduced to this case if the point of intersection of the lines of action of the forces of the system is taken as the center of reduction (since the moments of forces relative to this point are equal to zero).

If, after bringing the spatial system of forces to the selected center O, the main vector and the main moment are not equal to zero, and their directions make a right angle, i.e.

then such a system of forces can also be reduced to a resultant, but passing through another center of reduction - the point. To perform this operation, we first consider the equivalent force systems shown in Fig. 4.2.b and fig. 4.1. Obviously, if we change the notation (point B is called the center O, point A is called the center), the task facing us requires performing the operation inverse to that performed in the lemma on parallel transfer of force. Taking into account the above, the point must, firstly, be located in a plane perpendicular to the vector of the main moment passing through the center O, and, secondly, lie on a line parallel to the line of action of the main vector of forces and separated from it at a distance h equal to

Of the two lines found, you should choose the one for the points of which the vector of the main moment is equal to zero (the moment of the main vector of forces relative to the new center should be equal in magnitude and opposite in direction to the main moment of the system of forces relative to point O).

In the general case, after bringing the spatial system of forces to the selected center O, the main vector and the main moment, which are not equal to zero, do not form a right angle with each other (Fig. 4.5.a).

If the main moment is decomposed into two components - along the main vector of forces and perpendicular to it, then, in accordance with (4.5), a reduction center can be found for which the perpendicular component of the main moment becomes equal to zero, and the magnitudes and directions of the main vector and the first components of the main moment remain the same (Fig. 4.5.b). The collection of vectors is called power screw or dynamo.

Further simplification is not possible.

Since with such a change in the center of reduction, only the projection of the main moment changes to the direction perpendicular to the main vector of the system of forces, the value of the scalar product of these vectors remains unchanged, i.e.

This expression is called second invariant

statics.

Example 4.1. The vertices of a rectangular parallelepiped with sides and are acted upon by forces and (see Fig. 4.6). Taking the origin of coordinates of the Cartesian coordinate system indicated in the figure as the center of reduction of the force system, write down expressions for the projections of the main vector and the main moment.

Let's write down trigonometric relations to determine angles:

Now we can write expressions for the projections of the main vector and the main moment of forces of the system:

Note: knowledge of the vector projections onto the coordinate axes will allow, if necessary, to calculate its magnitude and direction cosines.

As was proven above, an arbitrary system of forces, arbitrarily located in space, can be reduced to a single force equal to the main vector of the system and applied at an arbitrary reduction center ABOUT, and one pair with a moment equal to the main moment of the system relative to the same center. Therefore, in the future, an arbitrary system of forces can be replaced by an equivalent set of two vectors - force and moment applied at a point ABOUT. When changing the position of the center of reduction ABOUT the main vector will maintain magnitude and direction, but the main moment will change. Let us prove that if the main vector is nonzero and is perpendicular to the main moment, then the system of forces is reduced to one force, which in this case we will call the resultant (Fig. 8). The main moment can be represented by a pair of forces ( , ) with a shoulder , then the forces and the main vector form a system of two

forces equivalent to zero, which can be discarded. There will remain one force acting along a straight line parallel to the main

Figure 8 to the vector and passing at a distance

h= from the plane formed by the vectors and . The considered case shows that if from the very beginning we choose the center of reduction on the straight line L, then the system of forces would immediately be brought to the resultant, the main moment would be equal to zero. Now we will prove that if the main vector is non-zero and not perpendicular to the main moment, then such a point can be chosen as the reduction center ABOUT* that the main moment relative to this point and the main vector will be located on the same straight line. To prove this, let us decompose the moment into two components - one directed along the main vector, and the other perpendicular to the main vector. Thus, the pair of forces is decomposed into two pairs with moments: and , and the plane of the first pair is perpendicular to , then the plane of the second pair, perpendicular to the vector (Fig. 9) contains the vector . The combination of a couple with a moment and a force forms a system of forces, which can be reduced to one force (Fig. 8) passing through the point O*. Thus (Fig. 9), the combination of the main vector and the main moment at the point ABOUT reduced to the force passing through a point ABOUT*, and a pair with a moment parallel to this line, which was what needed to be proven. The combination of a force and a couple, the plane of which is perpendicular to the line of action of the force, is called dynamism (Fig. 10). A pair of forces can be represented by two forces ( , ) of equal magnitude, located as shown in Fig. 10. But by adding the two forces and , we obtain their sum and the remaining force , from which it follows (Fig. 10) that the combination of the main vector and the main moment in point ABOUT, can be reduced to two non-intersecting forces and .

Let us consider some cases of reduction of a system of forces.

1. Flat system of forces. For definiteness, let all forces be in the plane OXY. Then in the most general case

The main vector is not zero, the main moment is not zero, their dot product is zero, indeed

therefore, the main vector is perpendicular to the main moment: the plane system of forces is reduced to the resultant.

2. System of parallel forces. For definiteness, let all forces be parallel to the axis OZ. Then in the most general case

Here also the main vector is not equal to zero, the main moment is not equal to zero, and their scalar product is equal to zero, indeed

therefore, in this case, the main vector is perpendicular to the main moment: the system of parallel forces is reduced to the resultant. In the particular case, if 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. Let us now systematize the cases considered. Let us recall: 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 (center of reduction), and a pair of forces with a moment equal to the main moment of the system of forces relative to the specified center of reduction.