As was proven above, an arbitrary system of forces, arbitrarily located in space, can be reduced to a single force equal to the main vector of the system and applied at an arbitrary reduction center ABOUT, and one pair with a moment equal to the main moment of the system relative to the same center. By
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Let us consider some cases of reduction of a system of forces. 
1. Flat system of forces. For definiteness, let all forces be in the plane OXY. Then in the most general case
The main vector is not zero, the main moment is not zero, their dot product is zero, indeed
therefore, the main vector is perpendicular to the main moment: the plane system of forces is reduced to the resultant.
2. System of parallel forces. For definiteness, let all forces be parallel to the axis OZ. Then in the most general case
Here also the main vector is not equal to zero, the main moment is not equal to zero, and their scalar product is equal to zero, indeed
therefore, the main vector is perpendicular to the main moment: the system of parallel forces is reduced to the resultant. In a particular case, if equal to zero, then the main vector of forces is equal to zero, and the system of forces is reduced to a pair of forces, the moment vector of which is in the plane OXY. Let us now systematize the cases considered. Let us recall: an arbitrary spatial system of forces applied to a rigid body is statically equivalent to a force equal to the main vector applied at an arbitrary point of the body (center of reduction), and a pair of forces with a moment equal to the main moment of the system of forces relative to the specified center of reduction.
1) Let =0,≠0. This is the case when a system of forces is reduced to one force, which we will call the resultant of the system of forces. An example of such a system of forces can be considered a converging system of forces, for which the lines of action of all forces intersect at one point.
2) ≠0,=0 . A system of forces is equivalent to a pair of forces.
3) ≠0,≠0, but. The main vector is not equal to zero, the main moment is not equal to zero, their scalar product is equal to zero, i.e. the principal vector and principal moment are orthogonal. Any system of vectors in which the main vector and the main moment are not equal to zero and they are perpendicular, is equivalent to a resultant whose line of action passes through the point ABOUT*(Figure 8). An example of such a system of forces can be considered a flat system of forces or a system of parallel forces.
4) ≠0,≠0, and the main vector and the main moment are non-orthogonal. In this case, the system of forces is reduced to dynamism or two non-intersecting forces.
Cases of reduction to the simplest form
Bringing to a pair
Let, as a result of bringing the forces to the center O, it turns out that the main vector is equal to zero, and the main moment is different from zero: . Then, by virtue of the fundamental theorem of statics, we can write
This means that the original system of forces in this case is equivalent to a pair of forces with a moment.
The couple's moment does not depend on which point is chosen as the moment center when calculating the couple's moment. Consequently, in this case the main point should not depend on the choice of the center of reduction. But this is precisely the conclusion that the relation leads to
connecting the main points regarding two different centers. When the additional term is also equal to zero, we get
Reduction to resultant
Let now the main vector is not equal to zero, and the main moment is equal to zero: . By virtue of the fundamental theorem of statics, we have
that is, the system of forces turns out to be equivalent to one force - the main vector. Consequently, in this case, the original system of forces is reduced to a resultant, and this resultant coincides with the main vector applied at the center of reduction: .
The system of forces is reduced to a resultant in the case when the main vector and the main moment are both not equal to zero, but mutually perpendicular: . The proof is carried out using the following sequence of actions.
Through the center of reduction O we draw a plane perpendicular to the main moment (Fig. 50, a). In the figure, this plane is combined with the drawing plane, and the main vector is located in it. In this plane we build a pair with a moment, and we choose the forces of the pair to be equal in magnitude to the main vector; then the pair's leverage will be equal to . Next, we move the pair in its plane so that one of the forces of the pair is applied at the center of reduction O opposite to the main one; the second force of the pair will be applied at point C, distant from the center O in the desired direction, determined by the direction, at a distance OS equal to the arm of the pair h (Fig. 50, b). Now discarding the balanced forces R and - applied at point O, we arrive at one force applied at point C (Fig. 50, c). It will serve as the resultant of this system of forces.
It can be seen that the reaction force is still equal to the main vector, but differs from the main vector in its point of application. If the main vector is applied at the reduction center O, then the resultant is at point C, the position of which requires a special definition. The geometric method of finding point C is visible from the construction done above.
For the moment of the resultant relative to the center of reduction O, we can write (see Fig. 50):
or, omitting intermediate values:
If we project this vector equality onto any axis passing through point O, we obtain the corresponding equality in projections:
Remembering that the projection of the moment of force relative to a point onto an axis passing through this point is the moment of force relative to the axis, we rewrite this equality as follows:
The resulting equalities express Varignon’s theorem in its general form (in Lecture 2 the theorem was formulated only for converging forces): if a system of forces has a resultant, then the moment of this resultant (relative to a point, relative to an axis) is equal to the sum of the moments of all given forces - components (relative to that same point, same axis). It is clear that in the case of a point the summation of moments is vectorial, in the case of an axis it is algebraic.
Reduction to dynamism
Dynamia or dynamic screw is the combination of a pair of forces and a force directed perpendicular to the plane of action of the pair. It can be shown that in the general case of reduction, when and is not perpendicular, the original system of forces is equivalent to some dynamism.
Let's consider some special cases of the previous theorem.
1. If for a given system forces R = 0, M 0 = 0, then it is in equilibrium.
2. If for a given system of forces R = 0, M 0 0, then it is reduced to one pair with a moment M 0 = m 0 (F i). In this case, the value of M 0 does not depend on the choice of center O.
3. If for a given system forces R 0, then it is reduced to one resultant, and if R 0 and M 0 = 0, then the system is replaced by one force, i.e. resultant R passing through center O; if R 0 and M 0 0, then the system is replaced by one force passing through a certain point C, and OC = d(OCR) and d = |M 0 |/R.
Thus, a flat system of forces, if it is not in equilibrium, is reduced either to one resultant (when R 0) or to one pair (when R = 0).
Example 2. Forces applied to the disk:
(Fig. 3.16) bring this system of forces to its simplest form.
Solution: choose the Oxy coordinate system. Let's choose point O as the reduction center. Main vector R:
R x = F ix = -F 1 cos30 0 – F 2 cos30 0 +F 4 cos45 0 = 0; Rice. 3.16
R y = F iy = -F 1 cos60 0 + F 2 cos60 0 – F 3 + F 4 cos45 0 = 0. Therefore R = 0.
The main moment of the M 0 system:
M 0: = m 0 (F i) = F 3 *a – F 4 *a*sin45 0 = 0, where a is the radius of the disk.
Answer: R = 0; M 0 = 0; the body is in balance.
Bring to its simplest form the system of forces F 1, F 2, F 3, shown in the figure (Fig. 3.17). Forces F 1 and F 2 are directed on opposite sides, and force F 3 is directed along the diagonal of rectangle ABCD, side AD of which is equal to a. |F 1 | = |F 2 | = |F 3 |/2 = F.
Solution: direct the coordinate axes as shown in the figure. Let us determine the projections of all forces on the coordinate axes:
The magnitude of the main vector R is equal to: 
;
.
The direction cosines will be: 
;
.
Hence: (x,R) = 150 0 ; (y, R) = 60 0 .
ABOUT 
let us determine the main moment of the system of forces relative to the center of reduction A. Then
m A = m A (F 1) + m A (F 2) + m A (F 3).
Considering that m A (F 1) = m A (F 3) = 0, since the direction of the forces passes through point A, then
m A = m A (F 2) = F*a.
Thus, the system of forces is reduced to a force R and a pair of forces with a moment m A directed counterclockwise (Fig. 3.18).
Answer: R = 2F; (x,^R) = 150 0 ; (y,^ R) = 60 0 ; m A = F*a.
Questions for self-control
What is the moment of force about the center?
What is a couple of forces?
Bringing an arbitrary plane system of forces to a given center?
Addition of parallel forces?
Literature: , , .
Lecture 4. Equilibrium conditions for an arbitrary plane system of forces
Basic form of equilibrium conditions. For the equilibrium of an arbitrary plane system of forces, it is necessary and sufficient that the sum of the projections of all forces on each of the two coordinate axes and the sum of their moments relative to any center lying in the plane of action of the forces are equal to zero:
F ix = 0; F iy = 0; m 0 (F i) = 0.
Second form of equilibrium conditions: For the equilibrium of an arbitrary plane system of forces, it is necessary and sufficient that the sum of the moments of all these forces relative to any two centers A and B and the sum of their projections onto the Ox axis not perpendicular to the line AB are equal to zero:
m A (F i) = 0; m B (F i) = 0; F ix = 0.
Third form of equilibrium conditions (three-moment equation): For the equilibrium of an arbitrary plane system of forces, it is necessary and sufficient that the sum of all these forces relative to any three centers A, B, C, not lying on the same straight line, be equal to zero:
m A (F i) = 0; m B (F i) = 0; m С (F i) = 0.
P 
example 1. Determine the embedding reactions of a cantilever beam under the action of a uniformly distributed load, one concentrated force and two pairs of forces (Fig. 4.1); load intensityq = 3*10 4 N/m; F = 4*10 4 H; m 1 = 2*10 4 H*m; m 2 = 3*10 4 H*m. BN = 3m; NC = 3m; CA = 4m.
R 
solution:
According to the principle of liberation from connections, we will replace connections with corresponding reactions. When a rigid embedding occurs in the wall, a reaction force R A of an unknown direction and an unknown moment m A arises (Fig. 4.2). Let's replace the distributed load with an equivalent concentrated force Q applied at point K (ВК = 1.5 m). Let us choose the coordinate system of the VCU and draw up the equilibrium conditions for the beam in the basic form:
projections of forces onto the X axis: - Fcos45 0 – R Ax = 0 (1)
projections of forces on the Y axis: -Q - Fsin45 0 + R Ax = 0 (2)
sum of moments: m A (F) = m 1 – m 2 + m A + Q*KA + F”*CA = 0 (3)
We will decompose the force F at point C into two mutually perpendicular components F” and F’; force F’ does not create a moment relative to point A, since the line of action of the force passes through point A. Force modulus F” = Fcos45 0 = F(2) 1/2 /2.
Substituting numerical values into equations (1), (2) and (3), we obtain:
In a given system of three equations there are three unknowns, so the system has a solution, and only a unique one.
4*10 4 *0.7 = R Ax R Ax = 2.8*10 4 H
3*10 4 *3 – 4*10 4 *0.7 + R Ay = 0 R Ay = 11.8*10 4 H
m A – 10 4 + 3*10 4 *3*8.5 + 4*10 4 *2.8 = 0 m A = - 86.8*10 4 H*m
Answer: R Ax = 2.8*10 4 H; R Ay = 11.8*10 4 H; m A = - 86.8*10 4 H*m.
Example 2. Determine the reactions of supports A, B, C and hinge D of a composite beam (Fig. 4.3).
q 
= 1.75*10 4 N/m; F = 6*10 4 H; P = 5*10 4 H.
Solution: Based on the principle of liberation from connections, we replace connections with corresponding reactions.
We will replace the distributed loadq with an equivalent concentrated force Q = q*KA applied at point M (AM = 2m). The number of unknown reaction forces: R Ax, R Ay, R B, R C and two pairs of reaction force components in the hinge D.
R 
Let's look separately at the reactions at hinge D. To do this, consider beams AD and DE separately (Fig. 4.5a, 4.5b).
According to Newton’s third law in the hinge D, a system of forces R Dx and R Dy acts on the beam KD, and the opposite system of forces acts on the beam DE: R’ Dx and R’ Dy, and the magnitudes of the forces are equal in pairs, i.e. R Dx = R Dx and R Dy = R Dy. These are the internal forces of the composite beam, so the number of unknown reaction forces is six. To determine them, it is necessary to create six independent equations of equilibrium states. The following options for composing equations of state are possible.
We draw up equilibrium conditions for the entire structure (3 equations) and for a separate element of this structure: beam KD or beam DE. When compiling equilibrium equations for the entire structure, internal forces are not taken into account, since upon summation they cancel each other out.
Equations of the equilibrium condition for the entire structure:
R Ax – Fcos60 0 = 0
Q - R Ay – Fsin60 0 + R B + R C – P = 0
m A (F) = Q*m A – Fsin60 0 *AN + R B *AB + R C *AC – P*AE = 0
Equations of the equilibrium condition for the element DE:
R’ Dy , + R C – P*DE = 0
M D (F) = R C *DC – P*DE = 0
In this way, six independent equations with six unknowns are compiled, so the system of equations has a solution, and only a unique one. By solving the system of equations, we will determine the unknown reaction forces.
As was proven above, an arbitrary system of forces, arbitrarily located in space, can be reduced to a single force equal to the main vector of the system and applied at an arbitrary center of reduction ABOUT, and one pair with a moment equal to the main moment of the system relative to the same center. Therefore, in the future, an arbitrary system of forces can be replaced by an equivalent set of two vectors - force and moment applied at a point ABOUT. When changing the position of the center of reduction ABOUT the main vector will maintain its magnitude and direction, but the main moment will change. Let us prove that if the main vector is nonzero and 
is perpendicular to the main moment, then the system of forces is reduced to one force, which in this case we will call the resultant (Fig. 8). The main moment can be represented by a pair of forces ( , ) with a shoulder , then the forces and the main vector form a system of two
forces equivalent to zero, which can be discarded. There will remain one force acting along a straight line parallel to the main
Figure 8 to the vector and passing at a distance
h= from the plane formed by the vectors and . The considered case shows that if from the very beginning we choose the center of reduction on the straight line L, then the system of forces would immediately be brought to a resultant force, the main moment would be equal to zero. Now we will prove that if the main vector is non-zero and not perpendicular to the main moment, then such a point can be chosen as the reduction center ABOUT* that the main moment relative to this point and the main vector will be located on the same straight line. To prove this, let us decompose the moment into two components - one directed along the main vector, and the other perpendicular to the main vector. Thus, the pair of forces is decomposed into two pairs with moments: and , and the plane of the first pair is perpendicular to , then the plane of the second pair, perpendicular to the vector (Fig. 9) contains the vector . The combination of a couple with a moment and a force forms a system of forces, which can be reduced to one force (Fig. 8), passing through the point O*. Thus (Fig. 9), the combination of the main vector and the main moment at the point ABOUT reduced to the force passing through a point ABOUT*, and a pair with a moment parallel to this line, which was what needed to be proven. The combination of a force and a couple, the plane of which is perpendicular to the line of action of the force, is called dynamism (Fig. 10). A pair of forces can be represented by two equal forces ( , ), located as shown in Fig. 10. But by adding the two forces and , we obtain their sum and the remaining force, from which it follows (Fig. 10) that the combination of the main vector and the main moment in point ABOUT, can be reduced to two non-intersecting forces and .
Let us consider some cases of reduction of a system of forces.
1. Flat system of forces. For definiteness, let all forces be in the plane OXY. Then in the most general case
The main vector is not zero, the main moment is not zero, their dot product is zero, indeed
therefore, the main vector is perpendicular to the main moment: the plane system of forces is reduced to the resultant.
2. System of parallel forces. For definiteness, let all forces be parallel to the axis OZ. Then in the most general case
Here also the main vector is not equal to zero, the main moment is not equal to zero, and their scalar product is equal to zero, indeed
therefore, in this case, the main vector is perpendicular to the main moment: the system of parallel forces is reduced to the resultant. In the particular case, if equal to zero, then the main vector of forces is equal to zero, and the system of forces is reduced to a pair of forces, the moment vector of which is in the plane OXY. Let us now systematize the cases considered. Let us recall: an arbitrary spatial system of forces applied to a rigid body is statically equivalent to a force equal to the main vector applied at an arbitrary point of the body (center of reduction), and a pair of forces with a moment equal to the main moment of the system of forces relative to the specified center of reduction.
Fundamental theorem of statics.An arbitrary system of forces acting on a rigid body can be replaced by an equivalent system consisting of a force and a pair of forces. The force is equal to the main vector of the force system and is applied at an arbitrarily selected point of the body (center of reduction), the moment of the couple is equal to the main moment of the force system relative to this point.
Main vector of the force system:
.
The main moment of the system of forces relative to the center O:
is determined by its projections on the coordinate axes:
, 
, 
,
.
The following cases of bringing a system of forces to the center are possible:
The system of forces is reduced to a resultant. The line of action of the resultant passes through the center of reduction.
A system of forces is reduced to a pair of forces.
3. , , - the system of forces has a resultant that does not pass through the center of reduction. Its line of action is determined by the equations
4. , , - the system of forces is reduced to a dynamic screw (force and a pair lying in a plane perpendicular to the force).
Moment of a couple of forces of a dynamic propeller
.
The axis of the dynamic screw is determined by the equations
5. , - a balanced system of forces.
Example 1.4.1. Bring the system of forces (Fig. 1.4.1) to its simplest form, if F 1 = 5 N, F 2 = 15 N, F 3 = 10 N, F 4 = 3 N, a= 2 m.
1. For the center of reduction, choose the origin of coordinates - the point O(Fig. 1.4.2) and indicate the angles a and b that determine the position of the force.
2. Find the projections of the main vector on the coordinate axes:
,
,
.
N.
3. Calculate the projections of the main moment relative to the point ABOUT on the coordinate axis:
,
,
,
Nm, 
Nm, Nm,
4. Find the value of the scalar product of the main vector and the main moment
Since , the system of forces is brought to the right dynamic screw. The moment vector of a pair of dynamic propellers and the main vector coincide in direction.
5. The equation of the dynamic propeller axis has the form:
or taking into account the found values:
To construct the axis of a dynamic propeller, we find the points A And B its intersections with coordinate planes Oxy And Oyz, respectively
–0.203 m 1.063 m
6. Let us determine the moment of a pair of forces of a dynamic propeller
Nm.
7. By coordinates of points A And B Let's depict the axis of the dynamic screw (Fig. 1.4.3). At an arbitrary point on this axis, we indicate a force equal to the main vector and the moment vector of the pair.
Problem 1.4.1. Does the resultant system of forces for which the main vector 
and the main moment relative to the center ABOUT 
.
Answer: yes.
Problem 1.4.2. Does the resultant system of forces for which the main vector and the main moment relative to the center ABOUT 
.
Answer: no.
Problem 1.4.3. Determine the distance from the center of reduction ABOUT the valley of action of the resultant system of forces (Fig. 1.4.4), if its main vector R= 15 N and main moment M O= 30 Nm.
Answer: 2 m.
Problem 1.4.4. Determine the angle between the main vector and the main moment of the system of forces shown in Figure 1.4.5, taking the point of reference as the center of reduction O, If F 1 = F 2 = 2 N, moment of a couple of forces M 1 = 3 Nm, OA= 1.5 m.
Answer: α = 0º.
Problem 1.4.5. Determine the angle between the main vector and the main moment of the force system shown in Figure 1.4.6, taking the point of reference as the center of reduction ABOUT, If F 1 = F 2 = F 3 = 10 N, a= 3 m.
Answer: α = 135º.
Problem 1.4.6. Find the main vector and main moment of the system of forces shown in Figure 1.4.7, if F 1 = F 2 = F 3 = 7 N, a OA = OB = OS= 2 m. Take the point as the reduction center ABOUT.
Answer: R = 0, M O= 17.146 Nm.
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| Rice. 1.4.6 | Rice. 1.4.7 |
Problem 1.4.7. Bring the system of forces applied to the vertices of the parallelepiped (Fig. 1.4.8) to its simplest form, if F 1 = 16 N, F 2 = 12 N, F 3 = 20 N, a = With= 2.4 m, b=1.8 m.
M= 48 Nm.
Problem 1.4.8. Bring the system of forces applied to the vertices of the cube (Fig. 1.4.9) to its simplest form, if F 1 = 15 N, F 2 = 40 N, F 3 = 25 N,
F 4 = F 5 = 20 N, a= 1.5 m.
Answer: the system of forces is reduced to a pair of forces with a moment M= 63.65 Nm.
Problem 1.4.9. Bring a system of forces applied to a regular quadrangular pyramid, as shown in Fig. 1.4.10, to the simplest form, if F 1 = F 2 = F 3 = F 4 = 1 N, F 5 = 2.83 N, AB = AS= 2 m.
Answer : the system of forces is balanced.
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| Rice. 1.4.8 | Rice. 1.4.9 |
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| Rice. 1.4.10 | Rice. 1.4.11 |
Problem 1.4.10. Bring the system of forces applied to the vertices of a rectangular parallelepiped (Fig. 1.4.11) to its simplest form, if F 1 = F 5 = 10 N, F 3 = 40 N, F 4 = 15 N, F 2 = 9 N, a= 2.4 m, b= 3.2 m, c= 1 m.
Answer: the system of forces is reduced to the resultant R= 32 N, the line of action of which is parallel to the axis Oy and passes through the point A (0,9; 0; 0).
Problem 1.4.11. Bring the system of forces applied to the vertices of a rectangular parallelepiped (Fig. 1.4.12) to its simplest form, if F 1 = F 3 = 3 N, F 2 = F 6 = 6 N, F 4 = F 5 = 9 N, a= 3 m, b= 2 m, c= 1 m.
Answer : the system of forces is balanced.
Problem 1.4.12. Bring the system of forces applied to the vertices of a rectangular parallelepiped (Fig. 1.4.13) to its simplest form, if F 1 = F 4 =F 5 = 50 N, F 2 = 120 N, F 3 = 30 N, a= 4 m, b= 3 m, c= 5 m.
R= 80 N, the line of action of which is parallel to the axis Oy and passes through the point A (0,0,10).
Problem 1.4.13. Bring the system of forces applied to the vertices of the cube (Fig. 1.4.14) to its simplest form, if a= 1 m, F 1 = 866 N, F 2 = F 3 = F 4 = F 5 = 500 N. When deciding to accept .
Answer: the system is reduced to the resultant R= 7.07 N.
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| Rice. 1.4.12 | Rice. 1.4.13 |
 |  |
| Rice. 1.4.14 | Rice. 1.4.15 |
Problem 1.4.14. Bring the system of forces applied to a regular triangular pyramid (Fig. 1.4.15) to its simplest form, if F 1 = F 2 = F 3 = F 4 = F 5 = F 6 = 1 N, AB = AS= 2 m.
Answer: the system of forces is driven to a dynamic screw with R= 1.41 N and M= 1.73 Nm, the axis of the power screw passes through the top S perpendicular to the base of the pyramid.
Problem 1.4.15. Radio mast weight with base G= 140 kN. Antenna tension force is applied to the mast F= 20 kN and resultant wind pressure forces P= 50 kN; both forces are horizontal and located in mutually perpendicular planes (Fig. 1.4.16). Determine the resulting reaction of the soil in which the base of the mast is laid.
Answer: a distributed system of ground reaction forces is driven to the left dynamic propeller with a force equal to 150 kN and a couple with a moment of 60 kN∙m. the equation of the central helical axis has the form
.
Center of gravity
The center of gravity of a solid body is the center of parallel forces of gravity of the particles of a given body.
,
To determine the position of the center of gravity of homogeneous bodies, the method of symmetry, the method of dividing into bodies of a simple shape with a known position of the centers of gravity, as well as the method of negative masses (lines, areas, volumes) are used.
Example 1.5.1. Determine the coordinates of the center of gravity of a flat truss (Fig. 1.5.1), composed of homogeneous rods with the same linear weight.
1. Let's apply the partitioning method, that is, imagine the truss as a set of seven rods.
2. Find the coordinates of the center of gravity of the truss using the formulas:
; 
,
where , , are the length and coordinates of the center of gravity of the rod with number .
Lengths and coordinates of the centers of gravity of the rods:
Then 
,
Example 1.5.2. The end wall of the hangar (Fig. 1.5.2) has the shape of a semicircle 1 radius with rectangular doorway 2 height and width Determine the coordinates of the wall’s center of gravity.
1. Let's apply the methods of symmetry and negative areas, considering a semicircle 1 and rectangular cutout 2 .
2. Find the coordinates of the wall’s center of gravity.
Because the axis Oh is the axis of symmetry, then the coordinate
The coordinate of the center of gravity of the plate is determined by the formula
where , , , are the areas and coordinates of the centers of gravity of the figures 1 And 2 .
Areas and coordinates of the centers of gravity of the figures:
Tasks 1.5.1 – 1.5.4. Determine the coordinates of the centers of gravity of flat trusses (Fig. 1.5.3 - 1.5.6), composed of homogeneous rods with the same linear weight.
Answers to problems 1.5.1 – 1.5.4:
| Task number | 1.5.1 | 1.5.2 | 1.5.3 | 1.5.4 |
| , m | 1,52 | 3,88 | 3,0 | 1,59 |
| , m | 0,69 | 1,96 | 1,73 | 0,17 |
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| Rice. 1.5.3 | Rice. 1.5.4 |
 | |
| Rice. 1.5.5 | Rice. 1.5.6 |
 |
|
| Rice. 1.5.7 | Rice. 1.5.8 |
Problems 1.5.5 – 1.5.7. Determine the coordinates of the centers of gravity of homogeneous composite lines (Fig. 1.5.7 – 1.5.9).
Answers to problems 1.5.5 – 1.5.7:
| Task number | 1.5.5 | 1.5.6 | 1.5.7 |
| , cm | –4,76 | ||
| , cm | 14,16 | 3,31 |
 |  |
| Rice. 1.5.9 | Rice. 1.5.10 |
 |  |
| Rice. 1.5.11 | Rice. 1.5.12 |
Problem 1.5.8. A homogeneous wire bent at a right angle is suspended on a thread (Fig. 1.5.10). Find the relationship between the lengths of sections AD And A.E., at which the area A.E. is in a horizontal position. AB = 0,3 l 1 .
Problem 1.5.9. Determine the coordinates of the center of gravity of a homogeneous wire (Fig. 1.5.11), if a= 3 m, b= 2 m, c= 1.5 m.
Answer: x C= 1.69 m, yC= 1.38 m, z C= 1.33 m.
Problem 1.5.10. A homogeneous closed contour bounding a semicircle is suspended on a thread (Fig. 1.5.12). Determine the angle α between the horizontal and the diameter of the semicircle.
Answer: α = 68.74º.
Problems 1.5.11 – 1.5.14. Determine the coordinates of the centers of gravity of homogeneous flat figures (Fig. 1.5.13 – 1.5.16).
Answers to problems 1.5.11 – 1.5.14:
| Task number | 1.5.11 | 1.5.12 | 1.5.13 | 1.5.14 |
| 37.07 cm | 32.38 cm | 2.31 m | ||
| 11.88 cm | 24.83 cm | 1.56 m |
 |  |
| Rice. 1.5.13 | Rice. 1.5.14 |
 |  |
| Rice. 1.5.15 | Rice. 1.5.16 |
 |  |
| Rice. 1.5.17 | Rice. 1.5.18 |
Problem 1.5.15. The support for the bearing journal is a part consisting of a support in the form of a parallelepiped and a key in the shape of a cube (Fig. 1.5.17). Determine the coordinates of the center of gravity of the stand. Dimensions are indicated in millimeters.
Answer:
Problem 1.5.16. The journal of a sliding bearing is a part consisting of a parallelepiped and a cylindrical support (Fig. 1.5.18). Determine the coordinates of the center of gravity of the axle. Dimensions are indicated in millimeters.
Answer: 
, 
,
Problem 1.5.17. A homogeneous body, the cross section of which is shown in Figure 1.5.19, consists of a hemisphere, a cylindrical part and a circular cone. Determine the coordinates of the body's center of gravity. Dimensions are indicated in millimeters.
Answer: , , 
Problem 1.5.18. The barrel of a tank gun has the shape of a truncated cone of length (Fig. 1.5.20). The outer diameter of the barrel at the point of attachment to the breech of the gun; the outer diameter in the section corresponding to the muzzle of the barrel; gun caliber d=100 mm. Determine the coordinate of the center of gravity of the trunk.
Answer:
Problem 1.5.19. Determine the coordinates of the center of gravity of a homogeneous body consisting of two rectangular parallelepipeds (Fig. 1.5.21). The lower parallelepiped has a cutout in the shape of a quarter cylinder with a base radius R= 10 cm. Dimensions in the picture are in cm.
Answer: x C= 17.1 cm, yC= 20.99 cm, z C= 7.84 cm.
Problem 1.5.20. Determine the coordinates of the center of gravity of a homogeneous body (Fig. 1.5.22), consisting of a triangular prism and a parallelepiped with a cutout. Dimensions in the picture are in cm.
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| Rice. 1.5.19 | Rice. 1.5.20 |
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| Rice. 1.5.21 | Rice. 1.5.22 |
Answer: x C= 20.14 cm, yC= 35.14 cm, z C= 5 cm.
Part 2. Kinematics
Kinematics of a point
There are three analytical ways to specify the movement of a point: vector, coordinate and natural.
With the vector method, the radius vector of a moving point is specified as a function of time. The velocity and acceleration vectors of a point are equal to the first and second time derivatives of the radius vector, respectively:
, 
.
The relationship between the radius vector and the Cartesian coordinates of a point is expressed by the equality: 
, where , , are unit vectors of the coordinate axes.
With the coordinate method, the law of motion of a point in a Cartesian coordinate system is given by specifying three functions: , , . Projections of velocity and acceleration on the coordinate axes, as well as the modules of velocity and acceleration of a point are determined by the formulas:
, , , 
,
In the natural method, the trajectory of a point and the law of motion of the point along the trajectory are specified, where the curvilinear coordinate is measured along an arc from some fixed point on the trajectory. The algebraic value of speed is determined by the formula, and the acceleration of a point is equal to the geometric sum of the tangential and normal accelerations, i.e. , , , 
, – radius of curvature of the trajectory at a given point.
Example 2.1.1. The projectile moves in a vertical plane according to the equations, 
(x,y– in m, t– in c). Find:
– trajectory equation;
– speed and acceleration at the initial moment;
– height and range of fire;
– radius of curvature at the initial and highest points of the trajectory.
1. Let us obtain the equations of the projectile trajectory, excluding the parameter t from the equations of motion
.
The trajectory of a projectile is a section of a parabola (Fig. 2.1.1), which has limiting points: the initial point with the coordinates X = 0, at= 0 and final, for which X = L(flight range), at = 0.
2. Determine the range of the projectile by substituting at= 0 into the trajectory equation. Where do we find it from? L= 24000 m.
3. We find the speed and acceleration of the projectile using projections on the coordinate axes:
At the initial moment of time v 0 = 500 m/s, A= 10 m/s 2.
4. To determine the flight altitude of the projectile, let’s find the time t 1 flight to this point. At the highest point, the projection of velocity onto the axis y equal to zero (Fig. 2.1.1), 
, where t 1 = 40 s. Substituting t 1 into the coordinate expression at, we get the height value N= 8000 m.
5. Radius of curvature of the trajectory
, Where 
.
m; 
m.
Example 2.1.2. In the crank-slider mechanism (Fig. 2.1.2) the crank 1 rotates at a constant angular velocity rad/s. Find the equations of motion, trajectory and speed of the midpoint M connecting rod 2 , If OA = AB= 80 cm.
1. Let's write down the equations of motion of a point M in coordinate form (Fig. 2.1.3)
2. We obtain the trajectory equation by eliminating time t from the equation of motion:
Point trajectory M– an ellipse with a center at the origin and semi-axes of 120 cm and 40 cm.
3. The speed of the point is determined by projections on the coordinate axes
Task 2.1.1. Given the equations of motion of a point, find the equation of its trajectory in coordinate form.
| Equation of motion | Answer |
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Task 2.1.2. Find the equation of the trajectory in coordinate form and the law of motion of a point along the trajectory if the equations of its motion in Cartesian coordinates are given. For the origin of the arc coordinate s accept the initial position of the point.
| Equation of motion | Answer |
| , ; | |
 | ; |
| ; | |
 ;
|
Task 2.1.3. The motion of a point is given by the equations , ( – in cm, – in s). Find the equation of the point’s trajectory in coordinate form, the speed and acceleration, the tangential and normal acceleration of the point, as well as the radius of curvature of the trajectory at time s. Draw the trajectory of the point and the found velocity and acceleration vectors in the drawing. , – in cm, if and when the angle is greatest.
Answer: 1) 
; 2) , , ; , , .
