The following methods are most often used to find the center of gravity of a body or figure:
· symmetry method;
· partitioning method;
· negative mass method.
Let's look at the techniques used in each of the listed methods.
Symmetry method
Let's imagine a homogeneous body that has a plane of symmetry. Let us choose a coordinate system such that the axes x And z lay in the plane of symmetry (see Figure 1).
In this case, each elementary particle by gravity G i with abscissa y i = +a corresponds to the same elementary particle with the abscissa y i = -a , Then:
y C = Σ(G i x i)/ΣG i = 0.
Hence the conclusion: if a homogeneous body has a plane of symmetry, then the center of gravity of the body lies in this plane.
The following propositions can be proven similarly:
· If a homogeneous body has an axis of symmetry, then the center of gravity of the body lies on this axis;
· If a homogeneous body has two axes of symmetry, then the center of gravity of the body is at the point of their intersection;
· The center of gravity of a homogeneous body of rotation lies on the axis of rotation.
Splitting method
This method consists in dividing the body into the smallest number of parts, the forces of gravity and the position of the centers of gravity of which are known, after which the previously given formulas are used to determine the overall center of gravity of the body.
Let's say we smashed the body with gravity G
into three parts G"
, G""
, G"""
, abscissas of the centers of gravity of these parts x" C , x"" C , x""" C
known.
Formula for determining the abscissa of the center of gravity of the whole body:
x C = Σ(G i x i)/ΣG i.
Let's rewrite it in the following form:
x C ΣG i = Σ(G i x i) or Gx C = Σ(G i x i) .
We write the last equality for each of the three parts of the body separately:
G"x" C = Σ(G"x" i), G""x"" C = Σ(G"" i x"" i), G"""x""" C = Σ(G""" i x""" i).
Adding the left and right sides of these three equalities, we get:
G"x" C + G"x"" C + G"""x""" C = Σ(G" i x" i) + Σ(G""x"" i) + Σ(G""" i x""" i) = Σ(G i x i).
But the right-hand side of the last equality is the product GxC , because
Gx C = Σ(G i x i),
Hence, x C = (G"x" C + G"x"" C + G"""x""" C)/G
, which was what needed to be proven.
The coordinates of the center of gravity on the coordinate axes are determined similarly y
And z
:
y C = (G"y" C + G""y"" C + G"""y""" C)/G
,
z C = (G"z" C + G""z"" C + G"""z""" C)/G
.
The resulting formulas are similar to the formulas for determining the coordinates of the center of gravity, derived above. Therefore, it is not possible to substitute the gravity forces of elementary particles into the original formulas G i , and the gravity forces of the final parts; under coordinates x i ,y i ,z i understand the coordinates of the centers of gravity of the parts into which the body is divided.
Negative mass method
This method is based on the fact that a body with free cavities is considered solid, and the mass of free cavities is considered negative. The form of the formulas for determining the coordinates of the center of gravity of the body does not change.
Thus, when determining the center of gravity of a body that has free cavities, the partitioning method should be used, but consider the mass of the cavities to be negative.
Practical methods for determining the center of gravity of bodies
In practice, to determine the center of gravity of flat bodies of complex shape, they are often used hanging method
, which consists in hanging a flat body on a thread from some point. A line is drawn along the thread, and the body is suspended from another point not located on the resulting line.
Then draw a line along the thread again.
The intersection point of the two lines will be the center of gravity of the flat body.
Another method of determining the center of gravity used in practice is called weighing method
. This method is often used to determine the center of gravity of large machines and products - cars, airplanes, wheeled tractors, etc., which have a complex volumetric shape and point support on the ground.
The method consists in applying equilibrium conditions, based on the fact that the sum of the moments of all forces acting on a stationary body is equal to zero.
In practice, this is done by weighing one of the machine’s supports (the rear or front wheels are mounted on the scales), while the readings of the scales are, in fact, the reaction of the support, which is taken into account when drawing up the equilibrium equation relative to the second point of support (located outside the scales).
Based on the known mass (respectively, weight) of the body, the reading of the scales at one of the support points, and the distance between the support points, you can determine the distance from one of the support points to the plane in which the center of gravity is located.
To find in this way the line (axis) on which the center of gravity of the machine is located, it is necessary to carry out two weighings according to the principle outlined above for the hanging method (see Fig. 1a).
Question 12
Moment of inertia of the body.
MOMENT OF INERTIA- a quantity that characterizes the distribution of masses in the body and is, along with mass, a measure of the inertia of the body when not moving. movement. In mechanics, there are M. and. axial and centrifugal. Osev M. and. body relative to the z-axis is called. quantity defined by equality
Where m i- masses of body points, h i- their distances from the z axis, r - mass density, V- body volume. Magnitude I z is a measure of the inertia of a body during its rotation around an axis (see Rotational motion ) . Axial M. and. can also be expressed through a linear quantity r z, called. radius of gyration relative to the z axis, according to f-le I z = M r 2 z, where M- body mass. Dimension M. and.- L 2 M; units of measurement - kg. m 2.
Centrifugal M. and. relative to the rectangular system. axes x, y, z, carried out at the point ABOUT, called quantities determined by equalities
or the corresponding volume integrals. These quantities are characteristics of the dynamic. imbalance of the body. For example, when rotating a body around the z axis from the values I xz And I yz The pressure forces on the bearings in which the axle is fixed depend.
M. and. relative to parallel axes z and z" are related by the relation (Huygens' theorem)
where z" is the axis passing through the center of mass of the body, d- distance between axes.
M. and. relative to any passing through the origin ABOUT axes Ol with direction cosines a, b, g is found according to the formula
Knowing six quantities I x , I y , I z , I xy , I yz , I zx, you can sequentially, using formulas (4) and (3), calculate the entire set of M. and. bodies relative to any axes. These six quantities determine the so-called. body inertia tensor. Through each point of the body you can draw 3 such mutually perpendicular axes, called. Ch. axes of inertia, for which I xy = I yz= I zx= 0. Then M. and. bodies relative to any axis can be determined by knowing Ch. axis of inertia and M. and. relative to these axes.
Before finding the center of gravity of simple figures, such as those that have a rectangular, round, spherical or cylindrical, as well as square shape, you need to know at what point the center of symmetry of a particular figure is located. Because in these cases, the center of gravity will coincide with the center of symmetry.
The center of gravity of a homogeneous rod is located at its geometric center. If you need to determine the center of gravity of a round disk of a homogeneous structure, then first find the point of intersection of the diameters of the circle. It will be the center of gravity of this body. Considering such figures as a ball, a hoop and a uniform rectangular parallelepiped, we can say with confidence that the center of gravity of the hoop will be in the center of the figure, but outside its points, the center of gravity of the ball is the geometric center of the sphere, and in the latter case, the center of gravity is considered to be the intersection diagonals of a rectangular parallelepiped.
Center of gravity of inhomogeneous bodies
To find the coordinates of the center of gravity, as well as the center of gravity of an inhomogeneous body itself, it is necessary to figure out on which segment of a given body the point is located at which all the gravitational forces intersect, acting on the figure if it is turned over. In practice, to find such a point, the body is suspended on a thread, gradually changing the points of attachment of the thread to the body. In the case when the body is in equilibrium, the center of gravity of the body will lie on a line that coincides with the line of the thread. Otherwise, gravity causes the body to move.
Take a pencil and a ruler, draw vertical straight lines that will visually coincide with the thread directions (threads attached to various points of the body). If the body shape is quite complex, then draw several lines that will intersect at one point. It will become the center of gravity for the body on which you performed the experiment.
Triangle center of gravity
To find the center of gravity of a triangle, you need to draw a triangle - a figure consisting of three segments connected to each other at three points. Before finding the center of gravity of the figure, you need to use a ruler to measure the length of one side of the triangle. Place a mark in the middle of the side, then connect the opposite vertex and the middle of the segment with a line called the median. Repeat the same algorithm with the second side of the triangle, and then with the third. The result of your work will be three medians that intersect at one point, which will be the center of gravity of the triangle.
If you are faced with a task regarding how to find the center of gravity of a body in the shape of an equilateral triangle, then you need to draw a height from each vertex using a rectangular ruler. The center of gravity in an equilateral triangle will be at the intersection of altitudes, medians and bisectors, since the same segments are simultaneously altitudes, medians and bisectors.
Coordinates of the center of gravity of the triangle
Before finding the center of gravity of the triangle and its coordinates, let’s take a closer look at the figure itself. This is a homogeneous triangular plate, with vertices A, B, C and, accordingly, coordinates: for vertex A - x1 and y1; for vertex B - x2 and y2; for vertex C - x3 and y3. When finding the coordinates of the center of gravity, we will not take into account the thickness of the triangular plate. The figure clearly shows that the center of gravity of the triangle is indicated by the letter E - to find it, we drew three medians, at the intersection of which we placed point E. It has its own coordinates: xE and yE.
One end of the median drawn from vertex A to segment B has coordinates x 1 , y 1 (this is point A), and the second coordinates of the median are obtained based on the fact that point D (the second end of the median) is in the middle of segment BC. The ends of this segment have coordinates known to us: B(x 2, y 2) and C(x 3, y 3). The coordinates of point D are denoted by xD and yD. Based on the following formulas:
x=(X1+X2)/2; y=(U1+U2)/2
Determine the coordinates of the middle of the segment. We get the following result:
xd=(X2+X3)/2; уd=(У2+У3)/2;
D *((X2+X3)/2, (U2+U3)/2).
We know what coordinates are typical for the ends of the segment AD. We also know the coordinates of point E, that is, the center of gravity of the triangular plate. We also know that the center of gravity is located in the middle of the segment AD. Now, using formulas and data known to us, we can find the coordinates of the center of gravity.
Thus, we can find the coordinates of the center of gravity of the triangle, or rather, the coordinates of the center of gravity of the triangular plate, given that its thickness is unknown to us. They are equal to the arithmetic mean of the homogeneous coordinates of the vertices of the triangular plate.
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 runs 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, 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 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 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 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, 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 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, it is necessary to simultaneously use different methods for 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
|
№ |
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.
|
How to find the center of gravity
Author: Let's take a body of arbitrary shape. Is it possible to hang it on a thread so that after hanging it retains its position (i.e. does not begin to turn) when any initial orientation (Fig. 27.1)?
In other words, is there a point relative to which the sum of the moments of gravity acting on various parts of the body would be equal to zero at any body orientation in space?
Reader: Yes, I think so. This point is called center of gravity of the body.
Proof. For simplicity, let us consider a body in the form of a flat plate of arbitrary shape, arbitrarily oriented in space (Fig. 27.2). Let's take the coordinate system X 0at with the beginning at the center of mass - point WITH, Then x C = 0, at C = 0.
Let us imagine this body as a collection of a large number of point masses m i, the position of each of which is specified by the radius vector.
By definition, the center of mass is , and the coordinate x C = .
Since in the coordinate system we adopted x C= 0, then . Let's multiply this equality by g and we get
As can be seen from Fig. 27.2, | x i| - this is the shoulder of strength. And if x i> 0, then the moment of force M i> 0, and if x j < 0, то M j < 0, поэтому с учетом знака можно утверждать, что для любого x i the moment of force will be equal M i = m i gx i . Then equality (1) is equivalent to equality , where M i– moment of gravity. This means that with an arbitrary orientation of the body, the sum of the moments of gravity acting on the body will be equal to zero relative to its center of mass.
In order for the body we are considering to be in equilibrium, it is necessary to apply to it at the point WITH force T = mg, directed vertically upward. The moment of this force relative to the point WITH equal to zero.
Since our reasoning did not depend in any way on how exactly the body is oriented in space, we proved that the center of gravity coincides with the center of mass, which is what we needed to prove.
Problem 27.1. Find the center of gravity of a weightless rod of length l, at the ends of which two point masses are fixed T 1 and T 2 .
| T 1 T 2 l | Solution. We will look not for the center of gravity, but for the center of mass (since these are the same thing). Let's introduce the axis X(Fig. 27.3).  Rice. 27.3 |
| x C =? | |
Answer: at a distance from the mass T 1 .
STOP! Decide for yourself: B1–B3.
Statement 1 . If a homogeneous flat body has an axis of symmetry, the center of gravity is on this axis.
Indeed, for any point mass m i, located to the right of the symmetry axis, there is the same point mass located symmetrically relative to the first one (Fig. 27.4). In this case, the sum of the moments of forces .
Since the entire body can be represented as divided into similar pairs of points, the total moment of gravity relative to any point lying on the axis of symmetry is equal to zero, which means that the center of gravity of the body is located on this axis. This leads to an important conclusion: if a body has several axes of symmetry, then the center of gravity lies at the intersection of these axes(Fig. 27.5).
Rice. 27.5 
Statement 2. If two bodies have masses T 1 and T 2 are connected into one, then the center of gravity of such a body will lie on a straight line segment connecting the centers of gravity of the first and second bodies (Fig. 27.6).
Rice. 27.6 
Rice. 27.7
Proof. Let us position the composite body so that the segment connecting the centers of gravity of the bodies is vertical. Then the sum of the moments of gravity of the first body relative to the point WITH 1 is equal to zero, and the sum of the moments of gravity of the second body relative to the point WITH 2 is equal to zero (Fig. 27.7).
notice, that shoulder gravity of any point mass t i the same with respect to any point lying on the segment WITH 1 WITH 2, and therefore the moment of gravity relative to any point lying on the segment WITH 1 WITH 2, the same. Consequently, the gravitational force of the entire body is zero relative to any point on the segment WITH 1 WITH 2. Thus, the center of gravity of the composite body lies on the segment WITH 1 WITH 2 .
An important practical conclusion follows from Statement 2, which is clearly formulated in the form of instructions.
Instructions,
how to find the center of gravity of a solid body if it can be broken
into parts, the positions of the centers of gravity of each of which are known
1. Each part should be replaced with a mass located at the center of gravity of that part.
2. Find center of mass(and this is the same as the center of gravity) of the resulting system of point masses, choosing a convenient coordinate system X 0at, according to the formulas:
In fact, let us arrange the composite body so that the segment WITH 1 WITH 2 was horizontal, and hang it on threads at points WITH 1 and WITH 2 (Fig. 27.8, A). It is clear that the body will be in equilibrium. And this balance will not be disturbed if we replace each body with point masses T 1 and T 2 (Fig. 27.8, b).
Rice. 27.8
STOP! Decide for yourself: C3.
Problem 27.2. Balls of mass are placed at two vertices of an equilateral triangle T every. A ball of mass 2 is placed at the third vertex T(Fig. 27.9, A). Triangle side A. Determine the center of gravity of this system.
| T 2T A |  Rice. 27.9 |
| x C = ? at C = ? | |
Solution. Let us introduce the coordinate system X 0at(Fig. 27.9, b). Then
,
.
Answer: x C = A/2; ; center of gravity lies at half height AD.
