Equilibrium equations for a spatial system of forces. Equilibrium equations for plane and spatial systems of forces. Test questions and assignments
That., for the equilibrium of an arbitrary spatial system forces are necessary and sufficient for the algebraic sum of the projections of all these forces onto each of the three arbitrarily chosen coordinate axes to be equal to zero and for the algebraic sum of their moments relative to each of these axes to also be zero.
Conditions (1.33) are called equilibrium conditions of an arbitrary spatial system of forces in analytical form.
Equilibrium conditions for a spatial system of parallel forces. If the lines of action of all the forces of a given system of forces are located in different planes and are parallel to each other, then such a system of forces is called spatial system of parallel forces.
Using the equilibrium conditions (1.33) of an arbitrary spatial system of forces, one can find the equilibrium conditions of a spatial system of parallel forces. (The equilibrium conditions we derived earlier for plane and spatial systems of converging forces, arbitrary flat system forces and a plane system of parallel forces could also be obtained using the equilibrium conditions (1.33) of an arbitrary spatial system of forces).
Let a spatial system of parallel forces act on a solid body (Figure 1.26). Since the choice of coordinate axes is arbitrary, it is possible to choose coordinate axes so that the axis z was parallel to the forces. With this choice of coordinate axes, the projections of each of the forces on the axis X And at and their moments about the axis z will be equal to zero, and therefore the equalities , and are satisfied regardless of whether this system forces are in equilibrium or not, and therefore cease to be conditions of equilibrium. Therefore, system (1.33) will give only three equilibrium conditions:
Hence, for the equilibrium of a spatial system of parallel forces, it is necessary and sufficient that the algebraic sum of the projections of all forces onto the axis parallel to these forces equals zero and that the algebraic sum of their moments relative to each of the two coordinate axes perpendicular to these forces also equals zero.
1. Select a body (or point) whose equilibrium should be considered in this problem.
2. Free the selected body from bonds and depict (arrange) all active forces and reaction forces of discarded bonds acting on this body (and only on this body). A body freed from connections, with a system of active and reaction forces attached to it, should be depicted separately.
3. Write equilibrium equations. To draw up equilibrium equations, you must first select the coordinate axes. This choice can be made arbitrarily, but the resulting equilibrium equations will be solved more easily if one of the axes is directed perpendicular to the line of action of some unknown reaction force. The solution of the resulting equilibrium equations should, as a rule, be carried out to the end in general view(algebraically). Then, for the required quantities, formulas will be obtained that allow one to analyze the results found; numerical values of the found quantities are substituted only into the final formulas. Equilibrium equations are compiled at analytical method solving problems on the equilibrium of a system of converging forces. However, if the number of converging forces whose equilibrium is considered is three, then it is convenient to apply the geometric method for solving these problems. The solution in this case comes down to the fact that instead of the equilibrium equations of all acting forces (active and reaction bonds), a force triangle is constructed, which, based on the geometric condition of equilibrium, must be closed (the construction of this triangle should begin with a given force). By solving the force triangle, we find the required quantities.
Dynamics
To understand the dynamics section, you need to know the following information. From mathematics - the scalar product of two vectors, differential equations. From physics – the laws of conservation of energy and momentum. Oscillation theory. It is recommended to review these topics.
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 of a rigid body 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 from general drawing It is difficult to see 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 (of which one is parallel to any coordinate axis), and then use Varignon’s theorem (see problem 36). In addition, you can calculate the 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 in the form of 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
20. Condition for equilibrium of a spatial system of forces:
21. Theorem about 3 non-parallel forces: The lines of action of three non-parallel mutually balancing forces lying in the same plane intersect at one point.
22. Statically definable problems- these are problems that can be solved using rigid body statics methods, i.e. problems in which the number of unknowns does not exceed the number of force equilibrium equations.
Statically indeterminate systems are systems in which the number of unknown quantities exceeds the number of independent equilibrium equations for a given system of forces
23. Equilibrium equations for a plane system of parallel forces:
AB is not parallel to F i
24. Cone and angle of friction: The limiting position of active forces under the influence of which equality can occur describes friction cone with angle (φ).
If the active force passes outside this cone, then equilibrium is impossible.
The angle φ is called the friction angle.
25. Indicate the dimension of the friction coefficients: the coefficients of static friction and sliding friction are dimensionless quantities, the coefficients of rolling friction and spinning friction have the dimension of length (mm, cm, m).m.
26. Basic assumptions made when calculating flat statically determined trusses:-truss rods are considered weightless; - fastening of rods in hinged truss nodes; -external load applied only at truss nodes; - the rod falls under the connection.
27. What is the relationship between the rods and nodes of a statically determinate truss?
S=2n-3 – simple statically definable truss, S-number of rods, n-number of nodes,
if S<2n-3 –не жесткая ферма, равновесие возможно, если внешние силы будут одинаково соотноситься
S>2n-3 – statically indeterminate truss, has extra connections, + calculation of deformation
28. A statically determinate truss must satisfy the condition: S=2n-3; S is the number of rods, n is the number of nodes.
29. Knot cutting method: This method consists of mentally cutting out the nodes of the truss, applying the corresponding external forces and reactions of the rods to them, and creating equilibrium equations for the forces applied to each node. It is conventionally assumed that all the rods are stretched (the reactions of the rods are directed away from the nodes).
30. Ritter method: We draw a secant plane that cuts the truss into 2 parts. The section must begin and end outside the truss. You can choose any part as an object of equilibrium. The section passes along the rods, and not through the nodes. The forces applied to an object of equilibrium form an arbitrary system of forces, for which 3 equilibrium equations can be drawn up. Therefore, we carry out the section so that no more than 3 rods are included in it, the forces in which are unknown.
A feature of the Ritter method is the choice of the form of the equation in such a way that each equilibrium equation includes one unknown quantity. To do this, we determine the positions of the Ritter points as the points of intersection of the lines of action of two unknown forces and write down the equations of moments rel. these points.
If the Ritter point lies at infinity, then as an equilibrium equation we construct equations of projections onto the axis perpendicular to these rods.
31. Ritter point- the point of intersection of the lines of action of two unknown forces. If the Ritter point lies at infinity, then as an equilibrium equation we construct equations of projections onto the axis perpendicular to these rods.
32. Center of gravity volumetric figure:
33. Center of gravity of a flat figure:
34. Center of gravity of the rod structure:
35. Center of gravity of the arc:
36. Center of gravity of a circular sector:
37. Center of gravity of the cone:
38. Center of gravity of the hemisphere:
39. Method of negative values: If a solid has cavities, i.e. cavities from which their mass is taken out, then we mentally fill these cavities to a solid body, and determine the center of gravity of the figure by taking the weight, volume, area of the cavities with the “-” sign.
40. 1st invariant: The 1st invariant of the force system is called the main vector of the force system. The main vector of the force system does not depend on the center of reduction R=∑ F i
41. 2nd invariant: Dot product of the principal vector by main point system of forces for any center of reduction is a constant value. 
42. In what case is a system of forces driven to a power screw? In the event that the main vector of the force system and its main moment relative to the center of reduction are not equal to zero and are not perpendicular to each other, given. the system of forces can be reduced to a power screw.
43. Equation of the central helical axis:
44.
M x - yR z + zR y = pR x ,
M y - zR x + xR z = pR y ,
M z - xR y + yR x = pR z
45. Moment of a couple of forces as a vector- this vector is perpendicular to the plane of action of the pair and is directed in the direction from where the rotation of the pair is visible counterclockwise. In modulus, the vector moment is equal to the product of one of the forces of the pair and the shoulder of the pair. Vector moment of a pair of phenomena. a free vector and can be applied to any point of a rigid body.
46. The principle of release from ties: If bonds are discarded, then they must be replaced by reaction forces from the bond.
47. Rope polygon- This is a construction of graphostatics, which can be used to determine the line of action of the resultant plane system of forces to find the reactions of supports.
48. What is the relationship between the rope and power polygon: To find unknown forces graphically in the force polygon we use an additional point O (pole), in the rope polygon we find the resultant, moving which into the force polygon we find the unknown forces
49. Condition for equilibrium of systems of pairs of forces: For equilibrium of pairs of forces acting on a solid body, it is necessary and sufficient that the moment of equivalent pairs of forces be equal to zero. Corollary: To balance a pair of forces, it is necessary to apply a balancing pair, i.e. a pair of forces can be balanced by another pair of forces with equal moduli and oppositely directed moments.
Kinematics
1. All methods of specifying the movement of a point:
natural way
coordinate
radius vector.
2. How to find the equation for the trajectory of a point’s movement using the coordinate method of specifying its movement? In order to obtain the trajectory equation for the motion of a material point, using the coordinate method of specifying, it is necessary to exclude the parameter t from the laws of motion.
3. Acceleration of a point at coordinates. method of specifying movement:
2 dots above the X
above y 2 dots
4. Acceleration of a point using the vector method of specifying motion:
5. Acceleration of a point at natural way movement tasks:
=
=
*
+v*
; a=
+
;
*
; v*
.
6. What is the normal acceleration equal to and how is it directed?– directed radially towards the center,
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 create 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 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.
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 
, 
That's why 
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 on 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) 
,
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) 
,
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) ,
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) 
,
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.
We find the projections of the main vector using formulas (4.4):
, 
, 
.
Its modulus is . The direction cosines will be
, ;
, ;
, 
.
The main vector is shown in Fig.
,
and the modulus of the main moment according to formula (4.8)
Now we determine the direction cosines of the main moment:
, 
;
, 
.
The main point is shown in Fig. The angle between vectors and is calculated using formula (4.11) and
We find the boundaries of the desired area from the conditions:
,
.
From here we find
,
.
In Fig. the desired region, constructed at , is shaded. The entire surface of the plate will be safe.
