Determination of the center of gravity of an irregularly shaped body. Methods for determining the coordinates of the center of gravity. Center of gravity of heterogeneous bodies
Lecture 4. Center of gravity.
This lecture addresses the following issues
1. The center of gravity of a rigid body.
2. Coordinates of the centers of gravity of heterogeneous bodies.
3. Coordinates of the centers of gravity of homogeneous bodies.
4. Methods for determining the coordinates of the centers of gravity.
5. Centers of gravity of some homogeneous bodies.
The study of these issues is necessary in the future to study the dynamics of motion of bodies, taking into account sliding and rolling friction, the dynamics of the center of mass of a mechanical system, kinetic moments, for solving problems in the discipline "Resistance of materials".
Bringing parallel forces.
After we have considered the reduction to the center of a flat system and an arbitrary spatial system of forces, we again return to considering the particular case of a system of parallel forces.
Bringing two parallel forces.
In the course of considering such a system of forces, the following three cases of reduction are possible.
1. System of two collinear forces. Consider a system of two forces parallel and directed in one direction P and Q applied at points BUT and IN... We will assume that the forces are perpendicular to this segment (Fig. 1, but).
FROM belonging to the segment AB and satisfying the condition:
AS/SV = Q/P.(1)
The main vector of the system R C = P + Q modulo is equal to the sum of these forces: R C = P + Q.
FROM taking into account (1) is equal to zero:MC = P ∙ AS- Q∙ SV = 0.
Thus, as a result of the cast, we got: R C ≠ 0, MC= 0. This means that the principal vector is equivalent to the resultant, passing through the center of reduction, that is:
The resultant of collinear forces is equal in modulus to their sum, and its line of action divides the segment connecting the points of their application, inversely proportional to the moduli of these forces in an internal way.
Note that the position of the point FROM will not change if the strength R and Q turn the cornerα. Point FROM with this property is called center of parallel forces.
2. System of two anticollinear and forces not equal in magnitude. Let the forces P and Q applied at points BUT and IN, are parallel, directed in opposite directions and are not equal in absolute value (Fig. 1, b).
Let us choose the point as the center of reduction FROM satisfying, as before, relation (1) and lying on the same straight line, but outside the segment AB.
The main vector of this system R C = P + Q the modulus will now be equal to the difference between the moduli of the vectors: R C = Q - P.
The main point relative to the center FROM is still zero:MC = P ∙ AS- Q∙ SV= 0, therefore
Resultant anticollinear and forces not equal in magnitude is equal to their difference, directed towards a greater force, and its line of action divides the segment connecting the points of their application, inversely proportional to the moduli of these forces outwardly.
Fig. 1
3. System of two anticollinear and forces equal in magnitude. Let us take as the initial previous case of reduction. We fix the force R and the strength Q let us strive modulo the force R.
Then at Q → R in formula (1), the ratio AS/SV → 1. This means that AS → SV, that is, the distance AS →∞ .
In this case, the module of the main vector R C → 0, and the modulus of the principal moment does not depend on the position of the reference center and remains equal to the initial value:
MC = P ∙ AS- Q∙ SV = P ∙ ( AS- SV) =P ∙ BUTB.
So, in the limit, we got a system of forces for which R C = 0, MC≠ 0, and the center of reduction is removed to infinity, which cannot be replaced by a resultant. In this system, it is not difficult to recognize a pair of forces, therefore a pair of forces has no resultant.
Center of the system of parallel forces.
Consider the system n forces P i applied at pointsA i (x i , y i , z i) and parallel to the axisOv with unit vector l(fig. 2).
If we exclude in advance the case of a system equivalent to a pair of forces, it is easy, on the basis of the previous section, to prove the existence of its resultantR.
Determine the coordinates of the centerC(x c, y c, z c) parallel forces, that is, the coordinates of the point of application of the resultant of this system.
For this purpose, we will use Varignon's theorem, on the basis of which:
M 0 (R) = Σ M 0(P i).
Fig. 2
The vector-moment of force can be represented as a vector product, therefore:
M 0 (R) = r c× R = Σ M0i(P i) = Σ ( r i× P i ).
Considering that R = R v ∙ l, but P i = P vi ∙ l and using the properties of the vector product, we get:
r c × R v ∙ l = Σ ( r i × P vi ∙ l),
r c ∙ R v × l = Σ ( r i ∙ P vi × l) = Σ ( r i ∙ P vi ) × l,
or:
[ r c R v - Σ ( r i P vi )] × l= 0.
The last expression is valid only if the expression in square brackets is equal to zero. Therefore, omitting the indexvand considering that the resultantR = Σ P i , from here we get:
r c = (Σ P i r i )/(Σ P i ).
Projecting the last vector equality on the coordinate axis, we obtain the required expression for the coordinates of the center of parallel forces:
x c = (Σ P i x i)/(Σ P i );
y c = (Σ P i y i )/(Σ P i );(2)
z c = (Σ P i z i )/(Σ P i ).
Center of gravity of bodies.
Coordinates of the centers of gravity of a homogeneous body.
Consider solid weight P and volume V in coordinate system Oxyz where the axes x and y connected to the surface of the earth, and the axis z aimed at the zenith.
If we break the body into elementary parts by volume∆ V i , then the force of attraction will act on each part of it∆ P idirected towards the center of the earth. Suppose that the dimensions of the body are much less than the dimensions of the Earth, then the system of forces applied to the elementary parts of the body can be considered not converging, but parallel (Fig. 3), and all the conclusions of the previous chapter are applicable to it.
Fig. 3
Definition ... The center of gravity of a rigid body is the center of parallel forces of gravity of the elementary parts of this body.
Recall that specific gravity an elementary part of the body is the ratio of its weight∆ P i to the volume ∆ V i : γ i = ∆ P i/ ∆ V i ... For a homogeneous body, this value is constant:γ i = γ = P/ V.
Substituting in (2) ∆ P i = γ i ∙∆ V i instead P i, taking into account the last remark and canceling the numerator and denominator byg, we get expressions for the coordinates of the center of gravity of a homogeneous body:
x c = (Σ ∆ V i∙ x i)/(Σ ∆ V i);
y c = (Σ ∆ V i∙ y i )/(Σ ∆ V i);(3)
z c = (Σ ∆ V i∙ z i )/(Σ ∆ V i).
Several theorems are useful in determining the center of gravity.
1) If a homogeneous body has a plane of symmetry, then its center of gravity is in this plane.
If the axes x and at position in this plane of symmetry, then for each point with coordinates... And the coordinate according to (3), will be equal to zero, because in total everything members with opposite signs are destroyed in pairs. So the center of gravity is located in the plane of symmetry.
2) If a homogeneous body has an axis of symmetry, then the center of gravity of the body is on this axis.
Indeed, in this case, if the axiszdraw along the axis of symmetry, for each point with coordinatesyou can find a point with coordinates and coordinates and calculated by formulas (3) will be equal to zero.
The third theorem is proved similarly.
3) If a homogeneous body has a center of symmetry, then the center of gravity of the body is at this point.
And a few more comments.
First. If the body can be divided into parts for which the weight and position of the center of gravity are known, then there is no need to consider each point, and in formulas (3) P i - defined as the weight of the part concerned, and- as the coordinates of its center of gravity.
Second. If the body is homogeneous, then the weight of its individual part is where - specific gravity the material from which the body is made, and V i - the volume of this part of the body. And formulas (3) will take on a more convenient form. For example,
And similarly, where - the volume of the whole body.
Third remark. Let the body have the form of a thin plate with area F and thick t lying in the plane Oxy... Substituting in (3)∆ V i =t ∙ ∆ F i , we obtain the coordinates of the center of gravity of a homogeneous plate:
x c = (Σ ∆ F i∙ x i) / (Σ ∆ F i);
y c = (Σ ∆ F i∙ y i ) / (Σ ∆ F i).
z c = (Σ ∆ F i∙ z i ) / (Σ ∆ F i).
Where - coordinates of the center of gravity of individual plates;- total body area.
Fourth remark. For a body in the form of a thin curved rod of length L with cross-sectional area a elementary volume∆ V i = a ∙∆ L i , so coordinates of the center of gravity of a thin curved bar will be equal:
x c = (Σ ∆ L i∙ x i)/(Σ ∆ L i);
y c = (Σ ∆ L i∙ y i )/(Σ ∆ L i);(4)
z c = (Σ ∆ L i∙ z i )/(Σ ∆ L i).
Where - coordinates of the center of gravityi-th site; ...
Note that, according to the definition, the center of gravity is a geometric point; it can lie outside the limits this body(for example, for a ring).
Note.
In this section of the course, we do not differentiate between gravity, gravity, and body weight. In reality, gravity is the difference between the Earth's gravity and the centrifugal force caused by its rotation.
Coordinates of the centers of gravity of heterogeneous bodies.
Center of gravity coordinates inhomogeneous solid(Fig. 4) in the selected frame of reference are determined as follows:
Fig. 4
Where - weight per unit volume of the body (specific gravity)
-the weight of the whole body.
irregular surface(Fig. 5), then the coordinates of the center of gravity in the selected frame of reference are determined as follows:
Fig. 5
Where - weight per unit area of the body,
-the weight of the whole body.
If the solid is non-uniform line(Fig. 6), then the coordinates of the center of gravity in the selected frame of reference are determined as follows:
Fig. 6
Where - unit weight of body length,
Whole body weight.
Methods for determining the coordinates of the center of gravity.
Based on the general formulas obtained above, specific methods can be indicated 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
S = S 1 + S 2.
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 cut is represented by a combination of a solid plate (without a cut) with an area S 1 and the area of the cut part S 2.
Fig. 9
S = S 1 - S 2.
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 a circular arc. Consider an arc AB radiusR with center angle... By virtue of symmetry, the center of gravity of this arc lies on the axisOx(fig. 10).
Fig. 10
Find the coordinate according to the formula ... To do this, select on the arc AB element MM ’ the length, the position of which is determined by the angle... Coordinate x element MM ’ will be... 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 is the length of the arc 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 Oh equal
where is the angle 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 BUT 1 BUT 2, we come to the conclusion that the center of gravity of the triangle should belong to the median BUT 3 M 3 (fig. 11).
Fig. 11
Breaking the triangle into strips parallel to the side BUT 2 BUT 3, you can make sure that it should lie on the median BUT 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 part from each median, counting from the corresponding side.
In particular, for the median BUT 1 M 1 we obtain, taking into account that the coordinates of the point M 1 - this is the arithmetic mean of the coordinates of the vertices BUT 2 and BUT 3 :
x c = x 1 + (2/3) ∙ (xM 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 corner 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 = (4 R) / (3 π) ≅ 0.4 R .
Example 1.Let us determine the center of gravity of the homogeneous body shown in Fig. 13.
Fig. 13
Decision.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
Decision. Center of gravity coordinates:
0.
Squares:
Therefore:
Example 3.
At a square sheet
see cut square hole
see (fig. 15). Find the center of gravity of the leaf. Example 4. Find the position of the center of gravity of the plate shown in Fig. 16. Dimensions are in centimeters.
Fig. 16
Decision. Divide the plate into figures (fig. 17), centers whose severities are known.
The areas of these figures and the coordinates of their centers of gravity:
1) a rectangle with sides 30 and 40 cm,S 1 =30 ∙ 40 = 1200 cm 2 ; x 1= 15 cm; at 1 = 20 cm.
2) right triangle with a base of 50 cm and a height of 40 cm;S 2 =0,5 ∙ 50 ∙ 40 = 1000 cm 2 ; x 2 = 30 + 50/3 = 46.7 cm; y 2 =40/3 = 13.3 cm;
3) half circle of a circle of radius r = 20 cm;S 3 =0,5 ∙π∙ 20 2 = 628 cm 2 ; x 3 =4 R /3 π = 8.5 cm; at
Decision. Recall that in physics, the density of a bodyρ and its sharegrelated by the ratio:γ = ρ g whereg - acceleration of gravity. To find the mass of such a homogeneous body, you need to multiply the density by its volume.
Fig. 19
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:
WhereL i lengthi -th truss rod, andx 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 itL 1 = 4 m,x 1 = 0 m,y 1 = 2 m. The second group of rods consists of five rods, for itL 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 FROM lies on the straight line connecting FROM 1 and FROM 2 and divides the segment FROM 1 FROM 2 regarding: FROM 1 FROM/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 equal to 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 to 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 the 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?
- What graphical construction can be used to find the center of gravity of a triangle?
- Write down the formula for the center of gravity of the circular sector.
- Using the formulas that determine the centers of gravity of a triangle and a circular sector, derive a similar formula for a circular segment.
- What formulas are used to calculate the coordinates of the centers of gravity of homogeneous bodies, 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?
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 types of 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 parts of the body, 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? 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 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 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 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 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 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 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 rear axle and the knee axle. 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 palms on the hips.
How will you know that 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.
Determining the center of gravity of an arbitrary body by successive addition of forces acting on its individual parts is a difficult task; it is facilitated only for bodies of a relatively simple shape.
Let the body consist of only two weights of mass and, connected by a rod (Fig. 125). If the mass of the rod is small in comparison with the masses and, then it can be neglected. Each of the masses is acted upon by the forces of gravity, equal respectively to and; both of them are directed vertically downward, that is, parallel to each other. As we know, the resultant of two parallel forces is applied at the point, which is determined from the condition
Fig. 125. Determination of the center of gravity of a body consisting of two weights
Consequently, the center of gravity divides the distance between two weights in a ratio inverse to the ratio of their masses. If this body is suspended at a point, it will remain in equilibrium.
Since two equal masses have a common center of gravity at a point that halves 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 circular disk divides it into two completely identical symmetrical parts (Fig. 127), the center of gravity must lie on each diameter of the disk, that is, at the point of intersection of the diameters - in the geometric center of the disk. Reasoning in a similar way, one 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 in its center. The last example shows that the center of gravity of the body can lie outside the body.
Fig. 126. The center of gravity of a homogeneous rod lies in its middle
Fig. 127. The center of a homogeneous disk lies in its geometric center
If the body has an irregular shape or if it is heterogeneous (for example, there are voids in it), then the calculation of the position of the center of gravity is often difficult and it is more convenient to find this position through experience. For example, suppose 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 suspension point, which would begin to rotate the body. Therefore, having drawn a straight line on our piece of plywood, representing the continuation of the thread, we can assert 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 straight 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 center of gravity of the aircraft is determined by rolling it with wheels onto weighing platforms. The resultant of the forces of weight falling on each wheel will be directed vertically, and the line along which it acts can be found according to the law of addition of parallel forces.
Fig. 128. The point of intersection of the vertical lines drawn through the suspension points is the center of gravity of the body
When masses change separate parts body or when the shape of the body changes, the position of the center of gravity changes. Thus, the center of gravity of an 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 changing the shape of the body, 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 weight on one of the bars, then the center of gravity will move towards this weight.
Fig. 129. a) The center of gravity of the bars connected by the hinge, located on one straight line, lies on the axis of the bars, b) The center of gravity of the bent system of bars lies outside the bars
81.1. Where is the center of gravity of two identical thin rods 12 cm long and held together in the shape of a T?
81.2. Prove that the center of gravity of a homogeneous triangular plate lies at the intersection of the medians.
Fig. 130. To exercise 81.3
81.3. A homogeneous board with a mass of 60 kg lies on two supports, as shown in fig. 130. Determine the forces acting on the supports.
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, a figure of 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 freely rotate around a horizontal axis passing through point O (Fig. 25.4, a). Then this figure can be considered as a lever with a fulcrum O.
Fig. 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).
Physics lesson summary for grade 7
Topic: Determination of the center of gravity
Physics teacher MOU Argayash secondary school №2
Khidiyatulina Z.A.
Laboratory work:
"Determination of the center of gravity of a flat plate"
purpose : finding the center of gravity of a flat plate.
Theoretical part:
All bodies have a center of gravity. The center of gravity of a body is the point relative to which the total moment of the forces of gravity acting on the body is zero. For example, if you hang an object by its center of gravity, then it will remain at rest. That is, its position in space will not change (it will not turn upside down or on its side). Why do some bodies topple over while others do not? If from the center of gravity of the body draw a line perpendicular to the floor, then in the case when the line goes beyond the boundaries of the body's support, the body will fall. The larger the support area, the closer the body's center of gravity is to the center point of the support area and the center line of the center of gravity, the more stable the body position will be. For example, the center of gravity of the famous Leaning Tower of Pisa is located just two meters from the center of its pillar. And the fall will happen only when this deviation is about 14 meters. The center of gravity of the human body is approximately 20.23 centimeters below the navel. An imaginary line drawn vertically from the center of gravity runs exactly between the feet. A tumbler doll also has a secret in the center of gravity of the body. Its stability is explained by the fact that the tumbler's center of gravity is at the very bottom, it actually stands on it. The condition for maintaining the balance of the body is the passage of the vertical axis of its general center of gravity inside the support area of the body. If the vertical of the center of gravity of the body leaves the area of support, the body loses its balance and falls. Therefore, the larger the area of support, the closer the center of gravity of the body is to the central point of the area of support and the center line of the center of gravity, the more stable the position of the body will be. The area of support in the vertical position of a person is limited by the space that is under the soles and between the feet. The center point of the plumb line of the center of gravity on the foot is 5 cm in front of the heel. The sagittal size of the support area always prevails over the frontal one, therefore, the displacement of the plumb line of the center of gravity is easier to the right and left than backward, and especially difficult - forward. In this regard, cornering stability during fast running is much less than in the sagittal direction (forward or backward). A foot in shoes, especially with a wide heel and a stiff sole, is more stable than without shoes, since it acquires a large support area.
Practical part:
Purpose of work: Using the proposed equipment, empirically find the position of the center of gravity of two figures made of cardboard and a triangle.
Equipment:A tripod, thick cardboard, a triangle from a school kit, a ruler, tape, thread, pencil ..
Task 1: Determine the center of gravity of a freeform planar shape
Using scissors, cut a free-form shape out of the cardboard. Use duct tape to attach the thread to it at point A. Hang the figure by the thread from the tripod leg. Using a ruler and pencil, mark the vertical line AB on the cardboard.
Move the thread attachment point to position C. Repeat the above steps
Point O of intersection of lines AB andCDgives the desired position of the figure's center of gravity.
Exercise 2: Using only a ruler and pencil, find the position of the center of gravity of a flat figure
Divide the shape into two rectangles using a pencil and ruler. Find the positions O1 and O2 of their centers of gravity by construction. Obviously, the center of gravity of the whole figure is on the O1O2 line
Break the shape into two rectangles in a different way. Find the positions of the centers of gravity O3 and O4 of each of them by construction. Connect points O3 and O4 with a line. The point of intersection of the lines О1О2 and О3О4 determines the position of the center of gravity of the figure
Task 2: Determine the position of the triangle's center of gravity
Use duct tape to secure one end of the thread at the apex of the triangle and hang it from the tripod foot. Use a ruler to mark the AB direction of the gravity line (mark on the opposite side of the triangle)
Repeat the same procedure, hanging the triangle by the vertex C. At the opposite vertex C side of the triangle, markD.
Using adhesive tape, attach the segments of the AB andCD... The point O of their intersection determines the position of the center of gravity of the triangle. In this case, the center of gravity of the figure is outside the body itself.
III ... Solving quality problems
1.For what purpose do circus performers hold heavy poles in their hands while walking on a tightrope?
2.Why does a person carrying a heavy load on their back lean forward?
3. Why can't you get up from the chair if you don't tilt your body forward?
4.Why does the crane not tip over towards the load to be lifted? Why does the crane not tip over towards the counterweight without load?
5. Why do cars and bicycles, etc. is it better to put the brakes on the rear rather than the front wheels?
6. Why is a truck loaded with hay easier to roll over than the same truck loaded with snow?
