01.02.2022

Theoretical mechanics. Statics

Necessary and sufficient conditions for the equilibrium of any system of forces are expressed by equalities (see § 13). But the vectors R and are equal only when, that is, when the acting forces, according to formulas (49) and (50), 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 onto each of the three coordinate axes and the sums of their moments relative to these axes are equal to zero.

Equalities (51) simultaneously express the equilibrium conditions solid, under the influence of any spatial system of forces.

If, in addition to the forces, a couple is also acting on the body, specified by its moment, then the form of the first three of the conditions (51) will not change (the sum of the projections of the forces of the couple on any axis is equal to zero), and the last three conditions will take the form:

The case of parallel forces. In the case when all the forces acting on the body are parallel to each other, you can choose the coordinate axes so that the axis is parallel to the forces (Fig. 96). Then the projections of each of the forces on the axis and their moments relative to the z axis will be equal to zero and system (51) will give three equilibrium conditions:

The remaining equalities will then turn into identities of the form

Consequently, for the equilibrium of a spatial system of parallel forces, it is necessary and sufficient that the sum of the projections of all forces onto the axis parallel to the forces and the sum of their moments relative to the other two coordinate axes are equal to zero.

Problem solving. The procedure for solving problems here remains the same as in the case of a plane system. Having established the equilibrium of which body (object) is being considered, it is necessary to depict all the external forces acting on it (both given and reaction connections) and draw up conditions for the equilibrium of these forces. From the resulting equations the required quantities are determined.

To get more simple systems equations, it is recommended to draw the axes so that they intersect more unknown forces or are perpendicular to them (unless this unnecessarily complicates the calculations of projections and moments of other forces).

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

In cases where it is difficult to see from the general drawing what the moment of a given force is relative to any axis, it is recommended to depict in an auxiliary drawing the projection of the body in question (along 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 arm of this projection, it is recommended to decompose the force into two mutually perpendicular components (one of which is parallel to some coordinate axis), and then use Varignon’s theorem (see. task 36). In addition, you can calculate moments analytically using formulas (47), as, for example, in problem 37.

Problem 39. There is a load on a rectangular plate with sides a and b. The center of gravity of the slab together with the load is located at point D with coordinates (Fig. 97). One of the workers holds the slab at corner A. At what points B and E should two other workers support the slab so that the forces applied by each of those holding the slab are equal.

Solution. We consider the equilibrium of a plate, which is a free body in equilibrium under the action of four parallel forces where P is the force of gravity. We draw up equilibrium conditions (53) for these forces, considering the plate horizontal and drawing the axes as shown in Fig. 97. We get:

According to the conditions of the problem, there should be Then from the last equation Substituting this value of P into the first two equations, we will finally find

The solution is possible when When and when will be When point D is in the center of the plate,

Problem 40. On a horizontal shaft lying in bearings A and B (Fig. 98), a pulley of radius cm and a drum of radius are mounted perpendicular to the shaft axis. The shaft is driven into rotation by a belt wrapped around a pulley; at the same time, a load weighing , tied to a rope, which is wound on a drum, is evenly lifted. Neglecting the weight of the shaft, drum and pulley, determine the reactions of bearings A and B and the tension of the driving branch of the belt, if it is known that it is twice the tension of the driven branch. Given: cm, cm,

Solution. In the problem under consideration, with uniform rotation of the shaft, the forces acting on it satisfy the equilibrium conditions (51) (this will be proven in § 136). Let's draw coordinate axes (Fig. 98) and depict the forces acting on the shaft: tension F of the rope, modulo equal to P, belt tension and components of bearing reactions.

To compile the equilibrium conditions (51), we first calculate and enter into the table the values of the projections of all forces onto the coordinate axes and their moments relative to these axes.

Now we create equilibrium conditions (51); since we get:

From equations (III) and (IV) we find immediately, taking into account that

Substituting the found values into the remaining equations, we find;

And finally

Problem 41. A rectangular cover with a weight forming an angle with the vertical is fixed on the horizontal axis AB at point B by a cylindrical bearing, and at point A by a bearing with a stop (Fig. 99). The lid is held in balance by rope DE and pulled back by a rope thrown over the block O with a weight at the end (line KO parallel to AB). Given: Determine the tension of the rope DE and the reactions of bearings A and B.

Solution. Consider the equilibrium of the lid. Let's draw coordinate axes, starting at point B (in this case, the force T will intersect the axes, which will simplify the form of the moment equations).

Then we depict all the given forces and reaction reactions acting on the cover: the force of gravity P applied at the center of gravity C of the cover, the force Q equal in magnitude to Q, the reaction T of the rope and the reaction of bearings A and B (Fig. 99; vector M k shown in dotted line not relevant to this task). To draw up the equilibrium conditions, we introduce an angle and denote the calculation of the moments of some forces is explained in the auxiliary fig. 100, a, b.

In Fig. 100, and the view is shown in projection onto the plane from the positive end of the axis

This drawing helps to calculate the moments of forces P and T relative to the axis. It can be seen that the projections of these forces onto the plane (plane perpendicular) are equal to the forces themselves, and the arm of the force P relative to point B is equal to; the shoulder of the force T relative to this point is equal to

In Fig. 100, b shows a view in projection onto a plane from the positive end of the y-axis.

This drawing (together with Fig. 100, a) helps to calculate the moments of forces P and relative to the y-axis. It shows that the projections of these forces onto the plane are equal to the forces themselves, and the arm of the force P relative to point B is equal to the arm of the force Q relative to this point is equal to or, as can be seen from Fig. 100, a.

Compiling the equilibrium conditions (51) taking into account the explanations made and assuming at the same time we obtain:

(I)

Considering what we find from equations (I), (IV), (V), (VI):

Substituting these values into equations (II) and (III), we obtain:

Finally,

Problem 42. Solve Problem 41 for the case when the lid is additionally acted upon by a pair located in its plane with a moment of rotation of the pair directed (when looking at the lid from above) counterclockwise.

Solution. In addition to the forces acting on the lid (see Fig. 99), we depict the moment M of the pair as a vector perpendicular to the lid and applied at any point, for example at point A. Its projections onto the coordinate axes: . Then, composing the equilibrium conditions (52), we find that equations (I) - (IV) will remain the same as in the previous problem, and the last two equations have the form:

Note that the same result can be obtained without composing an equation in the form (52), but by depicting the pair as two forces directed, for example, along the lines AB and KO (in this case, the moduli of the forces will be equal), and then using the usual equilibrium conditions.

Solving equations (I) - (IV), (V), (VI), we will find results similar to those obtained in problem 41, with the only difference that all formulas will include . Finally we get:

Problem 43. The horizontal rod AB is attached to the wall by a spherical hinge A and is held in a position perpendicular to the wall by braces KE and CD, shown in Fig. 101, a. A load with a weight is suspended from end B of the rod. Determine the reaction of hinge A and the tension of the guy wires if the Weight of the rod is neglected.

Solution. Let us consider the equilibrium of the rod. It is acted upon by force P and reactions. Let us draw coordinate axes and draw up equilibrium conditions (51). To find projections and moments of force, let us decompose it into components. Then, by Varignon’s theorem, since since

The calculation of moments of forces relative to the axis is explained by an auxiliary drawing (Fig. 101, b), which shows a view in projection onto a plane

Methods for solving equilibrium problems with an arbitrary spatial system of forces are considered. An example of solving the problem of equilibrium of a plate supported by rods in three-dimensional space is given. It is shown how, by choosing the axes when drawing up equilibrium equations, the solution to the problem can be simplified.

Content

The procedure for solving equilibrium problems with an arbitrary spatial system of forces

To solve the problem of equilibrium of a rigid body with an arbitrary spatial system of forces, it is necessary to select a rectangular coordinate system and, relative to it, draw up equilibrium equations.

The equilibrium equations for an arbitrary system of forces distributed in three-dimensional space are two vector equations:
the vector sum of the forces acting on the body is zero
(1) ;
the vector sum of the moments of forces, relative to the origin, is equal to zero
(2) .

Let Oxyz be the coordinate system we have chosen. By projecting equations (1) and (2) on the axis of this system, we obtain six equations:
the sums of force projections on the xyz axis are equal to zero
(1.x) ;
(1.y) ;
(1.z) ;
the sums of the moments of forces relative to the coordinate axes are equal to zero
(2.x) ;
(2.y) ;
(2.z) .
Here we assume that n forces act on the body, including reaction forces of the supports.

Let arbitrary force, with components, is applied to the body at point . Then the moments of this force relative to the coordinate axes are determined by the formulas:
(3.x) ;
(3.y) ;
(3.z) .

Thus, the procedure for solving the problem of equilibrium with an arbitrary spatial system of forces is as follows.

We discard the supports and replace them with reaction forces. If the support is a rod or thread, then the reaction force is directed along the rod or thread.
We choose the rectangular coordinate system Oxyz.
We find the projections of force vectors on the coordinate axes, , and points of their application, . The point of application of the force can be moved along a straight line drawn through the force vector. Such a movement will not change the values of the moments. Therefore, we select the most convenient points of application of forces for calculation.
We compose three equilibrium equations for forces (1.x,y,z).
For each force, using formulas (3.x,y,z), we find the projections of the moments of force on the coordinate axes.
We compose three equilibrium equations for moments of forces (2.x,y,z).
If the number of variables is greater than the number of equations, then the problem is statically indeterminate. It cannot be solved using static methods. It is necessary to use methods of resistance of materials.
We solve the resulting equations.

Simplify your calculations

In some cases, it is possible to simplify calculations if, instead of equation (2), we use the equivalent equilibrium condition.
The sum of the moments of forces about an arbitrary axis AA′ is equal to zero:
(4) .

That is, you can select several additional axes that do not coincide with the coordinate axes. And with respect to these axes, compose equations (4).

An example of solving a problem on the equilibrium of an arbitrary spatial system of forces

The equilibrium of the slab, in three-dimensional space, is maintained by a system of rods.

Find the reactions of the rods supporting a thin homogeneous horizontal plate in three-dimensional space. The rod fastening system is shown in the figure. The slab is acted upon by: gravity G; and the force P applied at point A, directed along side AB.

Given:
G= 28 kN; P= 35 kN; a = 7.5 m; b = 6.0 m; c = 3.5 m.

The solution of the problem

First we will solve this problem in a standard way, applicable for an arbitrary spatial system of forces. And then we will obtain a simpler solution, based on the specific geometry of the system, due to the choice of axes when drawing up the equilibrium equations.

Solving the problem in a standard way

Although this method will lead us to rather cumbersome calculations, it is applicable for an arbitrary spatial system of forces, and can be used in computer calculations.

Let's discard the connections and replace them with reaction forces. The connections here are rods 1-6. Instead, we introduce forces directed along the rods. We choose the directions of forces at random. If we do not guess the direction of any force, we will get a negative value for it.

We draw a coordinate system Oxyz with the origin at point O.

We find the projections of forces on the coordinate axes.

For strength we have:
.
Here α 1 - angle between LQ and BQ. From right triangle LQB:
m;
;
.

Forces , and are parallel to the z axis. Their components:
;
;
.

For strength we find:
.
Here α 3 - angle between QT and DT. From right triangle QTD:
m;
;
.

For strength:
.
Here α 5 - angle between LO and LA. From right triangle LOA:
m;
;
.

The force is directed diagonally across a rectangular parallelepiped. It has the following projections on the coordinate axes:
.
Here are the direction cosines of the diagonal AQ:
m;
;
;
.

We select the points of application of forces. Let's take advantage of the fact that they can be moved along lines drawn through the force vectors. So, as the point of application of force, you can take any point on the straight line TD. Let's take point T, since for it the x and z coordinates are equal to zero:
.
In a similar way, we select the points of application of the remaining forces.

As a result, we obtain the following values of the force components and their application points:
; (point B);
; (point Q);
; (point T);
; (point O);
; (point A);
; (point A);
; (point A);
; (point K).

We compose equilibrium equations for forces. The sums of the projections of forces on the coordinate axes are equal to zero.

;

;

.

We find the projections of the moments of forces on the coordinate axes.
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;

We compose equilibrium equations for moments of forces. The sums of the moments of forces about the coordinate axes are equal to zero.

;

;

;

So, we got the following system of equations:
(P1) ;
(P2) ;
(P3) ;
(P4) ;
(P5) ;
(P6) .

There are six equations and six unknowns in this system. Then you can substitute numerical values here and obtain a solution to the system using a mathematical program for calculating a system of linear equations.

But for this problem, you can get a solution without using funds computer technology.

An effective way to solve a problem

We will take advantage of the fact that equilibrium equations can be composed in more than one way. You can arbitrarily select the coordinate system and axes relative to which the moments are calculated. Sometimes, due to the choice of axes, it is possible to obtain equations that can be solved more simply.

Let us use the fact that, in equilibrium, the sum of the moments of forces about any axis is zero. Let's take the AD axis. The sum of the moments of forces about this axis is zero:
(P7) .
Next, we note that all forces except intersect this axis. Therefore their moments are equal to zero. Only one force does not cross the AD axis. It is also not parallel to this axis. Therefore, for equation (A7) to be satisfied, force N 1 should be equal to zero:
N 1 = 0 .

Now let's take the AQ axis. The sum of the moments of forces relative to it is zero:
(P8) .
This axis is crossed by all forces except . Since the force is not parallel to this axis, then to satisfy equation (A8) it is necessary that
N 3 = 0 .

Now let's take the AB axis. The sum of the moments of forces relative to it is zero:
(P9) .
This axis is crossed by all forces except , and . But N 3 = 0 . That's why
.
The moment of the force relative to the axis is equal to the product of the force arm by the magnitude of the projection of the force onto a plane perpendicular to the axis. The shoulder is equal to the minimum distance between the axis and the straight line drawn through the force vector. If the twist occurs in a positive direction, then the torque is positive. If it's negative, then it's negative. Then
.
From here
kN.

We will find the remaining forces from equations (A1), (A2) and (A3). From equation (A2):
N 6 = 0 .
From equations (A1) and (A3):
kN;
kN

Thus, solving the problem in the second way, we used the following equilibrium equations:
;
;
;
;
;
.
As a result, we avoided cumbersome calculations associated with calculating the moments of forces relative to the coordinate axes and obtained linear system equations with a diagonal matrix of coefficients, which was immediately resolved.

N 1 = 0 ; N 2 = 14.0 kN; N 3 = 0 ; N 4 = -2.3 kN; N 5 = 38.6 kN; N 6 = 0 ;

The minus sign indicates that the force N 4 directed in the direction opposite to that shown in the figure.

The case of such a balance of forces corresponds to two equilibrium conditions

M= Mo= 0, R* = 0.

Highlight Modules Mo and main vector R* of the system under consideration are determined by the formulas

Mo= (M x 2 + M y 2 + +M z 2) 1/2 ; R*= (X 2 + Y 2 +Z 2) 1/2.

They wound to zero only under the following conditions:

M x = 0, M y =0, M z = 0, X=0, Y=0, Z=0,

which correspond to the six basic equations of equilibrium of forces arbitrarily located in space

=0; =0;

=0; (5-17)

=0 ; =0.

The three equations of system (5-17) on the left are called equations of moments of forces relative to the coordinate axes, and the three on the right are equations of projections of forces on the axes.

Using these formulas, the moment equation can be represented as

å (y i Z i - z i Y i)=0; å(z i Х i - x i Z i)=0 ; å(x i Y i - y i X i)=0 .(5-18)

Where x i, y i, z i- coordinates of the points of application of force P; Y i , Z i , X i - projections of this force onto coordinate axes, which can have any direction.

There are other systems of six equations of equilibrium of forces, arbitrarily located in space.

Reducing a system of forces to a resultant force.

If the main vector of the force system R* is not equal to zero, but the main moment Mo or equal to zero, or directed perpendicular to the main vector, then the given system of forces is reduced to a resultant force.

There are 2 possible cases.

1st case.

Let R*¹ 0; Mo = 0 . In this case, the forces lead to a resultant, the line of action of which passes through the center of reduction O, and the force R* replaces a given system of forces, i.e. is its resultant.

2nd case.

R*¹0; Mo¹ 0 and Mo ⊥ R*. (Fig. 5.15).

After bringing the system of forces to the center O, the force is obtained R* , applied in this center and equal to the main vector of forces, and a pair of forces, the moment of which M equal to the main moment Mo all forces relative to the center of reduction, and Mo ⊥ R*.

Let's choose the strengths of this couple R' And R equal in modulus to the main vector R* , i.e. R= R' = R *. Then the leverage of this pair should be taken equal to OK = = M O/R * Let us draw plane I through point O, perpendicular to the moment of the pair of forces M . Couple of forces R' , R must be in this plane. Let's arrange this pair so that one of the forces of the pair R' was applied at point O and directed opposite to the force R * . Let us restore in plane I at point O a perpendicular to the line of action of the force R * , and at point K at a distance OK= M O/R * from point O we apply the second pair force R .

We put the segment OK in such a direction from point O that, looking towards the moment vector M, we see the pair tending to rotate its plane counterclockwise. Then strength R* And R' , applied at point O, will be balanced, and the force R pairs applied at point K will replace the given system of forces, i.e. will be its resultant. The straight line coinciding with the line of action of this force is the line of action of the resultant force. Rice. 5.15 shows the difference between the resultant force R and force R* , obtained by bringing forces to the center O.

Resultant R a system of forces applied at point K, having a certain line of action, is equivalent to a given system of forces, i.e. replaces this system.

The strength R* at point O replaces a given system of forces only in conjunction with a pair of forces with a moment M= Mo .

Strength R* can be applied at any point of the body to which forces are applied. Only the magnitude and direction of the main moment depend on the position of the point Mo .

Varignon's theorem. The moment of the resultant force about any point is equal to the geometric sum of the moments of the component forces about this point, and the moment of the resultant force about any axis is equal to the algebraic sum of the moments of the component forces about this axis.

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 couple with a moment. Reasoning in such a way that for the balance of this system of forces it is necessary and sufficient that at the same time there be R= 0 and M o = 0. But vectors and can vanish only when all their projections on the coordinate axes are equal to zero, i.e. when R x = R y = R z = 0 and M x = M y = M z = 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 Mz(P i) = 0.

Thus, for the equilibrium of a spatial system of forces, it is necessary and sufficient that the sum of the projections of all forces of the system onto 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, equals zero.

In special cases of a system of converging or parallel forces, these equations will be linearly dependent, and only three of the six equations will be linearly independent.

For example, the equilibrium equations for a system of forces, parallel to the axis Oz, have the form:

Σ Z i = 0;

Σ M x(P i) = 0;

Σ M y(P i) = 0.

Problems on body balance under the influence of a spatial system of forces.

The principle for solving problems in this section remains the same as for a plane system of forces. Having established the equilibrium of which body will be considered, they replace the connections imposed on the body with their reactions and draw up the conditions for the equilibrium of this body, considering it as free. From the resulting equations the required quantities are determined.

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

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

In cases where, when calculating the moment, difficulties arise in determining the projection of the force onto the corresponding plane or the arm 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 Varignon’s theorem.

Example 5. Frame AB(Fig. 45) is kept in balance by a hinge A and the rod Sun. On the edge of the frame there is a load weighing R. Let us determine the reactions of the hinge and the force in the rod.

Fig.45

We consider the equilibrium of the frame together with the load.

We build a calculation diagram, depicting the frame as a free body and showing all the forces acting on it: the reaction of the connections and the weight of the load R. These forces form a system of forces arbitrarily located on the plane.

It is advisable to create equations such that each contains one unknown force.

In our problem this is the point A, where the unknowns and are attached; dot WITH, where the lines of action of unknown forces and intersect; dot D– the point of intersection of the lines of action of forces and. Let's create an equation for the projection of forces onto the axis at(per axis X it is impossible to design, because it is perpendicular to the line AC).

And, before composing the equations, let's make one more useful remark. If in the design diagram there is a force located in such a way that its arm is not easy to locate, then when determining the moment, it is recommended to first decompose the vector of this force into two, more conveniently directed ones. In this problem we will decompose the force into two: and (Fig. 37) such that their modules

Let's make up 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 shelf weighing R held horizontally by two rods SE And CD, attached to the wall at a point E. Rods of equal length, AB=2 a,EO= a. Let us determine the forces in the rods and the reactions of the loops A And IN.

Fig.46

Consider the equilibrium of the plate. We build a design diagram (Fig. 46). Loop reactions are usually shown by two forces perpendicular to the loop axis: .

The forces form a system of forces arbitrarily located in space. We can create 6 equations. There are also six unknown people.

You need to think about what equations to create. It is desirable that they be simpler and that they contain fewer unknowns.

Let's make the following equations:

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

From (3): Y A =Y B and, according to (5), . This means From equation (6), because S 1 =S 2, follows Z A =Z B. Then according to (2) Z A =Z B =P/4.

From the triangle where , it follows ,

Therefore Y A =Y B =0.25P, Z A =Z B 0.25P.

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

The problem was solved correctly.

Self-test questions

What structure is called a truss?

Name the main components of a farm.

Which truss rod is called zero?

State the lemmas that determine the zero bar of the truss.

What is the essence of the method of cutting knots?

Based on what considerations, without calculations, can one determine the rods of spatial trusses in which, at a given load, the forces are equal to zero?

What is the essence of the Ritter method?

What is the relationship between the normal surface reaction and the normal pressure force?

What is frictional force?

Write down the Amonton-Coulomb law.

Formulate the basic law of friction. What is the coefficient of friction, the angle of friction and what does their value depend on?

The beam is in balance, resting on a smooth vertical wall and a rough horizontal floor; the center of gravity of the beam is in its middle. Is it possible to determine the direction of the overall sex response?

Name the dimension of the sliding friction coefficient.

What is the ultimate sliding friction force.

What characterizes a cone of friction?

Name the reason for the appearance of the rolling friction moment.

What is the dimension of the rolling friction coefficient?

Give examples of devices in which spinning friction occurs.

What is the difference between adhesion force and friction force?

What is a clutch cone called?

What are the possible directions of reaction of a rough surface?

What is the equilibrium region and what are the equilibrium conditions for the forces applied to a block resting on two rough surfaces?

What is the moment of a force about a point? What is the dimension of this quantity?

How to calculate the modulus of the moment of a force relative to a point?

Formulate a theorem about the moment of the resultant system of converging forces.

What is the moment of force about an axis?

Write down a formula connecting the moment of a force about a point with the moment of the same force about an axis passing through this point.

How is the moment of a force about an axis determined?

Why, when determining the moment of a force about an axis, is it necessary to project the force onto a plane perpendicular to the axis?

How should the axis be positioned so that the moment of a given force relative to this axis is equal to zero?

Give formulas for calculating moments of force about coordinate axes.

What is the direction of the force moment vector relative to the point?

How is the moment of a force relative to a point determined on a plane?

What area can determine the numerical value of the moment of force relative to a given point?

Does the moment of a force about a given point change when a force is transferred along the line of its action?

In what case is the moment of a force about a given point equal to zero?

Determine the geometric locus of points in space relative to which the moments of a given force are:

a) geometrically equal;

b) equal in modulus.

How are the numerical value and sign of the moment of force relative to the axis determined?

Under what conditions is the moment of force about the axis equal to zero?

In what direction of a force applied to a given point is its moment relative to a given axis greatest?

What relationship exists between the moment of a force about a point and the moment of the same force about an axis passing through this point?

Under what conditions is the modulus of the moment of a force relative to a point equal to the moment of the same force relative to an axis passing through this point?

What are the analytical expressions for moments of force about coordinate axes?

What are the main moments of a system of forces arbitrarily located in space relative to a point and relative to an axis passing through this point? What is the relationship between them?

What is the principal moment of a system of forces lying in one plane relative to any point in this plane?

What is the main moment of the forces composing the pair relative to any point in space?

What is the principal moment of a system of forces relative to a given pole?

How is the lemma on parallel force transfer formulated?

Formulate a theorem about bringing an arbitrary system of forces to the main vector and the main moment.

Write down formulas for calculating the projections of the main moment onto the coordinate axes.

Give a vector representation of the equilibrium conditions for an arbitrary system of forces.

Write down the equilibrium conditions for an arbitrary system of forces in projections onto rectangular coordinate axes.

How many independent scalar equilibrium equations can be written for a spatial system of parallel forces?

Write down the equilibrium equations for an arbitrary plane system of forces.

Under what condition are three non-parallel forces applied to a rigid body balanced?

What is the equilibrium condition for three parallel forces applied to a rigid body?

What are the possible cases of bringing arbitrarily located and parallel forces in space?

To what simplest form can a system of forces be reduced if it is known that the main moment of these forces relative to various points in space:

a) has the same value not equal to zero;

b) equal to zero;

c) has different values and is perpendicular to the main vector;

d) has different values and is not perpendicular to the main vector.

What are the conditions and equations of equilibrium of a spatial system of converging, parallel and arbitrarily located forces and how do they differ from the conditions and equations of equilibrium of the same type of forces on a plane?

What equations and how many of them can be composed for a balanced spatial system of converging forces?

Write down the system of equilibrium equations for a spatial system of forces?

What are the geometric and analytical conditions for reducing a spatial system of forces to a resultant?

Formulate a theorem about the moment of the resultant spatial system of forces relative to a point and an axis.

Write down equations for the line of action of the resultant.

What straight line in space is called the central axis of a system of forces?

Derive the equations for the central axis of the force system?

Show that two crossing forces can be driven to a force screw.

What formula is used to calculate the smallest principal moment of a given system of forces?

Write down the formulas for calculating the main vector of a spatial system of converging forces?

Write down the formulas for calculating the main vector of a spatial system of arbitrarily located forces?

Write down the formula for calculating the main moment of a spatial system of forces?

What is the dependence of the main moment of a system of forces in space on the distance of the center of reduction to the central axis of this system of forces?

Relative to which points in space do the main moments of a given system of forces have the same magnitude and make the same angle with the main vector?

Relative to what points in space are the main moments of the system of forces geometrically equal to each other?

What are the invariants of the force system?

What conditions are satisfied by the specified forces applied to a rigid body with one or two fixed points that is at rest?

Will there be a plane system of forces in equilibrium for which the algebraic sums of moments about three points located on the same straight line are equal to zero?

Let for a plane system of forces the sums of moments about two points be equal to zero. Under what additional conditions will the system be in equilibrium?

Formulate necessary and sufficient conditions for the equilibrium of a plane system of parallel forces.

What is a moment point?

What equations (and how many) can be composed for a balanced arbitrary plane system of forces?

What equations and how many of them can be composed for a balanced spatial system of parallel forces?

What equations and how many of them can be compiled for a balanced arbitrary spatial system of forces?

How is a plan for solving statics problems on the balance of forces formulated?

As was clarified in § 4.4, the necessary and sufficient conditions for the equilibrium of a spatial system of forces applied to a rigid body can be written in the form of three projection equations (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 where 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 cases when the body is only partially fixed, i.e. the connections that are imposed on the body do not guarantee the balance of the body. Four special cases can be indicated.

1. A solid body has one fixed point. In other words, it is attached to a fixed point using a perfect spherical joint.

Let us place the origin of the fixed coordinate system at this point. Action of connection at a point A Let's replace it with a reaction; since it is unknown in magnitude and direction, we will present 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) ,

6) . (7.16)

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

Thus, the condition for the equilibrium of a rigid body that has one fixed point is the equality to zero of each of the algebraic sums of the moments of all active forces of the system relative to three axes intersecting at a fixed point of the body .

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

Let us choose the origin of coordinates at the point A and direct the axis along the line passing through the points A And IN. Let us replace the action of bonds with reactions, directing the components of the reaction along the coordinate axes. Let us denote the distance between points A And IN through A; then the equilibrium equations (7.13) and (7.14) will be written in the following form:

1) ,

2) ,

3) ,

4) ,

5) ,

6) . (7.17)

The last equation does not contain reaction forces and establishes the connection between the active forces necessary for the balance of the body. Hence, the condition for the equilibrium of a rigid body that has 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 determined 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 If there is not a spherical, but a cylindrical hinge (i.e., a bearing), which does not interfere with the longitudinal sliding of the body 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 A And IN there are cylindrical hinges (bearings), and the components of their reactions along the axis of rotation are equal to zero. Consequently, the equilibrium equations will take the form:

1) ,

2) ,

4) ,

5) ,

6) . (7.18)

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

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

3) ,

4) ,

5) ,

6) . (7.19)

The third - fifth equations can serve to determine unknown reactions, and the first, second and sixth equations represent the conditions connecting the active forces and necessary for the equilibrium of the body. Of course, for the body to balance, the following conditions must be met: , , since at the support points only reactions of the direction accepted above can occur.

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

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