6.1. General information

Center of Parallel Forces
Consider two parallel, one-way forces, and applied to the body at the points A 1 and A 2 (Figure 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 around the points A 1 and A 2 in one direction and at the same angle, then we get a new system of parallel sals with the same modules. Moreover, their resultant will also pass through the point WITH... This point is called the center of parallel forces.
Consider a system of parallel and equally directed forces applied to a rigid body at points. This system has a resultant.
If each force of the system is rotated about the points of their application in the same direction and by the same angle, then new systems of equally directed parallel forces with the same modules and points of application will be obtained. The resultant of such systems will have the same module R but every time a different direction. Putting together forces F 1 and F 2 we find that their resultant R 1 that will always pass through the point WITH 1, the position of which is determined by equality. Adding 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 brought the process of addition of forces to the end, we come to the conclusion that the resultant of all forces will indeed always pass through the same point WITH, the position of which in relation to the points will be unchanged.
Dot WITH, through which the line of action of the resultant system of parallel forces passes at any rotations of these forces about the points of their application in the same direction at the same angle is called the center of parallel forces (Fig. 6.2).

Figure 6.2

Determine the coordinates of the center of parallel forces. Since the position of the point WITH with respect to the body is unchanged, then its coordinates do not depend on the choice of the coordinate system. Let us 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 since ,, we get

From here we find the coordinate of the center of parallel forces zc:

To determine the coordinates xc Let's compose the expression of the moment of forces about the axis Oz.

To determine the coordinates yc turn all 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 a rigid body is called a point invariably associated with this body WITH, through which the line of action of the resultant of the gravitational forces of the given body passes, for any position of the body in space.
The center of gravity is used in the study of the stability of the equilibrium positions of bodies and continuous media under the action 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 whole body and any of its parts is proportional to the volume P = Vγ, pk = vk γ, where γ - weight of a unit of volume, V- body volume. Substituting expressions P, pk into the formulas for determining the coordinates of the center of gravity and, canceling by a common factor γ , we get:

Dot WITH, whose coordinates are determined by the obtained formulas, is called the 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- the area of ​​the entire plate; sk- the area of ​​its part; xk, yk- coordinates of the center of gravity of the plate parts.
Dot WITH in this case is called center of gravity of the area.
The numerators of expressions that determine the coordinates of the center of gravity of plane figures are called with tatic moments of the square with respect 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 the dimensions of the cross-section, the center of gravity of the line is determined. The coordinates of the center of gravity of the line 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 the line parts.

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 the 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 method of splitting is used. The body is divided into parts, the center of gravity of which is found by the method of symmetry. The center of gravity of the whole body is determined by the formulas for the center of gravity of the volume (area).

Example... Determine the center of gravity of the plate shown in the figure below (Figure 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.

Figure 6.3

Answer: xc= 17.0cm; yc= 18.0cm.

3. Addition... This method is a special case of the splitting method. It is used when the body has cuts, cuts, etc., if the coordinates of the center of gravity of the body without a cut are known.

Example... Determine the center of gravity of a circular plate with a cutout radius r = 0,6 R(fig. 6.4).

Fig 6.4

The round plate has a center of symmetry. Place the origin at the center of the plate. The area of ​​the plate without a cut, the area of ​​the cut. Notched plate area; ...
Notched plate 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 pass to the limit, directing the elementary volumes to zero, i.e. by pulling the 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 center corner AOB= 2α (Fig. 6.5).

Rice. 6.5

The circular arc is symmetrical to the axis Oh, therefore, the center of gravity of the arc lies on the axis Oh, ys = 0.
According to the formula for the center of gravity of the line:

6.Experimental method... The centers of gravity of heterogeneous bodies of complex configuration can be determined experimentally: by the method of suspension and weighing. The first way is that the body is suspended on a rope at various points. The direction of the rope on which the body is suspended will give the direction of the force of gravity. The intersection point of these directions defines the center of gravity of the body.
The weighing method consists in first determining the weight of a body, such as a car. Then the balance determines the pressure of the rear axle of the car on the support. Having compiled the equation of equilibrium with respect to any 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).

Figure 6.6

Sometimes, when solving problems, it is necessary to apply simultaneously different methods for determining the coordinates of the center of gravity.

6.4. The centers of gravity of some of the simplest geometric shapes

To determine the centers of gravity of bodies of a frequently encountered shape (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

№	Figure name	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 is the radius of the circle.
	Homogeneous circular sector uc=0). where α is half of the central angle; R is the radius of the circle.
	Segment: the center of gravity is located on the axis of symmetry (coordinate uc=0). where α is half of the central angle; R is the radius of the circle.
	Semicircle:
	Triangle: the center of gravity of a homogeneous triangle is at the intersection of its medians. where x1, y1, x2, y2, x3, y3- coordinates of the vertices of the triangle
	Cone: the center of gravity of a homogeneous circular cone lies at its height and is at a distance of 1/4 of the height from the base of the cone.

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 newcomers to these activities face and one of the reasons why it is difficult to take their 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, awkward.

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 aggregate 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 ​​support 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, in women, it runs lower, and in men, higher. In men, it goes somewhere 4-5 fingers below the navel, and in women, approximately 10. Behind, the female line runs almost the dump, and the male line is about five fingers higher than it. It is also important to pay attention to the plumb line of the knee's center of gravity for stability while meditating. 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 the 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 described points 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 at a corresponding point at 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 rear axle and the knee axle. In this case, the knees should not be bent so that they go over the toes, 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 you have found your center of gravity?

You will feel a slight swaying, 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 should 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 through 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.

About axis x

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

About axis y

M y (R) = ΣM y (F k), - x C R = Σx kFk and x C = 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) = ΣM z (F k), - z C R = Σz kFk and z C = Σz kFk / Σ Fk .

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

r C = Σr kFk / Σ Fk.

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

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 with 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 shape, 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 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 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 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 = (s k x k) / S; Y C = (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 = s k - area of ​​the whole figure.

6. Center of gravity

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

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 = 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 = 0), we find only the coordinate Mustache.

2 How is the area of ​​the 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 to be 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 in this case.

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;

know formulas for determining the coordinates of the center of gravity of plane figures;

be able to determine 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 the formulas of the kinematics of a point are applicable. The formulas for determining the angular values ​​of a body rotating around a fixed axis have a complete semantic analogy with the formulas for determining the corresponding linear values ​​of 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 a 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 = s / t = const (it is assumed that the origin of the path and time coincide). In rectilinear motion, the speed 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 the modulus, but also by the 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 = 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 = lim a cp as t → 0 or lim Δv / Δt = dv / dt.

Considering that v = ds / dt, we get: a = dv / dt = 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) with respect to 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.
Path 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 measured 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 ​​of the distances traveled by the point for 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 of its positions serve as the axes. To set the law of motion of a point in a natural way, you need to:

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 = 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 point's true velocity 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 = f (t).

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

vav = Δs / Δt.

The average speed does not give an idea of ​​the true speed at any given moment in time (the true speed is otherwise called instantaneous). Obviously, the shorter the time interval over 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 = lim v cf as t → 0 or v = lim (Δs / Δt) = ds / dt.

Thus, the numerical value of the true speed is v = ds / dt.
The true (instantaneous) speed for any movement of a point is equal to the first derivative of the coordinate (that is, the distance from the origin of the movement) with respect to time.

As 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, 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 = - 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 motion is slow, then the direction of the tangential acceleration vector is opposite to the direction of the velocity vector.

6. What motion 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= const), the speed changes only in the direction. In this case, the tangential acceleration is zero, since v= 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 motion. 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 an oblique line.

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

have an idea about space, time, trajectory; average and true speed;

know ways of specifying the movement of a point; parameters of point movement along a given trajectory.

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

1. 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 the area are known.

Fig. 8

3.Negative area method. A special case of the partitioning method (Fig. 9). It applies to bodies with cutouts if the centers of gravity of the body without the cutout and the cutout part are known. The body in the form of a plate with a cutout is a combination of a solid plate (without a cutout) with an area S 1 and an area of ​​the cut part S 2.

Fig. 9

4.Grouping method. It is a good complement to the last two methods. After dividing the figure into its constituent elements, it can be 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) The center of gravity of the arc of a circle. Consider an arc AB radius R with a central corner. By virtue of 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 bearing in mind that the integral must be extended to the entire length of the arc, we get:

where L- arc length AB equal to.

Hence, we finally find that the center of gravity of the arc of a circle lies on its axis of symmetry at a distance from the center O equal to

where the angle is measured in radians.

2) Center of gravity of the area of ​​the triangle. 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 should belong to the median A 3 M 3 (fig. 11).

Fig. 11

Breaking the triangle into strips parallel to the side A 2 A 3, you can make sure that it should lie on the median A 1 M one . In this way, the center of gravity of a triangle lies at the intersection of its medians, which, as you know, separates a third 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 center of gravity of the triangle 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) The center of gravity of the area of ​​the circular sector. Consider a sector of a circle of radius R with a central angle 2α, located symmetrically about the axis Ox(fig. 12).

It's obvious that y c = 0, and the distance from the center of the circle from which this sector is cut to its center of gravity can be determined by the formula:

Fig. 12

The easiest way to calculate this integral is to divide the region of integration into elementary sectors with an angle dφ. Up to infinitesimal 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 get: x c = (4R) / (3π) ≅ 0.4 R .

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

Fig. 13

The body is homogeneous, consisting of two parts with a symmetrical shape. The coordinates of their centers of gravity:

Their volumes:

Therefore, the coordinates of the center of gravity of the body

Example 2. Find the center of gravity of the plate bent at right angles. Dimensions - in the drawing (Fig. 14).

Fig. 14

Center of gravity coordinates:

Squares:

Rice. 6.5.

Example 3. A square hole is cut out from a square sheet cm (Fig. 15). Find the center of gravity of the leaf.

Fig. 15

In this task, 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 staple (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:

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).

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 of gravity. 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 bar, the linear density must be multiplied by the length of this bar.

To solve the problem, you can use the splitting method. Representing a given truss as the sum of 6 individual rods, we get:

where L i length i-th truss rod, and x i, y i- coordinates of its center of gravity.

This problem can be simplified by grouping the last 5 truss members. 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 bars.

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 farm are found by 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 in relation to: WITH 1 WITH/SS 2 = (x c - x 1)/(x 2 - x c ) = L 2 /L 1 = 2,5/0,5.

Self-test questions

What is called the center of parallel forces?

How are the coordinates of the center of parallel forces determined?

How to determine the center of parallel forces, the resultant of which is zero?

What property does the center of parallel forces have?

What formulas are used to calculate the coordinates of the center of parallel forces?

What is called the center of gravity of the body?

Why the forces of gravity on the Earth, acting on a point of the body, can be taken as a system of parallel forces?

Write down the formula for determining the position of the center of gravity of inhomogeneous and homogeneous bodies, the formula for determining the position of the center of gravity of flat sections?

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

What is called the static moment of the square?

Give an example of a body whose center of gravity is outside the body.

How are the properties of symmetry used to determine the centers of gravity of bodies?

What is the essence of the negative weights method?

Where is the center of gravity of the circular arc?

How can you graphically find the center of gravity of a triangle?

Write down the formula for the center of gravity of the circular sector.

Using the formulas for 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, plane figures and lines?

What is called the static moment of the area of ​​a flat 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?

Grade 7 textbook

§ 25.3. How to find the center of gravity of the body?

Recall that the center of gravity is the point of application of the force of gravity. Let us consider how to experimentally find the position of the center of gravity of a flat body - say, an arbitrary shape cut out of cardboard (see laboratory work No. 12).

We suspend the cardboard figure with a pin or nail so that it can rotate freely around the horizontal axis passing through point O (Fig. 25.4, a). Then this figure can be considered as a lever with a fulcrum O.

Rice. 25.4. How to experimentally find the center of gravity of a flat figure

When a figure is in balance, the forces acting on it balance each other. This is the force of gravity F t, applied at the center of gravity of the figure T, and the elastic force F control, applied at point O (this force is applied from the side of a pin or nail).

These two forces balance each other only under the condition that the points of application of these forces (points T and O) lie on the same vertical (see Fig. 25.4, a). Otherwise, the force of gravity will rotate the figure around the point O (Fig. 25.4, b).

So, when the figure is in balance, the center of gravity lies on the same vertical line with the suspension point O. This allows us to determine the position of the figure’s center of gravity. Let us draw a vertical line with the help of a plumb line passing through the suspension point (blue line in Fig. 25.4, c). The center of gravity of the body lies on the drawn line. Let us repeat this experiment with a different position of the suspension point. As a result, we get the second line on which the center of gravity of the body lies (green line in Fig. 25.4, d). Consequently, at the intersection of these lines is the desired center of gravity of the body (red point D in Fig. 25.4, d).