Positions of the center of gravity of some figures. Determining the center of gravity of plane figures Measuring the center of gravity

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. When solving problems, the following methods are used:

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. method of negative areas: cavities (holes) are considered as part of a section with a negative area.

Examples of problem solving

Example1. Determine the position of the center of gravity of the figure shown in fig. 8.4.

Solution

We break the figure into three parts:

Similarly defined at C = 4.5 cm.

Example 2 Find the position of the center of gravity of a symmetrical rod truss ADBE(Fig. 116), the dimensions of which are as follows: AB = 6m, D.E.= 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) system of coordinate axes of the abscissa of the center of gravity of the farm

Unknown, therefore, is only the ordinate at C farm center of gravity. To determine it, we divide the farm into separate parts (rods). Their lengths are determined from the corresponding triangles.

From ∆AEF 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 farm are entered in the table and according to the formula

determine the ordinate u s the center of gravity of this flat truss.

Therefore, the center of gravity FROM the whole truss lies on the axis D.F. truss symmetry 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, creating the necessary design. Thus, metal consumption is reduced and a high-strength structure is formed.

For standard rolled sections, their own geometric characteristics are known. They are given in the relevant standards.

Solution

1. We denote 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\u003d 10.9 cm 2;

2 - I-beam No. 16 (GOST 8239-89); height 160 mm; shelf width 81 mm; sectional area A 2 - 20.2 cm 2;

3 - sheet 5x100; thickness 5 mm; width 100mm; sectional area A 3 \u003d 0.5 10 \u003d 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. Determining the center of gravity of a composite section:

Example 4 Determine the coordinates of the center of gravity of the section shown in fig. eight, a. The section consists of two corners 56x4 and channel No. 18. Check the correctness of determining the position of the center of gravity. Specify its position on the section.

Solution

1. : two corners 56 x 4 and channel No. 18. Let's denote them 1, 2, 3 (see Fig. 8, a).

2. Indicate the centers of gravity each profile using table. 1 and 4 adj. I, and denote them C 1, C 2, From 3 .

3. Let's choose 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 u s define by the formula

Using the application tables, we determine the areas of each profile and the coordinates of the centers of gravity:

Coordinates 1 and at 2 are equal to zero, since the axis X passes through the centers of gravity of the corners. Substitute the obtained values ​​into the formula to determine u s:

5. Let us indicate the center of gravity of the section in Fig. 8, and we will denote it by the letter C. We show the distance y C \u003d 2.43 cm from the axis X to point C.

Since the corners are symmetrically located, have the same area and coordinates, then A 1 \u003d A 2, y 1 = y 2 . Therefore, the formula for determining at C can be simplified:

6. Let's do a check. For this 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 profile areas will remain the same, but the coordinates of the centers of gravity of the corners and the channel will change. Let's write them out:

Finding the coordinate of the center of gravity:

According to the found coordinates x s and u s we put 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 at s, found in the first and second solution is: 6.51 - 2.43 \u003d 4.08 cm.

This is equal to the distance between the x-axes in the first and second solution: 5.6 - 1.52 = 4.08 cm.

Answer: at= 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 an I-beam No. 24 and a channel No. 24a. Show the position of the center of gravity on the section.

Solution

1.Let's break the section into rolled profiles: I-beam and channel. Let's call them 1 and 2.

3. We indicate the centers of gravity of each profile C 1 and C 2 using application tables.

4. Let's choose a system of coordinate axes. The x-axis is compatible with the axis of symmetry, and we draw the y-axis through the center of gravity of the I-beam.

5. Determine the coordinates of the center of gravity of the section. The y-coordinate c = 0, since the axis X coincides with the axis of symmetry. The x-coordinate with is determined by the formula

According to the table 3 and 4 app. I and the section scheme, we define

Substitute the numerical values ​​into the formula and get

5. Let's mark the point C (centre of gravity of the section) according to the found values ​​x c and y c (see Fig. 9, a).

Verification of the solution must be performed independently with the position of the axes, as shown in Fig. 9, b. As a result of the solution, we get x c \u003d 11.86 cm. The difference between the values ​​\u200b\u200bof x c for the first and second solutions is 11.86 - 6.11 \u003d 5.75 cm, which is equal to the distance between the y axes with the same solutions b dv / 2 = 5.75 cm.

Answer: x c \u003d 6.11 cm, if the y axis passes through the center of gravity of the I-beam; x c \u003d 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, which lies on the line KL of the intersection of the symmetry plane of the trolley with the drawing plane. The force of gravity of the crane winch Q l \u003d 10 kN is applied at the point D. The force of gravity of the counterweight G„=20 kN is applied at point E. The force of gravity of the boom G c = 5 kN is applied at point H. The crane overhang relative to the KL line 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 factor must be at least two.

Solution

1. In an unloaded state, the crane has a risk of tipping over when turning around the rail BUT. Therefore, with respect to the point BUT moment of stability

2. Overturning moment about a point BUT created by the 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 a load F there is a danger of the crane tipping over with a turn around the rail B. Therefore, with respect to the point AT moment of stability

5. Overturning moment relative to the rail AT

6. According to the condition of the problem, the operation of the crane is allowed with a stability coefficient k B ≥ 2, i.e.

Control questions and tasks

1. Why the forces of attraction to the Earth, acting on the points of the body, can 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: a rectangle, a triangle, a trapezoid and half a circle.

4.
What is called the static moment of the 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 figures 1 of the composite section (Fig. 8.8).

When deciding, use the reference data of the GOST tables "Hot-rolled steel" (see Appendix 1).

Determining the center of gravity of an arbitrary body by successively adding up the forces acting on its individual parts is a difficult task; it is facilitated only for bodies of comparatively simple form.

Let the body consist of only two weights of mass and connected by a rod (Fig. 125). If the mass of the rod is small compared to the masses and , then it can be neglected. Each of the masses is affected by gravity equal to and respectively; both of them are directed vertically down, that is, parallel to each other. As we know, the resultant of two parallel forces is applied at the point , which is determined from the condition

Rice. 125. Determination of the center of gravity of a body consisting of two loads

Therefore, the center of gravity divides the distance between two loads in a ratio inverse to the ratio of their masses. If this body is suspended at a point , it will remain in equilibrium.

Since two equal masses have a common center of gravity at a point that bisects the distance between these masses, it is immediately clear that, for example, the center of gravity of a homogeneous rod lies in the middle of the rod (Fig. 126).

Since any diameter of a homogeneous round disk divides it into two completely identical symmetrical parts (Fig. 127), the center of gravity must lie on each diameter of the disk, that is, at the point of intersection of the diameters - in the geometric center of the disk. Arguing in a similar way, we can find that the center of gravity of a homogeneous ball lies in its geometric center, the center of gravity of a homogeneous rectangular parallelepiped lies at the intersection of its diagonals, etc. The center of gravity of a hoop or ring lies in its center. The last example shows that the center of gravity of a body can lie outside the body.

Rice. 126. The center of gravity of a homogeneous rod lies in its middle

Rice. 127. The center of a homogeneous disk lies at its geometric center

If the body has an irregular shape or if it is inhomogeneous (for example, it has voids), then the calculation of the position of the center of gravity is often difficult and this position is more convenient to find through experience. Let, for example, it is required to find the center of gravity of a piece of plywood. Let's hang it on a thread (Fig. 128). Obviously, in the equilibrium position, the center of gravity of the body must lie on the continuation of the thread, otherwise the force of gravity will have a moment relative to the point of suspension, which would begin to rotate the body. Therefore, drawing a straight line on our piece of plywood, representing the continuation of the thread, we can assert that the center of gravity lies on this straight line.

Indeed, by suspending the body at different points and drawing vertical lines, we will make sure that they all intersect at one point. This point is the center of gravity of the body (since it must lie simultaneously on all such lines). In a similar way, one can determine the position of the center of gravity not only of a flat figure, but also of a more complex body. The position of the center of gravity of the aircraft is determined by rolling it with wheels onto the scale platform. The resultant of the weight forces on each wheel will be directed vertically, and you can find the line along which it acts by the law of addition of parallel forces.

Rice. 128. The point of intersection of vertical lines drawn through the points of suspension is the center of gravity of the body

When the masses of individual parts of the body change or when the shape of the body changes, the position of the center of gravity changes. So, the center of gravity of an aircraft moves when fuel is consumed from the tanks, when luggage is loaded, etc. For a visual experiment illustrating the movement of the center of gravity when the shape of the body changes, it is convenient to take two identical bars connected by a hinge (Fig. 129). In the case when the bars form a continuation of one another, the center of gravity lies on the axis of the bars. If the bars are bent at the hinge, then the center of gravity is outside the bars, on the bisector of the angle they form. If an additional load is put on one of the bars, then the center of gravity will move towards this load.

Rice. 129. a) The center of gravity of the bars connected by a hinge, located on one straight line, lies on the axis of the bars, b) The center of gravity of a bent system of bars lies outside the bars

81.1. Where is the center of gravity of two identical thin rods, having a length of 12 cm and fastened in the form of the letter T?

81.2. Prove that the centroid of a uniform triangular plate lies at the intersection of the medians.

Rice. 130. To exercise 81.3

81.3. A homogeneous board of mass 60 kg rests on two supports, as shown in Fig. 130. Determine the forces acting on the supports.

The center of gravity is the point through which the line of action of the resultant elementary forces of gravity passes. It has the property of the center of parallel forces (E. M. Nikitin, § 42). That's why formulas for determining the position of the center of gravity of various bodies look like:
x c = (∑ G i x i) / ∑ G i ;
(1) y c = (∑ G i y i) / ∑ G i ;
z c = (∑ G i z i) / ∑ G i .

If the body whose center of gravity needs to be determined can be identified with a figure made up of lines (for example, a closed or open contour made of wire, as in Fig. 173), then the weight G i of each segment l i can be represented as a product
G i \u003d l i d,
where d is the weight of a unit length of the material that is constant for the entire figure.

After substituting into formulas (1) instead of G i their values ​​l i d, the constant factor d in each term of the numerator and denominator can be taken out of brackets (outside the sign of the sum) and reduced. In this way, formulas for determining the coordinates of the center of gravity of a figure composed of line segments, will take the form:
x c = (∑ l i x i) / ∑ l i ;
(2) y c = (∑ l i y i) / ∑ l i ;
z c = (∑ l i z i) / ∑ l i .

If the body has the form of a figure composed of planes or curved surfaces located in various ways (Fig. 174), then the weight of each plane (surface) can be represented as follows:
G i = F i p,
where F i are the areas of each surface, and p is the weight per unit area of ​​the figure.

After substituting this value of G i into formulas (1), we obtain formulas for the coordinates of the center of gravity of a figure composed of areas:
x c = (∑ F i x i) / ∑ F i ;
(3) y c = (∑ F i y i) / ∑ F i ;
z c = (∑ F i z i) / ∑ F i .

If a homogeneous body can be divided into simple parts of a certain geometric shape (Fig. 175), then the weight of each part
G i = V i γ,
where V i is the volume of each part, and γ is the weight per unit volume of the body.

After substituting the values ​​of G i into formulas (1), we obtain formulas for determining the coordinates of the center of gravity of a body composed of homogeneous volumes:
x c = (∑ V i x i) / ∑ V i ;
(4) y c = (∑ V i y i) / ∑ V i ;
z c = (∑ V i z i) / ∑ V i .

When solving some problems to determine the position of the center of gravity of bodies, it is sometimes necessary to know where the center of gravity of an arc of a circle, a circular sector or a triangle is located.

If the radius of the arc r and the central angle 2α, contracted by the arc and expressed in radians, are known, then the position of the center of gravity C (Fig. 176, a) relative to the center of the arc O is determined by the formula:
(5) x c = (r sin α)/α.

If the chord AB=b of the arc is given, then in formula (5) it is possible to make the replacement
sinα = b/(2r)
and then
(5a) x c = b/(2α).

In a special case for a semicircle, both formulas will take the form (Fig. 176, b):
(5b) x c = OC = 2r/π = d/π.

The position of the center of gravity of the circular sector, if its radius r is given (Fig. 176, c), is determined using the formula:
(6) x c = (2r sin α)/(3α).

If the chord of the sector is given, then:
(6a) x c = b/(3α).

In a special case for a semicircle, both last formulas will take the form (Fig. 176, d)
(6b) x c = OC = 4r/(3π) = 2d/(3π).

The center of gravity of the area of ​​any triangle is located from any side at a distance equal to one third of the corresponding height.

In a right triangle, the center of gravity is at the intersection of perpendiculars raised to the legs from points located at a distance of one third of the length of the legs, counting from the top of the right angle (Fig. 177).

When solving problems for determining the position of the center of gravity of any homogeneous body, composed either of thin rods (lines), or of plates (areas), or of volumes, it is advisable to adhere to the following order:

1) draw a body, the position of the center of gravity of which needs to be determined. Since all body dimensions are usually known, scale must be observed;

2) break the body into component parts (line segments or areas, or volumes), the position of the centers of gravity of which is determined based on the size of the body;

3) determine either the lengths, or the areas, or the volumes of the constituent parts;

4) choose the location of the coordinate axes;

5) determine the coordinates of the centers of gravity of the constituent parts;

6) substitute the found values ​​of the lengths or areas or volumes of individual parts, as well as the coordinates of their centers of gravity, into the appropriate formulas and calculate the coordinates of the center of gravity of the whole body;

7) according to the found coordinates, indicate in the figure the position of the center of gravity of the body.

§ 23. Determining the position of the center of gravity of a body composed of thin homogeneous rods

§ 24. Determination of the position of the center of gravity of figures composed of plates

In the last problem, as well as in the problems given in the previous paragraph, the division of figures into component parts does not cause much difficulty. But sometimes the figure has such a form that allows you to divide it into its component parts in several ways, for example, a thin rectangular plate with a triangular cut (Fig. 183). When determining the position of the center of gravity of such a plate, its area can be divided into four rectangles (1, 2, 3 and 4) and one right triangle 5 in several ways. Two options are shown in Fig. 183, a and b.

The most rational is the way of dividing the figure into its component parts, in which the smallest number of them is formed. If the figure has cutouts, then they can also be included in the number of component parts of the figure, but the area of ​​the cut out part is considered negative. Therefore, this division is called the method of negative areas.

The plate in fig. 183, c is divided using this method into only two parts: rectangle 1 with the area of ​​​​the entire plate, as if it were whole, and triangle 2 with an area that we consider negative.

§ 26. Determination of the position of the center of gravity of a body composed of parts having a simple geometric shape

To solve problems of determining the position of the center of gravity of a body made up of parts that have a simple geometric shape, it is necessary to have the skills to determine the coordinates of the center of gravity of figures made up of lines or areas.

Center of gravity

a geometric point, invariably associated with a solid body, through which the resultant of all gravity forces acting on the particles of this body passes at any position of the latter in space; it may not coincide with any of the points of a given body (for example, near a ring). If a free body is suspended from threads that are attached sequentially to different points of the body, then the directions of these threads will intersect at the center of the body. The position of the center of gravity of a solid body in a uniform field of gravity coincides with the position of its center of mass. Breaking the body into pieces with weights p k , for which the coordinates x k , y k , z k their C. t. are known, you can find the coordinates of the C. t. of the whole body using the formulas:

Great Soviet Encyclopedia. - M.: Soviet Encyclopedia. 1969-1978 .

See what the "Center of Gravity" is in other dictionaries:

The center of mass (center of inertia, barycenter) in mechanics is a geometric point that characterizes the motion of a body or a system of particles as a whole. Contents 1 Definition 2 Centers of mass of homogeneous figures 3 In mechanics ... Wikipedia

A point invariably connected with a solid body through which the resultant of gravity forces acting on the particles of this body passes at any position of the body in space. For a homogeneous body with a center of symmetry (circle, ball, cube, etc.), ... ... encyclopedic Dictionary

Geom. a point, invariably connected with a solid body, through which passes the resultant force of all the forces of gravity acting on the particles of the body at any position in space; it may not coincide with any of the points of a given body (for example, at ... ... Physical Encyclopedia

A point invariably connected with a solid body through which the resultant of the forces of gravity acting on the particles of this body passes at any position of the body in space. For a homogeneous body with a center of symmetry (circle, ball, cube, etc.), ... ... Big Encyclopedic Dictionary

Center of gravity- CENTER OF GRAVITY, the point through which the resultant of gravity forces acting on particles of a solid body at any position of the body in space passes. For a homogeneous body with a center of symmetry (circle, ball, cube, etc.), the center of gravity is ... Illustrated Encyclopedic Dictionary

CENTER OF GRAVITY, the point at which the weight of a body is concentrated and around which its weight is distributed and balanced. A freely falling object rotates around its center of gravity, which in turn rotates along a trajectory that would be described by a point ... ... Scientific and technical encyclopedic dictionary

center of gravity- a rigid body; center of gravity The center of parallel forces of gravity acting on all particles of the body ... Polytechnic terminological explanatory dictionary

Centroid Dictionary of Russian synonyms. center of gravity n., number of synonyms: 12 main (31) spirit ... Synonym dictionary

CENTER OF GRAVITY- The human body does not have a permanent anat. location inside the body, but moves depending on changes in posture; its excursions relative to the spine can reach 20-25 cm. Experimental determination of the position of the central t. of the whole body with ... ... Big Medical Encyclopedia

The point of application of the resultant forces of gravity (weights) of all individual parts (details) that make up a given body. If the body is symmetrical with respect to a plane, a straight line, or a point, then in the first case, the center of gravity lies in the plane of symmetry, in the second, on ... ... Technical railway dictionary

center of gravity- The geometric point of a solid body through which the resultant of all gravity forces acting on the particles of this body passes at any position in space [Terminological dictionary for construction in 12 languages ​​(VNIIIS Gosstroy ... ... Technical Translator's Handbook

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• Center of gravity, A.V. Polyarinov. The novel by Alexei Polyarinov resembles a complex system of lakes. It has cyberpunk, and the majestic designs of David Mitchell, and Borges, and David Foster Wallace ... But his heroes are young journalists, ...

Based on the general formulas obtained above, it is possible to indicate specific methods for determining the coordinates of the centers of gravity of bodies.

1. If a homogeneous body has a plane, axis or center of symmetry, then its center of gravity lies respectively either in the plane of symmetry, or on the axis of symmetry, or in the center of symmetry.

Suppose, for example, that a homogeneous body has a plane of symmetry. Then, by this plane, it is divided into two such parts, the weights of which and are equal to each other, and the centers of gravity are at equal distances from the plane of symmetry. Consequently, the center of gravity of the body as a point through which the resultant of two equal and parallel forces passes will indeed lie in the plane of symmetry. A similar result is obtained in cases where the body has an axis or center of symmetry.

It follows from the properties of symmetry that the center of gravity of a homogeneous round ring, a round or rectangular plate, a rectangular parallelepiped, a ball and other homogeneous bodies with a center of symmetry lies in the geometric center (center of symmetry) of these bodies.

2. Partitioning. If the body can be divided into a finite number of such parts, for each of which the position of the center of gravity is known, then the coordinates of the center of gravity of the whole body can be directly calculated using formulas (59) - (62). In this case, the number of terms in each of the sums will be equal to the number of parts into which the body is divided.

Problem 45. Determine the coordinates of the center of gravity of the homogeneous plate shown in fig. 106. All measurements are in centimeters.

Solution. We draw the x, y axes and divide the plate into three rectangles (cut lines are shown in Fig. 106). We calculate the coordinates of the centers of gravity of each of the rectangles and their area (see table).

Whole plate area

Substituting the calculated quantities into formulas (61), we obtain:

The found position of the center of gravity C is shown in the drawing; point C is outside the plate.

3. Addition. This method is a special case of the partitioning method. It applies to bodies with cutouts if the centers of gravity of the body without the cutout and the cutout are known.

Problem 46. Determine the position of the center of gravity of a round plate of radius R with a radius cut (Fig. 107). Distance

Solution. The center of gravity of the plate lies on the line, since this line is the axis of symmetry. Draw coordinate axes. To find the coordinate, we supplement the area of ​​the plate to a full circle (part 1), and then subtract the area of ​​the cut circle from the resulting area (part 2). In this case, the area of ​​\u200b\u200bpart 2, as subtracted, should be taken with a minus sign. Then

Substituting the found values ​​into formulas (61), we obtain:

The found center of gravity C, as you can see, lies to the left of the point

4. Integration. If the body cannot be divided into several finite parts, the positions of the centers of gravity of which are known, then the body is first divided into arbitrary small volumes for which formulas (60) take the form

where are the coordinates of some point lying inside the volume. Then, in equalities (63), they pass to the limit, tending everything to zero, i.e., contracting these volumes into points. Then the sums in the equalities turn into integrals extended over the entire volume of the body, and formulas (63) give in the limit:

Similarly, for the coordinates of the centers of gravity of areas and lines, we obtain in the limit from formulas (61) and (62):

An example of applying these formulas to determining the coordinates of the center of gravity is considered in the next paragraph.

5. Experimental method. The centers of gravity of inhomogeneous bodies of complex configuration (aircraft, steam locomotive, etc.) can be determined experimentally. One of the possible experimental methods (suspension method) is that the body is suspended on a thread or cable at its various points. The direction of the thread on which the body is suspended will each time give the direction of gravity. The point of intersection of these directions determines the center of gravity of the body. Another possible way to experimentally determine the center of gravity is the weighing method. The idea behind this method is clear from the example below.