Let be a planar -gon, and the polygon is obtained from parallel translation to a vector not parallel to the plane. A polyhedron bounded by polygons and and parallelograms , , ... (Fig. 1) is called a -gonal prism (from the Greek word prisma - “sawn piece”) with bases and , side faces , , ... and side edges , , ... . If the side edges are perpendicular to the planes of the bases, then the prism is called straight, otherwise it is called inclined. Finally, a prism is called regular if it is straight and has regular polygons at its bases.

A regular angle-coal prism aligns with itself when rotated around its axis - a straight line passing through the centers of the bases and (Fig. 2). The planes of symmetry of the prism pass through the axis, and another plane of symmetry passes through the middle of the segment perpendicular to it. Exactly the same planes of symmetry have the dihedron, or bipyramid, dual to a regular angular prism - a polyhedron bounded by triangles with vertices at the centers of the bases and lateral faces of the prism (Fig. 3). Naturally occurring single crystals often take the form of regular, possibly truncated, prisms and dihedrals (due to crystallographic limitations, the number for crystal forms can only be 3, 4 or 6).

Another special case of symmetrical prisms is a parallelepiped, i.e. prism with parallelograms at the bases. A parallelepiped has 4 diagonals that intersect at one point - the center of symmetry of the parallelepiped. At this point the diagonals are divided in half (Fig. 4). Right parallelepipeds also have an axis of symmetry passing through the centers of the bases (Fig. 5). If the bases of a right parallelepiped are rectangles, then it is called rectangular. Rectangular parallelepipeds predominate among the polyhedral shapes around us: these are all kinds of boxes, rooms, buildings, etc. These parallelepipeds have three mutually perpendicular planes of symmetry, intersecting along three axes of symmetry (Fig. 6). Among rectangular parallelepipeds, regular quadrangular prisms (5 planes of symmetry) and a cube (9 planes of symmetry - Fig. 7 shows how they cut the surface of the cube) are even more symmetrical.

There is an interesting connection between parallelepipeds and tetrahedrons: if a pair of parallel planes are drawn through every two intersecting edges of the tetrahedron, then the resulting six planes will bound the parallelepiped described around the tetrahedron (Fig. 8). In this case, a regular tetrahedron corresponds to a cube, and isohedral tetrahedrons correspond to rectangular parallelepipeds.

The volume of an arbitrary prism is equal to the product of the area of ​​its base and its height, i.e. to the distance between the planes of the bases. There is another formula for the volume of a prism, where is the length of the side edge, a is the area of ​​the prism section perpendicular to the side edges.

Instructions

If in the conditions of the problem the volume (V) of the space bounded by the edges is given prisms, and the area of ​​its base (s), to calculate the height (H) use the formula common to the base of any geometric shape. Divide the volume by the area of ​​the base: H=V/s. For example, with a base of 1200 cm³ equal to 150 cm², the height prisms should be equal to 1200/150=8 cm.

If the quadrilateral at the base prisms, has the shape of any regular figure; instead of area, you can use edge lengths in calculations prisms. For example, with a square base, replace the area in the formula of the previous step with the second power of the length of its edge (a):H=V/a². And in the case of the same formula, substitute the product of the lengths of two adjacent edges of the base (a and b): H=V/(a*b).

To calculate height (H) prisms Knowing the total surface area (S) and the length of one base edge (a) may be sufficient. Since the total area is the sum of the areas of two bases and four side faces, and in such a polyhedron the base is , the area of ​​one side surface should be equal to (S-a²)/4. This face has two common edges with square edges of known size, which means to calculate the length of the other edge, divide the resulting area by the side of the square: (S-a²)/(4*a). Since the prism in question is rectangular, the edge of the length you calculated adjoins the bases at an angle of 90°, i.e. coincides with the height of the polyhedron: H=(S-a²)/(4*a).

In the correct height (H), knowing the length of the diagonal (L) and one edge of the base (a) is enough to calculate the height (H). Consider the triangle formed by this diagonal, the diagonal of the square base and one of the side edges. The edge here is an unknown quantity that coincides with the desired height, and the diagonal of the square, based on the Pythagorean theorem, is equal to the product of the length of the side and the root of two. In accordance with the same theorem, express the desired quantity (leg) in terms of the length of the diagonal prisms(hypotenuse) base (second leg): H=√(L²-(a*V2)²)=√(L²-2*a²).

Sources:

• quadrangular prism

A prism is a device that separates normal light into individual colors: red, orange, yellow, green, blue, indigo, violet. It is a translucent object, with a flat surface that refracts light waves depending on their wavelengths and thus allows light to be seen in different colors. Do prism It's pretty easy to do it yourself.

You will need

• Two sheets of paper
• Foil
• Cup
• CD
• Coffee table
• Flashlight
• Pin

Instructions

Adjust the position of the flashlight and paper until you see a rainbow on the sheets - this is how your beam of light is decomposed into spectra.

Video on the topic

A quadrangular pyramid is a pentahedron with a quadrangular base and a side surface of four triangular faces. The lateral edges of the polyhedron intersect at one point - the vertex of the pyramid.

Instructions

A quadrangular pyramid can be regular, rectangular or arbitrary. A regular pyramid has a regular quadrangle at its base, and its apex is projected into the center of the base. The distance from the top of the pyramid to its base is called the height of the pyramid. The side faces are isosceles triangles, and all edges are equal.

The base of a regular one can be a square or a rectangle. The height H of such a pyramid is projected to the point of intersection of the diagonals of the base. In a square and a rectangle, the diagonals d are the same. All lateral edges L of a pyramid with a square or rectangular base are equal to each other.

To find the edge of the pyramid, consider a right triangle with sides: the hypotenuse is the desired edge L, the legs are the height of the pyramid H and half the diagonal of the base d. Calculate the edge using the Pythagorean theorem: the square of the hypotenuse is equal to the sum of the squares of the legs: L²=H²+(d/2)². In a pyramid with a rhombus or parallelogram at the base, the opposite edges are equal in pairs and are determined by the formulas: L₁²=H²+(d₁/2)² and L₂²=H²+(d₂/2)², where d₁ and d₂ are the diagonals of the base.

Find the fourth edge L₃ of a rectangular pyramid using the Pythagorean theorem as the hypotenuse of a right triangle with legs H and d, where d is the diagonal of the base drawn from the base of the edge coinciding with the height of the pyramid H to the base of the desired edge L₃: L₃²= H²+d².

In an arbitrary pyramid, its vertex is projected to a random point on the base. To find the edges of such a pyramid, consider sequentially each of the right triangles, in which the hypotenuse is the desired edge, one of the legs is the height of the pyramid, and the second leg is the segment connecting the corresponding vertex of the base with the base of the height. To find the values ​​of these segments, it is necessary to consider the triangles formed at the base by connecting the projection point of the top of the pyramid and the corners of the quadrilateral.

Definition.

This is a hexagon, the bases of which are two equal squares, and the side faces are equal rectangles

Side rib- is the common side of two adjacent side faces

Prism height- this is a segment perpendicular to the bases of the prism

Prism diagonal- a segment connecting two vertices of the bases that do not belong to the same face

Diagonal plane- a plane that passes through the diagonal of the prism and its lateral edges

Diagonal section- the boundaries of the intersection of the prism and the diagonal plane. The diagonal cross section of a regular quadrangular prism is a rectangle

Perpendicular section (orthogonal section)- this is the intersection of a prism and a plane drawn perpendicular to its lateral edges

Elements of a regular quadrangular prism

The figure shows two regular quadrangular prisms, which are indicated by the corresponding letters:

• The bases ABCD and A 1 B 1 C 1 D 1 are equal and parallel to each other
• Side faces AA 1 D 1 D, AA 1 B 1 B, BB 1 C 1 C and CC 1 D 1 D, each of which is a rectangle
• Lateral surface - the sum of the areas of all lateral faces of the prism
• Total surface - the sum of the areas of all bases and side faces (sum of the area of ​​the side surface and bases)
• Side ribs AA 1, BB 1, CC 1 and DD 1.
• Diagonal B 1 D
• Base diagonal BD
• Diagonal section BB 1 D 1 D
• Perpendicular section A 2 B 2 C 2 D 2.

Properties of a regular quadrangular prism

• The bases are two equal squares
• The bases are parallel to each other
• The side faces are rectangles
• The side edges are equal to each other
• The side faces are perpendicular to the bases
• The lateral ribs are parallel to each other and equal
• Perpendicular section perpendicular to all side ribs and parallel to the bases
• Angles of perpendicular section - straight
• The diagonal cross section of a regular quadrangular prism is a rectangle
• Perpendicular (orthogonal section) parallel to the bases

Formulas for a regular quadrangular prism

Instructions for solving problems

When solving problems on the topic " regular quadrangular prism" means that:

Correct prism- a prism at the base of which lies a regular polygon, and the side edges are perpendicular to the planes of the base. That is, a regular quadrangular prism contains at its base square. (see properties of a regular quadrangular prism above) Note. This is part of a lesson with geometry problems (section stereometry - prism). Here are problems that are difficult to solve. If you need to solve a geometry problem that is not here, write about it in the forum. To denote the action of extracting the square root in solving problems, the symbol is used√ .

Task.

In a regular quadrangular prism, the base area is 144 cm 2 and the height is 14 cm. Find the diagonal of the prism and the total surface area.

Solution.
A regular quadrilateral is a square.
Accordingly, the side of the base will be equal

√144 = 12 cm.
From where the diagonal of the base of a regular rectangular prism will be equal to
√(12 2 + 12 2 ) = √288 = 12√2

The diagonal of a regular prism forms a right triangle with the diagonal of the base and the height of the prism. Accordingly, according to the Pythagorean theorem, the diagonal of a given regular quadrangular prism will be equal to:
√((12√2) 2 + 14 2 ) = 22 cm

Answer: 22 cm

Task

Determine the total surface of a regular quadrangular prism if its diagonal is 5 cm and the diagonal of its side face is 4 cm.

Solution.
Since the base of a regular quadrangular prism is a square, we find the side of the base (denoted as a) using the Pythagorean theorem:

A 2 + a 2 = 5 2
2a 2 = 25
a = √12.5

The height of the side face (denoted as h) will then be equal to:

H 2 + 12.5 = 4 2
h 2 + 12.5 = 16
h 2 = 3.5
h = √3.5

The total surface area will be equal to the sum of the lateral surface area and twice the base area

S = 2a 2 + 4ah
S = 25 + 4√12.5 * √3.5
S = 25 + 4√43.75
S = 25 + 4√(175/4)
S = 25 + 4√(7*25/4)
S = 25 + 10√7 ≈ 51.46 cm 2.

Answer: 25 + 10√7 ≈ 51.46 cm 2.

Prism– a polyhedron obtained from the intersection of a prismatic surface by two parallel planes. Equal polygons (faces) obtained in the section of a prismatic surface with parallel planes are called its reasons, and other faces (parallelograms) – side faces(Fig. 2.14).

The prism is called straight, if its lateral edges are perpendicular to the bases. The prism is called correct, if it is straight and its base is a regular polygon. The prism is called inclined, if its side edges (faces) are not perpendicular to the bases. The prism is called triangular, if its base is a triangle, quadrangular, if its base is a quadrilateral, and in general n-coal, if its base is n-gon. Prismatic surface- a surface formed by the movement of a straight line in space so that this straight line remains parallel to itself and intersects a given broken line. Rice. 2.14. Prism

The moving straight line is called generatrix prismatic surface, and this broken line is its guide.

2.6.5. Gravity and body weight

1. If the Earth didn't rotate , then onto a body of mass T, lying motionless on a support in a vacuum, the gravitational force would act
, directed towards the center of the Earth, as well as the ground reaction force , directed from the center of the Earth. In equilibrium

. (2.37)

Assuming for simplicity that the Earth has spherical symmetry (in shape and density), we can write the gravitational force as Newton’s law of universal gravitation:

, (2.38)

where is the gravitational constant;

M= 5.96·10 24 kg – mass of the Earth;

R= 6.37·10 6 m – the average radius of the Earth.

Being attracted to the Earth, the body acts on the stand with the force of weight . According to Newton's third law

. (2.39)

From equations (2.37)–(2.39) it follows that the weight force , acting in a vacuum from a body at rest on a support or suspension on a hypothetical non-rotating Earth, would be equal to the gravitational force

(2.40)

and would be directed towards the center of the Earth.

Without support there is no reaction force , no weight force . Then the body would fall freely in the field of one gravitational force with acceleration

, (2.41)

independent of body weight

(2.42)

and coinciding in magnitude and direction with the vector of the gravitational field strength at any point of the trajectory.

2. However, the Earth rotates in the system of fixed stars and is therefore a non-inertial frame of reference.

In a non-inertial frame of reference, each material point (body) is acted upon by inertial force
, which is not the result of the interaction of bodies, but the result of accelerated motion of the reference system. The inertial force is equal to the product of mass T material point (body) for acceleration reference systems:

. (2.43)

The minus sign shows that the inertial force is directed in the direction opposite to the acceleration vector reference systems. The inertial force in a rotating frame of reference is directed along the radiusrfrom the axis of rotation(Fig. 2.15, A).

The magnitude of the inertial force depends on the distance r to the axis of rotation. This distance depends on latitude

(2.44)

and at different latitudes it is different - at the equator it is greatest (  = 0), and at the pole it is equal to zero (
).

At latitude the inertial force is equal to

Where
– angular velocity of the Earth’s rotation.

Rice. 2.15. Inertial force in the rotating reference frame "Earth"

Let us consider in more detail the forces acting on a body that rests on the surface of the rotating Earth at a certain latitude in the absence of an environment. The gravitational force acts on the body
, directed towards the center of the Earth, and the force of inertia
, directed from the axis of its rotation. Ground reaction force keeps the body stationary relative to the Earth. Since the body is at rest in the reference frame, the forces acting on the body are compensated

. (2.46)

From equality (2.46) it follows that the reaction force balances the sum of gravitational forces
and inertia
. Reaction force line of action coincides with the line of action of the resultant two forces
+
and is directed from the Earth’s surface, forming with the O axis X drawn from the center of rotation (point O), a certain angle α , different from the angle geographical latitude ( α ≠ φ ).

Geometric sum of gravitational force
and inertia forces
,taking into account The daily rotation of the Earth is called gravity
, acting on a stationary body(Fig. 2.15, b):

. (2.47)

Then condition (2.46) for the equilibrium of the body has the form:

+ = 0. (2.48)

Body weight is the force with which any body located in the field of gravity acts on a support or suspension that prevents the free fall of the body (Fig. 2.16). Powers And – these are the forces of interaction between the body and the support. According to Newton's third law:

= – . (2.49)

Rice. 2.16. Forces acting on the body and support ( A); on the body and suspension ( b)

Thus, on a rotating Earth in the absence of a medium, the weight of a stationary body in magnitude and direction coincides with the force of gravity
(2.47):

=
, (2.50)

those. the weight is equal to the geometric sum of the gravitational force
and inertia forces
. Weight and gravity are applied to different objects (weight - to a support or suspension, gravity - to the body) and have a different physical nature (weight - elastic, i.e. essentially electromagnetic, and gravity - mainly gravitational). The weight of a body on a rotating planet is a static manifestation of gravity, as a result of which the support or suspension is deformed.

Let's define gravity and body weight, located at an arbitrary point on the earth's surface at latitude φ . From the triangle of forces (Fig. 2.17) it follows

. (2.51)

Rice. 2.17. Forces acting on the body and support

resting in a rotating reference frame "Earth"

Taking into account expressions (2.38) and (2.45), we obtain

Thus, body weight and gravity depend on body mass T, from the parameters characterizing the Earth ( M,ω), and from the position of the body on Earth ( R,φ ). At the poles, the weight of the body and the force of gravity are greatest and equal to the gravitational force

(2.53)

At the equator (
,
) body weight and gravity take the smallest value

If we take into account that the polar and equatorial radii of the Earth are not the same ( R floor= 6356.9 km, R eq= 6378.1 km), then

. (2.55)

After substituting the values ​​into formula (2.55) R floor , R eq , M, γ , and we also get

Thus, taking into account the difference in the polar and equatorial radii and the rotation of the Earth, the weight of the body and the force of gravity at the equator decrease by approximately 1.0% of the value at the pole!

Let us now define direction of gravity and body weight. The force of gravity and the weight of a body are directed towards the center of the Earth only at the poles and at the equator. There is no such coincidence at other points on the earth's surface. Deflection angle ∆α from the direction to the center of the Earth depends on the geographical latitude φ . Because the angle ∆α small, then from Fig. 2.17 it follows that for a spherical Earth

(2.56)

and for φ = 45° ∆α ≈ 0,1° .

Thus, if high accuracy is not required, then we can approximately assume that the force of gravity and the weight of the body are directed towards the center of the Earth and are equal in magnitude to the gravitational force.

3. Due to the relevance of weighing objects in motion, it is necessary to consider the influence of the Coriolis force . The Coriolis force is caused by the motion of bodies relative to a rotating reference frame. Coriolis force depends on speed body motion relative to the reference system and angular velocity reference systems.

The expression for the Coriolis force is:

Where
– vector product.

The magnitude of the Coriolis force is

, (2.58)

Where β – angle between vectors And . Vector
perpendicular to the plane in which the vectors lie And .

The Coriolis force is zero if the speed of the body is zero or the angle between the vectors And equal to zero or π (for example, when moving along the surface of the Earth near the equator along the geographic meridian). The Coriolis force takes its maximum value if the speed of the body is perpendicular to the axis of rotation of the Earth (for example,
when moving along the surface of the Earth along a parallel; in Fig. 2.18 a, b the case of a body moving to the east is presented).

Rice. 2.18. The Coriolis force acting on a body is

moving in a rotating reference frame "Earth"

If the body is at rest relative to the Earth, then
= 0 and the forces acting on it
,
,compensated. If a body is released from a support or suspension, it will begin to fall with the acceleration of free fall – dynamic manifestation of gravity on a rotating Earth. The equation of body motion has the form:

. (2.59)

Thus, gravity can be given a different interpretation by abandoning the expression (2.47)
, and replacing it with the more general

, (2.60)

. (2.61)

The resulting expression coincides with expression (2.47) for the static manifestation of gravity under the condition
, i.e. for the moment when a body begins to move or fall from a state of rest.

Acceleration of gravity bodies can be expressed from formula (2.59):

. (2.62)

Thus, acceleration of gravity body And gravity
,acting on a body moving relative to the rotating Earth are ambiguous quantities, depending on the speed of the body.

However, since the relative error

(2.63)

at body speed ≈ 67 m/s (≈ 240 km/h) and
does not exceed 0.1%, then usually use expressions for the static manifestation of gravity and body weight:

, (2.65)

Where – acceleration at the beginning of the free fall of a body from a state of rest, when the speed of movement is still very low
.

4. Body weight depends on the environment. The weight of a body in air (or liquid) is less than in airless space, since in these environments a buoyant force acts on the body. The occurrence of a buoyancy force can be explained by the fact that the contacting surfaces of the body and the support are not perfectly smooth - they have roughness (protrusions), the shape and size of which are different. Actually touching, i.e. real contact of the surfaces of two bodies occurs only in separate “spots” (Fig. 2.19).

The total actual contact area is 0.0010.01 of the nominal surface area and depends on the nature of the bodies and the nature of their surface treatment. Thus, a body located in an environment is actually surrounded by this environment.

Rice. 2.19. Buoyancy force
, acting on a body in the air

A body located in a gaseous medium is subject to a buoyant force equal to the product of its volume by the density of the medium and the acceleration of gravity:

, (2.66)

(2.67)

 density of the gas medium, which depends on gas pressure and its temperature T;  molar mass of gas; R = 8,31 J/(molTO) universal gas constant.

Ground reaction force is the resultant of all forces
, acting on the body in areas of actual contact.

In equilibrium

. (2.68)

Where does the body weight come from?

, (2.69)

. (2.70)

Thus, the weight of a body in the air is less than the force of gravity. Body weight is variable depends on the temperature, pressure and composition of the gas environment surrounding it, as well as on the volume of the body and the acceleration of gravity at the location of the body.

Calculations show that reducing the weight of a brass weight with a mass of T and density
in the air at temperature t = 20 0 C = 293 K and pressure R atm= 10 5 Pa is

from gravity. If such high accuracy is not required, then we can assume that the weight of the body in the air is equal to the force of gravity:

. (2.71)

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