Equilibrium conditions for a spatial system of arbitrarily located forces. Analytical conditions for the equilibrium of a spatial system of arbitrarily located forces

Equilibrium condition for the spatial system of converging forces: the algebraic sum of the projections of all forces on three mutually perpendicular coordinate axes must be equal to zero, i.e.

To find the moment of power about the axis z,need to design strength on the plane H perpendicular to the axis z(fig. 12), then find the projection moment F nrelative to point O, which is the point of intersection of the plane Hsuck z.Projection moment F nand will be a moment of power about the axis z:

Spatial system of arbitrarily placed forcesa system of forces is called, the lines of action of which do not lie in the same plane and do not intersect at one point. The resultant of such a system of forces is also equal to the geometric sum of these forces, but is depicted by the diagonal of complex volumetric figures (tetrahedron, octahedron, etc.).

Equilibrium condition for a spatial system of arbitrarily located forces:the algebraic sum of the projections of all forces on three mutually perpendicular axes of coordinates should be equal to zero and the algebraic sum of the moments of all forces relative to the same axes of coordinates should be equal to zero, i.e.

Friction

Frictioncalled resistance to body movement. The force with which the body resists movement is called friction force.

The friction force is always directed in the direction opposite to the movement. The friction force depends on the material of the rubbing bodies, the cleanliness of the processing and the presence of lubricant and does not depend on the size of the rubbing surfaces.

Friction happens: dry, semi-liquid, liquid.

Distinguish friction rest, movement, slidingand rolling.The static friction force is greater than the motion friction force.

The friction force is equal to the product of the normal pressure force and the sliding friction coefficient (Fig. 14):

F tr \u003d R n ƒ,

where R n \u003d mgcos a is the force of normal pressure;

ƒ - coefficient of sliding friction.

Sliding friction coefficientthe ratio of the friction force to the normal pressure force is called:

Materials with very low friction are called antifriction(babbitt, bronze, graphite). They are used for the manufacture of bearings, etc.

Materials with high friction are called frictional(special plastics with the use of asbestos and copper). They are used for brake pad linings, for clutch disc linings.

When the sliding surface is lubricated, the body begins to move with less friction.

Let's decompose the force of gravity G into components G 'and G "(Fig. 15)

Let's compose the equilibrium equation:

where h -distance from the surface to the line of action of the force;

k -rolling friction coefficient. It is equal to the OS segment (see fig. 16)

F dv \u003d F tr,

F tr \u003d R p k / h

If a h \u003d d,

F tr \u003d R p k / d

if h \u003d r,

F tr \u003d R p k / d

As was shown in § 4.4, the necessary and sufficient equilibrium conditions spatial system forces applied to a rigid body can be written in the form of three equations of projections (4.16) and three moments (4.17):

, , . (7.14)

If the body is completely fixed, then the forces acting on it are in equilibrium and equations (7.13) and (7.14) serve to determine the support reactions. Of course, there may be cases when these equations are not enough to determine the support reactions; we will not consider such statically indeterminate systems.

For a spatial system of parallel forces, the equilibrium equations take the form (§ 4.4 [‡]):

, , . (7.15)

Let us now consider the cases when the body is only partially fixed, i.e. the bonds that are placed on the body do not guarantee the balance of the body. Four special cases can be indicated.

1. A rigid body has one fixed point. In other words, it is attached to a fixed point by a perfect spherical hinge.

Place the origin of the fixed coordinate system at this point. Link action at point AND replace with a reaction; since it is unknown in absolute value and in direction, we will represent it in the form of three unknown components,, directed respectively along the axes,,.

Equilibrium equations (7.13) and (7.14) in this case will be written in the form:

1) ,

2) ,

3) ,

4) ,

5) ,

The last three equations do not contain reaction components, since the line of action of this force passes through the point AND... Consequently, these equations establish the relationship between the active forces necessary for the balance of the body, and the first three equations can be used to determine the components of the reaction.

Thus, equilibrium condition solidhaving one fixed point, is the equality to zero of each of the algebraic sums of the moments of all active forces of the system with respect to three axes intersecting at a fixed point of the body .

2. The body has two fixed points. This will be the case, for example, if it is attached to two fixed points by means of hinges.

Let's choose the origin at the point AND and direct the axis along the line passing through the points AND and IN... Let us replace the action of constraints with reactions, directing the components of the reaction along the coordinate axes. We denote the distance between the points AND and IN across and; then the equilibrium equations (7.13) and (7.14) are written in the following form:

1) ,

2) ,

3) ,

4) ,

5) ,

The last equation does not contain reaction forces and establishes the connection between the active forces necessary for the balance of the body. Consequently, the equilibrium condition of a rigid body having two fixed points is the equality to zero of the algebraic sum of the moments of all active forces applied to the body relative to the axis passing through the fixed points ... The first five equations are used to determine the unknown components of the reactions,,,,,.

Note that the components and cannot be defined separately. From the third equation, only the sum + is determined and, therefore, the problem with respect to each of these unknowns for a rigid body is statically indeterminate. However, if at the point IN there is not a spherical, but a cylindrical hinge (i.e., a bearing) that does not prevent the body from sliding along the axis of rotation, then the problem becomes statically definable.

The body has a fixed axis of rotation along which it can slide without friction. This means that at points AND and IN there are cylindrical joints (bearings), and their constituent reactions along the axis of rotation are equal to zero. Therefore, the equilibrium equations will take the form:

1) ,

2) ,

4) ,

5) ,

Two of the equations (7.18), namely the third and the sixth, impose restrictions on the system of active forces, and the rest of the equations serve to determine the reactions.

The body rests at three points on a smooth surface, and the support points do not lie on one straight line. Let us denote these points by AND, IN and FROM and is compatible with the plane ABC coordinate plane Ahu... Replacing the action of constraints with vertical reactions, and, we write the equilibrium conditions (7.14) in the following form:

3) ,

4) ,

5) ,

The third - fifth equations can serve to determine unknown reactions, and the first, second and sixth equations represent the conditions that connect the active forces and are necessary for the balance of the body. Of course, for the balance of the body, it is necessary to fulfill the conditions,,, since at the points of support only reactions of the direction adopted above can occur.

If the body rests on a horizontal plane at more than three points, then the problem becomes statically indeterminate, since in this case there will be as many reactions as there are points, and there will be only three equations for determining the reactions.

Task 7.3. Find the main vector and the main moment of the system of forces shown in Fig. Forces are applied to the vertices of the cube and are directed along its edges, and ,. The length of the edge of the cube is and.

We find the projections of the main vector by the formulas (4.4):

, , .

Its module is equal. The direction cosines will be

, ;

, .

The main vector is shown in Fig.

and the modulus of the principal moment by formula (4.8)

Now let's define the direction cosines of the main moment:

, ;

, .

The main point is shown in Fig. The angle between vectors and is calculated by the formula (4.11) and

We find the boundaries of the required area from the conditions:

From here we find

In fig. the sought-for region constructed for is shaded. The entire surface of the plate will be safe.

Theorem. For the balance of the spatial system of forces, it is necessary and sufficient that the main vector and the main moment of this system be equal to zero. Adequacy: for F o \u003d 0, the system of converging forces applied at the center of reduction O is equivalent to zero, and when Mo \u003d 0, the system of pairs of forces is equivalent to zero. Therefore, the original system of forces is equivalent to zero. Necessity: Let the given system of forces be equivalent to zero. Bringing the system to two forces, we note that the system of forces Q and P (Fig. 4.4) must be equivalent to zero, therefore, these two forces must have a common line of action and the equality Q \u003d –Р must be fulfilled. But this can be if the line of action of the force P passes through the point O, that is, if h \u003d 0. And this means that the main moment is equal to zero (M o \u003d 0). Because Q + P \u003d 0, a Q \u003d F o + P ", then F o + P" + P \u003d 0, and, therefore, F o \u003d 0. Necessary and sufficient conditions are equal to the spatial system of forces they are of the form: F o \u003d 0 , M o \u003d 0 (4.15),

or, in projections onto the coordinate axes, Fox \u003d åF kx \u003d F 1x + F 2x +… + F nx \u003d 0; F Oy \u003d åF ky \u003d F 1y + F 2y + ... + F ny \u003d 0; F oz \u003d åF kz \u003d F 1z + F 2z +… + F nz \u003d 0 (4.16). M Ox \u003d åM Ox (F k) \u003d M Ox (F 1) + M ox (F 2) + ... + M Ox (F n) \u003d 0, M Oy \u003d åM Oy (F k) \u003d M oy ( F 1) + M oy (F 2) +… + M oy (F n) \u003d 0, M oz \u003d åM O z (F k) \u003d M O z (F 1) + M oz (F 2) + .. . + M oz (F n) \u003d 0. (4.17)

So when solving problems, having 6 levels, you can find 6 unknowns. Note: a pair of forces cannot be reduced to a resultant. Special cases: 1) Equilibrium of the spatial system of parallel forces. Let the Z axis be parallel to the lines of force action (Figure 4.6), then the projections of forces on x and y are 0 (F kx \u003d 0 and F ky \u003d 0), and only F oz remains. As for the moments, only M ox and M oy remain, and M oz is absent. 2) Equilibrium of a plane system of forces. Ur-I F ox, F oy and moment M oz remain (Figure 4.7). 3) Equilibrium of a plane system of parallel forces. (Figure 4.8). Only 2 ur-I remain: F oy and M oz. When drawing up ur-th balance any point can be selected for the center of the ghost.

ABOUTR \u003d 0 and M R x \u003d R y \u003d R z \u003d 0 and M x \u003d M y \u003d M

Equilibrium conditions for an arbitrary spatial system of forces.

An arbitrary spatial system of forces, like a flat one, can be reduced to some center ABOUT and replace with one resultant force and a pair with moment. Reasoning that for the balance of this system of forces it is necessary and sufficient that at the same time R \u003d 0 and M o \u003d 0. But vectors can vanish only when all their projections on the coordinate axes are equal to zero, that is, when R x \u003d R y \u003d R z \u003d 0 and M x \u003d M y \u003d M z \u003d 0 or, when the acting forces satisfy the conditions

Thus, for the equilibrium of an arbitrary spatial system of forces, it is necessary and sufficient that the sums of the projections of all forces on each of the three coordinate axes and the sum of their moments about these axes are equal to zero.

Principles of solving problems on the balance of a body under the influence of a spatial system of forces.

The principle of solving the problems of this section remains the same as for a plane system of forces. Having established the balance of which body will be considered, they replace the constraints imposed on the body by their reactions and form the equilibrium conditions for this body, considering it as free. The sought values \u200b\u200bare determined from the obtained equations.

To obtain simpler systems of equations, it is recommended that the axes be drawn so that they intersect more unknown forces or are perpendicular to them (unless this unnecessarily complicates the calculation of projections and moments of other forces).

A new element in the formulation of equations is the calculation of the moments of forces about the coordinate axes.

In cases where it is difficult to see from the general drawing what the moment of a given force is about any axis, it is recommended to depict in the auxiliary drawing the projection of the body in question (together with the force) onto a plane perpendicular to this axis.

In cases where, when calculating the moment, difficulties arise in determining the projection of the force onto the corresponding plane or the shoulder of this projection, it is recommended to decompose the force into two mutually perpendicular components (of which one is parallel to some coordinate axis), and then use the Varignon theorem.

Example 5.

Frame AB (fig. 45) kept in balance by the hinge AND and rod Sun... There is a load on the edge of the frame R... Determine the reaction of the hinge and the force in the rod.

Fig. 45

Consider the balance of the frame together with the load.

We build a design scheme, depicting the frame as a free body and showing all the forces acting on it: the reactions of ties and the weight of the load R... These forces form a system of forces arbitrarily located on a plane.

It is advisable to draw up such equations so that each has one unknown force.

In our task, this is the point AND, where unknowns and are attached; dot FROMwhere the lines of action of unknown forces intersect and; dot D - the point of intersection of the lines of action of forces and. Let's compose the equation of the projections of forces on the axis at (per axis x it is impossible to design, because it is perpendicular to a straight line AS).

And, before composing the equations, we will make one more useful remark. If the design diagram has a force located so that its shoulder is not easy to find, then when determining the moment, it is recommended to first decompose the vector of this force into two, more conveniently directed. In this problem, we decompose the force into two: and (Fig. 37) such that their moduli

We compose the equations:

From the second equation we find ... From the third And from the first

So how did it happen S<0, то стержень Sun will be compressed.

An arbitrary spatial system of forces, like a flat one, can be brought to some center ABOUT and replace with one resultant force and a pair with a moment. Reasoning that for the balance of this system of forces it is necessary and sufficient that at the same time R \u003d 0 and M o \u003d 0. But vectors u can vanish only if all their projections on the coordinate axes are equal to zero, that is, when R x \u003d R y \u003d R z \u003d 0 and M x \u003d M y \u003d M z \u003d 0 or, when the acting forces satisfy the conditions:

Σ X i = 0; Σ M x(P i) = 0;

Σ Y i = 0; Σ M y(P i) = 0;

Σ Z i = 0; Σ M z(P i) = 0.

Thus, for the equilibrium of the spatial system of forces, it is necessary and sufficient that the sums of the projections of all forces of the system on each of the coordinate axes, as well as the sum of the moments of all forces of the system relative to each of these axes, equal zero.

A new element in the formulation of equations is the calculation of the moments of forces about the coordinate axes.

In cases where it is difficult to see from the general drawing what the moment of a given force is about any axis, it is recommended to depict on the auxiliary drawing the projection of the body in question (together with the force) onto a plane perpendicular to this axis.

Example 5. Frame AB (fig. 45) kept in balance by the hinge AND and rod Sun... There is a load on the edge of the frame R... Determine the reaction of the hinge and the force in the rod.

Fig. 45

Consider the balance of the frame together with the load.

It is desirable to compose such equations so that each has one unknown force.

In our task, this is the point AND, where unknowns and are attached; dot FROMwhere the lines of action of unknown forces intersect and; dot D - the point of intersection of the lines of action of the forces and. Let's compose the equation of projections of forces on the axis at (per axis x it is impossible to design, because it is perpendicular to a straight line AS).

And, before writing the equations, we will make one more useful remark. If the design diagram has a force located so that its shoulder is not easy to find, then when determining the moment, it is recommended to first decompose the vector of this force into two, more conveniently directed. In this problem, we decompose the force into two: and (Fig. 37) such that their moduli

We compose the equations:

From the second equation we find:

From the third

And from the first

So how did it happen S<0, то стержень Sun will be compressed.

Example 6. Rectangular weighing shelf R held horizontally by two rods CE and CD, attached to the wall at the point E... Rods of equal length, AB \u003d 2 a, EO \u003d a... Determine the efforts in the rods and the reaction of the loops AND and IN.

Fig. 46

Consider the balance of the slab. We build a calculation scheme (Fig. 46). Loop reactions are usually shown by two forces perpendicular to the loop axis:.

Forces form a system of forces, arbitrarily located in space. We can make 6 equations. There are also six unknowns.

What equations to make - you need to think. It is desirable that they be simpler and that there are fewer unknowns in them.

Let's compose the following equations:

From equation (1) we get: S 1 \u003d S 2. Then from (4):.

From (3): Y A \u003d Y B and, by (5),. Hence From equation (6), since S 1 \u003d S 2, it follows Z A \u003d Z B. Then by (2) Z A \u003d Z B \u003d P / 4.

From the triangle, where, it follows,

Therefore, Y A \u003d Y B \u003d 0.25P, Z A \u003d Z B 0.25P.

To check the solution, you can draw up another equation and see if it is satisfied with the found reaction values:

The problem was solved correctly.