Triangle is one of the basic geometric shapes. And only he has “delightful” points. These include, say, center gravity– the point to which the weight of each figure is brought. Where is this “delightful” point and how to find it?
You will need
- pencil, ruler
Instructions
1. Draw the triangle itself. To do this, take a ruler and draw a line with a pencil. Then draw another segment, starting from one of the ends of the previous one. Close the shape by combining the two remaining free points of the segments. The result is a triangle. It was his center gravity to be found.
2. Take a ruler and measure the length of one of the sides. Find the middle of this side and mark it with a pencil. Draw a segment from opposite vertex to the designated point. The resulting segment is called the median.
3. Proceed to side 2. Measure its length, divide it into two equal parts and draw the median from the opposite vertex.
4. Do the same with the third party. Please note that if you did everything correctly, the medians will intersect at one point. This is what will happen center gravity or, as it is also called, center triangle masses.
5. If you are faced with a task, discover center gravity equilateral triangle, then draw the height from the entire vertex of the figure. To do this, take a ruler with a right angle and one of the sides, lean it against the base of the triangle, and direct the other to the opposite vertex. Do the same with the other sides. The intersection point will be center ohm gravity. The specificity of equilateral triangles is that the same segments are medians, altitudes, and bisectors.
6. Center gravity of any triangle divides the medians into two segments. Their ratio is 2:1 when viewed from the top. If a triangle is placed on a pin in such a way that center If the oid is on its tip, it will not fall, but will be in balance. Also center gravity is the point to which each mass placed on the vertices of the triangle is brought. Perform this skill and make sure that this point is not called “delicious” for nothing.
Tip 2: How to Find the Height of an Equilateral Triangle
An equilateral triangle is a triangle whose sides are all equal, as its name suggests. This specificity greatly simplifies finding the remaining parameters triangle, including its height.
You will need
- Side length of an equilateral triangle
Instructions
1. In an equilateral triangle, all angles are also equal. Equilateral angle triangle, henceforth, is equal to 180/3 = 60 degrees. Apparently, because all sides and all angles of this triangle are equal, then all its heights will also be equal.
2. In an equilateral triangle ABC it is possible to draw, say, height A.E. Because an equilateral triangle is special case isosceles triangle, and AB = AC. Consequently, by the property of isosceles triangle the height AE will be the same time as the median (that is, BE = EC) triangle ABC and the angle bisector BAC (that is, BAE = CAE).
3. The height AE will be the side of the rectangular triangle BAE with hypotenuse AB. AB = a – length of the side of an equilateral triangle. Then AE = AB*sin(ABE) = a*sin(60o) = sqrt(3)*a/2. Consequently, to find the height of an equilateral triangle, it is enough to know only the length of its side.
4. Apparently, if the median or bisector of an equilateral triangle, then it will be its height.
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In an arbitrary triangle, you can select several segments, the lengths of which often need to be calculated. These segments connect the points lying at the vertices of the triangle, at the midpoints of its sides, at the centers of the inscribed and circumscribed circles, as well as other points important for the geometry of the triangle. Some options for calculating the lengths of such segments in Euclidean geometry are given below.
Instructions
1. If the segment that you want to detect connects any two vertices arbitrary triangle, then he is one of the sides of this geometric figure. If we know, say, the lengths of the other 2 sides (A and B) and the size of the angle that they form (?), then you can calculate the length of this segment (C) based on the cosine theorem. Add the squares of the lengths of the sides, subtract from the total two lengths of the same sides, multiplied by the cosine of the given angle, and then find square root from the resulting value: C=?(A?+B?-2*A*B*cos(?)).
2. If a segment begins at one of the vertices of the triangle, ends on the opposite side and is perpendicular to it, then such a segment is called height (h). It can be detected, say, by knowing the area (S) and length (A) of the side on which the height is lowered - divide the doubled area by the length of the side: h=2*S/A.
3. If a segment connects the midpoint of each side of an arbitrary triangle and a vertex lying opposite this side, then it is called this segment median (m). You can find its length, say, by knowing the lengths of all sides (A, B, C) - add the double squares of the lengths of 2 sides, subtract from the resulting value the square of the side in the middle of which the segment ends, and then find the square root of the quarter the resulting total: m=?((2*A?+2*B?-C?)/4).
4. If a segment connects the center of a circle inscribed in an arbitrary triangle and any of the tangent points of this circle with the sides of the triangle, then its length can be determined by calculating the radius (r) of the inscribed circle. To do this, say, divide the area (S) of the triangle by its perimeter (P): r=S/P.
5. If a segment connects the center of a circle circumscribed about an arbitrary triangle with each of the vertices of this figure, then its length can be calculated by finding the radius of the circumscribed circle (R). If we know, say, the length of one of the sides (A) in such a triangle and the angle (?) lying opposite it, then to calculate the length of the segment you need, divide the length of the side by the double sine of the angle: R=A/(2*sin(? )).
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Tip 4: How to find the median of an equilateral triangle
Median triangle- this is the segment that connects the vertex of the triangle with the middle of the opposite side. In an equilateral triangle, the median is both the bisector and the height. Thus, the required segment can be constructed using several methods.
You will need
- - pencil;
- - ruler;
- – protractor;
- - compass.
Instructions
1. Using a ruler and pencil, divide the side of the equilateral triangle in half. Draw a line connecting the detected point and opposite corner triangle. Set aside the next two pieces in the same way. You have drawn the medians of an equilateral triangle.
2. Draw the height of the equilateral triangle. Using a square, lower a perpendicular from the vertex of the triangle to opposite side. You have constructed the height of an equilateral triangle. It is also its median.
3. Construct the bisectors of an equilateral triangle. Every angle of an equilateral triangle is equal to 60°. Attach the protractor to one of the sides of the triangle so that the reference point coincides with the vertex of the triangle. One of its sides must follow the line of the measuring device correctly, the other side must intersect the semicircle at the point marked 60?.
4. Mark the 30? division with a dot. Draw a ray connecting the detected point and the vertex of the triangle. Find the point where the ray intersects the side of the triangle. The resulting segment is the bisector of an equilateral triangle, which is its median.
5. If an equilateral triangle is inscribed in a circle, draw a straight line connecting its vertex to the center of the circle. Mark the point of intersection of this line with the side of the triangle. The segment connecting the vertex of the triangle and its side will be the median of the equilateral triangle.
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Useful advice
Construct the bisector of an angle? an equilateral triangle is permitted using a compass. To do this, construct two circles with a center at the other 2 vertices of the triangle and a radius equal to side triangle. The circles will intersect at 2 points: at the vertex of the corner? and at point N. Connect these points with each other. Have you constructed the bisector of the angle?
The center of a figure can be detected by several methods, depending on what data is already known about it. It is worth considering finding the center of a circle, which is a community of points located on equal distance from the center, because this figure is one of the most common.
You will need
- – square;
- - ruler.
Instructions
1. A primitive method of finding the center of a circle is to bend the sheet of paper on which it is drawn, making sure, by looking at the gap, that it is folded correctly in half. After this, bend the sheet perpendicular to the first fold. This way you will get diameters, the intersection point of which is the center of the figure.
2. Undoubtedly, this method perfect only if the circle is drawn on paper that is quite thin, so that one can see through the light whether the sheet is correct.
3. It is possible that the figure in question was drawn on a hard, inflexible surface, or it is a separate part that also cannot be bent. In order to find the center of the circle in this case, you need a ruler.
4. The diameter is the longest line segment connecting 2 points on a circle. As you know, it passes through the center, therefore the task of finding the center of the circle comes down to finding the diameter and its midpoint.
5. Place a ruler on the circle, then fix the zero mark at every point of the figure. Attach the ruler to the circle, obtaining a secant, and then move towards the center of the figure. The length of the secant will increase until it reaches the peak point. You will get the diameter, and having found its middle, you will also find the center of the circle.
6. The center of the circumscribed circle for any triangle is located at the intersection of the median perpendiculars. If the triangle is right-angled, its center will invariably coincide with the middle of the hypotenuse. That is, the solution lies in building inside the circle right triangle with vertices lying on a circle.
7. Stencil for right angle A school or construction square, a ruler, or even a sheet of paper/cardboard can serve. Place the vertex of a right angle at any point on the circle, make marks in those places where the sides of the angle intersect the boundary of the circle, and combine them. You have a diameter - the hypotenuse.
8. Using the same method, find another diameter, the intersection of 2 such segments will be the center of the circle.
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Pay attention!
The tasks may indicate that you need to detect the center of gravity, center of mass, or centroid. All three names mean the same thing.
Center of gravity(or center of mass) of a certain body is a point that has the property that if the body is suspended from this point, it will maintain its position.
Below we consider two-dimensional and three-dimensional problems associated with the search for various centers of mass - mainly from the point of view of computational geometry.
In the solutions discussed below, two main ones can be distinguished: fact. The first is that the center of mass of a system of material points is equal to the average of their coordinates, taken with coefficients proportional to their masses. The second fact is that if we know the centers of mass of two non-intersecting figures, then the center of mass of their union will lie on the segment connecting these two centers, and it will divide it in the same ratio as the mass of the second figure relates to the mass of the first.
Two-dimensional case: polygons
In fact, when speaking about the center of mass of a two-dimensional figure, one can mean one of next three tasks:
- The center of mass of the system of points - i.e. all the mass is concentrated only at the vertices of the polygon.
- The center of mass of the frame - i.e. The mass of a polygon is concentrated on its perimeter.
- The center of mass of a solid figure - i.e. The mass of the polygon is distributed over its entire area.
Each of these tasks has independent decision, and will be discussed separately below.
Center of mass of the point system
This is the simplest of three tasks, and its solution is the well-known physical formula for the center of mass of a system of material points:
where are the masses of the points, are their radius vectors (specifying their position relative to the origin), and is the desired radius vector of the center of mass.
In particular, if all points have the same mass, then the coordinates of the center of mass are arithmetic mean coordinates of points. For triangle this point is called centroid and coincides with the intersection point of the medians:
For proof These formulas suffice to remember that equilibrium is achieved at a point at which the sum of the moments of all forces is equal to zero. IN in this case this turns into the condition that the sum of the radius vectors of all points relative to the point, multiplied by the masses of the corresponding points, equals zero:
and, expressing from here , we obtain the required formula.
Frame center of mass
But then each side of the polygon can be replaced by one point - the middle of this segment (since the center of mass of a homogeneous segment is the middle of this segment), with a mass equal to length this segment.
Now we have a problem about a system of material points, and applying the solution from the previous paragraph to it, we find:
where is the midpoint of the ith side of the polygon, is the length of the ith side, is the perimeter, i.e. the sum of the lengths of the sides.
For triangle can be shown next statement: this point is bisector intersection point a triangle formed by the midpoints of the sides of the original triangle. (to show this, you need to use the above formula, and then notice that the bisectors divide the sides of the resulting triangle in the same ratios as the centers of mass of these sides).
Center of mass of a solid figure
We believe that the mass is distributed uniformly over the figure, i.e. the density at each point of the figure is equal to the same number.
Triangle case
It is argued that for a triangle the answer will be the same centroid, i.e. the point formed by the arithmetic mean of the coordinates of the vertices:
Triangle case: proof
Here we give an elementary proof that does not use the theory of integrals.
Archimedes was the first to give such a purely geometric proof, but it was very complicated, with a large number geometric constructions. The proof given here is taken from the article "Finding Centroids the Easy Way" by Apostol, Mnatsakanian.
The proof boils down to showing that the center of mass of the triangle lies on one of the medians; By repeating this process twice more, we will thereby show that the center of mass lies at the point of intersection of the medians, which is the centroid.
Let's break it down given triangle into four, connecting the middles of the sides, as shown in the figure:
The four resulting triangles are similar to the triangle with the coefficient .
Triangles No. 1 and No. 2 together form a parallelogram, the center of mass of which lies at the point of intersection of its diagonals (since this is a figure that is symmetrical with respect to both diagonals, and, therefore, its center of mass must lie on each of the two diagonals). The point is in the middle common side triangles No. 1 and No. 2, and also lies on the median of the triangle:
Let now the vector be the vector drawn from the vertex to the center of mass of triangle No. 1, and let the vector be the vector drawn from to the point (which, recall, is the middle of the side on which it lies):
Our goal is to show that the vectors and are collinear.
Let us denote by and the points that are the centers of mass of triangles No. 3 and No. 4. Then, obviously, the center of mass of the set of these two triangles will be the point, which is the middle of the segment. Moreover, the vector from point to point coincides with the vector.
The desired center of mass of the triangle lies in the middle of the segment connecting points and (since we divided the triangle into two parts of equal areas: No. 1-No. 2 and No. 3-No. 4):
Thus, the vector from the vertex to the centroid is . On the other hand, because triangle No. 1 is similar to a triangle with a coefficient, then the same vector is equal to . From here we get the equation:
where we find:
Thus, we have proven that the vectors and are collinear, which means that the desired centroid lies on the median emanating from the vertex.
Moreover, along the way we proved that the centroid divides each median in the ratio, counting from the vertex.
Polygon case
Let us now move on to the general case - i.e. to the occasion polygon. For him, such reasoning is no longer applicable, so we reduce the problem to a triangular one: namely, we divide the polygon into triangles (i.e., triangulate it), find the center of mass of each triangle, and then find the center of mass of the resulting centers of mass of the triangles.
The final formula is as follows:
where is the centroid of the th triangle in the triangulation given polygon, is the area of the th triangulation triangle, and is the area of the entire polygon.
Triangulation convex polygon- a trivial task: for this, for example, you can take triangles, where .
Polygon case: alternative way
On the other hand, the use of the above formula is not very convenient for non-convex polygons, since triangulating them is not an easy task in itself. But for such polygons, you can come up with a simpler approach. Namely, let’s draw an analogy with how you can search for the area of an arbitrary polygon: choose arbitrary point, and then the sign areas of the triangles formed by this point and the points of the polygon are summed up: . A similar technique can be used to find the center of mass: only now we will sum up the centers of mass of triangles taken with coefficients proportional to their areas, i.e. The final formula for the center of mass is:
where is an arbitrary point, is the points of the polygon, is the centroid of the triangle, is the signed area of this triangle, is the signed area of the entire polygon (i.e.).
Three-dimensional case: polyhedra
Similar to the two-dimensional case, in 3D we can immediately talk about four possible formulations of the problem:
- The center of mass of the system of points - vertices of the polyhedron.
- The center of mass of the frame is the edges of the polyhedron.
- The center of mass of the surface - i.e. the mass is distributed over the surface area of the polyhedron.
- The center of mass of a solid polyhedron - i.e. the mass is distributed throughout the polyhedron.
Center of mass of the point system
As in two-dimensional case, we can apply physical formula and get the same result:
which in case equal masses turns into the arithmetic mean of the coordinates of all points.
Center of mass of the polyhedron frame
Similar to the two-dimensional case, we simply replace each edge of the polyhedron with a material point located in the middle of this edge, and with a mass equal to the length of this edge. Having received the problem of material points, we easily find its solution as a weighted sum of the coordinates of these points.
Center of mass of the polyhedron surface
Each face of the surface of a polyhedron is a two-dimensional figure, the center of mass of which we know how to look for. Having found these centers of mass and replacing each face with its center of mass, we get a problem with material points, which is already easy to solve.
Center of mass of a solid polyhedron
The case of the tetrahedron
As in the two-dimensional case, let us first solve simplest task- a problem for a tetrahedron.
It is stated that the center of mass of a tetrahedron coincides with the point of intersection of its medians (the median of a tetrahedron is a segment drawn from its vertex to the center of mass opposite face; Thus, the median of the tetrahedron passes through the vertex and through the point of intersection of the medians of the triangular face).
Why is this so? Here, reasoning similar to the two-dimensional case is valid: if we cut a tetrahedron into two tetrahedra using a plane passing through the vertex of the tetrahedron and some median of the opposite face, then both resulting tetrahedra will have the same volume (since triangular face will be divided by the median into two triangles equal area, and the height of the two tetrahedra will not change). Repeating these arguments several times, we find that the center of mass lies at the point of intersection of the tetrahedron medians.
This point - the point of intersection of the medians of the tetrahedron - is called its centroid. It can be shown that it actually has coordinates equal to the arithmetic mean of the coordinates of the tetrahedron vertices:
(this can be inferred from the fact that the centroid divides the medians in the ratio)
Thus, there is no fundamental difference between the cases of a tetrahedron and a triangle: the point equal to the arithmetic mean of the vertices is the center of mass in two formulations of the problem: both when the masses are located only at the vertices, and when the masses are distributed over the entire area/volume. In fact, this result generalizes to an arbitrary dimension: the center of mass of an arbitrary simplex(simplex) is the arithmetic mean of the coordinates of its vertices.
The case of an arbitrary polyhedron
Let us now move on to the general case—the case of an arbitrary polyhedron.
Again, as in the two-dimensional case, we reduce this problem to an already solved one: we divide the polyhedron into tetrahedrons (i.e., we tetrahedronize it), find the center of mass of each of them, and obtain the final answer to the problem in the form of a weighted sum of the found centers wt.
Determining the location of the barycenter by integration
Barycenter subset X space R n (\displaystyle \mathbb (R) ^(n)) can be calculated using the integral
G = ∫ x g (x) d x ∫ g (x) d x , (\displaystyle G=(\frac (\int xg(x)\;dx)(\int g(x)\;dx)),)Another formula for calculating the coordinates of the barycenter:
G k = ∫ z S k (z) d z ∫ S k (z) d z , (\displaystyle G_(k)=(\frac (\int zS_(k)(z)\;dz)(\int S_(k )(z)\;dz)),)Where G k is k th coordinate G, A S k (z) - measure of intersection X with the hyperplane defined by the equation x k = z. Again the denominator is a measure of the set X.
For flat figure the coordinates of the barycenter will be
G x = ∫ x S y (x) d x A ; (\displaystyle G_(\mathrm (x) )=(\frac (\int xS_(\mathrm (y) )(x)\;dx)(A));) G y = ∫ y S x (y) d y A , (\displaystyle G_(\mathrm (y) )=(\frac (\int yS_(\mathrm (x) )(y)\;dy)(A)) ,)Where A- area of the figure X, S y ( x) - intersection length [unknown term ] X with a vertical line with an abscissa x, S x( y) - a similar value when exchanging axes.
Determining the location of the barycenter for an area bounded by graphs of continuous functions
Barycenter coordinates (x ¯ , y ¯) (\displaystyle ((\bar (x)),\;(\bar (y)))) areas, limited by schedules continuous functions f (\displaystyle f) And g (\displaystyle g), such that f (x) ≥ g (x) (\displaystyle f(x)\geq g(x)) on the interval [ a , b ] (\displaystyle ), a ≤ x ≤ b (\displaystyle a\leq x\leq b), are given by the expressions
x ¯ = 1 A ∫ a b x [ f (x) − g (x) ] d x (\displaystyle (\bar (x))=(\frac (1)(A))\int _(a)^(b) x\left\;dx) . y ¯ = 1 A ∫ a b [ f (x) + g (x) 2 ] [ f (x) − g (x) ] d x , (\displaystyle (\bar (y))=(\frac (1)( A))\int _(a)^(b)\left[(\frac (f(x)+g(x))(2))\right]\left\;dx,)Where A (\displaystyle A)- area of the region (calculated by the formula ∫ a b [ f (x) − g (x) ] d x (\displaystyle \int _(a)^(b)\left\;dx)) .
Determining the location of the barycenter of an L-shaped object
A method for finding the barycenter of an L-shaped figure.
Barycenters of a triangle and tetrahedron
G = 1 a: 1 b: 1 c = b c: c a: a b = csc A: csc B: csc C (\displaystyle G=(\frac (1)(a)):(\frac (1) (b)):(\frac (1)(c))=bc:ca:ab=\csc A:\csc B:\csc C) = cos A + cos B ⋅ cos C: cos B + cos C ⋅ cos A: cos C + cos A ⋅ cos B (\displaystyle =\cos A+\cos B\cdot \cos C:\cos B+\cos C\cdot \cos A:\cos C+\cos A\cdot \cos B) = sec A + sec B ⋅ sec C: sec B + sec C ⋅ sec A: sec C + sec A ⋅ sec B . (\displaystyle =\sec A+\sec B\cdot \sec C:\sec B+\sec C\cdot \sec A:\sec C+\sec A\cdot \sec B.)The barycenter is also physically the center of mass of a triangle made of uniform sheet material, and also if all the mass is concentrated at the vertices and equally divided between them. If the mass is distributed evenly along the perimeter, then the center of mass lies at the Spieker point (incenter of the middle triangle), which (at general case) does not coincide with the centroid of the entire triangle.
The area of a triangle is equal to 3/2 the length of any side multiplied by the distance from the centroid to the side.
The centroid of a triangle lies on the Euler straight line between its orthocenter H and the center of its circumcircle O, exactly twice as close to the second as to the first:
G H = 2 G O . (\displaystyle GH=2GO.)In addition, for incenter I and the center of nine points N, we have
G H = 4 G N , (\displaystyle GH=4GN,) G O = 2 G N , (\displaystyle GO=2GN,) I G< H G , {\displaystyle IGHas similar properties
The median of a triangle is the diameter that bisects the chords parallel to the base, therefore the center of gravity (n° 217) of the area of the triangle lies on it. Consequently, the three medians of the triangle, intersecting, determine the center of gravity of the area of the triangle.
Elementary considerations show that the medians of a triangle intersect at a point two-thirds the length of each of them from the corresponding vertex. Therefore, the center of gravity of the area of a triangle lies on any of its medians at a distance of two-thirds of its length from the vertex.
219. Quadrangle.
The center of gravity of the area of the quadrilateral is determined by the intersection of two straight lines, which we obtain by applying the distribution property of centers of gravity (item 213).
First, divide the quadrilateral diagonally into two triangles. The center of gravity of the quadrilateral lies on the straight line connecting the centers of gravity of these triangles. This straight line is the first of the two required straight lines.
We obtain the second straight line in the same way, dividing the quadrilateral into two triangles (different from the previous ones) using another diagonal.
220. Polygon.
We know how to find the centers of gravity of the area of a triangle and a quadrilateral. To determine the centroid of the area of a polygon with an arbitrary number of sides, assume that we know how to find the centroid of the area of a polygon with a smaller number of sides.
Then you can do the same as in the case of a quadrilateral. The area of a given polygon is divided into two parts in two different ways by drawing diagonals. In each of the two cases, the centers of gravity of the individual parts are connected directly. These two lines intersect at the desired center of gravity.
221. Arc of a circle.
Let it be necessary to determine the center of gravity of a circular arc AB of length s. Let us relate the circle to two mutually perpendicular diameters OX and OY, of which the first passes through the middle C of the arc AB. The center of gravity lies on the OX axis, which is the axis of symmetry. It is therefore sufficient to determine 5. For this we have the formula:
Let them be: a - the radius of the circle, c - the length of the chord AB, - the angle between the OX axis and the radius drawn to the element values , corresponding to the ends of the arc AB. We have:
Then, taking B as the integration variable and performing integration along the arc AB, we obtain:
Consequently, the center of gravity of a circular arc lies on the radius drawn through the middle of the arc, at a point whose distance from the center of the circle is a fourth proportional to the length of the arc, the radius and the chord.
222. Circular sector.
The sector enclosed between the arc of a circle and two radii OA and OB can be decomposed by intermediate radii into infinitesimal sectors equal to each other. These elementary sectors can be considered as infinitely narrow triangles; the center of gravity of each of them, according to the previous one, lies on the radius drawn through the middle of the elementary arc of this sector, at a distance of two-thirds of the length of the radius from the center of the circle. The equal masses of all elementary triangles, concentrated at their centers of gravity, form a uniform arc of a circle, the radius of which is equal to two-thirds of the radius of the sector arc. The case under consideration is thus reduced to finding the center of gravity of this homogeneous arc, i.e., to the problem solved in the previous paragraph.
223. Tetrahedron.
Let us determine the center of gravity of the volume of the tetrahedron. The plane passing through one of the edges and through the middle of the opposite edge is the diametrical plane which bisects the chords parallel to this last edge: it therefore contains the center of gravity of the volume of the tetrahedron. Consequently, the six planes of the tetrahedron, each of which passes through one of the edges and through the middle of the opposite edge, intersect at one point, which represents the center of gravity of the volume of the tetrahedron.
Consider the tetrahedron ABCD (Fig. 37); connect vertex A to the center of gravity I of the base BCD; straight line AI is the intersection of the diametrical planes passing
through the edges AB and therefore it contains the desired center of gravity. The point is located at a distance of two-thirds of the median BH from the vertex B. In the same way, let’s take a point K on the median AH at a distance of two-thirds of its length from the vertex. Line B K will intersect line A at the center of gravity of the tetrahedron. Let us draw from the similarity of triangles ABN and JN, it is clear that IK is the third part of AB), then from the similarity of triangles and BGA we conclude that there is a third part.
Consequently, the center of gravity of the volume of the tetrahedron lies on the segment connecting any vertex of the tetrahedron with the center of gravity of the opposite face, at a distance of three quarters of the length of this segment from the vertex.
Let us also note that the straight line connecting the midpoints R and L of two opposite edges (Fig. 38) is the intersection of the diametrical planes passing through these edges; it also passes through the center of gravity of the tetrahedron. Thus, three straight lines connecting the midpoints of opposite edges of the tetrahedron intersect at its center of gravity.
Let H and be the midpoints of one pair of opposite edges (Fig. 38) and M, N the midpoints of two other opposite edges. The figure HNLM is a parallelogram whose sides are respectively parallel to the others
two ribs. The straight lines HL and MN, connecting the midpoints of two opposite edges, are the diagonals of this parallelogram, which means they are divided in half at the point of intersection. Thus, the center of gravity of the tetrahedron lies in the middle of the segment connecting the midpoints of two opposite edges of the tetrahedron.
224. Pyramid with a polygonal base.
The center of gravity of the pyramid lies on the segment connecting the top of the pyramid with the center of gravity of the base at a distance of three quarters of the length of this segment from the top.
To prove this theorem, we decompose the pyramid into tetrahedrons by planes drawn through the top of the pyramid and through the diagonals of the base ABCD (for example, BD in Fig. 39).
Let's draw a plane intersecting the edges at a distance of three quarters of their length from the top. This plane contains the centers of gravity of the tetrahedrons, and therefore the pyramids. The masses of tetrahedrons, which we assume to be concentrated at their centers of gravity, are proportional to their volumes, and therefore to the areas of the bases (Fig. 39) or also to the areas of triangles bad, bed,..., similar to the previous ones and located in the secant plane abcd... Thus Thus, the desired center of gravity coincides with the center of gravity of the polygon abcd. The latter lies on the straight line connecting the vertex S of the pyramid with the center of gravity (similarly located) of the base polygon.
225. Prism. Cylinder. Cone.
Based on symmetry, the centers of gravity of the prism and the cylinder lie in the middle of the segment connecting the centers of gravity of the bases.
Considering the cone as the limit of a pyramid inscribed in it with the same vertex, we are convinced that the center of gravity of the cone lies on the segment connecting the vertex of the cone with the center of gravity of the base, at a distance of three quarters of the length of this segment from the vertex. We can also say that the center of gravity of the cone coincides with the center of gravity of the section of the cone by a plane parallel to the base and drawn at a distance of one quarter of the height of the cone from the base.
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 area are known.
Fig.8
3.Negative area method. A special case of the partitioning method (Fig. 9). It applies to bodies that have cutouts if the centers of gravity of the body without the cutout and the cutout part are known. A body in the form of a plate with a cutout is represented by a combination of a solid plate (without a cutout) with an area S 1 and an area of the cut out part S 2 .
Fig.9
4.Grouping method. It is a good complement to the last two methods. After dividing a figure into its component elements, it is 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) Center of gravity of a circular arc. Consider the arc AB radius R with a central angle. Due to 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 over the entire length of the arc, we obtain:
Where L- arc length AB, equal to .
From here we finally find that the center of gravity of a circular arc lies on its axis of symmetry at a distance from the center ABOUT, equal
where the angle is measured in radians.
2) Center of gravity of the triangle's area. 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 must belong to the median A 3 M 3 (Fig. 11).
Fig.11
Breaking a triangle into strips parallel to the side A 2 A 3, we can verify that it must lie on the median A 1 M 1. Thus, the center of gravity of a triangle lies at the point of intersection of its medians, which, as is known, separates a third part 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 triangle’s center of gravity 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) Center of gravity of the area of a circular sector. Consider a sector of a circle with radius R with a central angle of 2α, located symmetrically relative to 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 by dividing the integration domain into elementary sectors with an angle dφ. Accurate to infinitesimals of the 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 obtain: x c = (4R)/(3π) ≅ 0.4 R .
Example 1. Let us determine the center of gravity of the homogeneous body shown in Fig. 13.
Fig.13
The body is homogeneous, consisting of two parts with a symmetrical shape. Coordinates of their centers of gravity:
Their volumes:
Therefore, the coordinates of the center of gravity of the body
Example 2. Let us find the center of gravity of a plate bent at a right angle. Dimensions are in the drawing (Fig. 14).
Fig.14
Coordinates of the centers of gravity:
Areas:
|
Fig.15
In this problem, 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 bracket (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 are:
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).
Let us 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- free fall acceleration. 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 rod, the linear density must be multiplied by the length of this rod.
To solve the problem, you can use the partitioning method. Representing a given truss as a sum of 6 individual rods, we obtain:
Where L i length i th truss rod, and x i, y i- coordinates of its center of gravity.
The solution to this problem can be simplified by grouping the last 5 bars of the truss. 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 rods.
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 truss are found using 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 regarding: 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 the center of parallel forces called?
How are the coordinates of the center of parallel forces determined?
How to determine the center of parallel forces whose resultant is zero?
What properties does the center of parallel forces have?
What formulas are used to calculate the coordinates of the center of parallel forces?
What is the center of gravity of a body?
Why can the gravitational forces of the Earth acting on a point on a body 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 the static moment of area?
Give an example of a body whose center of gravity is located outside the body.
How are the properties of symmetry used in determining the centers of gravity of bodies?
What is the essence of the negative weights method?
Where is the center of gravity of a circular arc?
What graphical construction can be used to find the center of gravity of a triangle?
Write down the formula that determines the center of gravity of a circular sector.
Using 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, flat figures and lines?
What is called the static moment of the area of a plane 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?
