Determination of the center of mass of an irregularly shaped body. Determining the coordinates of the center of gravity of plane figures

The ability to stay in balance without making any effort is very important for effective meditation, yoga, qigong and also for belly dancing. This is the first requirement that newcomers to these activities face and one of the reasons why it is difficult to take the first steps without an instructor. A question suggesting that a person does not know his center of gravity may look somewhat different. In qigong, for example, a person will ask how to be relaxed and at the same time perform movements while standing, a beginner oriental dancer will not understand how to separate and coordinate the movements of the lower and upper torso, and also in both cases people will overextend and often lose stability. Their movements will be uncertain, clumsy.

Therefore, it is important to understand how to find your center of gravity yourself, this requires both mental work and dexterity, but over time the skill moves to an instinctive level.

What you need to do in order not to strain your muscles and at the same time do not use external supports. The answer is obvious, you need to move the support inward. More precisely, rely on a conventional internal axis. Where does this axis run? The concept of the center of gravity is conditional, but nevertheless it is used in physics. There it is customary to define it as the point of application of the resultant gravity forces. The resultant force of gravity is the combination of all the forces of gravity, taking into account the direction of their action.

Difficult yet? Please be patient.

That is, we are looking for a point in our body that will allow us not to fall, without consciously struggling with earthly attraction. This means that the force of gravity of the earth must be directed so that it converges with the rest of the acting forces somewhere in the center of our body.

This direction of forces creates a conditional axis in the very center of our body, the vertical surface is the vertical of the center of gravity. That part of the body which we rest against the ground is our support area (we rest against the ground with our feet) In the place where this vertical rests against the surface on which we stand, that is, we rest against the ground, this is the point of the center of gravity inside the support area. If the vertical is displaced from this place, we will lose our balance and fall. The larger the area of \u200b\u200bsupport itself, the easier it is for us to stay close to its center, and therefore we will all instinctively take a wide step while standing on an unstable surface. That is, the support area is not only the feet themselves, but also the space between them.

It is also important to know that the width of the support area affects more than the length. In the case of a person, this means that we have more chances to fall on our side than back, and even more so forward. Therefore, when running, it is harder for us to maintain balance, the same can be said about heels. But in wide, stable shoes, on the contrary, it is easier to resist, even easier than completely barefoot. However, the activities mentioned at the beginning involve very soft, light shoes or no shoes at all. Therefore, we will not be able to help ourselves with shoes.

Therefore, it is very important to find the center point of the vertical line on your foot. Usually it is not located in the center of the foot, as some automatically assume, but closer to the heel, somewhere half way from the center of the foot to the heel.
But that's not all.

In addition to the vertical line of the center of gravity, there is also a horizontal one, as well as a separate one for the limbs.
The horizontal line for women and men runs slightly differently.

In front of women, it runs lower, and higher in men. For men, it goes somewhere 4-5 fingers below the navel, and for women, about 10. Behind, the female line runs almost over the bark, and the male line is about five fingers higher than it. In addition, for stability during meditation, it is important to pay attention to the plumb line of the knee's center of gravity. It is located slightly above the bone (lower leg), but two or three fingers below the cartilage.

During meditation, as during belly dancing, it is not very good to spread your feet wide, the maximum width usually corresponds to the width of the shoulders.

Therefore, you need to help yourself a little with your knees trying to build the vertical axis as straight as possible. Stand in front of the mirror, find all the points described on yourself. Place your feet shoulder-width apart. Relax the muscles in your legs and body. Then, straighten your back without straining your body, relax your legs by bending your knees slightly. Imagine three vertical lines, each of which runs at a corresponding point in the back of the torso, in the front of the torso, and around the knees. Try to position the points so that the front axle of the torso is about halfway between the back and knee axles. In this case, the knees should not be bent so that they go over the toe, they should only be slightly bent and well relaxed. Preferably above the center of gravity inside the support area that we found on the foot. In this case, hands can be freely positioned along the gods or put your palms on your hips.

How will you know that you have found your center of gravity?

You will feel a slight wiggle, but at the same time you will definitely know that you will not fall.

The topic is relatively easy to learn, but extremely important when studying the course on strength of materials. The main attention here must be paid to solving problems with both flat and geometric shapes and with standard rolled profiles.

Questions for self-control

1. What is the center of parallel forces?

The center of parallel forces is a point through which the line of the resultant system of parallel forces applied at given points passes, with any change in the direction of these forces in space.

2. How to find the coordinates of the center of parallel forces?

To determine the coordinates of the center of parallel forces, we use the Varignon theorem.

Axis x

M x (R) \u003d ΣM x (F k), - y C R \u003d Σy kFk and y C \u003d Σy kFk / Σ Fk .

Axis y

M y (R) \u003d ΣM y (F k), - x C R \u003d Σx kFk and x C \u003d Σx kFk / Σ Fk .

To determine the coordinate z C , turn all forces 90 ° so that they become parallel to the axis y (Figure 1.5, b). Then

M z (R) \u003d ΣM z (F k), - z C R \u003d Σz kFk and z C \u003d Σz kFk / Σ Fk .

Therefore, the formula for determining the radius vector of the center of parallel forces takes the form

r C \u003d Σr kFk / Σ Fk.

3. What is the center of gravity of the body?

The center of gravity - a point invariably associated with a rigid body through which the resultant of the gravity forces acting on the particles of this body passes at any position of the body in space. For a homogeneous body having a center of symmetry (circle, ball, cube, etc.), the center of gravity is at the center of symmetry of the body. The position of the center of gravity of a rigid body coincides with the position of its center of mass.

4. How to find the center of gravity of a rectangle, triangle, circle?

To find the center of gravity of a triangle, you need to draw a triangle - a figure consisting of three line 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. Put 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, which intersect at one point, which will be the center of gravity of the triangle. If you need to determine the center of gravity of a circular disc of a homogeneous structure, then first find the point of intersection of the diameters of the circle. She will be the center of gravity of this body. Considering such figures as a ball, a hoop and a uniform rectangular parallelepiped, we can confidently say 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 the intersection diagonals of a rectangular parallelepiped.

5. How to find the coordinates of the center of gravity of a planar composite section?

Splitting method: if a flat figure 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 entire figure are determined by the formulas:

X C \u003d (s k x k) / S; Y C \u003d (s k y k) / S,

where x k, y k - coordinates of the centers of gravity of parts of the figure;

s k - their areas;

S \u003d s k - area of \u200b\u200bthe whole figure.

6. Center of gravity

1. In what case to determine the center of gravity is it enough to determine one coordinate by calculation?

In the first case, to determine the center of gravity, it is enough to determine one coordinate. The body is divided into a finite number of parts, for each of which the position of the center of gravity C and area S are known. For example, the projection of the body onto the plane xOy (Figure 1.) can be represented as two flat figures with areas S 1 and S 2 (S \u003d S 1 + S 2 ). The centers of gravity of these figures are at points C 1 (x 1, y 1) and C 2 (x 2, y 2) ... Then the coordinates of the center of gravity of the body are

Since the centers of the figures lie on the ordinate axis (x \u003d 0), we find only the coordinate Mustache.

2 How is the area of \u200b\u200bthe hole in figure 4 taken into account in the formula to determine the center of gravity of the figure?

Negative mass method

This method consists in the fact that a body with free cavities is considered solid, and the mass of free cavities is considered negative. The form of 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 with free cavities, the method of partitioning should be used, but the mass of the cavities should be considered negative.

have an idea about the center of parallel forces and its properties;

knowformulas for determining the coordinates of the center of gravity of plane figures;

be able todetermine the coordinates of the center of gravity of flat figures of simple geometric figures and standard rolled profiles.

ELEMENTS OF KINEMATICS AND DYNAMICS
Having studied the kinematics of a point, pay attention to the fact that the rectilinear movement of a point, both uneven and uniform, is always characterized by the presence of normal (centripetal) acceleration. With the translational motion of the body (characterized by the motion of any of its points), all formulas of the kinematics of a point are applicable. Formulas for determining the angular values \u200b\u200bof a body rotating around a fixed axis have a complete semantic analogy with formulas for determining the corresponding linear values \u200b\u200bof a translationally moving body.

Topic 1.7. Point kinematics
When studying the topic, pay attention to the basic concepts of kinematics: acceleration, speed, path, distance.

Questions for self-control

1. What is the relativity of the concepts of rest and motion?

Mechanical movement is a change in the movement of a body, or (its parts) in space relative to other bodies over time. The flight of a thrown stone, the rotation of a wheel are examples of mechanical movement.

2. Give the definition of the basic concepts of kinematics: trajectory, distance, path, speed, acceleration, time.

Velocity is a kinematic measure of a point's movement, which characterizes the rate at which its position in space changes. Velocity is a vector quantity, that is, it is characterized not only by the modulus (scalar component), but also by the direction in space.

As is known from physics, with uniform motion, the speed can be determined by the length of the path traversed per unit of time: v \u003d s / t \u003d const (it is assumed that the origin of the path and time coincide). In rectilinear motion, the velocity is constant both in absolute value and in direction, and its vector coincides with the trajectory.

System speed unit SI is determined by the length / time ratio, i.e. m / s.

Acceleration is a kinematic measure of the change in the speed of a point in time. In other words, acceleration is the rate of change in speed.
Like speed, acceleration is a vector quantity, that is, it is characterized not only by modulus, but also by direction in space.

In rectilinear motion, the velocity vector always coincides with the trajectory and therefore the vector of the velocity change also coincides with the trajectory.

It is known from the physics course that acceleration is the change in speed per unit of time. If for a short period of time Δt the speed of the point changed by Δv, then the average acceleration over this period of time was: a cf \u003d Δv / Δt.

Average acceleration does not provide an indication of the true magnitude of the change in speed at any given time. In this case, it is obvious that the shorter the considered period of time, during which the speed change occurred, the closer the acceleration value will be to the true (instantaneous) one.
Hence the definition: true (instantaneous) acceleration is the limit to which the average acceleration tends when Δt tends to zero:

a \u003d lim a cp as t → 0 or lim Δv / Δt \u003d dv / dt.

Considering that v \u003d ds / dt, we get: a \u003d dv / dt \u003d d 2 s / dt 2.

The true acceleration in rectilinear motion is equal to the first derivative of the velocity or the second derivative of the coordinate (distance from the origin of the displacement) in time. The unit of acceleration is a meter divided by a second squared (m / s 2).

Trajectory - a line in space along which a material point moves.
Way is the length of the trajectory. The traversed path l is equal to the arc length of the trajectory traversed by the body in some time t. Path is a scalar.

Distance determines the position of a point on its trajectory and is counted from a certain origin. Distance is an algebraic quantity, since depending on the position of the point relative to the origin and on the accepted direction of the distance axis, it can be both positive and negative. Unlike distance, the path traveled by a point is always a positive number. The path coincides with the absolute value of the distance only if the movement of the point starts from the origin and follows the path in one direction.

In the general case of a point's movement, the path is equal to the sum of the absolute values \u200b\u200bof the distances traveled by the point during a given period of time:

3. In what ways can the law of motion of a point be specified?

1. The natural way to define the movement of a point.

With the natural method of specifying the movement, it is assumed to determine the parameters of the movement of a point in a moving frame of reference, the origin of which coincides with the moving point, and the tangent, normal and binormal to the trajectory of the point in each position serve as the axes. To set the law of motion of a point in a natural way, you must:

1) know the trajectory of movement;

2) set the origin on this curve;

3) establish a positive direction of movement;

4) give the law of motion of a point along this curve, i.e. express the distance from the origin to the position of a point on the curve at a given time ∪OM \u003d S (t) .

2.Vector way of specifying point movement

In this case, the position of a point on a plane or in space is determined by a vector function. This vector is plotted from a fixed point selected as the origin, its end defines the position of the moving point.

3.Coordinate way of specifying point movement

In the selected coordinate system, the coordinates of the moving point are set as a function of time. In a rectangular Cartesian coordinate system, these will be the equations:

4. How is the vector of the true speed of a point directed during curvilinear motion?

With an uneven movement of a point, the modulus of its speed changes over time.
Imagine a point whose motion is given in a natural way by the equation s \u003d f (t).

If in a short time interval Δt the point has passed the path Δs, then its average speed is equal to:

vav \u003d Δs / Δt.

The average speed does not give an idea of \u200b\u200bthe true speed at any given moment in time (the true speed is otherwise called instantaneous). It is obvious that the shorter the time interval for which the average speed is determined, the closer its value will be to the instantaneous speed.

The true (instantaneous) speed is the limit to which the average speed tends as Δt tends to zero:

v \u003d lim v cf as t → 0 or v \u003d lim (Δs / Δt) \u003d ds / dt.

Thus, the numerical value of the true speed is v \u003d ds / dt.
The true (instantaneous) speed for any movement of a point is equal to the first derivative of the coordinate (i.e., the distance from the origin of the movement) in time.

When Δt tends to zero, Δs also tends to zero, and, as we have already found out, the velocity vector will be tangential (i.e., it coincides with the true velocity vector v). It follows from this that the limit of the conditional velocity vector v p, which is equal to the limit of the ratio of the point displacement vector to an infinitely small time interval, is equal to the point's true velocity vector.

5. How are the tangential and normal accelerations of a point directed?

The direction of the acceleration vector coincides with the direction of the speed change Δ \u003d - 0

The tangential acceleration at a given point is directed tangentially to the trajectory of the point; if the motion is accelerated, then the direction of the tangential acceleration vector coincides with the direction of the velocity vector; if the movement is slow, then the direction of the tangential acceleration vector is opposite to the direction of the velocity vector.

6. What movement does a point make if the tangential acceleration is zero, and the normal acceleration does not change over time?

Uniform curvilinear movement characterized by the fact that the numerical value of the speed is constant ( v \u003d const), the speed changes only in the direction. In this case, the tangential acceleration is zero, since v \u003d const (fig.b),

and the normal acceleration is not zero, since r is the final value.

7. What do kinematic graphs look like with uniform and equally variable motion?

With uniform movement, the body traverses equal paths for any equal intervals of time. For the kinematic description of uniform rectilinear motion, the coordinate axis OX conveniently positioned along the line of movement. The position of the body during uniform movement is determined by specifying one coordinate x... The displacement vector and the velocity vector are always directed parallel to the coordinate axis OX... Therefore, displacement and speed in straight-line motion can be projected onto the axis OX and consider their projections as algebraic quantities.

With uniform movement, the path changes according to a linear relationship. In coordinates. The graph is a slanted line.

As a result of studying the topic, the student must:

have an ideaabout space, time, trajectory; average and true speed;

knowways of specifying point movement; parameters of point movement along a given trajectory.

Note. The center of gravity of a symmetrical figure is on the axis of symmetry.

The center of gravity of the rod is at mid-height. When solving problems, the following methods are used:

1.method of symmetry: the center of gravity of symmetrical figures is on the axis of symmetry;

2. method of separation: complex sections are divided into several simple parts, the position of the centers of gravity of which is easy to determine;

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

Examples of problem solving

Example 1. Determine the position of the center of gravity of the figure shown in Fig. 8.4.

Decision

We break the figure into three parts:

Similarly, at C \u003d 4.5 cm.

Example 2. Find the position of the center of gravity of a symmetrical bar truss ADBE (Fig. 116), the dimensions of which are as follows: AB \u003d 6m, DE \u003d 3 m and EF \u003d 1m.

Decision

Since the truss is symmetrical, its center of gravity lies on the axis of symmetry DF. With the selected (Fig. 116) coordinate system of the abscissa of the center of gravity of the truss

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

Of ΔAEF we have

Of 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 into the table and using the formula

define the ordinate withcenter of gravity of this flat truss.

Hence the center of gravity FROM the whole farm lies on the axis DFsymmetry of the truss at a distance of 1.59 m from the point F.

Example 3. Determine the coordinates of the center of gravity of the composite section. The section consists of a sheet and rolled profiles (Fig. 8.5).

Note. Frames are often welded from different profiles to create the required structure. Thus, metal consumption is reduced and a high strength structure is formed.

The intrinsic geometric characteristics are known for standard rolled sections. They are listed in the relevant standards.

Decision

1. Let's mark the figures with numbers and write out the necessary data from the tables:

1 - channel No. 10 (GOST 8240-89); height h \u003d 100 mm; shelf width b \u003d 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; cross-sectional area And 2 - 20.2 cm 2;

3 - 5x100 sheet; thickness 5 mm; width 100mm; cross-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 compound section is symmetrical, so the center of gravity is on the axis of symmetry and the coordinate x C \u003d 0.

3. Determination of the center of gravity of a composite section:

Example 4. Determine the coordinates of the center of gravity of the section shown in Fig. 8, a. The section consists of two corners 56x4 and channel No. 18. Check the correctness of determining the position of the center of gravity. Indicate its position in the section.

Decision

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

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

3. Choose a coordinate system. Axis at compatible with the axis of symmetry, and the axis x will lead 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 with \u003d 0. Coordinate with defined by the formula

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

Coordinates at 1 and at 2equal to zero, since the axis xpasses through the centers of gravity of the corners. Substitute the obtained values \u200b\u200binto the formula to determine with:

5. We indicate the center of gravity of the section in Fig. 8, a and denote it by the letter C.Let us show the distance at 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 \u003d y 2. Therefore, the formula for determining at C can be simplified:

6. Let's check.For this, the axis x draw along the lower edge of the corner shelf (Fig. 8, b). Axis at leave as in the first solution. Formulas for determining x C and at C do not change:

The profile areas will remain the same, and the coordinates of the centers of gravity of the corners and the channel will change. Let's write them out:

Find the coordinate of the center of gravity:

According to the found coordinates x with and with we draw point C on the drawing. The position of the center of gravity found in two ways is at the same point. Let's check it out. Difference between coordinates with, found in the first and second solutions is: 6.51 - 2.43 \u003d 4.08 cm.

This is equal to the distance between the x-axes for the first and second solutions: 5.6 - 1.52 \u003d 4.08 cm.

Answer: with \u003d 2.43 cm, if the x-axis passes through the centers of gravity of the corners, or y with \u003d 6.51 cm if the x-axis runs along the bottom edge of the corner shelf.

Example 5. Determine the coordinates of the center of gravity of the section shown in Fig. nine, a. The section consists of an I-beam No. 24 and a channel No. 24a. Show the position of the center of gravity in the section.

Decision

1. We divide the section into rolled profiles: I-beam and channel. Let's designate them with numbers 1 and 2.

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

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

5. Determine the coordinates of the center of gravity of the section. The coordinate y c \u003d 0, since the axis x coincides with the axis of symmetry. We define the x coordinate by the formula

According to the table. 3 and 4 adj. I and the section diagram, we define

Substitute numerical values \u200b\u200bin the formula and get

5. Draw point C (the center of gravity of the section) according to the found values \u200b\u200bof xc and yc (see Fig. 9, a).

The solution must be checked independently with the position of the axes, as shown in Fig. 9, b. As a result of the solution, we obtain 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 for the same solutions b dv / 2 \u003d 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 \u003d 1.5 m (Fig. 1.102). The gravity force of the crane trolley is G r \u003d 30 kN, the center of gravity of the trolley is at point C, which lies on the line KL of intersection of the trolley symmetry plane 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 „\u003d 20 kN is applied at point E. The force of gravity of the boom G c \u003d 5 kN is applied at point H. The outreach of the crane relative to the line KL is 2 m. Determine the stability coefficient of the crane in an unloaded state and what kind of load F can be lifted with this crane, provided that the stability factor must be at least two.

Decision

1. In an unloaded state, the crane has a risk of tipping over when turning around the rail A. Therefore, relative to the point Amoment of stability

2. Overturning moment about a point A is created by the gravity of the counterweight, i.e.

3. Hence the coefficient of stability of the crane in an unloaded state

4. When the crane is loaded with a load Fthere is a danger of the crane overturning with a turn near rail B. Therefore, relative to the point IN moment of stability

5. Tipping moment relative to the rail IN

6. According to the condition of the problem, the operation of the crane is allowed with the 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 the formulas for determining the position of the center of gravity of inhomogeneous and homogeneous bodies, the formulas for determining the position of the center of gravity of plane sections.

3. Repeat the formulas to determine the position of the center of gravity of simple geometric shapes: rectangle, triangle, trapezoid and half circle.

4.
What is called the static moment of the square?

5. Calculate the static moment of the given figure about the axis Ox. h \u003d 30 cm; b \u003d 120 cm; from \u003d 10 cm (Figure 8.6).

6. Determine the coordinates of the center of gravity of the shaded figure (Fig. 8.7). Dimensions are in mm.

7. Determine the coordinate at figure 1 of the composite section (Fig. 8.8).

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

Before you can find the center of gravity of simple shapes, such as those that have a rectangular, round, spherical or cylindrical or square shape, you need to know where the center of symmetry of a particular shape is. Since in these cases, the center of gravity will coincide with the center of symmetry.

The center of gravity of a homogeneous bar is located in its geometric center. If it is necessary to determine the center of gravity of a circular disc of a homogeneous structure, then first find the point of intersection of the diameters of the circle. She will be the center of gravity of this body. Considering such figures as a ball, a hoop and a homogeneous rectangular parallelepiped, it is safe to say 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 the intersection diagonals of a rectangular parallelepiped.

Center of gravity of heterogeneous bodies

To find the coordinates of the center of gravity, as well as the center of gravity of an inhomogeneous body, it is necessary to figure out on which segment of this body the point is located at which all the gravity forces acting on the figure intersect 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 sets the body in motion.

Take a pencil and a ruler, draw vertical lines that will visually coincide with the thread directions (threads that are fixed at various points on the body). If the shape of the body is complex enough, then draw several lines that will intersect at one point. It will become the center of gravity for the body you were experimenting with.

Center of gravity of the triangle

If you are faced with a task concerning how to find the center of gravity of a body in the shape of an equilateral triangle, then it is necessary 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 heights, medians and bisectors, since the same segments are simultaneously heights, 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 vertices A - x1 and y1; for vertex В - x2 and y2; for vertex С - 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 designated by the letter E - to find it, we drew three medians, at the intersection of which we put the 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 the point D are denoted by xD and yD. Based on the following formulas:

x \u003d (X1 + X2) / 2; y \u003d (Y1 + Y2) / 2

Determine the coordinates of the midpoint of the segment. We get the following result:

xd \u003d (X2 + X3) / 2; yd \u003d (Y2 + Y3) / 2;

D * ((X2 + X3) / 2, (Y2 + Y3) / 2).

We know which coordinates are characteristic for the ends of the blood pressure segment. 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 BP segment. Now, using the formulas and the data we know, 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.

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 got the wrong answer. You may have measured distances from different reference points. Repeat measurements.

For example, if children are sitting on a swing, the center of gravity will be somewhere between the children, 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.
This reasoning is true in two-dimensional space. Draw a square that will fit all the objects in the system. The center of gravity should be inside this square.

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

Correct the error if you find multiple centers of gravity. Each system has only one center of gravity. If you find multiple centers of gravity, chances are you haven't added all the points. The center of gravity is equal to the ratio of the "total" moment to the "total" weight. You don't have to divide “every” moment by “every” weight: this is how you find the position of each object.

Check the starting point if the answer differs by some integer value. In our example, the answer is 3.4 m. Let's say you received an answer of 0.4 m or 1.4 m, or another number ending in ", 4". This is because you did not choose the left end of the board as a reference point, but a point that is located to the right by a whole amount. In fact, your answer is correct, no matter which starting point you choose! Just remember: the origin is always at x \u003d 0. Here's an example:

In our example, the origin was at the left end of the board, and we found that the center of gravity is 3.4 m from this origin.
If you choose a point as a reference point that is located 1 m to the right of the left end of the board, you get the answer 2.4 m.That is, the center of gravity is at a distance of 2.4 m from the new reference point, which, in turn, located at a distance of 1 m from the left end of the board. So the center of gravity is 2.4 + 1 \u003d 3.4 m from the left end of the board. That's the old answer!
Note: when measuring distance, remember that the distances to the "left" reference point are negative and to the "right" are positive.

Measure distances in straight lines. Suppose there are two children on the swing, but one child is much taller than the other, or one child is hanging under the board instead of sitting on it. Ignore this difference and measure the straight line distances. Measuring distances at angles will yield close but not entirely accurate results.

For a swing board problem, remember that the center of gravity is between the right and left ends of the board. Later, you will learn how to calculate the center of gravity of more complex two-dimensional systems.