Note. The center of gravity of a symmetrical figure is on the axis of symmetry.
The center of gravity of the rod is at the middle of the height. The following methods are used to solve problems:
1. symmetry method: the center of gravity of symmetrical figures is on the axis of symmetry;
2. separation method: complex sections are divided into several simple parts, the position of the centers of gravity of which is easy to determine;
3. negative area method: cavities (holes) are considered as part of a section with a negative area.
Examples of problem solving
Example 1. Determine the position of the center of gravity of the figure shown in Fig. 8.4.
Solution
We divide the figure into three parts:
Defined similarly at C = 4.5 cm.
Example 2. Find the position of the center of gravity of a symmetrical bar truss ADBE(Fig. 116), the dimensions of which are as follows: AB = 6m, DE = 3 m and EF = 1m.
Solution
Since the truss is symmetrical, its center of gravity lies on the axis of symmetry D.F. With the selected (Fig. 116) coordinate axes system, the abscissa of the center of gravity of the truss
Consequently, only the ordinate is unknown at C center of gravity of the farm. To determine it, we divide the truss into separate parts (rods). Their lengths are determined from the corresponding triangles.
From ΔAEF we have
From ΔADF we have
The center of gravity of each rod lies in its middle; the coordinates of these centers are easily determined from the drawing (Fig. 116).
The found lengths and ordinates of the centers of gravity of individual parts of the truss are entered into the table and according to the formula
determine the ordinate y s the center of gravity of a given flat truss.
Therefore, the center of gravity WITH the entire truss lies on the axis DF symmetry of the truss at a distance of 1.59 m from the point F.
Example 3. Determine the coordinates of the center of gravity of the composite section. The section consists of a sheet and rolled profiles (Fig. 8.5).
Note. Often frames are welded from different profiles to create the required structure. Thus, metal consumption is reduced and a high-strength structure is formed.
For standard rolled profiles, their own geometric characteristics are known. They are given in the relevant standards.
Solution
1. Let’s designate the figures by numbers and write out the necessary data from the tables:
1 - channel No. 10 (GOST 8240-89); height h = 100 mm; shelf width b= 46 mm; cross-sectional area A 1= 10.9 cm 2;
2 - I-beam No. 16 (GOST 8239-89); height 160 mm; shelf width 81 mm; cross-sectional area A 2 - 20.2 cm 2;
3 - sheet 5x100; thickness 5 mm; width 100mm; cross-sectional area A 3 = 0.5 10 = 5 cm 2.
2. The coordinates of the centers of gravity of each figure can be determined from the drawing.
The composite section is symmetrical, so the center of gravity is on the axis of symmetry and the coordinate X C = 0.
3. Determination of the center of gravity of a composite section:
Example 4. Determine the coordinates of the center of gravity of the section shown in Fig. 8, A. The section consists of two angles 56x4 and channel No. 18. Check the correctness of determining the position of the center of gravity. Indicate its position on the section.
Solution
1. : two corners 56 x 4 and channel No. 18. Let us denote them 1, 2, 3 (see Fig. 8, A).
2. We indicate the centers of gravity each profile, using table 1 and 4 adj. I, and denote them C 1, C 2, C 3.
3. Select a system of coordinate axes. Axis at compatible with the axis of symmetry, and the axis X draw through the centers of gravity of the corners.
4. Determine the coordinates of the center of gravity of the entire section. Since the axis at coincides with the axis of symmetry, then it passes through the center of gravity of the section, therefore x s= 0. Coordinate y s we will determine by the formula
Using the tables in the appendix, we determine the areas of each profile and the coordinates of the centers of gravity:
Coordinates at 1 And at 2 are equal to zero, since the axis X passes through the centers of gravity of the corners. Let's substitute the obtained values into the formula to determine y s:
5. Let us indicate the center of gravity of the section in Fig. 8, a and denote it by the letter C. Let us show the distance y C = 2.43 cm from the axis X to point C.
Since the corners are symmetrically located and have the same area and coordinates, then A 1 = A 2, y 1 = y 2. Therefore, the formula for determining at C can be simplified:
6. Let's check. For this purpose the axis X Let's draw along the lower edge of the corner shelf (Fig. 8, b). Axis at Let's leave it as in the first solution. Formulas for determining x C And at C do not change:
The areas of the profiles will remain the same, but the coordinates of the centers of gravity of the angles and channels will change. Let's write them down:
Find the coordinate of the center of gravity:
According to the found coordinates x s And y s draw point C on the drawing. The position of the center of gravity found in two ways is at the same point. Let's check it out. Difference between coordinates y s, found in the first and second solutions is: 6.51 - 2.43 = 4.08 cm.
This is equal to the distance between the x-axis in the first and second solution: 5.6 - 1.52 = 4.08 cm.
Answer: s= 2.43 cm if the x axis passes through the centers of gravity of the corners, or y c = 6.51 cm if the x-axis runs along the bottom edge of the corner flange.
Example 5. Determine the coordinates of the center of gravity of the section shown in Fig. 9, A. The section consists of I-beam No. 24 and channel No. 24a. Show the position of the center of gravity on the section.
Solution
1.Let's divide the section into rolled profiles: I-beam and channel. Let's denote them by numbers 1 and 2.
3. We indicate the centers of gravity of each profile C 1 and C 2 using application tables.
4. Select a system of coordinate axes. The x axis is compatible with the axis of symmetry, and the y axis is drawn through the center of gravity of the I-beam.
5. Determine the coordinates of the center of gravity of the section. Coordinate y c = 0, since the axis X coincides with the axis of symmetry. We determine the x coordinate with the formula
According to the table 3 and 4 adj. I and the cross-sectional diagram we determine
Let's substitute the numerical values into the formula and get
5. Let's plot point C (the center of gravity of the section) using the found values of x c and y c (see Fig. 9, a).
The solution must be checked independently with the axes positioned as shown in Fig. 9, b. As a result of the solution, we obtain x c = 11.86 cm. The difference between the values of x c for the first and second solutions is 11.86 - 6.11 = 5.75 cm, which is equal to the distance between the y axes for the same solutions b dv /2 = 5.75 cm.
Answer: x c = 6.11 cm, if the y-axis passes through the center of gravity of the I-beam; x c = 11.86 cm if the y-axis passes through the left extreme points of the I-beam.
Example 6. The railway crane rests on rails, the distance between which is AB = 1.5 m (Fig. 1.102). The force of gravity of the crane trolley is G r = 30 kN, the center of gravity of the trolley is at point C, lying on the line KL of the intersection of the plane of symmetry of the trolley with the plane of the drawing. The gravity force of the crane winch Q l = 10 kN is applied at the point D. The gravity force of the counterweight G„=20 kN is applied at point E. The gravity force of the boom G c = 5 kN is applied at point H. The outreach of the crane relative to the line KL is 2 m. Determine the stability coefficient of the crane in an unloaded state and what load F can be lifted with this crane, provided that the stability coefficient must be at least two.
Solution
1. When unloaded, the crane is in danger of tipping over when turning around the rail A. Therefore, relative to the point A moment of stability
2. Overturning moment relative to a point A is created by the force of gravity of the counterweight, i.e.
3. Hence the stability coefficient of the crane in an unloaded state
4. When loading the crane boom with cargo F there is a danger of the crane overturning when turning near rail B. Therefore, relative to the point IN moment of stability
5. Overturning moment relative to the rail IN
6. According to the conditions of the problem, operation of the crane is permitted with a stability coefficient k B ≥ 2, i.e.
Test questions and assignments
1. Why can the forces of attraction to the Earth acting on points of the body be taken as a system of parallel forces?
2. Write down formulas for determining the position of the center of gravity of inhomogeneous and homogeneous bodies, formulas for determining the position of the center of gravity of flat sections.
3. Repeat the formulas for determining the position of the center of gravity of simple geometric shapes: rectangle, triangle, trapezoid and half circle.
4. 
What is the static moment of area?
5. Calculate the static moment of this figure about the axis Ox. h= 30 cm; b= 120 cm; With= 10 cm (Fig. 8.6).
6. Determine the coordinates of the center of gravity of the shaded figure (Fig. 8.7). Dimensions are given in mm.
7. Determine the coordinate at figure 1 of the composite section (Fig. 8.8).
When deciding, use the reference data from the GOST tables “Hot-rolled steel” (see Appendix 1).
Physics lesson notes, grade 7
Topic: Determining the center of gravity
Physics teacher, Argayash Secondary School No. 2
Khidiyatulina Z.A.
Laboratory work:
"Determination of the center of gravity of a flat plate"
Target : finding the center of gravity of a flat plate.
Theoretical part:
All bodies have a center of gravity. The center of gravity of a body is the point relative to which the total moment of gravity acting on the body is zero. For example, if you hang an object by its center of gravity, it will remain at rest. That is, its position in space will not change (it will not turn upside down or on its side). Why do some bodies tip over while others don't? If you draw a line perpendicular to the floor from the center of gravity of the body, then if the line goes beyond the boundaries of the body’s support, the body will fall. The larger the area of support, the closer the center of gravity of the body is to the central point of the area of support and the center line of the center of gravity, the more stable the position of the body will be. For example, the center of gravity of the famous Leaning Tower of Pisa is located just two meters from the middle of its support. And the fall will happen only when this deviation is about 14 meters. The center of gravity of the human body is approximately 20.23 centimeters below the navel. An imaginary line drawn vertically from the center of gravity passes exactly between the feet. For a tumbler doll, the secret also lies in the center of gravity of the body. Its stability is explained by the fact that the center of gravity of the tumbler is at the very bottom; it actually stands on it. The condition for maintaining the balance of a body is the passage of the vertical axis of its common center of gravity within the area of the body’s support. If the vertical center of gravity of the body leaves the support area, the body loses balance and falls. Therefore, the larger the area of support, the closer the center of gravity of the body is located to the central point of the area of support and the central line of the center of gravity, the more stable the position of the body will be. The area of support when a person is in a vertical position is limited by the space that is under the soles and between the feet. The center point of the vertical line of the center of gravity on the foot is 5 cm in front of the heel tubercle. The sagittal size of the support area always prevails over the frontal one, therefore the displacement of the vertical line of the center of gravity occurs more easily to the right and left than backward, and is especially difficult forward. In this regard, stability during turns during fast running is significantly less than in the sagittal direction (forward or backward). A foot in shoes, especially with a wide heel and a hard sole, is more stable than without shoes, as it acquires a larger area of support.
Practical part:
Purpose of the work: Using the proposed equipment, experimentally find the position of the center of gravity of two figures made of cardboard and a triangle.
Equipment:Tripod, thick cardboard, triangle from a school kit, ruler, tape, thread, pencil...
Task 1: Determine the position of the center of gravity of a flat figure of arbitrary shape
Using scissors, cut out a random shape from cardboard. Attach the thread to it at point A with tape. Hang the figure by the thread to the tripod leg. Using a ruler and pencil, mark the vertical line AB on the cardboard.
Move the thread attachment point to position C. Repeat the above steps.
Point O of the intersection of lines AB andCDgives the desired position of the center of gravity of the figure.
Task 2: Using only a ruler and pencil, find the position of the center of gravity of a flat figure
Using a pencil and ruler, divide the shape into two rectangles. By construction, find the positions O1 and O2 of their centers of gravity. It is obvious that the center of gravity of the entire figure is on the O1O2 line
Divide the figure into two rectangles in another way. By construction, find the positions of the centers of gravity O3 and O4 of each of them. Connect points O3 and O4 with a line. The intersection point of lines O1O2 and O3O4 determines the position of the figure’s center of gravity
Task 2: Determine the position of the center of gravity of the triangle
Using tape, secure one end of the thread at the top of the triangle and hang it from the tripod leg. Using a ruler, mark the direction AB of the gravity line (make a mark on the opposite side of the triangle)
Repeat the same procedure, hanging the triangle from vertex C. On the opposite vertex C side of the triangle, make a markD.
Using tape, attach pieces of thread AB andCD. Point O of their intersection determines the position of the center of gravity of the triangle. In this case, the center of gravity of the figure is outside the body itself.
III . Solving quality problems
1.For what purpose do circus performers hold heavy poles in their hands when walking on a tightrope?
2. Why does a person carrying a heavy load on his back lean forward?
3. Why can’t you get up from a chair unless you tilt your body forward?
4.Why does the crane not tip towards the load being lifted? Why, without a load, does the crane not tip towards the counterweight?
5. Why do cars and bicycles, etc. Is it better to put brakes on the rear wheels rather than the front wheels?
6. Why does a truck loaded with hay overturn more easily than the same truck loaded with snow?
Rectangle.
Since a rectangle has two axes of symmetry, its center of gravity is at the intersection of the axes of symmetry, i.e. at the point of intersection of the diagonals of the rectangle. 
Triangle.
The center of gravity lies at the point of intersection of its medians. From geometry it is known that the medians of a triangle intersect at one point and are divided in a ratio of 1:2 from the base. 
Circle. Since a circle has two axes of symmetry, its center of gravity is at the intersection of the axes of symmetry.
Semicircle.
A semicircle has one axis of symmetry, then the center of gravity lies on this axis. Another coordinate of the center of gravity is calculated by the formula: . 
Many structural elements are made from standard rolled products - angles, I-beams, channels and others. All dimensions, as well as geometric characteristics of rolled profiles, are tabular data that can be found in reference literature in tables of normal assortment (GOST 8239-89, GOST 8240-89).
Example 1. Determine the position of the center of gravity of the figure shown in the figure.
Solution:
We select the coordinate axes so that the Ox axis goes along the bottommost overall dimension, and the Oy axis goes along the leftmost overall dimension.
We break a complex figure into a minimum number of simple figures:
rectangle 20x10;
triangle 15x10;
circle R=3 cm.
We calculate the area of each simple figure and its coordinates of the center of gravity. The calculation results are entered into the table
|
Figure no. |
Area of figure A, |
Center of gravity coordinates |
|
|
| |||
Answer: C(14.5; 4.5)
Example 2
.
Determine the coordinates of the center of gravity of a composite section consisting of a sheet and rolled sections. 
Solution.
We select the coordinate axes as shown in the figure.
Let's designate the figures by numbers and write out the necessary data from the table:
|
Figure no. |
Area of figure A, |
Center of gravity coordinates |
|
|
|
|||
|
|
|||
We calculate the coordinates of the center of gravity of the figure using the formulas:
Answer: C(0; 10)
Laboratory work No. 1 “Determination of the center of gravity of composite flat figures”
Target: Determine the center of gravity of a given flat complex figure using experimental and analytical methods and compare their results.
Work order
Divide the figure into the minimum number of figures whose centers of gravity we know how to determine.
Indicate the area numbers and coordinates of the center of gravity of each figure.
Calculate the coordinates of the center of gravity of each figure.
Calculate the area of each figure.
Calculate the coordinates of the center of gravity of the entire figure using the formulas (the position of the center of gravity is plotted on the drawing of the figure):
Draw your flat figure in your notebooks in size, indicating the coordinate axes.
Determine the center of gravity analytically.
The installation for experimentally determining the coordinates of the center of gravity using the hanging method consists of a vertical stand 1
(see figure) to which the needle is attached 2
. Flat figure 3
Made of cardboard, which is easy to punch holes in. Holes A
And IN
pierced at randomly located points (preferably at the furthest distance from each other). A flat figure is suspended on a needle, first at a point A
, and then at the point IN
. Using a plumb line 4
, attached to the same needle, draw a vertical line on the figure with a pencil corresponding to the thread of the plumb line. Center of gravity WITH
the figure will be located at the intersection point of the vertical lines drawn when hanging the figure at the points A
And IN
.
6.1. General information
Center of Parallel Forces
Let us consider two parallel forces directed in one direction, and , applied to the body at points A 1 and A 2 (Fig.6.1). This system of forces has a resultant, the line of action of which passes through a certain point WITH. Point position WITH can be found using Varignon's theorem:
If you turn the forces and near the points A 1 and A 2 in one direction and at the same angle, then we get a new system of parallel salas having the same modules. In this case, their resultant will also pass through the point WITH. This point is called the center of parallel forces.
Let's consider a system of parallel and identically directed forces applied to a solid body at points. This system has a resultant.
If each force of the system is rotated near the points of their application in the same direction and at the same angle, then new systems of identically directed parallel forces with the same modules and points of application will be obtained. The resultant of such systems will have the same modulus R, but every time a different direction. Having folded my strength F 1 and F 2 we find that their resultant R 1, which will always pass through the point WITH 1, the position of which is determined by the equality . Folding further R 1 and F 3, we find their resultant, which will always pass through the point WITH 2 lying on a straight line A 3
WITH 2. Having completed the process of adding forces to the end, we will come to the conclusion that the resultant of all forces will indeed always pass through the same point WITH, whose position relative to the points will be unchanged.
Dot WITH, through which the line of action of the resultant system of parallel forces passes for any rotation of these forces near the points of their application in the same direction at the same angle is called the center of parallel forces (Fig. 6.2).
Fig.6.2
Let us determine the coordinates of the center of parallel forces. Since the position of the point WITH relative to the body is unchanged, then its coordinates do not depend on the choice of coordinate system. Let's turn all the forces around their application so that they become parallel to the axis OU and apply Varignon’s theorem to rotated forces. Because R" is the resultant of these forces, then, according to Varignon’s theorem, we have 
, because , , we get
From here we find the coordinate of the center of parallel forces zc:
To determine the coordinates xc Let's create an expression for the moment of forces about the axis Oz.
To determine the coordinates yc let's turn all the forces so that they become parallel to the axis Oz.
The position of the center of parallel forces relative to the origin (Fig. 6.2) can be determined by its radius vector:
6.2. Center of gravity of a rigid body
Center of gravity of a rigid body is a point invariably associated with this body WITH, through which the line of action of the resultant forces of gravity of a given body passes, for any position of the body in space.
The center of gravity is used in studying the stability of equilibrium positions of bodies and continuous media under the influence of gravity and in some other cases, namely: in the strength of materials and in structural mechanics - when using Vereshchagin's rule.
There are two ways to determine the center of gravity of a body: analytical and experimental. The analytical method for determining the center of gravity directly follows from the concept of the center of parallel forces.
The coordinates of the center of gravity, as the center of parallel forces, are determined by the formulas:
Where R- whole body weight; pk- weight of body particles; xk, yk, zk- coordinates of body particles.
For a homogeneous body, the weight of the entire body and any part of it is proportional to the volume P=Vγ, pk =vk γ, Where γ
- weight per unit volume, V- body volume. Substituting expressions P, pk into the formula for determining the coordinates of the center of gravity and, reducing by a common factor γ
, we get:
Dot WITH, whose coordinates are determined by the resulting formulas, is called center of gravity of the volume.
If the body is a thin homogeneous plate, then the center of gravity is determined by the formulas:
Where S- area of the entire plate; sk- area of its part; xk, yk- coordinates of the center of gravity of the plate parts.
Dot WITH in this case it is called center of gravity area.
The numerators of expressions that determine the coordinates of the center of gravity of plane figures are called with static moments of area relative to the axes at And X:
Then the center of gravity of the area can be determined by the formulas:
For bodies whose length is many times greater than the cross-sectional dimensions, the center of gravity of the line is determined. The coordinates of the line's center of gravity are determined by the formulas:
Where L- line length; lk- the length of its parts; xk, yk, zk- coordinate of the center of gravity of parts of the line.
6.3. Methods for determining the coordinates of the centers of gravity of bodies
Based on the formulas obtained, it is possible to propose practical methods for determining the centers of gravity of bodies.
1. Symmetry. If a body has a center of symmetry, then the center of gravity is at the center of symmetry.
If the body has a plane of symmetry. For example, the XOU plane, then the center of gravity lies in this plane.
2. Splitting. For bodies consisting of bodies with simple shapes, the splitting method is used. The body is divided into parts, the center of gravity of which is determined by the method of symmetry. The center of gravity of the entire body is determined by the formulas for the center of gravity of volume (area).
Example. Determine the center of gravity of the plate shown in the figure below (Fig. 6.3). The plate can be divided into rectangles in various ways and the coordinates of the center of gravity of each rectangle and their area can be determined.
Fig.6.3
Answer: xc=17.0cm; yc=18.0cm.
3. Addition. This method is a special case of the partitioning method. It is used when the body has cutouts, cuts, etc., if the coordinates of the center of gravity of the body without the cutout are known.
Example. Determine the center of gravity of a circular plate having a cutout radius r = 0,6 R(Fig. 6.4).
Fig.6.4
A round plate has a center of symmetry. Let's place the origin of coordinates at the center of the plate. Plate area without cutout, cutout area. Square plate with cutout; .
The plate with a cutout has an axis of symmetry О1 x, hence, yc=0.
4. Integration. If the body cannot be divided into a finite number of parts, the positions of the centers of gravity of which are known, the body is divided into arbitrary small volumes, for which the formula using the partitioning method takes the form: 
.
Then they go to the limit, directing the elementary volumes to zero, i.e. contracting volumes into points. The sums are replaced by integrals extended to the entire volume of the body, then the formulas for determining the coordinates of the center of gravity of the volume take the form:
Formulas for determining the coordinates of the center of gravity of an area:
The coordinates of the center of gravity of the area must be determined when studying the equilibrium of plates, when calculating the Mohr integral in structural mechanics.
Example. Determine the center of gravity of a circular arc of radius R with central angle AOB= 2α (Fig. 6.5).
Rice. 6.5
The arc of a circle is symmetrical to the axis Oh, therefore, the center of gravity of the arc lies on the axis Oh, yс = 0.
According to the formula for the center of gravity of a line:
6.Experimental method. The centers of gravity of inhomogeneous bodies of complex configuration can be determined experimentally: by the method of hanging and weighing. The first method is to suspend the body on a cable at various points. The direction of the cable on which the body is suspended will give the direction of gravity. The point of intersection of these directions determines the center of gravity of the body.
The weighing method involves first determining the weight of a body, such as a car. Then the pressure of the vehicle's rear axle on the support is determined on the scales. By drawing up an equilibrium equation relative to a point, for example, the axis of the front wheels, you can calculate the distance from this axis to the center of gravity of the car (Fig. 6.6).
Fig.6.6
Sometimes, when solving problems, different methods for determining the coordinates of the center of gravity should be used simultaneously.
6.4. Centers of gravity of some simple geometric figures
To determine the centers of gravity of bodies of frequently occurring shapes (triangle, circular arc, sector, segment), it is convenient to use reference data (Table 6.1).
Table 6.1
Coordinates of the center of gravity of some homogeneous bodies
|
№ |
Name of the figure |
Drawing |
|
Arc of a circle: the center of gravity of an arc of a uniform circle is on the axis of symmetry (coordinate uc=0).
R- radius of the circle. |
|
|
|
Homogeneous circular sector uc=0).
where α is half the central angle; R- radius of the circle. |
|
|
|
Segment: the center of gravity is located on the axis of symmetry (coordinate uc=0). where α is half the central angle; R- radius of the circle. |
|
|
|
Semicircle:
|
|
|
|
Triangle: the center of gravity of a homogeneous triangle is at the point of intersection of its medians. Where x1, y1, x2, y2, x3, y3- coordinates of the triangle vertices |
|
|
|
Cone: the center of gravity of a uniform circular cone lies at its height and is located at a distance of 1/4 of the height from the base of the cone.
|
Lecture 4. Center of gravity.
This lecture covers the following issues
1. Center of gravity of a solid body.
2. Coordinates of the centers of gravity of inhomogeneous bodies.
3. Coordinates of the centers of gravity of homogeneous bodies.
4. Methods for determining the coordinates of centers of gravity.
5. Centers of gravity of some homogeneous bodies.
The study of these issues is necessary in the future to study the dynamics of the movement of bodies taking into account sliding and rolling friction, the dynamics of the movement of the center of mass of a mechanical system, kinetic moments, to solve problems in the discipline “Strength of Materials”.
Bringing parallel forces.
After we have considered bringing a flat system and an arbitrary spatial system of forces to the center, we again return to considering the special case of a system of parallel forces.
Bringing two parallel forces.
In the course of considering such a system of forces, the following three cases of reduction are possible.
1. System of two collinear forces. Let us consider a system of two parallel forces directed in one direction P And Q, applied at points A And IN. We will assume that the forces are perpendicular to this segment (Fig. 1, A).
WITH, belonging to the segment AB and satisfying the condition:
AC/NE = Q/P.(1)
Main vector of the system R C = P + Q is equal in modulus to the sum of these forces: R C = P + Q.
WITH taking into account (1) is equal to zero:MC = P ∙ AC- Q∙ NE = 0.
Thus, as a result of the casting we got: R C ≠ 0, MC= 0. This means that the main vector is equivalent to the resultant passing through the center of reduction, that is:
The resultant of collinear forces is equal in modulus to their sum, and its line of action divides the segment connecting the points of their application, in inverse proportion to the moduli of these forces in an internal manner.
Note that the position of the point WITH will not change if the forces R And Q turn an angleα. Dot WITH, which has this property is called center of parallel forces.
2. System of two anticollinear and forces not equal in magnitude. May the strength P And Q, applied at points A And IN, parallel, directed in opposite directions and unequal in magnitude (Fig. 1, b).
Let us choose a point as the reduction center WITH, which still satisfies relation (1) and lies on the same line, but outside the segment AB.
The main vector of this system R C = P + Q the modulus will now be equal to the difference between the moduli of the vectors: R C = Q - P.
The main point regarding the center WITH is still zero:MC = P ∙ AC- Q∙ NE= 0, so
Resultant anticollinear and forces that are not equal in magnitude are equal to their difference, directed towards the greater force, and its line of action divides the segment connecting the points of their application, in inverse proportion to the external moduli of these forces.
Fig.1
3. System of two anticollinear and forces equal in magnitude. Let's take the previous case of reduction as the initial one. Let's fix the force R, and strength Q let us direct the modulus to the force R.
Then at Q → R in formula (1) the relation AC/NE → 1. This means that AC → NE, that is, the distance AC →∞ .
In this case, the module of the main vector R C → 0, and the modulus of the main moment does not depend on the position of the center of reduction and remains equal to the original value:
MC = P ∙ AC- Q∙ NE = P ∙ ( AC- NE) =P ∙ AB.
So, in the limit we have obtained a system of forces for which R C = 0, MC≠ 0, and the center of reduction is removed to infinity, which cannot be replaced by the resultant. It is not difficult to recognize a couple of forces in this system, so a pair of forces has no resultant.
Center of the system of parallel forces.
Consider the system n strength P i, applied at pointsA i (x i , y i , z i) and parallel to the axisOv with orth l(Fig. 2).
If we exclude in advance the case of a system equivalent to a pair of forces, it is not difficult, based on the previous paragraph, to prove the existence of its resultantR.
Let's determine the coordinates of the centerC(x c, y c, z c) parallel forces, that is, the coordinates of the point of application of the resultant of this system.
For this purpose, we use Varignon’s theorem, based on which:
M0 (R) = Σ M0(P i).
Fig.2
The vector-moment of a force can be represented as a vector product, therefore:
M 0 (R) = r c× R = Σ M0i(P i) = Σ ( r i× P i ).
Considering that R = Rv ∙ l, A P i = Pvi ∙ l and using the properties of the vector product, we get:
r c × Rv ∙ l = Σ ( r i × Pvi ∙ l),
r c ∙ R v× l = Σ ( r i ∙ Pvi × l) = Σ ( r i ∙ Pvi ) × l,
or:
[ r c R v - Σ ( r i Pvi )] × l= 0.
The last expression is valid only if the expression in square brackets is equal to zero. Therefore, omitting the indexvand taking into account that the resultantR = Σ P i , from here we get:
r c = (Σ P i r i )/(Σ P i ).
Projecting the last vector equality on the coordinate axis, we obtain the required expression for the coordinates of the center of parallel forces:
x c = (Σ P i x i)/(Σ P i );
y c = (Σ P i y i )/(Σ P i );(2)
z c = (Σ P i z i )/(Σ P i ).
Center of gravity of bodies.
Coordinates of the centers of gravity of a homogeneous body.
Consider a rigid body weighing P and volume V in the coordinate system Oxyz, where are the axes x And y connected to the surface of the earth, and the axis z aimed at the zenith.
If you break the body into elementary parts with a volume∆ V i , then the force of attraction will act on each part of it∆ P i, directed towards the center of the Earth. Let us assume that the dimensions of the body are significantly smaller than the dimensions of the Earth, then the system of forces applied to the elementary parts of the body can be considered not converging, but parallel (Fig. 3), and all the conclusions of the previous chapter are applicable to it.
Fig.3
Definition . The center of gravity of a solid body is the center of parallel forces of gravity of the elementary parts of this body.
Let us recall that specific gravity of an elementary part of the body is called the ratio of its weight∆ P i to volume ∆ V i : γ i = ∆ P i/ ∆ V i . For a homogeneous body this value is constant:γ i = γ = P/ V.
Substituting ∆ into (2) P i = γ i ∙∆ V i instead of P i, taking into account the last remark and reducing the numerator and denominator byg, we get expressions for the coordinates of the center of gravity of a homogeneous body:
x c = (Σ ∆ V i∙ x i)/(Σ ∆ V i);
y c = (Σ ∆ V i∙ y i )/(Σ ∆ V i);(3)
z c = (Σ ∆ V i∙ z i )/(Σ ∆ V i).
Several theorems are useful in determining the center of gravity.
1) If a homogeneous body has a plane of symmetry, then its center of gravity is in this plane.
If the axes X And at located in this plane of symmetry, then for each point with coordinates. And the coordinate according to (3), will be equal to zero, because in total All members with opposite signs are destroyed in pairs. This means that the center of gravity is located in the plane of symmetry.
2) If a homogeneous body has an axis of symmetry, then the center of gravity of the body is on this axis.
Indeed, in this case, if the axiszdraw along the axis of symmetry, for each point with coordinatesyou can find a point with coordinates and coordinates and , calculated using formulas (3), will be equal to zero.
The third theorem is proved in a similar way.
3) If a homogeneous body has a center of symmetry, then the center of gravity of the body is at this point.
And a few more comments.
First. If the body can be divided into parts for which the weight and position of the center of gravity are known, then there is no need to consider each point, and in formulas (3) P i – determined as the weight of the corresponding part and– as the coordinates of its center of gravity.
Second. If the body is homogeneous, then the weight of an individual part of it, Where - specific gravity of the material from which the body is made, and V i - the volume of this part of the body. And formulas (3) will take a more convenient form. For example,
And similarly, where - volume of the whole body.
Third note. Let the body have the form of a thin plate with an area F and thickness t, lying in the plane Oxy. Substituting in (3)∆ V i =t ∙ ∆F i , we obtain the coordinates of the center of gravity of a homogeneous plate:
x c = (Σ ∆ F i∙ x i) / (Σ ∆ F i);
y c = (Σ ∆ F i∙ y i ) / (Σ ∆ F i).
z c = (Σ ∆ F i∙ z i ) / (Σ ∆ F i).
Where – coordinates of the center of gravity of individual plates;– total body area.
Fourth note. For a body in the form of a thin curved rod of length L with cross-sectional area a elementary volume∆ V i = a ∙∆ L i , That's why coordinates of the center of gravity of a thin curved rod will be equal:
x c = (Σ ∆ L i∙ x i)/(Σ ∆ L i);
y c = (Σ ∆ L i∙ y i )/(Σ ∆ L i);(4)
z c = (Σ ∆ L i∙ z i )/(Σ ∆ L i).
Where – coordinates of the center of gravityi-th section; .
Note that, according to the definition, the center of gravity is a geometric point; it can also lie outside the boundaries of a given body (for example, for a ring).
Note.
In this section of the course we do not differentiate between gravity, gravity and body weight. In reality, gravity is the difference between the gravitational force of the Earth and the centrifugal force caused by its rotation.
Coordinates of the centers of gravity of inhomogeneous bodies.
Center of gravity coordinates inhomogeneous solid(Fig.4) in the selected reference system are determined as follows:
Fig.4
Where - weight per unit volume of a body (specific gravity)
- whole body weight.
non-uniform surface(Fig. 5), then the coordinates of the center of gravity in the selected reference system are determined as follows:
Fig.5
Where - weight per unit body area,
- whole body weight.
If the solid is non-uniform line(Fig. 6), then the coordinates of the center of gravity in the selected reference system are determined as follows:
Fig.6
Where - weight per body length,
Whole body weight.
Methods for determining the coordinates of the center of gravity.
Based on the general formulas obtained above, it is possible to indicate specific methods determining the coordinates of the centers of gravity of bodies.
1. Symmetry. If a homogeneous body has a plane, axis or center of symmetry (Fig. 7), then its center of gravity lies, respectively, in the plane of symmetry, axis of symmetry or in the center of symmetry.
Fig.7
2. Splitting. The body is divided into a finite number of parts (Fig. 8), for each of which the position of the center of gravity and area are known.
Fig.8
S =S 1 +S 2.
3.Negative area method. A special case of the partitioning method (Fig. 9). It applies to bodies that have cutouts if the centers of gravity of the body without the cutout and the cutout part are known. A body in the form of a plate with a cutout is represented by a combination of a solid plate (without a cutout) with an area S 1 and the area of the cut part S2.
Fig.9
S = S 1 - S 2.
4.Grouping method. It is a good complement to the last two methods. After dividing a figure into its component elements, it is convenient to combine some of them again in order to then simplify the solution by taking into account the symmetry of this group.
Centers of gravity of some homogeneous bodies.
1) Center of gravity of a circular arc. Consider the arc AB radiusR with central angle. Due to symmetry, the center of gravity of this arc lies on the axisOx(Fig. 10).
Fig.10
Let's find the coordinate according to the formula . To do this, select on the arc AB element MM ’ length, whose position is determined by the angle. Coordinate X element MM' will. Substituting these values X And d l and keeping in mind that the integral must be extended over the entire length of the arc, we obtain:
where L is the length of arc AB, equal to .
From here we finally find that the center of gravity of a circular arc lies on its axis of symmetry at a distance from the center O equal
where is the angle measured in radians.
2) Center of gravity of the triangle's area. Consider a triangle lying in the plane Oxy, the coordinates of the vertices of which are known: A i (x i,y i ), (i= 1,2,3). Breaking the triangle into narrow strips parallel to the side A 1 A 2, we come to the conclusion that the center of gravity of the triangle must belong to the median A 3 M 3 (Fig. 11).
Fig.11
Breaking a triangle into strips parallel to the side A 2 A 3, we can verify that it must lie on the median A 1 M 1 . Thus, the center of gravity of a triangle lies at the point of intersection of its medians, which, as is known, separates a third part from each median, counting from the corresponding side.
In particular, for the median A 1 M 1 we obtain, taking into account that the coordinates of the point M 1 - this is the arithmetic mean of the coordinates of the vertices A 2 and A 3 :
x c = x 1 + (2/3) ∙ (xM 1 - x 1 ) = x 1 + (2/3) ∙ [(x 2 + x 3 )/2 - x 1 ] = (x 1 + x 2 + x 3 )/3.
Thus, the coordinates of the triangle’s center of gravity are the arithmetic mean of the coordinates of its vertices:
x c =(1/3) Σ x i ; y c =(1/3) Σ y i .
3) Center of gravity of the area of a circular sector. Consider a sector of a circle with radius R with central angle 2α , located symmetrically about the axis Ox (Fig. 12) .
It's obvious that y c = 0, and the distance from the center of the circle from which this sector is cut to its center of gravity can be determined by the formula:
Fig.12
The easiest way to calculate this integral is by dividing the integration domain into elementary sectors with an angle dφ . Accurate to infinitesimals of the first order, such a sector can be replaced by a triangle with a base equal to R × dφ and height R. The area of such a triangle dF =(1/2)R 2 ∙ dφ , and its center of gravity is at a distance of 2/3 R from the vertex, therefore in (5) we put x = (2/3)R∙ cosφ. Substituting in (5) F= α R 2, we get:
Using the last formula, we calculate, in particular, the distance to the center of gravity semicircle.
Substituting α = π /2 into (2), we obtain: x c = (4 R)/(3π) ≅ 0.4 R .
Example 1.Let us determine the center of gravity of the homogeneous body shown in Fig. 13.
Fig.13
Solution.The body is homogeneous, consisting of two parts with a symmetrical shape. Coordinates of their centers of gravity:
Their volumes:
Therefore, the coordinates of the center of gravity of the body
Example 2. Let us find the center of gravity of a plate bent at a right angle. Dimensions are in the drawing (Fig. 14).
Fig.14
Solution. Coordinates of the centers of gravity:
0.
Areas:
That's why:
Example 3.
On a square sheet
cm square hole cut
cm (Fig. 15). Let's find the center of gravity of the sheet. Example 4. Find the position of the center of gravity of the plate shown in Fig. 16. Dimensions are given in centimeters.
Fig.16
Solution. Let's divide the plate into figures (Fig. 17), centers the severity of which is known.
The areas of these figures and the coordinates of their centers of gravity:
1) a rectangle with sides 30 and 40 cm,S 1 =30 ∙ 40=1200 cm 2 ; x 1=15 cm; at 1 =20 cm.
2) a right triangle with a base of 50 cm and a height of 40 cm;S 2 =0,5 ∙ 50 ∙ 40= 1000 cm 2 ; X 2 =30+50/3=46.7 cm; y 2 =40/3 =13.3 cm;
3) half circle radius circle r = 20 cm;S 3 =0,5 ∙π∙ 20 2 =628 cm 2 ; X 3 =4 R /3 π =8.5 cm; at
Solution. Recall that in physics the density of a bodyρ and its specific gravitygrelated by the relation:γ = ρ g , Whereg - acceleration of gravity. To find the mass of such a homogeneous body, you need to multiply the density by its volume.
Fig.19
The term “linear” or “linear” density means that to determine the mass of a truss rod, the linear density must be multiplied by the length of this rod.
To solve the problem, you can use the partitioning method. Representing a given truss as a sum of 6 individual rods, we obtain:
WhereL i lengthi th truss rod, andx i , y i - coordinates of its center of gravity.
The solution to this problem can be simplified by grouping the last 5 bars of the truss. It is easy to see that they form a figure with a center of symmetry located in the middle of the fourth rod, where the center of gravity of this group of rods is located.
Thus, a given truss can be represented by a combination of only two groups of rods.
The first group consists of the first rod, for itL 1 = 4 m,x 1 = 0 m,y 1 = 2 m. The second group of rods consists of five rods, for itL 2 = 20 m,x 2 = 3 m,y 2 = 2 m.
The coordinates of the center of gravity of the truss are found using the formula:
x c = (L 1 ∙ x 1 + L 2 ∙ x 2 )/(L 1 + L 2 ) = (4∙0 + 20∙3)/24 = 5/2 m;
y c = (L 1 ∙ y 1 + L 2 ∙ y 2 )/(L 1 + L 2 ) = (4∙2 + 20∙2)/24 = 2 m.
Note that the center WITH lies on the straight line connecting WITH 1 and WITH 2 and divides the segment WITH 1 WITH 2 regarding: WITH 1 WITH/SS 2 = (x c - x 1 )/(x 2 - x c ) = L 2 / L 1 = 2,5/0,5.
Self-test questions
- What is called the center of parallel forces?
- How are the coordinates of the center of parallel forces determined?
- How to determine the center of parallel forces whose resultant is zero?
- What properties does the center of parallel forces have?
- What formulas are used to calculate the coordinates of the center of parallel forces?
- What is the center of gravity of a body called?
- Why can the gravitational forces of the Earth acting on a point on a body be taken as a system of parallel forces?
- Write down the formula for determining the position of the center of gravity of inhomogeneous and homogeneous bodies, the formula for determining the position of the center of gravity of flat sections?
- Write down the formula for determining the position of the center of gravity of simple geometric shapes: rectangle, triangle, trapezoid and half a circle?
- What is called the static moment of area?
- Give an example of a body whose center of gravity is located outside the body.
- How are the properties of symmetry used to determine the centers of gravity of bodies?
- What is the essence of the method of negative weights?
- Where is the center of gravity of a circular arc located?
- What graphical construction can be used to find the center of gravity of a triangle?
- Write down the formula that determines the center of gravity of a circular sector.
- Using formulas that determine the centers of gravity of a triangle and a circular sector, derive a similar formula for a circular segment.
- What formulas are used to calculate the coordinates of the centers of gravity of homogeneous bodies, flat figures and lines?
- What is called the static moment of the area of a plane figure relative to the axis, how is it calculated and what dimension does it have?
- How to determine the position of the center of gravity of an area if the position of the centers of gravity of its individual parts is known?
- What auxiliary theorems are used to determine the position of the center of gravity?