equalness of the body with a diversified system of forces. Write the rіvnyannya rіvnovagi dovіlnoї prostorovoї system of forces. Center of gravity of shear structure
As a result, the system of forces is changing from equal, її head vector and head moment to zero:
Number of vector equalities is produced up to the next six scalar equalities:
yaki are called the minds of the equal expanse of a vast system of forces.
The first three minds show the equality of zero of the head vector, the next three - the equality of zero to the head moment of the system of forces.
In the minds of a jealous person, everything is to be blamed fiery forces- as active (set), and reactions of calls. Rest far behind the unknown, and the minds of the jealous become the equals of the appointment of the unknown - the equals of the jealous.
If the maximum number of equals is more than six, then the task of equalizing the body under a sufficiently large system of forces can be assigned to six unknown reactions. For a large number of unknown leaders, they become statically insignificant.
And one more respect. If the head vector and head moment if the center O reaches zero, then the stench reaches zero unless there is some other center. Tse directly exclaims the material about changing the center of the attraction (to bring it independently). Otzhe, as you know equal bodies are fixed in one coordinate system, the stench is fixed in any other non-robust coordinate system. Otherwise, it seems that the choice of coordinate axes pіd hour ordering rіvnyan іvnovagi as a whole is more than enough.
Rectangular slab (Fig. 51 a) is flattened in a horizontal position by a spherical hinge O, a bearing A and a cable BE, and the points are on the same vertical. At point D to the slab, a force was applied perpendicular to the side OD and heeled up to the slab area under the cut of 45 °. Calculate the tension of the cable and the reaction of the supports at the points of Vin A, as well as i.
To complete the task, we look at the equal plate. Before the active forces P, G, we add the reaction of the ties - the warehouse reaction of the spherical hinge, the reaction of the bearing, the reaction of the cable. The coordinate axes Oxyz are entered at once (Fig. 51 b). It can be seen that the collection of forces has been cut off, establishing a fairly spacious system, de force is unknown.
For the signing of the uninhabited, we become jealous of the jealousy.
We start from equal projections of forces on the whole:
It is understandable that the projection of the calculation is calculated in two steps; the projection of the force T on the plane, distance, projecting on the axis (more on the axis, parallel), is known (div. Fig. 51,b):
In this way of underwire design, it is manually corristuvatsya, if the line of force and that all do not change. We add up:
Rivnyannya momentіv forces shdo osі maє vglyad:
Moments of forces at equal days, shards of force or force change all x (), or parallel ї th. In both vibrations, the moment of force should be equal to zero (div. p. 41).
Calculation of the moment of force is often easier, as the force is laid out in the same order of warehouses and speeded up by the Varignon theorem. At to this particular type tse manually zrobiti for strength. Laying out її on horizontal and vertical warehouses, we can write.
Let's take a look at the vast expanse of the system of forces, like a solid body. Let's direct the system of forces to the given center and succumb to that drop, if the head vector and moment of the system of forces are equal to zero, then.
(1) Such a system of forces is equivalent to zero, that is. vrіvnovazhen. Otzhe, equanimity (1) є enough minds jealous. Ale tsi mind so necessary, tobto. if the system of forces is known in equivalence, then equivalence (1) is also victorious. 
then the given system would be pinned to a level of equality in the center of the given system, and there would be no equanimity. Yakbi alé Mo =**Oh, given the system was cleaved to a bet, and it was not possible for a bet to bet against one another. In this way, we have brought, that for a fairly sufficient space system of forces it is necessary and sufficient, that the head vector and the head moment of the system should be equal to zero for a sufficiently chosen center of reduction. Washes (1) are called the minds of equal vector form. For otrimannya zruchnіshoї for practical purposes of the analytical form of the minds of equal projected equality (1) on the axis of the Cartesian coordinate system. As a result, we take:
(2)wash the equal systems of parallel forces in space For a fairly spacious system of forces to be equal, it is necessary and sufficient that the sum of the projections of all forces on the x, y and z coordinate axes, as well as the sum of the moments of all forces of all these axes, is equal to zero. Let it be on firm body diє space system parallel forces. Scales of the choice of axes are sufficient, you can choose the coordinate system so that one of the axes is parallel to the forces, and two
others are perpendicular (Fig. 138). With such a choice of coordinate axes, the projection of the skin forces on the x and y axes and the x moments should be equal to zero. Tse means what 
Numbers of equivalence are also victorious, regardless of what, the system of forces is changed from equivalence to that. cease to be jealous minds. Therefore, think jealously to get rid of like this:
In this way, for the equalization of the system of parallel forces in the space, it is necessary and sufficient that the sum of the projections of the forces on the whole, parallel to the forces, equals zero and that the sum of the sulima of the moment in the skin of the two coordinate axes perpendicular to the forces is also equal to zero.
17, Theorem about the equivalence of 2 pairs of space force.
Bringing the force to the given center (Poinsot's method) - the force can be transferred parallel to itself to a point on the plane, in order to add a pair of forces, the moment of which is closer to the moment of the force to the point that is being looked at. Dodamo to the system at point A, the motion of force, equal to the value between itself and the value of the given force, straightened along one straight line at the opposite side of that parallel to the given force: The external force, that one of the additional forces, is directly directed to establish a pair of forces. The moment of the wager is numerically equal to the moment of the external force like the center of the reduction. In rich vipads, a couple of forces are visualized by hand with an arc arrow. Bringing a flat enough system of forces to a predetermined center - we choose a sufficient point on the plane and the skin forces are transferred by the Poinsot method to that point. The replacement of the vyhіdnoї dovіlnoї system is taken away by the system of forces and the system of pairs. The system of forces to go down is reduced to one force applied in the center of the reduction, as previously it was called equal, but now the force does not replace the external system of forces, the shards after the reduction were called the system of pairs. The system of pairs can be reduced to one pair (theorem about folding pairs), the moment is equal to the sum of the algebra of momentum of the external forces to the center of the reduction. In a zagalny slope of a flat, a fairly system of forces is reduced to one force, as it is called the head vector i to a bet with a moment equal to the head moment of all forces of the system at the center of the reduction: - head vector, - head moment. A. A. Mindfulness of the plane doubly system of forces є one-hour reversal of the head vector and the head moment of the system to zero: The equalization of the force (I form) appears at the sight of the system of three equalities from the minds of the equalities of the victories of the head vector for the projections of the head vector: that III form)
17. 
27-28. Fall between the main moments of forces for two fairly chosen centers of reduction. Invariant systems of forces
Let the open space system be brought to the center Oh, tobto.
de 
The head moment is controlled directly by the head vector deaky Kut (Fig. 1.32)
Let's take a new center of reduction O1 and bring all the forces to the center. As a result, we take a new head vector, which is equal to the head vector R, and a new head moment, which is defined by the formula de pk - the radius vector of the point of reporting the force Fk, passing from the new center of reduction O1 (div. Fig. 1.32). The head moment Mo1 is similar to the new center of reduction and now satisfies the direct head vector R and cut a1. Let us establish a link between the moments Mo and Mo1. From the little one 1.32 it is clear that (3) Substituting (3) for equanimity (2), we take 
4
(- Projections of the head moment about the point Pro on the coordinate axis).
Bringing the force to the given center.
In order to bring the force applied at any point of a solid body to the given center, it is necessary:
1) Transfer the force parallel to yourself to the task center without changing the force module.
2) At the given center, report a pair of forces, the vector moment of which is equal to the vector moment of force, which is transferred to the new center. Qiu pair of forces is called an advent couple.
Diya forces on a solid body do not change when transferring it parallel to itself to the other point of a solid body, just to add a couple of forces.
33 
32
34. For a flat system of parallel forces, two equal equals can be added. if the forces are parallel to the Y axis, then the equalness of the equal may look.
Another equal can be put together any way you like.
35 for the equal of an absolutely free body, on the yak of space there is a sufficient system of forces, necessary and sufficient, so that six equal equals were victorious. Although the body is fixed in one point, it has three steps of freedom. Step by step, such a body cannot collapse, but it can only turn around on some kind of axis, that is, on some coordinate axes. In order for such a body to be in equal time, it is necessary, so that it does not turn around, and for whom it is enough to crave equalness to zero three equal moments
Also, in order for the body to be absolutely rigid with one fixed point, on the yak, there is a fairly spacious system of forces, it was equal, it was necessary and sufficient, so that the sum of the moments in the forces of three mutually perpendicular axes was equal to zero.
Three other levels serve to determine the storage reaction of the hinge at the point of fixing Nx, Ny, Nz
37. A body that can have two fixed points, can have one step of freedom. It can only wrap around the axis, so that it can pass through two fixed points. It is enough for jealousy to be thirsty, so that the sum of the moments of all forces, which is on the body, which is the axis, which can pass through two fixed points, is equal to zero: ∑Mxx(Fi)=0 
38 / System tіl є kіlka tіl, z'ednah mizh themselves like a rank. The forces that are on the body of the system are subdued on the outside and inside. The internal forces are called mutually between the bodies of one and the same system, and the external forces are called the forces that are on the body of a given system to develop a body, but do not enter before it.
As the system of bodies is rebuying from the equal, then we look at the equalness of the skin body of the okremo, protecting the internal forces between the bodies. As a flat full system is given N tіl, then the lines of the system can be folded into 3N equal equals. When rozv'yazannі tasks on the equalization of the system tіl, it is also possible to look at the rіvnovag like the system tіl zagalom, so for whether or not they are better. At a glance, the system’s equivalence with a flame of internal forces in mutual modality between bodies does not stand up against the axioms about the equivalence of the forces of di and opposing forces. In this order, there are 2 types of knowledge of the rіvnovagi systems tіl ... 1sp In the first line, the whole structure is considered. and then let's look at the whole system and look at it. jealous in the new. 2sp. razchlenovuєmo sis-mu on the rim of the body and comp.
Static primary systemic systems, in which the number of unknown values does not outweigh the number of independent equalities of the given system of forces.
statically undefined. Systems are systems in which the number of unknown quantities exceeds the number of independent equals equal to given systems of forces Kct=R-Y de R-number of reactions. Y-number of independent regions
41. When the body exits from the position of the equal, the strength of the rub is calmly changing and in Russia it is called the force of the rub of forging, so that the coefficient of the rub of the forge is less for the coefit of the rub of peace. In technical rozrahunka, it is accepted that these coefficients are equal. W for more materials, the coeficient of forging is changed. The coefficient of the tertya forging is determined experimentally.
The strength of the forging is straightened out to the body's ability to move.
The strength of rubbing does not lie on the surface of the surface, which stick together.
Maximum strength rubbing is proportional to the normal vice. Under the normal grip, a new grip is formed over the entire area of the surface to be rubbed: Fmax=fN
43. For the obviousness of rubbing the surface, the reaction of the short surface of the breath is normal to the surface on the deky kut<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 kuta is increasing the coefficient of the coefficient.
A cone is called a rubbing cone, descriptions of the overall reaction R are directly normal reactions. If the rubbing coefficient f is the same for all the straight lines, then the rubbing cone will be circular
For the equalization of the body on the short surface, it is necessary and sufficient, so that the equally active force was in the middle of the cone, rubbing or passed along the satisfying cone
30. Head vector module Ro=√Rx^2+Ry^2 de Rx= Fkx Ry = Fky
Kuti utavlenі head vector іz vіdpovіdnoyu vіssyu coordinates Сos(x^Ro)=Rx/Ro Сos(y^Ro)=Ry/Ro
Modulus of the head moment inverse to the center of the reduction Pro Mo√Mox^2+Moy^2 de Mox=∑Mx(Fk) Moy=∑My(Fk)
Kuti utvorenі head moment іz corresponding axes of coordinates Сos(x^Mo)=Mox/Mo Сos(y^Mo)=Moy/Mo
Where 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
Ronot=0 Mo not=0 ale Ro perpendicularlyMo is replaced by one force that does not pass through the center of the reduction
31. Flat system of forces. All forces of the system lie at the same plane. Come on, for example, tse will be the XAY area, de A is a pretty center of the reduction. The forces of the system on the entire AZ are not designed and unless the axes AX and AY are created, the rocks lie at the XAY plane (p. 13). At what point does jealousy win
Vrakhovuychi tse, it is necessary to wash the mind for a flat system of forces:
In this way, for the equalization of a solid body under a plane system of forces, it is necessary and sufficient, so that two sums of projection forces on the coordinate axes and the sum of moments in the algebra of all forces can be equal to zero points of the plane.
39. different names given oblige or given parts are superficial, or lines. Ras boundary forces are characterized by intensity q, tobto by force, fall down per unit volume, the surface of the line. Rozpodіlenі forces ring out to replace the serendipity.
As if the forces were divided near the flat on a straight line, they will be replaced by the guarded force in such a way.
Progressively razpodіlene navantazhennia _intensivn_styu q zamenyuyu zoseredzhennoy force Q = qL as applied in the middle of the plot. Step by step razpodіlenim navantazhennyam name the forces that may have the same magnitude and directives on the task of the body.
Yakshcho rozpodіlenі forces zminyuyutsya for linear law
(according to the tricutnik), then the power Q = qmaxL / 2- is applied at the center of gravity of the tricutnik, ruffled with a winder - on the first base……………….
44. Tertya kochennya - opir ruhu, which is to blame for the migration of bodies one by one. Viyavlyaєєєєєєєєєє, for example, between the elements of the bearing nodules, between the tire of the wheel of the car and the roadbed. As a rule, the amount of rubbish is less than the amount of rubbish forged, and therefore the rub is a broader type of movement in technology.
Rubbing the frostbite, weeping between two bodies, and to that it is classified as a good-looking rubbish.
45. Rubbed wrap. It is acceptable that on a horizontal plane there is an important sack, the center of the spool is significant through O, and the point of the spool of the spool with the flat through S. The spool wrap is almost straight and is called the turns. Dosvid shows that as a moment of betting, if it is guilty to bring a sack from the wrapper, even more small, then the sack will not come to the wrapper. It sounds like a fluffy bet is paralyzed by another couple, like a rubbish wrap.
One of the methods for calculating the moment of wear of the bearing of the stiffness is based on the fact that the moment of wear is divided into such ranks as the independent moment M0 and the residual moment M1, which then add up and give the total moment:
Two parallels and straight lines in one beam of force are induced to one force - equal force, applied at the point, which divides straight on the line, is wrapped in proportion to the magnitude of the forces. Sequentially adding up in pairs parallel forces, one also comes to one force - equal R: If the force can be transferred along the line її dії, then the point of reporting the force (even equal) is essentially not assigned. If all forces turn to the same corner and again spend additional forces, then we will take another straight line of divinity. The point of crossing between two lines of two equal equals can be seen, like a point of reporting of equal equals, which does not change its position with a one-hour turn of all forces on the same kut. Such a point is called the center of parallel forces. The center of parallel forces is a point of supplementation equal, does not change its position with an one-hour turn of all forces on one and the same point
47 The radius-vector of a point is called a vector, the cob of which moves with the cob of the coordinate system, and the end - with the center point.
In this way, especially the radius-vector, which introduces it into other vectors, these are those whose cob must always be located at the point of the cob of coordinates (Fig. 17).
The center of parallel forces, a point, through the yak to pass a line of a diversified system of parallel forces Fk, with any turn of all these forces, the points were reported in one and the same direction and on the same kut. The coordinates of the center of parallel forces are determined by the formulas:
de xk, yk, zk – coordinate point of force reporting.
48Vaga Center of a solid body - a point, invariably tied to this body, through which to pass a line of divinity equal to the forces of gravity of the particles of the body for any position of the body in space. In each field of gravity it is important to be the same, tobto. the forces of gravity of the particles of the body are parallel, one and the same, and they take a constant value for any turns of the body. Center of gravity coordinates:
; ; , de P = åp k, x k, y k, z k - Coordinates of the gravity reporting points p k. The center of gravity is a geometric point and can lie behind the boundaries of the body (for example, ring). Center of gravity of a flat figure:
DF k - elementary maidan, F - figure area. If the area is not possible to break the kіlka of the last parts, then. Even though the body can be all symmetrical, the center of the body's body is located on this axis.
49 The distribution of tasks on the assignment of the position (coordinates) to the center of gravity of a homogeneous plate, the system of bodies, which are located on the plane or space, is brought up to the folding of the alignment and the distant subset to the new number of numerical data and the calculation of the result:
Tobto. It is necessary to break up the system in the warehouse, to determine the position of the center of gravity of these warehouse elements. Calculate the mass of storage parts, showing through the following thickness - linear, volume surface, fallow according to the type of system presented. For example, the solution of the pet is fast, then it’s not varto її soromity to introduce (sound out it is given, but in the text of the task it is indicated that the plate, shears, plate are the same). From the features of this plant, the following two words should be mentioned: 1) the center of gravity of a warehouse is straight-cut, square, or a shear, the stake does not create difficulties - the center of gravity of such figures is located in the center.
50. circular sector: ; Trikutnik. To the beats of the tricout on a thin line,
Parallel skin from the yoga side determines what the center of the
severity of the skin line lie on the її geometric center (near the center
symmetry), then the center of gravity of the tricot lies on the yoga
median. Krapka peretina median dilit їх at spіvvіdnoshnі (2:1).
Circular sector (Figure 54). The center of gravity lies on the axis
symmetry. To the beats of the circular sector on elementary tricoutniks
signify an arc, studded with centers of heaviness trikutnikov. Radius
arcs are 2/3 of the sector radius. In this rank, the coordinate to the center
the severity of the circular sector is determined
virase xC = sin α.
51Pivkul. The center of the vaga lies on the axis of symmetry with the wind
3/8 view of the base.
Pyramid (cone) (Fig. 55).
The center of gravity lies on the line,
what is the bottom vertex from the center
heaviness of the base on the steel ¾ of the
The arc of the stake The center of gravity lies on the axis of symmetry
coordinates xC = sin α; uC = 0.
Kinematics
1Kinematics, Razdіl theoretical mechanics, vvchaє ruh material tіl not tsіkavlyachisya reasons that call or change tsey ruh. For her, it is more important than physical priming and mathematical rigor within the framework of accepted models Head of kinematics Set the ruh of the material point (system) - tse means to give a way to determine the position of the point (all points that make up the system) at some point in time.
The task of kinematics is based on the development of methods for the development of the point (system) and the methods of determining the speed, accelerating the point and other kinematic values of the point to establish a mechanical system. point trajectory
Set the ruh point means set the position of the skin moment to the hour. The camp may be assigned, as it was intended, to the coordinate system. However, for whom it is not mandatory to put the coordinates themselves; you can win the values, but they are connected with them. Below are three main ways to set a ruhu point.
1. Natural method. In this way, they are koristuyutsya, as if the trajectory of the movement of the point is visible. The trajectory is called the confluence of points of space, through the yak pass the material part that collapses. The whole line, as if out of sight in the open space. With the natural method, it is necessary to set (Fig. 1):
a) the trajectory of the movement (whatever the coordinate system);
b) hit a point on its zero, in a way that the winding up of S to the particle, so that the trajectory collapses;
c) positive straight line to S (when the point M is shifted, the opposite straight line S is negative);
d) cob at the hour t;
e) the function S(t), as it is called the law of rotation**) of the point.
2. Coordinate method. The most universal and last way to describe the movement. Vin transfer date:
a) coordinate systems (not necessarily Cartesian) q1, q2, q3;
b) cob according to the hour t;
c) the law of rucu points, tobto. functions q1(t), q2(t), q3(t).
Speaking about the coordinates of a point, we must always use the coordinates of the Cartesian coordinates.
3. Vector method. The position of a point near the space can also be determined by the radius vector, we draw from the last cob of the point (Fig. 2). In this way, for the description of the flow, it is necessary to ask:
a) the ear of the radius vector r;
b) cob according to the hour t;
c) the ruhu law of the point r(t).
Oskilki zavdannya one vector quantity r is equivalent to zavdannya three її projections x, y, z on the coordinate axes, the vector way is easy to go to the coordinate one. If we introduce single vectors i, j, k (i = j = k = 1), straightening the x, y and z axes (Fig. 2), then, obviously, the law of rotation can be
r(t) = x(t)i + y(t)j+z(t)k. (one)
The advantage of the vector form of the record before the coordinate compactness (replacement of three quantities is operated on from one) and often in greater accuracy.
butt. There is a small ring M on the unruly drotyan, and a straight rod AB (Fig. 3) passes through the yak, which evenly wraps around the point A (= t, de = const). Know the law of the kіltsya M vzdovzh shear AB i shdo pіvkola.
To complete the first part of the task, we speed it up in a coordinate way, directing the entire Cartesian system of the shear and picking the cob at point A. The scales of the AMC entries are straight (like spiraling on the diameter),
x(t) = AM = 2Rcos = 2Rcoswt,
de R is the radius of the pivcol. The omission of the law of movement is called harmonic kolvannyam (the kolvanya ce trivatime, obviously, is less than doti, until the ringlet reaches point A).
The other part of the plant is vicarious, natural way. Viberally positive direction in the direction of the flare of the trajectory (pivkola AS) against the arrow of the year (Fig. 3), and zero runs from the point C. Then the length of the arc CM as a function of the hour will give the law of the movement of the point M
S(t) = R2 = 2Rt,
tobto. The ring will be evenly collapsing along the stake with radius R with the apex sweep 2 . How screeching from the conducted review,
zero for the hour at both points in the moment, if you change the ring at point Z.
2.Vector way to set the ruhu point
The speed of the point is straightened to the trajectory (Figure 2.1) that is calculated, zgіdno (1.2), according to the formula
turn around Folding ruh points (tila)- such a move, at which point (body) at once takes the fate of a number of moves (for example, a passenger who moves around the car, who collapses). In this way, the roaming coordinate system (Oxyz) is introduced, as a way to set the tasks of the rohodo non-robust (basic) coordinate system (O 1 x 1 y 1 z 1). Absolute Rush sound points ruh by extension to a non-robust coordinate system. Vidnosny Rukh- Rukh according to the standard to the Rukhoma system of coordinates. (Rukh on the car). portable roc- Rukh Rukhlivy syst. coordinates of the schodo nerukhomoy (ruh wagon). The folding theorem: , ; -orti (alone vectors) of a ruhomo coordinate system, the ort wraps around the mitt axis, so the speed of the end, etc., Þ: ,
; - Vіdnosna shvidkіst. ; portable speed: 
Therefore, the absolute flexibility of the point = the geometric sum of the figurative (v e) and the visual (v r) flexibility , modulus: . : 
and etc. Warehouse virazi, which signifies the acceleration: 1) - the acceleration of the pole; 2) 3) - visible accelerated point; 4) , otrimuєmo: . 
The first three additions are quickening points in figurative Russian: - quickening of the pole O; - wraparound usk., 
- Guard usk., Tobto. 
. Quick folding theorem (Corioles theorem): 
, de 
– Coriole acceleration (Coriolis acceleration) – in the case of a non-transferable portable rush, absolutely acceleration = geometric sum of a portable, visual and Coriolis acceleration. Korіolisové priskrennya charac- terizes є: 1) change of the modulus and directness of figurative portability of a point through її vіdnosny ruh; 2) changing the straight line of the point through the wraparound portable hand. Coriolis acceleration modulus: a z = 2×|w e ×v r |×sin(w e ^ v r), directly of the vector follows the vector creation rule, or Zhukovsky’s rule: 90 about direct wrapping. 
Coriolisov usk. = 0 three times: 1) w e =0, then. at the time of the progressive portable ruhu chi, the moment of the beast is kut. speed 0; 2) v r =0; 3) sin (w e ^ v r) = 0, then. Ð(w e ^ v r) = 0, if the visibility v r is parallel to the axis of the portable wrap. At different times in one plane - cut between v r i vector w e \u003d 90 o, sin90 o \u003d 1, a \u003d 2 × w e × v r. Folding ruh solid body
With the addition of two forward flow, the resulting flow is also forward and the speed of the resulting flow is more than the sum of the flow rates of the warehouse flow. Foldable wrap tb. the body is close to the axes, which are shifted. All the wrapping, the camp that is in the expanse, is changed by the call of the year. mitteva veil body wrap. The vector of the apex shvidkost is a forged vector, straightening the mitt axis of the wrap. Absolute top winding of the body = geometric sum of windings of warehouse wraps - the rule of parallelogram of windings. 
. Yakshcho tіlo take a fate at once in mittevih wrappers for a number of axes, which intertwine in one point, then. With a spherical Russia of a solid body, one of the points of which the whole hour of the ruh is filled with unbreakable, perhaps equal to the spherical ruh: Y \u003d f 1 (t); q=f 2 (t); j = f 3 (t). Y - kut pretsії, q - kut nutatsі, j - kut of your wrapping - Euler's kuti. Kutova swidkіst pretsії, kut. swidkіst nutatsії, kut. sk. wet wrap. 
, - The module of the apex tightness of the body is near the mitten axis. Through projections on non-violent coordinate axes: - Euler's kinematic alignment. Foldable wrap around 2 parallel axles.
1) 
The wrapping was sent in one Bik. w=w 2 + w 1 
, 
. 2) Wrapping straight from the different side. 
w \u003d w 2 -w 1 
Z - inst. center that instant all wrapping, . The vectors of the apex slips when wrapped around ||-their axes add up the same way, like the vectors of parallel forces. 3) Pair of wraps- Wrapping around | |-their axes are directed in different directions and the apex of the shvidkost modulo equal (- a pair of apex of the frills). For this swing v A = v B, the resulting movement of the body is translational (or mittevian translational) movement v = w 1 × AB - the moment of parity of the winding movements (the translational movement of the bicycle pedal is carried out by rams). Instant. the center of shvidkost is known to the indistinct. Folding forward and wrapping ruhіv. 1) Swiftness of the forward movement ^ to the axis wrapping - plane-parallel movement - mittve wrapping around the axis Рр іz the apical swivel w=w". 2) Gvintovy Rukh- The movement of the body is folded from the overt movement on the axis Aa from the corner. w that progressive zі shvidkіstyu v||Aa. All Aa - all gvinta. Like v and w in one beck, gwent - right, like in different - levi. Look, how to pass an hour of one turn, be some point of the body, which lies on the axis of the screw, sound. crochet gwent - h. How v and w are constant, h = = const, with constant crochet be-like (×)M, without lying on the axis of the gwent, describe the gwent line. 
directed along the dotichny gvintovіy line. 3) Swiftness of the forward movement makes a pretty wrap around it, in which direction you can see how it is composed of a series of mittevyh screwed ruhіv, like the screwed axes, which constantly change - the mittevo-gvintovy ruh. Let's move on the cob of coordinates with the point of the line of the line of the force of the system. All forces are projected on the coordinate axes and subsuming the projections (Fig. 7.4). We take projections that are equal on the coordinate axis:
The modulus of an equal and equal system of similar forces is significant behind the formula
Directly the vector is equal to the cuts.
Pretty spacious system of forces
Bringing a fairly spacious system to the center of Pro.
Given the space system of forces (Fig. 7.5, a). Navigate її to the center of O.
Forces must be moved in parallel, a system of pairs of forces is established in one's own. The moment of skin s s tsih pairs is more expensive to increase the power module on the way to the center of the reduction.
At the center of the given there is a bundle of forces, which can be replaced by the total force (head vector) F GL (Fig. 7.5, b).
The moment of pairs of forces can be summed up by taking away the total moment of the system M goal (head moment).
In this rank, quite a spacious system of forces is brought to the head vector and head moment.
The head vector was taken into three warehouses, straightened out by the coordinate axes (Fig. 7.5 c).
Sound the total moment of the warehouse: three moments according to the coordinate axes.
The absolute value of the head vector (Fig. 7.5b) is more
The absolute value of the head moment is attributed to such a formula.
Rivnyannya rіvnovagi prostorovoї system of forces
When rivnovazi F Goal = 0; M goal = 0. We take six equal equals:
Six equal equals of the space system of forces give six independent possible displacements of the body in space: three displacements of coordinate axes and three wrappings around these axes.
Apply the solution of tasks
example 1. On the body in the form of a cube with an edge a\u003d 10 cm for three forces (Fig. 7.6). Calculate the moments and forces of the coordinate axes that run along the edges of the cube.
Solution
1. Moments of forces Oh:
2. Moments of forces schodo axis OU.
butt 2. Two wheels are fixed on a horizontal shaft, r 1 = 0.4 m; d 2 = 0.8 m. 7.7. Power added to wheel 1 F1, up to wheel 2 - forces F2= 12 kN, F3= 4kN.
Signify strength F1 that reaction at the hinges BUTі At at the station of jealousy.
Guessing:
1. When equal, six equal equals are victorious.
R_vnyannya momentіv slid fold schodo supports And that St.
2. Sealy F 2 \\O x; F 2 \\Oy;F 3 \\Oy.
The moment of these forces should be equal to zero.
3. 
The rozrahunok should be completed with a re-verification, having become a supplementary equalizer.
Solution
1. Significant force F\, having combined the equal moment of the forces on the axis Oz:
2. Significant reactions in the support BUT. On the support there are two warehouse reactions ( Y A ; X A ).
We add up the equal moment of the forces of the axis Oh"(in support U).
Rotate about axis Oh" does not apply:
The sign "minus" means that the reaction is direct from the protile bed.
Rotate about axis OU" does not change, we add equal moments of forces to the axis OU"(in support AT):
3. Significantly, the reaction at the U support. On the support, there are two warehouse reactions ( XB , Y B ). We add up the equal moment of the forces of the axis Oh(support BUT):
We store equal moments at any axis OU(support BUT):
4. Recheck. Vikoristovuemo alignment of projections:
Rozrahunok vykonaniy correctly.
example 3. Calculate the numerical value of the force P1 , for which the shaft ND(Fig. 1.21, a) perebuvatime at Rivnovazi. With the known value of the force R 1 designate the reference reactions.
Dyuchi on the gears of the wheel of power R
і R 1
directions according to dotichnyh to cob kіl kolіs; forces T
і T 1
- according to the radii of wheels; forces A 1 parallel to the axis of the shaft. T = 0.36P, 7T1 = P1; A1 \u003d 0.12P 1.
Solution
Support the shaft, shown in fig. 1.21 a, it is necessary to look at how spacious the hinged supports are, which allow the linear movements of the straight lines of the axes іі v(The selected coordinate system is shown in Fig. 1.21, b).
It is necessary to change the shaft in the form of bonds and replace them with reactions V B, H B, V C , NS (Fig. 1.21, b). They took away the expanse of the system of forces, on the basis of the equalization of the equal, the equalization of the coordinate system (Fig. 1.21.6):
de A 1* 1,25D/2 - moment of a wide axis і forces A 1 , applied to the right gear wheel.
Moments are welcome і forces T 1і A 1(addition to the middle gear wheel), P 1 (addition to the right gear wheel) and P add to zero, so the forces P, T 1, P 1 are parallel to the axis і, and the force A 1 peretinaє all in.
stars V C \u003d 0.37P;
stars VB=0.37P.
father, reactions V Bі V Z assigned correctly;
de A 1* 1,25D/2- moment v forces A 1 , applied to the middle gear wheel.
Moments are welcome v forces T, R 1 (added to the middle gear wheel), A 1і T 1(advance to the right gear wheel) add to zero, so how strong T, R 1 , T 1 parallel axis v, strength A 1 rethink everything v.
stars H C = 0.81Р;
stars H С = 1.274Р
Warehouse rechecking:
father, reactions H Bі N C assigned correctly.
At the end, it was significant that the supporting reactions turned out to be a plus sign. Tse point to those who take direct V B , H B, V C і N C zbіgayutsya z dijsnimi direct reactions zv'yazkіv.
butt 4. The pressure force of the connecting rod of the steam engine P = 25 kN is transmitted to the middle of the neck of the crankshaft at the point D under the hood α
\u003d 30 ° to the horizon with a vertical expansion of the neck of the knee (Fig. 1.22). On the end of the shaft of planting the pulleys of the belt transmission. The tightness of the wire pins of the double belt is greater, lower, tobto. S1 = 2S2. The force of the flywheel shaft G = 10 kN.
Calculate the tightness of the shafts of the belt drive and the reaction of the bearings BUTі AT, nehtuyuchi masoyu shaft.
Solution
Considering the alignment of the horizontal crankshaft with a pulley. Applied visibly to the mind of the given task of force P, S 1 , S 2 і G . It is necessary to change the shaft in the form of supporting fastenings and replace them with reactions V A , H A, V Bі N St. The coordinate axes are chosen as shown in Fig. 1.22. At the hinges BUTі At do not blame the reaction of the axis w, so that the tightness of the waist belt and all other forces are felt at the planes perpendicular to the center of the axis.
Warehouse equalization:
In addition, for the mind's task, there may be one more equal
In this rank, there are six nevіdomih zusil S 1, S 2, H A, V A, H B і V B and six ties of ties.
Alignment of projections on the whole w at the butt, it turns into the same 0 = 0, so all the forces lie at the planes perpendicular to the axis w.
Substituting equal equal S 1 \u003d 2S 2 and virishuyuchi їh, we know:
Reaction value H B veyshlo zі minus sign. Tse means that it is actually directly opposite to that taken in fig. 1.22.
Control nutrition and task
1. Write down the formulas for the distribution of the head vector of the space system of forces that converge.
2. Write down the formula for the expansion of the head vector of the space system of sufficient expansion of forces.
3. Write down the formula for the head moment of the space system of forces.
4. Write down the system of equalities of equalities of the spacious system of forces.
5. How is it necessary to vicorate for the purpose of the reaction of the shear R 1 (Fig. 7.8)?
6. Calculate the head moment of the system of forces (Fig. 7.9). The point of reduction is the cob of coordinates. The coordinate axes run along the edges of the cube, the edge of the cube is 20 cm long; F 1 - 20kN; F 2 - 30kN.
7. Determine the reaction Xv (Fig. 7.10). The vertical weight is driven by a pulley by two horizontal forces. seely F1 і F2 parallel axis Oh. AT = 0.3 m; OV= 0.5 m; F 1 = 2kN; F 2 = 3.5 books
 |
Recommendation. Fold equal moment at any time OU" at the point BUT.
8. Give feedback on the supply of the test task.
20. Umova equal space system of forces:
21. Theorem about 3 non-parallel forces: Lines of three non-parallel forces, which are mutually equal, lie in the same plane, overlap in one point.
22. Static fixed tasks- tse zavdannya, yakі can be decoupled by the methods of statics of a solid body, tobto. zavdannya, among them, the number of nevidomyh does not exceed the number of equal equal forces.
Static non-original systems, in which a number of unknown values outweigh the number of independent equals of a given system of forces
23. Rivnyannya rіvnovagy flat system of parallel forces:
AB is not parallel to F i
24. Cone ta kut tertya: Border camp of active forces cone rubbing with cut (φ).
If the force is active to pass the pose with a cone, then even equal is impossible.
Kut φ is called kut tertya.
25. Specify the expansion of the coefficients of the rub: coefficients of rubbing calmness and tertya forging-bezrozmirnі values, coefficients of rubbing stiffness and tertya wrapping may rozmirnіst dozhini (mm, cm, m).m
26. The main allowances, which are accepted during the raising of flat statically defined trusses:- swift fermi vvazhayut nevagomimi; - fastening of the shear at the nodes of the fermi-hinge; -zvnіshnє navantazhennya superimposed less at fermi knots; - shearer is wearing a bell
27. What is the link between strands and knots of a statically assigned fermi?
S = 2n-3 - simple statically initial farm, S-number of shearers, n-number of knots,
Yakscho S<2n-3 –не жесткая ферма, равновесие возможно, если внешние силы будут одинаково соотноситься
S>2n-3 – the truss is not statically defined, ties can be added, +deformation expansion
28. A statically appointed farm is responsible for satisfying the mind: S=2n-3; S-number of shears, n-number of knots.
29. Method of visualization of nodes: This method is based on the fact that thoughts see fermi knots, apply strong forces to them, and the reactions of shearing and become even equal forces that reach the skin knot. Mentally allow that all the shears are stretched (reactions of shears in the direction of the nodes).
30. Ritter method: We carry out a sіchnu ploschina, scho rozsіkaє farm for 2 parts. Peretin can start and finish beyond the borders of the fermi. As an object of jealousy, you can choose whether or not a part. Peretin pass with shears, not knots. Forces, applied to the object of equivalence, establish a sufficient system of forces, on which 3 equivalences of equivalence can be added. To that, the retin is carried out in such a way that no more than 3 haircuts have been consumed by the new one, there are no such cases.
The peculiarity of Ritter's method is to choose the form of equalization in such a way that one unknown value was included in the skin equalization. For which position, the Ritter point is the point of the cross line of the line dividing two nevіdomih zusil and it is recorded equal to the moment rel. tsich point.
If the Ritter point lies at the inconsistency, then as equal equalization of the equal projections on the whole, perpendicular to these shears.
31. Krapka Ritter- the point of the cross line is the line of two nevidomyh zusil. If the Ritter point lies at the inconsistency, then as equal equalization of the equal projections on the whole, perpendicular to these shears.
32. Center of gravity of volumetric figure:
33. Center of gravity of a flat figure:
34. Center of gravity of shear structure:
35. Center of gravity of the arc:
36. Center of gravity of the circular sector:
37. Cone gravity center:
38. Center of heaviness pіvkulі:
39. Method of negative values: How hard. body may be empty, tobto. empty from some vinnyat їх masses, we remember the thoughts of the empty to the suctile body, and determine the center of gravity of the figure, taking the vag, hugging, the area of the empty zі sign "-".
40. 1st invariant: The first invariant of the system of forces is the head vector of the system of forces. The head vector of the system of forces lie in the center of the reduction R=∑ F i
41. 2nd invariant: The scalar dobutok of the head vector at the moment of the system of forces for the center of the reduced value is constant. 
42. How many times does a system of forces aim up to a power gwent? In time, as the head vector of the system of forces and її the head moment to the center of the reduction is not equal to zero and not perpendicular to each other, tasks. the power system can be reduced to a power gwent.
43. Alignment of the central screw 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 bet forces yak vector the whole vector is perpendicular to the plane of the pari and the straight lines of the bet, the stars can be seen wrapping the bet against the arrow of the year. Behind the module, the vector moment is more profitable for one of the forces of the bet on the shoulder of the bet. Vector moment of the bet yavl. Vіlnym vector i mozhe but dodany to be-yakoy ї point of a solid body.
46. The principle of calling from calls: If the links are seen, it is necessary to replace them with the forces of reactions in the form of a link.
47. Motuzkovy bagatokutnik- tse pobudova graphostatics, which can be used to designate a line of a equal plane system of forces for the significance of the reactions of the supports.
48. What a mutual connection between a motuzyanim and a powerful bagatokutnik: For the knowledge of unknown forces, graphically in the power bagatokutnik we know the additional point O (pole), at the motuzkovy bagatokutnik we know equally, moving the yak into the power bagatokutnik we know the unknown force
49. Umova equalization of systems of pairs of forces: For the equality of pairs of forces, which are on a solid body, it is necessary and sufficient, so that the moment of equivalent pairs of forces reaches zero. Naslidok: In order to restore a couple of forces, it is necessary to report a worthy couple, tobto. a pair of forces can be combined with another pair of forces with equal modules and parallel straightening moments.
Kinematics
1. All ways to set the point flow:
natural way
coordinate
vector radius.
2. How to determine the alignment of the trajectory of the movement of a point with the coordinate method of specifying the movement? In order to take the alignment of the trajectory of the movement of a material point, with the coordinate method of setting, it is necessary to turn on the parameter t from the laws of movement.
3. Accelerated point in coordination. ways to set the pace:
above x 2 dots
above y 2 points
4. Accelerating points with the vector method of setting the speed:
5. Accelerating points in a natural way
=
=
*
+v*
; a=
+
;
*
; v*
.
6. Why is it even and how is it straightened normally?– straightened along the radius to the center,
Enthusiasm...