The center of gravity of a symmetrical figure is determined by the formula. Determining the coordinates of the center of gravity of plane figures

Draw a diagram of the system and mark the center of gravity on it. If the found center of gravity is outside the object system, you received an incorrect answer. You may have measured the distances from different points countdown. Repeat the measurements.

• For example, if children are sitting on a swing, the center of gravity will be somewhere between the children, and not to the right or left of the swing. Also, the center of gravity will never coincide with the point where the child is sitting.
• These arguments are valid in two-dimensional space. Draw a square that will contain all the objects of the system. The center of gravity should be inside this square.

Check mathematical calculations, if you get a small result. If the reference point is at one end of the system, a small result places the center of gravity near the end of the system. This may be the correct answer, but in the vast majority of cases this result indicates an error. When you calculated the moments, did you multiply the corresponding weights and distances? If instead of multiplying you added the weights and distances, you would get a much smaller result.

Correct the error if you found multiple centers of gravity. Each system has only one center of gravity. If you found multiple centers of gravity, you most likely did not add up all the moments. Center of gravity equal to the ratio“total” moment to “total” weight. There is no need to divide “every” moment by “every” weight: this way you will find the position of each object.

Goal of the work – determine the center of gravity of a complex figure analytically and experimentally.

Theoretical background. Material bodies consist of elementary particles, whose position in space is determined by their coordinates. The forces of attraction of each particle to the Earth can be considered a system parallel forces, the resultant of these forces is called the force of gravity of the body or the weight of the body. The center of gravity of a body is the point of application of gravity.

The center of gravity is geometric point, which can be located outside the body (for example, a disk with a hole, a hollow ball, etc.). Big practical significance has a definition of the center of gravity of thin flat homogeneous plates. Their thickness can usually be neglected and the center of gravity can be assumed to be located in a plane. If coordinate plane xOy is aligned with the plane of the figure, then the position of the center of gravity is determined by two coordinates:

where is the area of ​​part of the figure, ();

– coordinates of the center of gravity of the parts of the figure, mm (cm).

Section of a figure	A, mm 2	X c ,mm	Yc, mm
	bh	b/2	h/2
	bh/2	b/3	h/3
	R 2 a
At 2α = π πR 2 /2

Work procedure.

Draw a figure complex shape, consisting of 3-4 simple figures(rectangle, triangle, circle, etc.) on a scale of 1:1 and indicate its dimensions.

Draw the coordinate axes so that they cover the entire figure, break the complex figure into simple parts, determine the area and coordinates of the center of gravity of each simple figure relative to the selected coordinate system.

Calculate the coordinates of the center of gravity of the entire figure analytically. Cut this figure from thin cardboard or plywood. Drill two holes, the edges of the holes should be smooth, and the diameter of the holes should be slightly larger than the diameter of the needle for hanging the figure.

First hang the figure at one point (hole), draw a line with a pencil that coincides with the plumb line. Repeat the same when hanging the figure at another point. The center of gravity of the figure, found experimentally, must coincide.

Determine the coordinates of the center of gravity of a thin homogeneous plate analytically. Check experimentally

Solution algorithm

1. Analytical method.

a) Draw the drawing on a scale of 1:1.

b) Break a complex figure into simple ones

c) Select and draw coordinate axes (if the figure is symmetrical, then along the axis of symmetry, otherwise along the figure’s contour)

d) Calculate the area of ​​simple figures and the entire figure

e) Mark the position of the center of gravity of each simple figure in the drawing

f) Calculate the coordinates of the center of gravity of each figure

(x and y axis)

g) Calculate the coordinates of the center of gravity of the entire figure using the formula

h) Mark the position of the center of gravity on drawing C (

2. Experimental determination.

The correctness of the solution to the problem can be verified experimentally. Cut out this figure from thin cardboard or plywood. Drill three holes, the edges of the holes should be smooth, and the diameter of the holes should be slightly larger than the diameter of the needle for hanging the figure.

First hang the figure at one point (hole), draw a line with a pencil that coincides with the plumb line. Repeat the same when hanging the figure at other points. The value of the coordinates of the center of gravity of the figure, found when hanging the figure at two points: . The center of gravity of the figure, found experimentally, must coincide.

3. Conclusion about the position of the center of gravity during analytical and experimental determination.

Exercise

Determine the center of gravity of a flat section analytically and experimentally.

Execution example

Task

Determine the coordinates of the center of gravity of a thin homogeneous plate.

I Analytical method

1. The drawing is drawn to scale (dimensions are usually given in mm)

2. We break a complex figure into simple ones.

1- Rectangle

2- Triangle (rectangle)

3- Area of ​​the semicircle (it doesn’t exist, minus sign).

We find the position of the center of gravity of simple figures of points, and

3. Draw the coordinate axes as convenient and mark the origin of coordinates.

4. Calculate the areas of simple figures and the area of ​​the entire figure. [size in cm]

(3. no, sign -).

Area of ​​the entire figure

5. Find the coordinate of the central point. , and in the drawing.

6. Calculate the coordinates of points C 1, C 2 and C 3

7. Calculate the coordinates of point C

8. Mark a point on the drawing

II Experienced

Coordinates of the center of gravity experimentally.

Control questions.

1. Is it possible to consider the force of gravity of a body as a resultant system of parallel forces?

2. Can the center of gravity of the entire body be located?

3. What is the essence experimental determination center of gravity of a flat figure?

4. How is the center of gravity of a complex figure consisting of several simple figures determined?

5. How should a figure of complex shape be rationally divided into simple figures when determining the center of gravity of the entire figure?

6. What sign does the area of ​​the holes have in the formula for determining the center of gravity?

7. At the intersection of which lines of the triangle is its center of gravity located?

8. If a figure is difficult to break down into a small number of simple figures, what method of determining the center of gravity can provide the fastest answer?

Practical work №6

“Solving complex problems”

Goal of the work: be able to solve complex problems (kinematics, dynamics)

Theoretical background: Velocity is a kinematic measure of the movement of a point, characterizing the speed of change in its position. The speed of a point is a vector characterizing the speed and direction of movement of a point in this moment time. When specifying the motion of a point by the equations of velocity projection on the axis Cartesian coordinates are equal:

The velocity modulus of a point is determined by the formula

The direction of speed is determined by direction cosines:

The characteristic of the speed of change of speed is acceleration a. The acceleration of a point is equal to the time derivative of the velocity vector:

When specifying the motion of a point, the equations for the projection of acceleration onto the coordinate axes are equal to:

Acceleration module:

Module full acceleration

The tangential acceleration module is determined by the formula

The normal acceleration modulus is determined by the formula

where is the radius of curvature of the trajectory at a given point.

The direction of acceleration is determined by the direction cosines

The equation rotational movement solid around fixed axis looks like

Angular velocity of the body:

Sometimes angular velocity is characterized by the number of revolutions per minute and is designated by the letter . The dependence between and has the form

Angular acceleration of the body:

A force equal to the product of the mass of a given point by its acceleration and the direction in the direction directly opposite to the acceleration of the point is called the force of inertia.

Power is the work done by a force per unit time.

Basic dynamics equation for rotational motion

– the moment of inertia of the body relative to the axis of rotation, is the sum of the products of the masses of material points by the square of their distances to this axis

Exercise

A body of mass m, with the help of a cable wound on a drum of diameter d, moves up or down along inclined plane with inclination angle α. Equation of body motion S=f(t), equation of drum rotation, where S is in meters; φ - in radians; t – in seconds. P and ω are, respectively, the power and angular velocity on the drum shaft at the moment of the end of acceleration or the beginning of braking. Time t 1 – acceleration time (from rest to a given speed) or braking (from a given speed to a stop). The coefficient of sliding friction between the body and the plane is –f. Neglect friction losses on the drum, as well as the mass of the drum. When solving problems, take g=10 m/s 2

Execution example

P, ω Problem 1

(picture 1). Solution 1. Rectilinear movement (Figure 1, a). A point moving uniformly at some point in time received new law movement, and after a certain period of time stopped. Define everything kinematic characteristics point movements for two cases; a) movement along straight path ; b) movement along curvilinear trajectory

constant radius of curvature r=100cm

Figure 1(a).

Law of change in point velocity

We find the initial speed of the point from the condition:

We find the braking time to stop from the condition:

at , from here .

Law of motion of a point during a period of uniform motion

The distance traveled by the point along the trajectory during the braking period,

Law of change in tangential acceleration of a point

whence it follows that during the period of braking the point moved equally slow, since the tangential acceleration is negative and constant in value.

The normal acceleration of a point on a straight trajectory is zero, i.e. . Solution 2.

Curvilinear movement (Figure 1, b).

Figure 1(b) In this case, compared to the case rectilinear movement

All kinematic characteristics remain unchanged, with the exception of normal acceleration.

Law of change in normal acceleration of a point Normal acceleration of a point at starting moment

braking The numbering of point positions on the trajectory accepted in the drawing: 1 – points in uniform motion before braking begins; 2 – position of the point at the moment of braking; 3 – current position of the point during the braking period; 4 – final position of the point.

Task 2.

The load (Fig. 2, a) is lifted using a drum winch. The diameter of the drum is d=0.3m, and the law of its rotation is .

The drum acceleration lasted until angular velocity. Determine all kinematic characteristics of the movement of the drum and load.

Solution. Law of change in drum angular velocity. We find the initial angular velocity from the condition: ; therefore, acceleration began from a state of rest. We will find the acceleration time from the condition: . Drum rotation angle during acceleration period.

Law of Change angular acceleration drum, it follows that during the acceleration period the drum rotated uniformly accelerated.

The kinematic characteristics of the load are equal to the corresponding characteristics of any point of the traction rope, and therefore point A lying on the rim of the drum (Fig. 2, b). As is known, the linear characteristics of a point of a rotating body are determined through its angular characteristics.

The distance traveled by the load during the acceleration period, . Velocity of the load at the end of acceleration.

Acceleration of cargo.

Law of cargo movement.

The distance, speed and acceleration of the load could be determined in another way, through the found law of motion of the load:

Task 3. The load, moving uniformly upward along an inclined support plane, at some point in time received braking in accordance with the new law of motion , where s is in meters and t is in seconds. Mass of the load m = 100 kg, coefficient of sliding friction between the load and the plane f = 0.25. Determine the force F and power on the traction rope for two moments of time: a) uniform motion before braking begins;

b) initial moment of braking. When calculating, take g=10 m/.

Solution. We determine the kinematic characteristics of the movement of the load.

Law of change in speed of load

starting speed load (at t=0)

Cargo acceleration

Since the acceleration is negative, the movement is slow.

1. Uniform movement of the load.

For determining driving force F we consider the equilibrium of the load, which is acted upon by a system of converging forces: the force on the cable F, the gravitational force of the load G=mg, normal reaction supporting surface N and friction force directed towards the movement of the body. According to the law of friction, . Choosing a direction coordinate axes, as shown in the drawing, and draw up two equilibrium equations for the load:

The power on the cable before braking begins is determined by well-known formula

Where is m/s.

2. Slow movement of cargo.

As is known, with uneven forward movement body, the system of forces acting on it in the direction of movement is not balanced. According to d'Alembert's principle (kinetostatic method), the body in this case can be considered to be in conditional equilibrium if we add to all the forces acting on it an inertial force, the vector of which is directed opposite to the acceleration vector. The acceleration vector in our case is directed opposite to the velocity vector, since the load moves slowly. We create two equilibrium equations for the load:

Power on the cable at the start of braking

Control questions.

1. How to determine numerical value and the direction of the point's velocity at the moment?

2. What characterizes the normal and tangential components of total acceleration?

3. How to move from expressing angular velocity in min -1 to expressing it in rad/s?

4. What is body weight called? Name the unit of measurement of mass

5. At what movement material point does inertial force arise? What is its numerical value and what is its direction?

6. State d'Alembert's principle

7. Does the force of inertia arise with uniform curvilinear movement material point?

8. What is torque?

9. How is the relationship between torque and angular velocity expressed for a given transmitted power?

10. Basic dynamics equation for rotational motion.

Practical work No. 7

"Calculation of structures for strength"

Goal of the work: determine strength, cross-sectional dimensions and permissible load

Theoretical background.

Knowing the force factors and geometric characteristics of the section during tensile (compression) deformation, we can determine the stress using the formulas. And to understand whether our part (shaft, gear, etc.) will withstand external load. It is necessary to compare this value with the permissible voltage.

So, the static strength equation

Based on it, 3 types of problems are solved:

1) strength test

2) determination of section dimensions

3) determination of permissible load

So, the equation of static stiffness

Based on it, 3 types of problems are also solved

Equation of static tensile (compressive) strength

1) First type - strength test

,

i.e. we decide left side and compare with the permissible voltage.

2) Second type - determination of section dimensions

from the right side cross-sectional area

Section circle

hence the diameter d

Rectangle section

Section square

A = a² (mm²)

Semicircle section

Sections: channel, I-beam, angle, etc.

Area values ​​- from the table, accepted according to GOST

3) The third type is determining the permissible load;

taken to the smaller side, integer

EXERCISE

Task

A) Strength check (test calculation)

For a given beam, construct a diagram of longitudinal forces and check the strength in both sections. For timber material (steel St3) accept

B) Selection of section (design calculation)

For a given beam, construct a diagram of longitudinal forces and determine the cross-sectional dimensions in both sections. For timber material (steel St3) accept

B) Determination of permissible longitudinal force

For a given beam, determine the permissible values ​​of loads and ,

construct a diagram of longitudinal forces. For timber material (steel St3) accept . When solving the problem, assume that the type of loading is the same on both sections of the beam.

Example of completing a task

P, ω Problem 1

Check the strength of a column made of I-profiles of a given size. For the column material (steel St3), accept the permissible tensile stresses and during compression . In the event of overloading or significant underloading, select I-beam sizes that ensure optimal column strength.

Solution.

A given beam has two sections 1, 2. The boundaries of the sections are the sections in which the external forces. Since the forces loading the beam are located along its central longitudinal axis, only one internal force factor arises in the cross sections - longitudinal force, i.e. there is tension (compression) of the beam.

To determine the longitudinal force, we use the section method. Mentally drawing a section within each section, we will discard the lower fixed part of the beam and leave it for consideration top part. In section 1, the longitudinal force is constant and equal to

The minus sign indicates that the beam is compressed in both sections.

We build a diagram of longitudinal forces. Having drawn the base (zero) line of the diagram parallel to the axis of the beam, we plot the obtained values ​​perpendicular to it on an arbitrary scale. As you can see, the diagram turned out to be outlined by straight lines parallel to the base one.

We check the strength of the timber, i.e. We determine the calculated stress (for each section separately) and compare it with the permissible one. To do this, we use the compressive strength condition

where area is a geometric characteristic of the strength of the cross section. From the table of rolled steel we take:

for I-beam
for I-beam

Strength test:

The values ​​of longitudinal forces are taken in absolute value.

The strength of the beam is ensured, however, there is a significant (more than 25%) underload, which is unacceptable due to excessive consumption of material.

From the strength condition, we determine the new dimensions of the I-beam for each section of the beam:
Hence the required area

According to the GOST table, we select I-beam No. 16, for which;

Hence the required area

According to the GOST table, we select I-beam No. 24, for which ;

With the selected I-beam sizes, underload also occurs, but it is insignificant (less than 5%)

Task No. 2.

For a beam with given cross-sectional dimensions, determine the permissible load values ​​and . For timber material (steel St3), accept permissible tensile stresses and during compression .

Solution.

The given beam has two sections 1, 2. There is tension (compression) of the beam.

Using the method of sections, we determine the longitudinal force, expressing it through the required forces and. Carrying out a section within each section, we will discard the left part of the beam and leave it for consideration right side. In section 1, the longitudinal force is constant and equal to

In section 2, the longitudinal force is also constant and equal to

The plus sign indicates that the beam is stretched in both sections.

We build a diagram of longitudinal forces. The diagram is outlined by straight lines parallel to the base one.

From the condition of tensile strength, we determine the permissible load values ​​and, having previously calculated the areas of the given cross sections:

Control questions.

1. What internal force factors arise in the section of a beam during tension and compression?

2. Write down the condition for tensile and compressive strength.

3. How are the signs of longitudinal force and normal stress assigned?

4. How will the voltage change if the cross-sectional area increases by 4 times?

5. Are the strength conditions different for tensile and compressive calculations?

6. In what units is voltage measured?

7. Which one mechanical characteristics selected as the ultimate stress for ductile and brittle materials?

8. What is the difference between limiting and permissible stress?

Practical work No. 8

“Solving problems to determine the main central moments of inertia of flat geometric shapes»

Goal of the work: determine analytically the moments of inertia flat bodies complex shape

Theoretical background. The coordinates of the center of gravity of the section can be expressed in terms of the static moment:

where relative to the Ox axis

relative to the Oy axis

The static moment of the area of ​​a figure relative to an axis lying in the same plane, equal to the product the area of ​​a figure by the distance of its center of gravity to this axis. The static moment has a dimension. The static moment can be a positive, negative or equal to zero(relative to any central axis).

The axial moment of inertia of a section is the sum of the products or integral of elementary areas taken over the entire section by the squares of their distances to a certain axis lying in the plane of the section under consideration.

The axial moment of inertia is expressed in units - . The axial moment of inertia is a quantity that is always positive and not equal to zero.

The axes passing through the center of gravity of the figure are called central. The moment of inertia about the central axis is called central point inertia.

The moment of inertia about any axis is equal to the center

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, we get new system 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 to study the stability of equilibrium positions of bodies and continuum, under the influence of gravity and in some other cases, namely: in the resistance 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 common multiplier 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 V in this case is called center of gravity of the area.
Numerators of expressions defining the coordinates of the center of gravity flat 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, determine the center of gravity of the line. 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 obtained formulas, we can propose practical ways 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 of simple shape, 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 different ways and determine the coordinates of the center of gravity of each rectangle and their area.

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, slices, 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 final number parts, the position 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 you should use simultaneously different methods determining the coordinates of the center of gravity.

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

IN engineering practice It happens that it becomes necessary to calculate the coordinates of the center of gravity of a complex flat figure consisting of simple elements for which the location of the center of gravity is known. This task is part of the task of determining...

Geometric characteristics of composite cross sections of beams and rods. Often with similar questions design engineers of cutting dies have to deal with when determining the coordinates of the center of pressure, and developers of loading schemes various transport when placing loads, for designers of building metal structures when selecting sections of elements and, of course, for students when studying disciplines “ Theoretical mechanics" and "Strength of materials".

Library of elementary figures.

For symmetrical plane figures, the center of gravity coincides with the center of symmetry. The symmetrical group of elementary objects includes: circle, rectangle (including square), parallelogram (including rhombus), regular polygon.

Of the ten figures presented in the figure above, only two are basic. That is, using triangles and sectors of circles, you can combine almost any figure of practical interest. Any arbitrary curves can be divided into sections and replaced with circular arcs.

The remaining eight figures are the most common, which is why they were included in this unique library. In our classification, these elements are not basic. A rectangle, parallelogram and trapezoid can be formed from two triangles. A hexagon is the sum of four triangles. A circle segment is the difference between a sector of a circle and a triangle. The annular sector of a circle is the difference between two sectors. A circle is a sector of a circle with an angle α=2*π=360˚. A semicircle is, accordingly, a sector of a circle with an angle α=π=180˚.

Calculation in Excel of the coordinates of the center of gravity of a composite figure.

It is always easier to convey and perceive information by considering an example than to study the issue using purely theoretical calculations. Let's consider the solution to the problem “How to find the center of gravity?” using the example of the composite figure shown in the figure below this text.

The composite section is a rectangle (with dimensions a1 =80 mm, b1 =40 mm), to which was added at the top left isosceles triangle(with base size a2 =24 mm and height h2 =42 mm) and from which a semicircle was cut out from the top right (with the center at the point with coordinates x03 =50 mm and y03 =40 mm, radius r3 =26 mm).

We will use a program to help you perform the calculations MS Excel or program OOo Calc . Any of them will easily cope with our task!

In cells with yellow we will fill it auxiliary preliminary calculations .

We calculate the results in cells with a light yellow fill.

Blue font is initial data .

Black font is intermediate calculation results .

Red font is final calculation results .

We begin solving the problem - we begin the search for the coordinates of the center of gravity of the section.

Initial data:

1. We will write the names of the elementary figures forming a composite section accordingly

to cell D3: Rectangle

to cell E3: Triangle

to cell F3: Semicircle

2. Using the “Library of Elementary Figures” presented in this article, we will determine the coordinates of the centers of gravity of the elements of the composite section xci And yci in mm relative to arbitrarily selected axes 0x and 0y and write

to cell D4: =80/2 = 40,000

xc 1 = a 1 /2

to cell D5: =40/2 =20,000

yc 1 = b 1 /2

to cell E4: =24/2 =12,000

xc 2 = a 2 /2

to cell E5: =40+42/3 =54,000

yc 2 = b 1 + h 2 /3

to cell F4: =50 =50,000

xc 3 = x03

to cell F5: =40-4*26/3/PI() =28,965

yc 3 = y 03 -4* r3 /3/ π

3. Let's calculate the areas of the elements F 1 , F 2 , F3 in mm2, again using the formulas from the section “Library of elementary figures”

in cell D6: =40*80 =3200

F1 = a 1 * b1

in cell E6: =24*42/2 =504

F2 = a2 *h2 /2

in cell F6: =-PI()/2*26^2 =-1062

F3 =-π/2*r3 ^2

The area of ​​the third element - the semicircle - is negative because it is a cutout - an empty space!

Calculation of center of gravity coordinates:

4. Let's define total area final figure F0 in mm2

in merged cell D8E8F8: =D6+E6+F6 =2642

F0 = F 1 + F 2 + F3

5. Let's calculate the static moments of a composite figure Sx And Sy in mm3 relative to the selected axes 0x and 0y

in merged cell D9E9F9: =D5*D6+E5*E6+F5*F6 =60459

Sx = yc1 * F1 + yc2 *F2 + yc3 *F3

in the merged cell D10E10F10: =D4*D6+E4*E6+F4*F6 =80955

Sy = xc1 * F1 + xc2 *F2 + xc3 *F3

6. And finally, let’s calculate the coordinates of the center of gravity of the composite section Xc And Yc in mm in the selected coordinate system 0x - 0y

in merged cell D11E11F11: =D10/D8 =30,640

Xc = Sy / F0

in merged cell D12E12F12: =D9/D8 =22,883

Yc =Sx /F0

The problem has been solved, the calculation in Excel has been completed - the coordinates of the center of gravity of the section, compiled using three simple elements, have been found!

Conclusion.

The example in the article was chosen to be very simple in order to make it easier to understand the methodology for calculating the center of gravity of a complex section. The method is that any complex figure should be divided into simple elements With famous places location of the centers of gravity and make final calculations for the entire section.

If the section is made up of rolled profiles - angles and channels, then there is no need to divide them into rectangles and squares with cut out circular “π/2” sectors. The coordinates of the centers of gravity of these profiles are given in the GOST tables, that is, both the angle and the channel will be basic in your calculations of composite sections elementary elements(there is no point in talking about I-beams, pipes, rods and hexagons - these are centrally symmetrical sections).

The location of the coordinate axes, of course, does not affect the position of the figure’s center of gravity! Therefore, choose a coordinate system that simplifies your calculations. If, for example, I were to rotate the coordinate system 45˚ clockwise in our example, then calculating the coordinates of the centers of gravity of a rectangle, triangle and semicircle would turn into another separate and cumbersome stage of calculations that cannot be performed “in the head”.

The calculation shown below Excel file in this case it is not a program. Rather, it is a sketch of a calculator, an algorithm, a template that follows in each specific case create your own sequence of formulas for cells with a bright yellow fill.

So, you now know how to find the center of gravity of any section! The complete calculation of all geometric characteristics of arbitrary complex composite sections will be considered in one of the upcoming articles in the “” section. Follow the news on the blog.

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A few words about the glass, coin and two forks, which are depicted in the “illustration icon” at the very beginning of the article. Many of you are certainly familiar with this “trick”, which evokes admiring glances from children and uninitiated adults. The topic of this article is the center of gravity. It is he and the fulcrum, playing with our consciousness and experience, who are simply fooling our minds!

The center of gravity of the “fork+coin” system is always located on fixed distance vertically down from the edge of the coin, which in turn is the fulcrum. This position stable equilibrium! If you shake the forks, it immediately becomes obvious that the system is striving to take its previous stable position! Imagine a pendulum - a fixing point (= the point of support of a coin on the edge of a glass), a rod - the axis of the pendulum (= in our case, the axis is virtual, since the mass of the two forks is separated by different sides space) and the load at the bottom of the axle (=the center of gravity of the entire “fork + coin” system). If you begin to deflect the pendulum from the vertical in any direction (forward, backward, left, right), then it will inevitably return to its original position under the influence of gravity. steady state of equilibrium(the same thing happens with our forks and coin)!

If you don’t understand, but want to understand, figure it out yourself. It’s very interesting to “get there” yourself! I will add that the same principle of using stable equilibrium is also implemented in the toy Vanka-stand-up. Only the center of gravity of this toy is located above the fulcrum, but below the center of the hemisphere of the supporting surface.

I am always glad to see your comments, dear readers!!!

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Based on the above general formulas, you can specify specific methods for 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

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 an area of ​​the cut out part S 2 .

Fig.9

4.Grouping method. Is good addition 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 radius R with a central angle. Due to symmetry, the center of gravity of this arc lies on the axis Ox(Fig. 10).

Fig.10

Let's find the coordinate using the formula. To do this, select on the arc AB element MM' length, the position of which 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- arc length 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 ABOUT, equal

where the angle is 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 is the arithmetic mean of the coordinates of the vertices A 2 and A 3:

x c = x 1 + (2/3)∙(x M 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 circular sector. Consider a sector of a circle with radius R with a central angle of 2α, located symmetrically relative to 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 = (4R)/(3π) ≅ 0.4 R .

Example 1. Let us determine the center of gravity of the homogeneous body shown in Fig. 13.

Fig.13

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

Coordinates of the centers of gravity:

Areas:

Fig.15

In this problem, it is more convenient to divide the body into two parts: a large square and a square hole. Only the area of ​​the hole should be considered negative. Then the coordinates of the center of gravity of the sheet with the hole:

coordinate since the body has an axis of symmetry (diagonal).

Example 4. The wire bracket (Fig. 16) consists of three sections of equal length l.

Fig.16

Coordinates of the centers of gravity of the sections:

Therefore, the coordinates of the center of gravity of the entire bracket are:

Example 5. Determine the position of the center of gravity of the truss, all the rods of which have the same linear density (Fig. 17).

Let us recall that in physics the density of a body ρ and its specific gravity g are related by the relation: γ= ρ g, Where g- acceleration free fall. To find the mass of such a homogeneous body, you need to multiply the density by its volume.

Fig.17

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:

Where L i length i th truss rod, and x 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 it L 1 = 4 m, x 1 = 0 m, y 1 = 2 m. The second group of rods consists of five rods, for it L 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 the center of parallel forces called?

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?

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 flat sections?

Write down the formula for determining the position of the center of gravity of simple geometric shapes: rectangle, triangle, trapezoid and half circle?

What is 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 in determining the centers of gravity of bodies?

What is the essence of the negative weights method?

Where is the center of gravity of a circular arc?

What graphical construction can you 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?