Equilibrium equations for a converging spatial system of forces. Equilibrium of an arbitrary spatial system of forces is the solution to the problem. Points without taking into account the forces applied to them
Vector equilibrium conditions for an arbitrary system of forces: for the equilibrium of a system of forces applied to a solid body, it is necessary and sufficient that the main vector of the force system be equal to zero and main point system of forces relative to any center of reduction was also equal to zero. Otherwise: in order for ~0, the following conditions are necessary and sufficient:
,
or 
,
. (19)
Equilibrium conditions for a spatial system of forces in analytical form
For the equilibrium of a spatial system of forces applied to a solid body, it is necessary and sufficient that the three sums of the projections of all forces on the axis Cartesian coordinates were equal to zero and the three sums of the moments of all forces relative to the three coordinate axes were also equal to zero.
. (20)
Equilibrium conditions for a spatial system of converging forces
For the equilibrium of a spatial system of converging forces applied to a solid body, it is necessary and sufficient that the sums of the projections of forces on each of the three rectangular coordinate axes be equal to zero:
;
;
, (21)
In the case of a plane system of converging forces, one of the coordinate axes, usually 
, is chosen perpendicular to the forces, and the other two axes are chosen, respectively, in the plane of the forces. D for balance flat system converging forces acting on a solid body are necessary and sufficient so that the sum of the projections of these forces onto each of the two rectangular coordinate axes, lying in the plane of forces, were equal to zero:
;
, (22)
Equilibrium conditions for a spatial system of parallel forces
Let's direct the axis 
parallel to the forces: for the equilibrium of a spatial system of parallel forces applied to a solid body, it is necessary and sufficient that the algebraic sum of these forces be equal to zero and the sum of the moments of forces relative to two coordinate axes perpendicular to the forces are also equal to zero:
Equilibrium conditions for a plane system of forces
Let's position the axes 
And 
in the plane of action of forces.
Equilibrium conditions for a plane system of forces in the first form: for the equilibrium of a plane system of forces acting on a solid body, it is necessary and sufficient that the sums of the projections of these forces onto each of the two rectangular coordinate axes located in the plane of action of the forces are equal to zero and the sum of the algebraic moments of forces relative to any point located in the plane of action forces was also zero:
(24)
For the equilibrium of a plane system of parallel forces applied to a solid body, it is necessary and sufficient that the algebraic sum of the forces be equal to zero and the sum of the algebraic moments of the forces relative to any point located in the plane of the forces is also equal to zero:
(25)
Three-moment theorem (second form of equilibrium conditions): for the equilibrium of a plane system of forces applied to a rigid body, it is necessary and sufficient that the sums of the algebraic moments of the forces of the system relative to any three points located in the plane of action of the forces and not lying on the same straight line are equal to zero:
Third form of equilibrium conditions: for the equilibrium of a plane system of forces applied to a solid body, it is necessary and sufficient that the sums of the algebraic moments of forces relative to any two points lying in the plane of action of the forces are equal to zero and the algebraic sum of the projections of these forces onto any axis of the plane that is not perpendicular to the straight line , passing through two moment points, was also equal to zero, i.e.
Let us consider an arbitrary spatial system of forces acting on a rigid body. Let us bring this system of forces to a given center and focus on the case when the main vector and the main moment of this system of forces are equal to zero, i.e.
(1) Such a system of forces is equivalent to zero, i.e. balanced. Consequently, equalities (1) are sufficient equilibrium conditions. But these conditions are also necessary, i.e. if the system of forces is in equilibrium, then equalities (1) are also satisfied. Indeed, if the system were in equilibrium, but, for example 
That this system would have been grafted onto the resultant at the center of reduction and there would have been no equilibrium. If but Mo =**O, this system would be grafted onto the pair and there would be no balance either. The pair cannot balance each other. Thus, we have proven that for the equilibrium of an arbitrary spatial system of forces it is necessary and sufficient that the main vector and the main moment of this system relative to an arbitrarily chosen center of reduction are equal to zero. Conditions (1) are called equilibrium conditions in vector form. To obtain an analytical form of equilibrium conditions that is more convenient for practical purposes, let us project equalities (1) onto the axes of the Cartesian coordinate system. As a result we get:
(2)equilibrium conditions for a system of parallel forces in space For the equilibrium of an arbitrary spatial system of forces, it is necessary and sufficient that the sum of the projections of all forces on the coordinate axes x, y and z, as well as the sum of the moments of all forces relative to the same axes, equal zero. Let a rigid body be acted upon by spatial system parallel forces. Since the choice of axes is arbitrary, it is possible to choose a coordinate system so that one of the axes is parallel to the forces, and two
others are perpendicular to them (Fig. 1.38). With this choice of coordinate axes, the projections of each of the forces on the x and y axes and their moments relative to the z axis will always be equal to zero. This means that 
These equalities are identically satisfied, regardless of whether a given system of forces is in equilibrium or not, i.e. cease to be conditions of equilibrium. Therefore, the following equilibrium conditions will remain:
Thus, for the equilibrium of a system of parallel forces in space, it is necessary and sufficient that the sum of the projections of all forces onto the axis parallel to these forces is equal to zero and that the promises of their moments relative to each of the two coordinate axes perpendicular to the forces are also equal to zero.
17, Theorem on the equivalence of two pairs of forces in space.
Bringing a force to a given center (Poinsot method) - a force can be transferred parallel to itself to any point in the plane if you add the appropriate pair of forces, the moment of which is equal to the moment of this force relative to the point in question. Let us add two forces to the system at point A, equal in magnitude to each other and to the magnitude of the given force, directed along one straight line in opposite sides and parallel to a given force: The kinematic state has not changed (attachment axiom). The original force and one of the added forces in the opposite direction form a pair of forces. The moment of this pair is numerically equal to the moment of the initial force relative to the center of reduction. In many cases, it is convenient to represent a pair of forces with an arc arrow. Bringing a plane arbitrary system of forces to a given center - we select an arbitrary point on the plane and transfer each of the forces using the Poinsot method to this point. Instead of the original arbitrary system, we obtain a convergent system of forces and a system of pairs. The converging system of forces is reduced to a single force applied at the center of reduction, which was previously called the resultant, but now this force does not replace the original system of forces, since after the reduction a system of pairs has arisen. A system of pairs is reduced to one pair (the theorem on the addition of pairs), the moment of which is equal to the algebraic sum of the moments of the original forces relative to the center of reduction. IN general case a flat arbitrary system of forces is reduced to one force, called the main vector, and to a pair with a moment equal to the main moment of all forces of the system relative to the center of reduction: - the main vector, - the main moment. A. A. The condition for equilibrium of a flat arbitrary system of forces is the simultaneous reversal of the main vector and the main moment of the system to zero: Equilibrium equations (I form) are obtained in the form of a system of three equations from equilibrium conditions using expressions for projections of the main vector: There are two more forms of Equilibrium equations (II and III forms)
17. 
27-28. Dependence between the main moments of forces relative to two arbitrarily chosen centers of reduction. Invariants of the force system
Let this spatial system be brought to the center O, i.e.
Where 
The main moment forms a certain Angle a with the direction of the main vector (Fig. 1.32)
Let us now take a new center of reduction O1 and bring all forces to this center. As a result, we again obtain a main vector equal to the main vector R, and a new main moment determined by the formula where pk is the radius vector of the point of application of force Fk, drawn from the new center of reduction O1 (see Fig. 1.32). Main moment Mo1 relative to the new center the reduction has changed and now forms a certain angle a1 with the direction of the main vector R. Let us establish a connection between the moments Mo and Mo1. From Figure 1.32 it is clear that (3) Substituting (3) into equality (2), we obtain 
(4)Next, opening the parentheses on the right side of equality (4) and taking the common factor O1O beyond the sum sign, we have
( - projections of the main moment relative to point O onto the coordinate axes).
Bringing force to a given center.
To bring a force applied at any point of a solid body to a given center it is necessary:
1) Transfer the force parallel to itself to a given center without changing the modulus of the force.
2) At a given center, apply a pair of forces, the vector moment of which is equal to the vector moment of the transferred force relative to the new center. This pair of forces is called an adjoint pair.
The action of a force on a rigid body does not change when it is transferred parallel to itself to another point of the rigid body, if a couple of forces are added.
33 
32
34. For a plane system of parallel forces, two equilibrium equations can be drawn up. If the forces are parallel to the Y axis, then the equilibrium equations have the form.
The second equation can be constructed for any point.
35 for the equilibrium of a completely free body on which a spatial arbitrary system of forces acts, it is necessary and sufficient that the six equilibrium equations be satisfied. If a body is fixed at one point, then it has three degrees of freedom. Such a body cannot move translationally, but can only rotate around any axis, that is, around the coordinate axes. In order for such a body to be in equilibrium, it is necessary that it does not rotate, and for this it is sufficient to require that the three moment equations be equal to zero
So, in order for an absolutely rigid body with one fixed point, on which an arbitrary spatial system of forces acts, to be in equilibrium, it is necessary and sufficient that the sum of the moments of all forces relative to three mutually perpendicular axes equals zero.
Three other equations are used to determine the components of the hinge reaction at the attachment point Nx, Ny, Nz
37. A body that has two fixed points has one degree of freedom. It can only rotate around an axis passing through these two fixed points. Equilibrium will exist if the body does not rotate around this axis. Therefore, for equilibrium it is enough to require that the sum of the moments of all forces acting on the body relative to the axis passing through two fixed points is equal to zero: ∑Mxx(Fi)=0 
38/A system of bodies is several bodies connected to each other in some way. The forces acting on the bodies of the system are divided into external and internal. Internal are the forces of interaction between bodies of the same system, and external are the forces with which the bodies of a given system are acted upon by bodies that are not part of it.
If a system of bodies is in equilibrium, then we consider the equilibrium of each body separately, taking into account the internal forces of interaction between the bodies. If a flat arbitrary system is given N bodies, then 3N equilibrium equations can be compiled for this system. When solving problems on the equilibrium of a system of bodies, one can also consider the equilibrium of both the system of bodies as a whole and for any combination of bodies. When considering the equilibrium of the system as a whole, the internal forces of interaction between bodies are not taken into account on the basis of the axiom of the equality of the forces of action and reaction. Thus, there are 2 types of finding the equilibrium of systems of bodies...1sp First of all, we consider the entire structure, and then we disconnect any body from this system and consider. There is balance in it. 2sp. We divide the system into separate bodies and the composition of the equilibrium equation for each body.
Statically definable systems are systems in in which the number of unknown quantities does not exceed the number of independent equilibrium equations for a given system of forces.
Statically undefined Systems are systems in which the number of unknown quantities exceeds the number of independent equilibrium equations for a given system of forces Kst = R-Y where R is the number of reactions. Y-number of independent equations
41.After the body leaves the equilibrium position, the static friction force decreases and during movement it is called the sliding friction force, i.e., the sliding friction coefficient is slightly less than the static friction coefficient. In technical calculations, these coefficients are assumed to be equal. WITH By increasing the speed of movement, the coefficient of sliding friction for most materials decreases. The sliding friction coefficient is determined experimentally.
The sliding friction force is directed opposite to the possible movement of the body.
The friction force does not depend on the area of contacting surfaces.
The maximum frictional force is proportional to the normal pressure. Under normal pressure understand the total pressure on the entire contact area of the rubbing surfaces: Fmax=fN
43. In the presence of friction, the total reaction of a rough surface is deviated from the normal to the surface by a certain angle<р, который в случае выхода тела из равновесия достигает максимума и называется углом трения tgφ=Fmax/N Fmax=fN тогда tgφ=f
The tangent of the friction angle is equal to the friction coefficient.
A cone of friction is a cone described by the total reaction R around the direction of the normal reaction. If the coefficient of friction f is the same in all directions, then the cone of friction will be circular
For a body to balance on a rough surface, it is necessary and sufficient that the resultant of the active forces be inside the friction cone or pass along the generatrix of the cone
30. Modulus of the main vector Ro=√Rx^2+Ry^2 where Rx= ƩFkx Ry= ƩFky (Rx,Ry projections of the main vector onto the corresponding coordinate axes)
Angles formed by the main vector with the corresponding coordinate axis Сos(x^Ro)=Rx/Ro Сos(y^Ro)=Ry/Ro
Modulus of the main moment relative to the selected center of reduction O Mo√Mox^2+Moy^2 where Mox=∑Mx(Fk) Moy=∑My(Fk) Mox Moy-projections of the main moment relative to the point O on the coordinate axes)
Angles formed by the principal moment with the corresponding coordinate axes Сos(x^Mo)=Mox/Mo Сos(y^Mo)=Moy/Mo
If Ro is not=0 Mo=0 the system of forces can be replaced by one force
Ro=0 Mo not=0 the system of forces is replaced by a pair of forces
Rone=0 Mo not=0 but Ro perpendicular to Mo is replaced by one force not passing through the center of reduction
31.Flat system of forces. All forces of this system lie in one plane. Let, for example, this be the XAY plane, where A is an arbitrary reduction center. The forces of this system are not projected onto the AZ axis and do not create moments relative to the AX and AY axes, since they lie in the XAY plane (section 13). In this case the equality
Taking this into account, we obtain equilibrium conditions for a plane system of forces:
Thus, for the equilibrium of a rigid body under the action of a plane system of forces, it is necessary and sufficient that two sums of the projections of forces on the coordinate axes and the sum of the algebraic moments of all forces relative to any point in the plane be equal to zero.
39.forces acting on all points are called distributed given volume or a given part of a surface or line. Ras limited forces are characterized by intensity q, i.e. by force, due per unit volume, surface or line length. Distributed forces are usually replaced by concentrated ones.
If distributed forces act in a plane on a straight line, then they are replaced by a concentrated force as follows.
A uniformly distributed load of intensity q is replaced by a concentrated force Q =qL which is applied in the middle of the section. A uniformly distributed load refers to forces that have the same magnitudes and directions on a given area of the body.
If the distributed forces change according to a linear law
(along the triangle), then the concentrated force Q = qmaxL/2- is applied at the center of gravity of the triangle, located at a distance - from its base……………….
44.Rolling friction is the resistance to movement that occurs when bodies roll over each other. It appears, for example, between the elements of rolling bearings, between the tire of a car wheel and the road surface. As a rule, the value of rolling friction is much less than the value of sliding friction, and therefore rolling is a common type of movement in technology.
Rolling friction occurs at the interface of two bodies and is therefore classified as a type of external friction.
45.spin friction. Suppose that a heavy ball lies on a horizontal plane, we denote the center of the ball by O, and the point of contact of the ball with the plane by C. The rotation of the ball around the straight line CO is called spinning. Experience shows that if the moment of the couple that should cause the ball to spin is very small, then the ball will not spin. It follows that the action of the driving pair is paralyzed by some other pair, on the presence of which the spinning friction depends.
One method for calculating the frictional torque of a rolling bearing is to divide the frictional torque into the so-called load-independent torque M0 and the load-dependent torque M1, which are then added together to give the total torque:
46two parallel forces directed in the same direction are reduced to one force - a resultant force applied at a point dividing a straight line into distances inversely proportional to the magnitudes of the forces. Consistently adding parallel forces in pairs, we also arrive at one force - the resultant R: Since the force can be transferred along the line of its action, the point of application of the force (resultant) is essentially undefined. If all the forces are rotated by the same angle and the addition of the forces is carried out again, then we obtain a different direction of the line of action of the resultant. The point of intersection of these two lines of action of the resultants can be considered as the point of application of the resultant, which does not change its position when all forces simultaneously rotate through the same angle. This point is called the center of parallel forces. The center of parallel forces is the point of application of the resultant, which does not change its position when all forces simultaneously rotate through the same angle
47The radius vector of a point is a vector whose beginning coincides with the origin of the coordinate system, and the end with the given point.
Thus, a feature of the radius vector that distinguishes it from all other vectors is that its origin is always located at the origin point (Fig. 17).
The center of parallel forces, the point through which the line of action of the resultant system of parallel forces Fk passes for any rotation of all these forces near their points of application in the same direction and at the same angle. The coordinates of the Center of parallel forces are determined by the formulas:
where xk, yk, zk are the coordinates of the points of application of forces.
48Center of gravity of a rigid body - a point invariably associated with this body, through which the line of action of the resultant forces of gravity of the particles of the body passes at any position of the body in space. In this case, the gravity field is considered homogeneous, i.e. the gravitational forces of the particles of the body are parallel to each other and remain constant during any rotation of the body. Center of gravity coordinates:
; ; , where Р=åр k, x k,y k,z k – coordinates of points of application of gravity forces р k. The center of gravity is a geometric point and can lie outside the body (for example, a ring). Center of gravity of a flat figure:
DF k – elementary area, F – area of the figure. If the area cannot be divided into several finite parts, then . If a homogeneous body has an axis of symmetry, then the center of gravity of the body is on this axis.
49 Solving problems to determine the position (coordinates) of the center of gravity of a homogeneous plate, a system of bodies located on a plane or space comes down to drawing up equations and further inserting known numerical data into it and calculating the result:
Those. it is necessary to break the system into components and find the positions of the center of gravity of these component elements. Calculate the mass of the components, expressing it through the specific density - linear, volumetric or surface, depending on the type of system presented. At the end of the solution, the specific density will be reduced, so do not be embarrassed to enter it (as a rule, it is not given, but the text of the problem indicates that the plate, rods, and slab are homogeneous). Of the features of this task, two things should be noted: 1) determining the center of gravity of a component of a rectangular, square shape or rod, circle is not difficult - the center of gravity of such figures is in the center.
50. circular sector: ; Triangle. Dividing the triangle into thin lines,
parallel to each of its sides determine that since the center
the gravity of each line lies on its geometric center (at the center
symmetry), then the center of gravity of the triangle lies at the intersection of its
median The intersection point of the medians divides them in the ratio (2:1).
Circular sector (Figure 54). The center of gravity lies on the axis
symmetry. By dividing a circular sector into elementary triangles
determine the arc formed by the centers of gravity of the triangles. Radius
arc is equal to 2/3 of the sector radius. Thus, the coordinate of the center
the gravity of the circular sector is determined
expression xC = sin α.
51Hemisphere. The center of gravity lies on the axis of symmetry at a distance
3/8 from the base.
Pyramid (cone) (Figure 55).
The center of gravity lies on the line
connecting the vertex to the center
gravity of the base at a distance of ¾ from
Arc of a circle The center of gravity lies on the axis of symmetry and has
coordinates xC = sin α ; уС = 0 .
Kinematics
1Kinematics, a branch of theoretical mechanics, studies the movement of material bodies without being interested in the reasons that cause or change this movement. For it, only physical validity and mathematical rigor within the framework of the accepted models are important. Kinematics problems To set the movement of a material point (system) means to give a way to determine the position of a point (all points forming a system) at any moment in time.
The tasks of kinematics are to develop methods for specifying the movement of a point (system) and methods for determining the speed, acceleration of a point and other kinematic quantities of the points that make up a mechanical system. point trajectory
To specify the movement of a point means to specify its position at each moment of time. This position must be determined, as already noted, in some coordinate system. However, for this it is not always necessary to specify the coordinates themselves; you can use quantities that are somehow related to them. Below are three main ways to specify the movement of a point.
1. The natural way. This method is used if the trajectory of the point is known. A trajectory is a set of points in space through which a moving material particle passes. This is the line that she draws in space. With the natural method, you need to set (Fig. 1):
a) trajectory of movement (relative to any coordinate system);
b) an arbitrary point on it, zero, from which the distance S to the moving particle along the trajectory is measured;
c) positive direction of reference S (when point M is shifted in the opposite direction, S is negative);
d) start of time t;
e) function S(t), which is called the law of motion**) of the point.
2. Coordinate method. This is the most universal and comprehensive way to describe movement. It assumes the task:
a) coordinate systems (not necessarily Cartesian) q1, q2, q3;
b) start of time t;
c) the law of motion of a point, i.e. functions q1(t), q2(t), q3(t).
When talking about the coordinates of a point, we will always mean (unless otherwise stated) its Cartesian coordinates.
3. Vector method. The position of a point in space can also be determined by a radius vector drawn from a certain origin to a given point (Fig. 2). In this case, to describe the movement you need to set:
a) the origin of the radius vector r;
b) start of time t;
c) the law of motion of the point r(t).
Since specifying one vector quantity r is equivalent to specifying its three projections x, y, z on the coordinate axes, it is easy to move from the vector method to the coordinate one. If we introduce unit vectors i, j, k (i = j = k = 1), directed respectively along the x, y and z axes (Fig. 2), then, obviously, the law of motion can be represented in the form*)
r(t) = x(t)i +y(t)j+z(t)k. (1)
The advantage of the vector form of recording over the coordinate form is compactness (instead of three quantities one operates with one) and often greater clarity.
Example. A small ring M is put on a fixed wire semicircle, through which another straight rod AB passes (Fig. 3), uniformly rotating around point A (= t, where = const). Find the laws of motion of the ring M along the rod AB and relative to the semicircle.
To solve the first part of the problem, we will use the coordinate method, directing the x-axis of the Cartesian system along the rod and choosing its origin at point A. Since the inscribed AMS is a straight line (as based on the diameter),
x(t) = AM = 2Rcos = 2Rcoswt,
where R is the radius of the semicircle. The resulting law of motion is called a harmonic oscillation (this oscillation will obviously continue only until the moment the ring reaches point A).
We will solve the second part of the problem using the natural method. Let us choose the positive direction of counting the distance along the trajectory (semicircle AC) counterclockwise (Fig. 3), and zero coinciding with point C. Then the length of the arc SM as a function of time will give the law of motion of point M
S(t) = R2 = 2R t,
those. the ring will move uniformly around a circle of radius R with an angular velocity of 2. As is clear from the examination,
the zero of the time count in both cases corresponded to the moment when the ring was at point C.
2.Vector method of specifying the movement of a point
The speed of the point is directed tangentially to the trajectory (Fig. 2.1) and is calculated, according to (1.2), using the formula
We combine the origin of coordinates with the point of intersection of the lines of action of the forces of the system. We project all forces onto the coordinate axes and sum up the corresponding projections (Fig. 7.4). We obtain the projections of the resultant on the coordinate axes:
The module of the resultant system of converging forces is determined by the formula
The direction of the resultant vector is determined by the angles
Arbitrary spatial system of forces
Bringing an arbitrary spatial system of forces to the center of O.
A spatial system of forces is given (Fig. 7.5, a). Let's bring it to the center O.
Forces must be moved in parallel, and a system of pairs of forces is formed. The moment of each of these pairs is equal to the product of the force modulus and the distance to the center of reduction.
A beam of forces arises at the reduction center, which can be replaced by the total force (main vector) F GL (Fig. 7.5, b).
The moments of pairs of forces can be added, obtaining the total moment of the system M ch (main moment).
Thus, an arbitrary spatial system of forces is reduced to the main vector and the main moment.
The main vector is usually decomposed into three components directed along the coordinate axes (Fig. 7.5, c).
Usually the total moment is decomposed into components: three moments relative to the coordinate axes.
The absolute value of the main vector (Fig. 7.5b) is equal to
The absolute value of the main moment is determined by the formula.
Equilibrium equations for a spatial system of forces
At equilibrium F ch = 0; M ch = 0. We obtain six equilibrium equations:
The six equilibrium equations of the spatial system of forces correspond to six independent possible movements of the body in space: three movements along the coordinate axes and three rotations around these axes.
Examples of problem solving
Example 1. On a cube-shaped body with an edge A= 10 cm three forces act (Fig. 7.6). Determine the moments of forces relative to the coordinate axes coinciding with the edges of the cube.
Solution
1. Moments of forces about the axis Oh:
2. Moments of forces about the axis Oh.
Example 2. Two wheels are fixed on a horizontal shaft, g 1 = 0.4 m; g 2 = 0.8 m. Other dimensions are in Fig. 7.7. Force is applied to wheel 1 F 1, to wheel 2 - power F 2= 12 kN, F 3= 4kN.
Define strength F 1 and reactions in the hinges A And IN in a state of balance.
Let us remind you:
1. In equilibrium, six equilibrium equations are satisfied.
Moment equations should be written relative to the supports A and B.
2. Powers F 2 \\O x; F 2\\Oy;F 3\\Oy.
The moments of these forces relative to the corresponding axes are equal to zero.
3. 
The calculation should be completed by verification using additional equilibrium equations.
Solution
1. Determine strength F\, having composed the equation of moments of forces relative to the Oz axis:
2. Determine reactions in support A. There are two reaction components acting on the support ( Y A ; X A ).
We compose the equation of moments of forces about the axis Oh"(in support IN).
Rotation around an axis Oh" doesn't happen:
The minus sign means that the reaction is directed in the opposite direction.
Rotation around an axis Oh" does not happen, we draw up an equation for the moments of forces relative to the axis Oh"(in support IN):
3. Determine the reactions in support B. Two components of the reaction act on the support ( X B , Y B ). We compose the equation of moments of forces about the axis Oh(support A):
We compose the equation of moments about the axis Oh(support A):
4.Check. We use projection equations:
The calculation was done correctly.
Example 3. Determine the numerical value of the force P 1 , at which the shaft Sun(Fig. 1.21, A) will be in equilibrium. At the found force value P 1 determine support reactions.
Forces acting on gears R
And P 1
directed tangentially to the initial circles of the wheels; strength T
And T 1
- according to the radii of the wheels; strength A 1 parallel to the shaft axis. T = 0.36P, 7T 1 = P 1; A 1 = 0.12P 1.
Solution
The shaft supports shown in Fig. 1.21, a, should be considered as spatial hinge supports that prevent linear movements in the directions of the axes And And v(the selected coordinate system is shown in Fig. 1.21, b).
We free the shaft from the connections and replace their action with reactions V V, N V, V C, N C (Fig. 1.21, b). We have obtained a spatial system of forces, for which we draw up equilibrium equations using the selected coordinate system (Fig. 1.21.6):
Where A 1*1.25D/2 - moment about the axis And strength A 1, applied to the right gear.
Moments about the axis And strength T 1 And A 1(applied to the middle gear), P 1 (applied to the right gear) and P are equal to zero, since the forces P, T 1, P 1 are parallel to the axis And, and force A 1 crosses the axis And.
where V C = 0.37P;
where V B =0.37P.
hence the reactions V B And V C defined correctly;
Where A 1 * 1.25D/2- moment about the axis v strength A 1, applied to the middle gear.
Moments about the axis v forces T, P 1 (applied to the middle gear), A 1 And T 1(applied to the right gear) are equal to zero, since the forces T, R 1, T 1 parallel to the axis v, strength A 1 crosses the axis v.
whence H C = 0.81P;
from where H C = 1.274P
Let's create a verification equation:
hence the reactions N V And N S defined correctly.
In conclusion, we note that the support reactions turned out to have a plus sign. This indicates that the selected directions V B, N B, V C And N S coincide with the actual directions of bond reactions.
Example 4. The pressure force of the steam engine connecting rod P = 25 kN is transmitted to the middle of the crankshaft journal at the point D at an angle α
= 30° to the horizontal with the knee cheeks vertical (Fig. 1.22). A belt drive pulley is mounted on the end of the shaft. The tension of the driving branch of the belt is two times greater than that of the driven branch, i.e. S 1 = 2S 2 . Flywheel gravity G = 10 kN.
Determine the tension of the belt drive branches and the reactions of the bearings A And IN, neglecting the mass of the shaft.
Solution
We consider the equilibrium of a horizontal crankshaft with a pulley. We apply the specified forces in accordance with the problem conditions P, S 1, S 2 And G . We free the shaft from the supporting fasteners and replace their action with reactions V A, N A, V B And N V. We select the coordinate axes as shown in Fig. 1.22. Hinged A And IN no reactions occur along the axis w, since the tension of the belt branches and all other forces act in planes perpendicular to this axis.
Let's create the equilibrium equations:
In addition, according to the conditions of the problem, we have another equation
So there are six unknown forces here S 1, S 2, N A, V A, N B And V B and six equations connecting them.
Equation of projections onto an axis w in the example under consideration turns into the identity 0 = 0, since all forces lie in planes perpendicular to the axis w.
Substituting S 1 =2S 2 into the equilibrium equations and solving them, we find:
Reaction value N V It turned out with a minus sign. This means that in reality its direction is opposite to that assumed in Fig. 1.22.
Test questions and assignments
1. Write down the formulas for calculating the main vector of a spatial system of converging forces.
2. Write down the formula for calculating the main vector of a spatial system of arbitrarily located forces.
3. Write down the formula for calculating the main moment of a spatial system of forces.
4. Write down the system of equilibrium equations for the spatial system of forces.
5. Which equilibrium equation should be used to determine the reaction of the rod R 1 (Fig. 7.8)?
6. Determine the main moment of the force system (Fig. 7.9). The reference point is the origin of coordinates. The coordinate axes coincide with the edges of the cube, the edge of the cube is 20 cm; F 1 - 20kN; F 2 - 30kN.
7. Determine the Xb reaction (Fig. 7.10). The vertical axis with the pulley is loaded by two horizontal forces. Powers F 1 And F 2 parallel to the axis Oh. AO = 0.3 m; OB= 0.5 m; F 1 = 2kN; F 2 = 3.5 kN.
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Recommendation. Create an equation for moments about the axis Oh" at the point A.
8. Answer the test questions.
It was established above (6.5, case 6) that
Considering that, 
, let us project formulas (6.18) onto Cartesian coordinate axes. We have analytical form of equilibrium equations for an arbitrary spatial system of forces:
(6.19)
The last three equations occur due to the fact that the projection of the moment of force relative to a point onto the axis that passes through this point is equal to the moment of force relative to the axis (formula (6.9)).
Conclusion arbitrary spatial system of forces, which is applied to a solid body, we must compose six equilibrium equations(6.19), therefore we have the opportunity to determine using these equations six unknown quantities.
Consider the case spatial system of parallel forces. We choose the coordinate system so that the axis Oz was parallel to the lines of action of the forces (Fig. 6.11).
This leaves three equations:
Conclusion. When solving balance problems parallel spatial system of forces, which is applied to a solid body, we must compose three equilibrium equations and with the help of these equations we have the opportunity determine three unknown quantities.
At the first lecture in the “Statics” section, we found out that there are six types of force systems, which may be encountered in your practice of engineering calculations. In addition, there are two possibilities for arranging pairs of forces: in space and in a plane. Let us summarize all the equilibrium equations for forces and for pairs of forces into one table (Table 6.2), in which in the last column we note the number of unknown quantities that the system of equilibrium equations will allow us to determine.
Table 6.2 – Equilibrium equations for different systems of forces
| Type of force system | Equilibrium equations | Number of unknowns to be determined | |
Convergent flat  | |||
Parallel flat (axis 0 at)
 |  t. A 0xy
| ||
Arbitrary flat (in the 0xy plane)  |  t. A– arbitrary, belonging to the plane 0xy
|
Continuation of Table 6.2
Continuation of Table 6.2
Questions for self-control on topic 6
1. How to find the moment of force about an axis?
2. What relationship exists between the moment of a force relative to a point and the moment of the same force relative to the axis that passes through this point?
3. In what cases is the moment of force about the axis equal to zero? And when is it greatest?
4. In what cases is a system of forces reduced to a resultant?
5. In what case is the spatial system of forces given:
– to a pair of forces;
– to the dynamic screw?
6. What is called an invariant of statics? What static invariants do you know?
7. Write down the equilibrium equations for an arbitrary spatial system of forces.
8. Formulate a necessary and sufficient condition for the equilibrium of a parallel spatial system of forces.
9. Will the main vector of the force system change when the center of gravity changes? And the main point?
Topic 7. FARMS. DEFINITION OF EFFORT
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 equilibrium 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.
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.
