R in theoretical mechanics. Basic mechanics for dummies. Introduction

Statics is a branch of theoretical mechanics in which the conditions of equilibrium of material bodies under the influence of forces are studied.

In statics, a state of equilibrium is understood as a state in which all parts of a mechanical system are at rest (relative to a fixed coordinate system). Although the methods of statics are also applicable to moving bodies, and with their help it is possible to study problems of dynamics, the basic objects of study of statics are stationary mechanical bodies and systems.

Force is a measure of the influence of one body on another. Force is a vector that has a point of application on the surface of the body. Under the influence of a force, a free body receives an acceleration proportional to the force vector and inversely proportional to the mass of the body.

Law of equality of action and reaction

The force with which the first body acts on the second is equal in absolute value and opposite in direction to the force with which the second body acts on the first.

Hardening principle

If a deformable body is in equilibrium, then its equilibrium will not be disturbed if the body is considered absolutely solid.

Statics of a material point

Let us consider a material point that is in equilibrium. And let n forces act on it, k = 1, 2, ..., n.

If a material point is in equilibrium, then the vector sum of the forces acting on it is equal to zero:
(1) .

In equilibrium, the geometric sum of the forces acting on a point is zero.

Geometric interpretation. If you place the beginning of the second vector at the end of the first vector, and place the beginning of the third at the end of the second vector, and then continue this process, then the end of the last, nth vector will be aligned with the beginning of the first vector. That is, we get a closed geometric figure, the lengths of the sides are equal to the modules of the vectors. If all vectors lie in the same plane, then we get a closed polygon.

It is often convenient to choose rectangular coordinate system Oxyz. Then the sums of the projections of all force vectors on the coordinate axes are equal to zero:

If you choose any direction specified by some vector, then the sum of the projections of the force vectors onto this direction is equal to zero:
.
Let's multiply equation (1) scalarly by the vector:
.
Here - scalar product vectors and .
Note that the projection of the vector onto the direction of the vector is determined by the formula:
.

Rigid body statics

Moment of force about a point

Determination of moment of force

A moment of power, applied to the body at point A, relative to the fixed center O, is called a vector equal to the vector product of vectors and:
(2) .

Geometric interpretation

The moment of force is equal to the product of force F and arm OH.

Let the vectors and be located in the drawing plane. According to property vector product, the vector is perpendicular to the vectors and , that is, perpendicular to the plane of the drawing. Its direction is determined by the right screw rule. In the figure, the torque vector is directed towards us. Absolute torque value:
.
Since then
(3) .

Using geometry, we can give a different interpretation of the moment of force. To do this, draw a straight line AH through the force vector. From the center O we lower the perpendicular OH to this straight line. The length of this perpendicular is called shoulder of strength. Then
(4) .
Since , then formulas (3) and (4) are equivalent.

Thus, absolute value of the moment of force relative to the center O is equal to product of force per shoulder this force relative to the selected center O.

When calculating torque, it is often convenient to decompose the force into two components:
,
Where . The force passes through point O. So it's her moment equal to zero. Then
.
Absolute torque value:
.

Moment components in a rectangular coordinate system

If we choose a rectangular coordinate system Oxyz with a center at point O, then the moment of force will have the following components:
(5.1) ;
(5.2) ;
(5.3) .
Here are the coordinates of point A in the selected coordinate system:
.
The components represent the values ​​of the moment of force about the axes, respectively.

Properties of the moment of force relative to the center

The moment about the center O, due to the force passing through this center, is equal to zero.

If the point of application of the force is moved along a line passing through the force vector, then the moment, with such movement, will not change.

The moment from the vector sum of forces applied to one point of the body is equal to the vector sum of moments from each of the forces applied to the same point:
.

The same applies to forces whose continuation lines intersect at one point. In this case, their intersection point should be taken as the point of application of forces.

If the vector sum of forces is zero:
,
then the sum of the moments from these forces does not depend on the position of the center relative to which the moments are calculated:
.

Couple of forces

Couple of forces- these are two forces equal in absolute magnitude and having opposite directions, applied to different points of the body.

A pair of forces is characterized by the moment they create. Since the vector sum of the forces entering the pair is zero, the moment created by the pair does not depend on the point relative to which the moment is calculated. From the point of view of static equilibrium, the nature of the forces involved in the pair does not matter. A couple of forces is used to indicate that a moment of force of a certain value acts on a body.

Moment of force about a given axis

There are often cases when we do not need to know all the components of the moment of a force about a selected point, but only need to know the moment of a force about a selected axis.

The moment of force about an axis passing through point O is the projection of the vector of the moment of force, relative to point O, onto the direction of the axis.

Properties of the moment of force about the axis

The moment about the axis due to the force passing through this axis is equal to zero.

The moment about an axis due to a force parallel to this axis is equal to zero.

Calculation of the moment of force about an axis

Let a force act on the body at point A. Let's find the moment of this force relative to the O′O′′ axis.

Let's construct a rectangular coordinate system. Let the Oz axis coincide with O′O′′. From point A we lower the perpendicular OH to O′O′′. Through points O and A we draw the Ox axis. We draw the Oy axis perpendicular to Ox and Oz. Let us decompose the force into components along the axes of the coordinate system:
.
The force intersects the O′O′′ axis. Therefore its moment is zero. The force is parallel to the O′O′′ axis. Therefore, its moment is also zero. Using formula (5.3) we find:
.

Note that the component is directed tangentially to the circle whose center is point O. The direction of the vector is determined by the right screw rule.

Conditions for the equilibrium of a rigid body

In equilibrium, the vector sum of all forces acting on the body is equal to zero and the vector sum of the moments of these forces relative to an arbitrary fixed center is equal to zero:
(6.1) ;
(6.2) .

We emphasize that the center O, relative to which the moments of forces are calculated, can be chosen arbitrarily. Point O can either belong to the body or be located outside it. Usually the center O is chosen to make calculations simpler.

The equilibrium conditions can be formulated in another way.

In equilibrium, the sum of the projections of forces on any direction specified by an arbitrary vector is equal to zero:
.
The sum of the moments of forces relative to an arbitrary axis O′O′′ is also equal to zero:
.

Sometimes such conditions turn out to be more convenient. There are cases when, by selecting axes, calculations can be made simpler.

Body center of gravity

Let's consider one of the most important forces - gravity. Here the forces are not applied at certain points of the body, but are continuously distributed throughout its volume. For every area of ​​the body with an infinitesimal volume ΔV, the force of gravity acts. Here ρ is the density of the body’s substance, and is the acceleration of gravity.

Let be the mass of an infinitely small part of the body. And let point A k determine the position of this section. Let us find the quantities related to gravity that are included in the equilibrium equations (6).

Let us find the sum of gravity forces formed by all parts of the body:
,
where is body mass. Thus, the sum of the gravitational forces of individual infinitesimal parts of the body can be replaced by one vector of the gravitational force of the entire body:
.

Let us find the sum of the moments of gravity, in a relatively arbitrary way for the selected center O:

.
Here we have introduced point C, which is called center of gravity bodies. The position of the center of gravity, in a coordinate system centered at point O, is determined by the formula:
(7) .

So, when determining static equilibrium, the sum of the gravity forces of individual parts of the body can be replaced by the resultant
,
applied to the center of mass of the body C, the position of which is determined by formula (7).

Center of gravity position for different geometric shapes can be found in the relevant reference books. If a body has an axis or plane of symmetry, then the center of gravity is located on this axis or plane. Thus, the centers of gravity of a sphere, circle or circle are located at the centers of the circles of these figures. Centers of gravity rectangular parallelepiped, rectangle or square are also located at their centers - at the points of intersection of the diagonals.

Uniformly (A) and linearly (B) distributed load.

There are also cases similar to gravity, when forces are not applied at certain points of the body, but are continuously distributed over its surface or volume. Such forces are called distributed forces or .

(Figure A). Also, as in the case of gravity, it can be replaced by a resultant force of magnitude , applied at the center of gravity of the diagram. Since the diagram in Figure A is a rectangle, the center of gravity of the diagram is at its center - point C: | AC| = | CB|.

(Figure B). It can also be replaced by the resultant. The magnitude of the resultant is equal to the area of ​​the diagram:
.
The application point is at the center of gravity of the diagram. The center of gravity of a triangle, height h, is located at a distance from the base. That's why .

Friction forces

Sliding friction. Let the body be on a flat surface. And let be the force perpendicular to the surface with which the surface acts on the body (pressure force). Then the sliding friction force is parallel to the surface and directed to the side, preventing the movement of the body. Its greatest value is:
,
where f is the friction coefficient. The friction coefficient is a dimensionless quantity.

Rolling friction. Let a round shaped body roll or be able to roll on the surface. And let be the pressure force perpendicular to the surface from which the surface acts on the body. Then a moment of friction forces acts on the body, at the point of contact with the surface, preventing the movement of the body. The greatest value of the friction moment is equal to:
,
where δ is the rolling friction coefficient. It has the dimension of length.

References:
S. M. Targ, Short course theoretical mechanics, " graduate School", 2010.

Within any training course The study of physics begins with mechanics. Not from theoretical, not from applied or computational, but from good old classical mechanics. This mechanics is also called Newtonian mechanics. According to legend, a scientist was walking in the garden, saw an apple falling, and it was this phenomenon that prompted him to discover the law universal gravity. Of course, the law has always existed, and Newton only gave it a form understandable to people, but his merit is priceless. In this article we will not describe the laws of Newtonian mechanics in as much detail as possible, but we will outline the fundamentals, basic knowledge, definitions and formulas that can always play into your hands.

Mechanics is a branch of physics, a science that studies the movement of material bodies and the interactions between them.

The word itself has Greek origin and translates as “the art of building machines.” But before we build machines, we are still like the Moon, so let’s follow in the footsteps of our ancestors and study the movement of stones thrown at an angle to the horizon, and apples falling on our heads from a height h.

Why does the study of physics begin with mechanics? Because this is completely natural, shouldn’t we start with thermodynamic equilibrium?!

Mechanics is one of the oldest sciences, and historically the study of physics began precisely with the foundations of mechanics. Placed within the framework of time and space, people, in fact, could not start with something else, no matter how much they wanted. Moving bodies are the first thing we pay attention to.

What is movement?

Mechanical motion is a change in the position of bodies in space relative to each other over time.

It is after this definition that we quite naturally come to the concept of a frame of reference. Changing the position of bodies in space relative to each other. Key words here: relative to each other . After all, a passenger in a car moves relative to the person standing on the side of the road at a certain speed, and is at rest relative to his neighbor in the seat next to him, and moves at some other speed relative to the passenger in the car that is overtaking them.

That is why, in order to normally measure the parameters of moving objects and not get confused, we need reference system - rigidly interconnected reference body, coordinate system and clock. For example, the earth moves around the sun in a heliocentric frame of reference. In everyday life, we carry out almost all our measurements in a geocentric reference system associated with the Earth. The earth is a body of reference relative to which cars, planes, people, and animals move.

Mechanics, as a science, has its own task. The task of mechanics is to know the position of a body in space at any time. In other words, mechanics builds a mathematical description of motion and finds connections between physical quantities, which characterize it.

In order to move further, we need the concept “ material point " They say that physics is an exact science, but physicists know how many approximations and assumptions have to be made in order to agree on this very accuracy. No one has ever seen a material point or smelled an ideal gas, but they exist! They are simply much easier to live with.

A material point is a body whose size and shape can be neglected in the context of this problem.

Sections of classical mechanics

Mechanics consists of several sections

• Kinematics
• Dynamics
• Statics

Kinematics from a physical point of view, it studies exactly how a body moves. In other words, this section deals with the quantitative characteristics of movement. Find speed, path - typical kinematics problems

Dynamics solves the question of why it moves the way it does. That is, it considers the forces acting on the body.

Statics studies the balance of bodies under the influence of forces, that is, answers the question: why doesn’t it fall at all?

Limits of applicability of classical mechanics.

Classical mechanics no longer claims to be a science that explains everything (at the beginning of the last century everything was completely different), and has a clear framework of applicability. In general, the laws of classical mechanics are valid in the world we are accustomed to in size (macroworld). They stop working in the case of the particle world, when quantum mechanics replaces classical mechanics. Also, classical mechanics is not applicable to cases when the movement of bodies occurs at a speed close to the speed of light. In such cases, relativistic effects become pronounced. Roughly speaking, within the framework of quantum and relativistic mechanics, classical mechanics is special case, when the body size is large and the speed is low. You can learn more about it from our article.

Generally speaking, quantum and relativistic effects never go away; they also occur during the ordinary motion of macroscopic bodies at a speed much lower than the speed of light. Another thing is that the effect of these effects is so small that it does not go beyond the most accurate measurements. Classical mechanics will thus never lose its fundamental importance.

We will continue to study physical foundations mechanics in the following articles. For a better understanding of the mechanics, you can always refer to individually will shed light on the dark spot of the most difficult task.

In all its beauty and elegance. With its help, Newton once derived his law of universal gravitation based on Kepler's three empirical laws. The subject, in general, is not that complicated and is relatively easy to understand. But passing is difficult, since teachers are often terribly picky (like Pavlova, for example). When solving problems, you need to be able to solve diffuses and calculate integrals.

Key Ideas

In essence, theoretical mechanics in this course is the application of the variational principle to calculate the “motion” of various physical systems. Calculus of variations is covered briefly in the course Integral Equations and Calculus of Variations. Lagrange's equations are Euler's equations, which are the solution to a problem with fixed ends.

One problem can usually be solved by 3 different methods at once:

• Lagrange method (Lagrange function, Lagrange equations)
• Hamilton method (Hamilton function, Hamilton equations)
• Hamilton-Jacobi method (Hamilton-Jacobi equation)

It is important to choose the simplest one for a specific task.

Materials

First semester (test)

Basic formulas

View in large size!

Theory

Videos

Lectures by V.R. Khalilova - Attention! Not all lectures are recorded

Second semester (exam)

We need to start with the fact that different groups The exam goes differently. Usually Examination ticket consists of 2 theoretical questions and 1 problem. Questions are required for everyone, but you can either get rid of a task (for excellent work in the semester + written tests) or grab an extra one (and more than one). Here you will be told about the rules of the game at seminars. In the groups of Pavlova and Pimenov, theormin is practiced, which is a kind of admission to the exam. It follows that this theory must be known perfectly.

Exam in Pavlova groups goes something like this: First, a ticket with 2 term questions. There is little time to write, and the key here is to write it absolutely perfectly. Then Olga Serafimovna will be kind to you and the rest of the exam will go very pleasantly. Next is a ticket with 2 theory questions + n problems (depending on your work in the semester). Theory in theory can be written off. Solve problems. Having a lot of problems in an exam is not the end if you know how to solve them perfectly. This can be turned into an advantage - for each exam point you get a +, +-, -+ or -. The rating is given “based on the overall impression” => if in theory everything is not perfect for you, but then you get 3+ for the tasks, then the overall impression is good. But if you had no problems in the exam and the theory is not ideal, then there is nothing to smooth it out.

Theory

• Julia. Lecture notes (2014, pdf) - both semesters, 2nd stream
• Second stream tickets part 1 (lecture notes and part for tickets) (pdf)
• Second stream tickets and table of contents for all these parts (pdf)
• Answers to tickets for the 1st stream (2016, pdf) - in printed form, very convenient
• Recognized theory for the exam for Pimenov groups (2016, pdf) - both semesters
• Answers to theorymin for Pimenov groups (2016, pdf) - neat and seemingly error-free

Tasks

• Pavlova's seminars 2nd semester (2015, pdf) - neat, beautifully and clearly written
• Problems that may be on the exam (jpg) - once in some shaggy year they were in the 2nd stream, may also be relevant for V.R. groups. Khalilov (he gives similar problems in kr)
• Problems for tickets (pdf)- for both streams (on the 2nd stream these tasks were in A.B. Pimenov’s groups)

Search the library by authors and keywords from the book title:

Theoretical and analytical mechanics

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• Perelman Ya.I. Entertaining mechanics (4th edition). M.-L.: ONTI, 1937 (djvu)
• Planck M. Introduction to Theoretical Physics. Part one. General mechanics (2nd edition). M.-L.: GTTI, 1932 (djvu)
• Polak L.S. (ed.) Variational principles of mechanics. Collection of articles by classics of science. M.: Fizmatgiz, 1959 (djvu)
• Poincare A. Lectures on celestial mechanics. M.: Nauka, 1965 (djvu)
• Poincare A. New mechanics. Evolution of laws. M.: Contemporary issues: 1913 (djvu)
• Rose N.V. (ed.) Theoretical mechanics. Part 1. Mechanics of a material point. L.-M.: GTTI, 1932 (djvu)
• Rose N.V. (ed.) Theoretical mechanics. Part 2. Mechanics of material systems and solids. L.-M.: GTTI, 1933 (djvu)
• Rosenblat G.M. Dry friction in problems and solutions. M.-Izhevsk: RHD, 2009 (pdf)
• Rubanovsky V.N., Samsonov V.A. Stability of stationary motions in examples and problems. M.-Izhevsk: RHD, 2003 (pdf)
• Samsonov V.A. Lecture notes on mechanics. M.: MSU, 2015 (pdf)
• Sugar N.F. Course of theoretical mechanics. M.: Higher. school, 1964 (djvu)
• Collection of scientific and methodological articles on theoretical mechanics. Issue 1. M.: Higher. school, 1968 (djvu)
• Collection of scientific and methodological articles on theoretical mechanics. Issue 2. M.: Higher. school, 1971 (djvu)
• Collection of scientific and methodological articles on theoretical mechanics. Issue 3. M.: Higher. school, 1972 (djvu)
• Collection of scientific and methodological articles on theoretical mechanics. Issue 4. M.: Higher. school, 1974 (djvu)
• Collection of scientific and methodological articles on theoretical mechanics. Issue 5. M.: Higher. school, 1975 (djvu)
• Collection of scientific and methodological articles on theoretical mechanics. Issue 6. M.: Higher. school, 1976 (djvu)
• Collection of scientific and methodological articles on theoretical mechanics. Issue 7. M.: Higher. school, 1976 (djvu)
• Collection of scientific and methodological articles on theoretical mechanics. Issue 8. M.: Higher. school, 1977 (djvu)
• Collection of scientific and methodological articles on theoretical mechanics. Issue 9. M.: Higher. school, 1979 (djvu)
• Collection of scientific and methodological articles on theoretical mechanics. Issue 10. M.: Higher. school, 1980 (djvu)
• Collection of scientific and methodological articles on theoretical mechanics. Issue 11. M.: Higher. school, 1981 (djvu)
• Collection of scientific and methodological articles on theoretical mechanics. Issue 12. M.: Higher. school, 1982 (djvu)
• Collection of scientific and methodological articles on theoretical mechanics. Issue 13. M.: Higher. school, 1983 (djvu)
• Collection of scientific and methodological articles on theoretical mechanics. Issue 14. M.: Higher. school, 1983 (djvu)
• Collection of scientific and methodological articles on theoretical mechanics. Issue 15. M.: Higher. school, 1984 (djvu)
• Collection of scientific and methodological articles on theoretical mechanics. Issue 16. M.: Vyssh. school, 1986

Examples of solving problems in theoretical mechanics

Statics

Problem conditions

Kinematics

Kinematics of a material point

The task

Determining the speed and acceleration of a point by given equations her movements.
Using the given equations of motion of a point, establish the type of its trajectory and for the moment of time t = 1 s find the position of the point on the trajectory, its speed, total, tangential and normal acceleration, as well as the radius of curvature of the trajectory.
Equations of motion of a point:
x = 12 sin(πt/6), cm;
y = 6 cos 2 (πt/6), cm.

Kinematic analysis of a flat mechanism

The task

The flat mechanism consists of rods 1, 2, 3, 4 and a slider E. The rods are connected to each other, to the sliders and fixed supports using cylindrical hinges. Point D is located in the middle of rod AB. The lengths of the rods are equal, respectively
l 1 = 0.4 m; l 2 = 1.2 m; l 3 = 1.6 m; l 4 = 0.6 m.

The relative arrangement of the mechanism elements in a specific version of the problem is determined by the angles α, β, γ, φ, ϑ. Rod 1 (rod O 1 A) rotates around a fixed point O 1 counterclockwise with a constant angular velocityω 1.

For a given position of the mechanism it is necessary to determine:

• linear velocities V A, V B, V D and V E of points A, B, D, E;
• angular velocities ω 2, ω 3 and ω 4 of links 2, 3 and 4;
• linear acceleration a B of point B;
• angular acceleration ε AB of link AB;
• positions of instantaneous speed centers C 2 and C 3 of links 2 and 3 of the mechanism.

Determination of absolute speed and absolute acceleration of a point

The task

The diagram below considers the motion of point M in the trough of a rotating body. Using the given equations of portable motion φ = φ(t) and relative motion OM = OM(t), determine the absolute speed and absolute acceleration of a point at a given point in time.

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Dynamics

Integration of differential equations of motion of a material point under the influence of variable forces

The task

A load D of mass m, having received an initial speed V 0 at point A, moves in a curved pipe ABC located in a vertical plane. In a section AB, the length of which is l, the load is acted upon by a constant force T (its direction is shown in the figure) and a force R of the medium resistance (the modulus of this force R = μV 2, the vector R is directed opposite to the speed V of the load).

The load, having finished moving in section AB, at point B of the pipe, without changing the value of its speed module, moves to section BC. In section BC, the load is acted upon by a variable force F, the projection F x of which on the x axis is given.

Considering the load to be a material point, find the law of its motion in section BC, i.e. x = f(t), where x = BD. Neglect the friction of the load on the pipe.

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Theorem on the change in kinetic energy of a mechanical system

The task

The mechanical system consists of weights 1 and 2, a cylindrical roller 3, two-stage pulleys 4 and 5. The bodies of the system are connected by threads wound on the pulleys; sections of threads are parallel to the corresponding planes. The roller (a solid homogeneous cylinder) rolls along the supporting plane without sliding. The radii of the stages of pulleys 4 and 5 are respectively equal to R 4 = 0.3 m, r 4 = 0.1 m, R 5 = 0.2 m, r 5 = 0.1 m. The mass of each pulley is considered to be uniformly distributed along its outer rim . The supporting planes of loads 1 and 2 are rough, the sliding friction coefficient for each load is f = 0.1.

Under the action of a force F, the modulus of which changes according to the law F = F(s), where s is the displacement of the point of its application, the system begins to move from a state of rest. When the system moves, pulley 5 is acted upon by resistance forces, the moment of which relative to the axis of rotation is constant and equal to M 5 .

Determine the value of the angular velocity of pulley 4 at the moment in time when the displacement s of the point of application of force F becomes equal to s 1 = 1.2 m.

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Application of the general equation of dynamics to the study of the motion of a mechanical system

The task

For a mechanical system, determine the linear acceleration a 1 . Assume that the masses of blocks and rollers are distributed along the outer radius. Cables and belts should be considered weightless and inextensible; there is no slippage. Neglect rolling and sliding friction.

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Application of d'Alembert's principle to determining the reactions of the supports of a rotating body

The task

The vertical shaft AK, rotating uniformly with an angular velocity ω = 10 s -1, is fixed by a thrust bearing at point A and a cylindrical bearing at point D.

Rigidly attached to the shaft are a weightless rod 1 with a length of l 1 = 0.3 m, at the free end of which there is a load with a mass of m 1 = 4 kg, and a homogeneous rod 2 with a length of l 2 = 0.6 m, having a mass of m 2 = 8 kg. Both rods lie in the same vertical plane. The points of attachment of the rods to the shaft, as well as the angles α and β are indicated in the table. Dimensions AB=BD=DE=EK=b, where b = 0.4 m. Take the load as a material point.

Neglecting the mass of the shaft, determine the reactions of the thrust bearing and the bearing.