Finding the center of gravity of a flat body write down the experiment. Determining the coordinates of the center of gravity of plane figures. Center of gravity of inhomogeneous bodies

Determining the center of gravity of an arbitrary body by sequential addition of forces acting on its individual parts is a difficult task; it becomes easier only for bodies of relatively simple shape.

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

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

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

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

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

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

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

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

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

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

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

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

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

81.2. Prove that the center of gravity of a homogeneous triangular plate lies at the intersection of the medians.

Rice. 130. For exercise 81.3

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

The 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 general center body gravity.

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 right part the last equality is the product Gx C , 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 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 movement) . 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 axles.

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= Izx= 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.

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

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

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

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

Based on the above general formulas, you can specify specific methods for determining the coordinates of the centers of gravity of bodies.

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

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

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

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

Problem 45. Determine the coordinates of the center of gravity of the homogeneous plate shown in Fig. 106. All dimensions are given in centimeters.

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

Area of ​​the entire plate

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

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

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

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

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

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

The found center of gravity C, as can be seen, lies to the left of the point

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

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

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

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

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

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 books 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 passes along the lowest overall size, and the Oy axis is 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

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:

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

Draw your flat figure in your notebooks in size, indicating the coordinate axes.

Determine the center of gravity analytically.

1. 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):

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 .