Theoretical mechanics course of lectures. Basic mechanics for dummies. Introduction. Limits of applicability of classical mechanics

1 slide

Course of lectures on theoretical mechanics Dynamics (I part) Bondarenko A.N. Moscow - 2007 The electronic training course was written on the basis of lectures given by the author for students studying in the specialties of SZhD, PGS and SDM at NIIZhT and MIIT (1974-2006). The educational material corresponds to the calendar plans in the amount of three semesters. To fully implement animation effects during a presentation, you must use a Power Point viewer no lower than the one built into Microsoft Office of the Windows-XP Professional operating system. Comments and suggestions can be sent by e-mail: [email protected]. Moscow State University of Railway Engineering (MIIT) Department of Theoretical Mechanics Scientific and Technical Center of Transport Technologies

2 slide

Contents Lecture 1. Introduction to dynamics. Laws and axioms of material point dynamics. Basic equation of dynamics. Differential and natural equations of motion. Two main tasks of dynamics. Examples of solving the direct problem of dynamics Lecture 2. Solving the inverse problem of dynamics. General instructions for solving the inverse problem of dynamics. Examples of solving the inverse problem of dynamics. The motion of a body thrown at an angle to the horizon, without taking into account air resistance. Lecture 3. Rectilinear oscillations of a material point. The condition for the occurrence of oscillations. Classification of oscillations. Free vibrations without taking into account the forces of resistance. damped vibrations. Oscillation decrement. Lecture 4. Forced oscillations of a material point. Resonance. Influence of resistance to motion during forced vibrations. Lecture 5. Relative motion of a material point. Forces of inertia. Particular cases of movement for various types of portable movement. Influence of the Earth's rotation on the balance and motion of bodies. Lecture 6. Dynamics of a mechanical system. mechanical system. External and internal forces. Center of mass of the system. The theorem on the motion of the center of mass. Conservation laws. An example of solving the problem of using the theorem on the movement of the center of mass. Lecture 7. Impulse of force. The amount of movement. Theorem on the change in momentum. Conservation laws. Euler's theorem. An example of solving the problem on the use of the theorem on the change in momentum. moment of momentum. The theorem on changing the angular momentum. Lecture 8. Conservation laws. Elements of the theory of moments of inertia. Kinetic moment of a rigid body. Differential equation of rotation of a rigid body. An example of solving the problem of using the theorem on the change in the angular momentum of the system. Elementary theory of the gyroscope. Recommended literature 1. Yablonsky A.A. Course of theoretical mechanics. Part 2. M.: Higher school. 1977. 368 p. 2. Meshchersky I.V. Collection of problems in theoretical mechanics. M.: Science. 1986. 416 p. 3. Collection of assignments for term papers /Ed. A.A. Yablonsky. M.: Higher school. 1985. 366 p. 4. Bondarenko A.N. “Theoretical mechanics in examples and tasks. Dynamics” (electronic manual www.miit.ru/institut/ipss/faculties/trm/main.htm), 2004

3 slide

Lecture 1 Dynamics is a section of theoretical mechanics that studies mechanical motion from the most general point of view. The movement is considered in connection with the forces acting on the object. The section consists of three sections: Dynamics of a material point Dynamics Dynamics of a mechanical system Analytical mechanics ■ Dynamics of a point - studies the motion of a material point, taking into account the forces that cause this movement. The main object is a material point - a material body with a mass, the dimensions of which can be neglected. Basic assumptions: - there is an absolute space (it has purely geometric properties that do not depend on matter and its movement. - there is an absolute time (does not depend on matter and its movement). It follows from this: - there is an absolutely immobile frame of reference. - time does not depend on motion of the frame of reference. - the masses of moving points do not depend on the motion of the frame of reference. These assumptions are used in classical mechanics created by Galileo and Newton. It still has a fairly wide scope, since the mechanical systems considered in applied sciences do not have such large masses and speeds of motion, for which it is necessary to take into account their influence on the geometry of space, time, motion, as is done in relativistic mechanics (the theory of relativity) ■ The basic laws of dynamics - first discovered by Galileo and formulated by Newton form the basis of all methods for describing and analyzing the motion of mechanical systems and their dynamic interaction action under the influence of various forces. ■ Law of inertia (Galileo-Newton law) - An isolated material point of a body retains its state of rest or uniform rectilinear motion until the applied forces force it to change this state. This implies the equivalence of the state of rest and motion by inertia (the law of relativity of Galileo). The frame of reference, in relation to which the law of inertia is fulfilled, is called inertial. The property of a material point to strive to keep the speed of its movement (its kinematic state) unchanged is called inertia. ■ The law of proportionality of force and acceleration (Basic equation of dynamics - Newton's II law) - The acceleration imparted to a material point by force is directly proportional to the force and inversely proportional to the mass of this point: or Here m is the mass of the point (a measure of inertia), measured in kg, numerically equal to weight divided by the gravitational acceleration: F is the acting force, measured in N (1 N imparts an acceleration of 1 m / s2 to a point with a mass of 1 kg, 1 N \u003d 1/9. 81 kg-s). ■ Dynamics of a mechanical system - studies the movement of a set of material points and solid bodies, united by the general laws of interaction, taking into account the forces that cause this movement. ■ Analytical mechanics - studies the motion of non-free mechanical systems using general analytical methods. one

4 slide

Lecture 1 (continued - 1.2) Differential equations of motion of a material point: - differential equation of motion of a point in vector form. - differential equations of point motion in coordinate form. This result can be obtained by formal projection of the vector differential equation (1). After grouping, the vector relation is decomposed into three scalar equations: In coordinate form: We use the relationship of the radius-vector with coordinates and the force vector with projections: differential equation of motion on natural (moving) coordinate axes: or: - natural equations of motion of a point. ■ Basic equation of dynamics: - corresponds to the vector way of specifying the movement of a point. ■ The law of independence of the action of forces - The acceleration of a material point under the action of several forces is equal to the geometric sum of the accelerations of a point from the action of each of the forces separately: or The law is valid for any kinematic state of bodies. The forces of interaction, being applied to different points (bodies) are not balanced. ■ The law of equality of action and reaction (Newton's III law) - Every action corresponds to an equal and oppositely directed reaction: 2

5 slide

Two main problems of dynamics: 1. Direct problem: Motion is given (equations of motion, trajectory). It is required to determine the forces under the action of which a given movement occurs. 2. Inverse problem: The forces under the action of which the movement occurs are given. It is required to find motion parameters (motion equations, motion trajectory). Both problems are solved using the basic equation of dynamics and its projection onto the coordinate axes. If the motion of a non-free point is considered, then, as in statics, the principle of release from bonds is used. As a result of the reaction, the bonds are included in the composition of the forces acting on the material point. The solution of the first problem is connected with differentiation operations. The solution of the inverse problem requires the integration of the corresponding differential equations, and this is much more difficult than differentiation. The inverse problem is more difficult than the direct problem. The solution of the direct problem of dynamics - let's look at examples: Example 1. A cabin with a weight G of an elevator is lifted by a cable with an acceleration a . Determine cable tension. 1. Select an object (the elevator car moves forward and can be considered as a material point). 2. We discard the connection (cable) and replace it with the reaction R. 3. Compose the basic equation of dynamics: Determine the reaction of the cable: Determine the cable tension: With a uniform movement of the cab ay = 0 and the cable tension is equal to the weight: T = G. When the cable breaks T = 0 and the acceleration of the cabin is equal to the acceleration of free fall: ay = -g. 3 4. We project the basic equation of dynamics onto the y axis: y Example 2. A point of mass m moves along a horizontal surface (the Oxy plane) according to the equations: x = a coskt, y = b coskt. Determine the force acting on the point. 1. Select an object (material point). 2. We discard the connection (plane) and replace it with the reaction N. 3. Add an unknown force F to the system of forces. 4. Compose the basic equation of dynamics: 5. Project the basic equation of dynamics on the x,y axes: Determine the force projections: Force modulus: Direction cosines : Thus, the magnitude of the force is proportional to the distance of the point to the center of coordinates and is directed towards the center along the line connecting the point to the center. The trajectory of the point movement is an ellipse centered at the origin: O r Lecture 1 (continued - 1.3)

6 slide

Lecture 1 (continuation 1.4) Example 3: A load of weight G is suspended on a cable of length l and moves along a circular path in a horizontal plane with some speed. The angle of deviation of the cable from the vertical is equal to. Determine the tension of the cable and the speed of the load. 1. Select an object (cargo). 2. Discard the connection (rope) and replace it with the reaction R. 3. Compose the main equation of dynamics: From the third equation, determine the reaction of the cable: Determine the tension of the cable: Substitute the value of the reaction of the cable, normal acceleration into the second equation and determine the speed of the load: 4. Project the main equation axle dynamics,n,b: Example 4: A car of weight G moves on a convex bridge (radius of curvature is R) with a speed V. Determine the pressure of the car on the bridge. 1. We select an object (a car, we neglect the dimensions and consider it as a point). 2. We discard the connection (rough surface) and replace it with the reactions N and the friction force Ffr. 3. We compose the basic equation of dynamics: 4. We project the basic equation of dynamics onto the n axis: From here we determine the normal reaction: We determine the pressure of the car on the bridge: From here we can determine the speed corresponding to zero pressure on the bridge (Q = 0): 4

7 slide

Lecture 2 After substituting the found values of the constants, we obtain: Thus, under the action of the same system of forces, a material point can perform a whole class of movements determined by the initial conditions. The initial coordinates take into account the initial position of the point. The initial velocity, given by the projections, takes into account the influence on its movement along the considered section of the trajectory of the forces that acted on the point before arriving at this section, i.e. initial kinematic state. The solution of the inverse problem of dynamics - In the general case of the movement of a point, the forces acting on the point are variables that depend on time, coordinates and speed. The motion of a point is described by a system of three second-order differential equations: After integrating each of them, there will be six constants C1, C2,…., C6: The values of the constants C1, C2,…., C6 are found from six initial conditions at t = 0: Example 1 of the solution inverse problem: A free material point of mass m moves under the action of a force F, which is constant in magnitude and magnitude. . At the initial moment, the speed of the point was v0 and coincided in direction with the force. Determine the equation of motion of a point. 1. We compose the basic equation of dynamics: 3. We lower the order of the derivative: 2. We choose the Cartesian reference system, directing the x axis along the direction of the force and project the main equation of dynamics onto this axis: or x y z 4. Separate the variables: 5. Calculate the integrals from both parts of the equation : 6. Let's represent the velocity projection as a derivative of the coordinate with respect to time: 8. Calculate the integrals of both parts of the equation: 7. Separate the variables: 9. To determine the values of the constants C1 and C2, we use the initial conditions t = 0, vx = v0 , x = x0: As a result, we obtain the equation of uniformly variable motion (along the x axis): 5

8 slide

General instructions for solving direct and inverse problems. Solution procedure: 1. Compilation of the differential equation of motion: 1.1. Choose a coordinate system - rectangular (fixed) with an unknown trajectory of movement, natural (moving) with a known trajectory, for example, a circle or a straight line. In the latter case, one rectilinear coordinate can be used. The reference point should be combined with the initial position of the point (at t = 0) or with the equilibrium position of the point, if it exists, for example, when the point fluctuates. 6 1.2. Draw a point at a position corresponding to an arbitrary moment in time (for t > 0) so that the coordinates are positive (s > 0, x > 0). We also assume that the velocity projection in this position is also positive. In the case of oscillations, the velocity projection changes sign, for example, when returning to the equilibrium position. Here it should be assumed that at the considered moment of time the point moves away from the equilibrium position. The implementation of this recommendation is important in the future when working with resistance forces that depend on speed. 1.3. Release the material point from bonds, replace their action with reactions, add active forces. 1.4. Write down the basic law of dynamics in vector form, project onto selected axes, express given or reactive forces in terms of time, coordinates or speed variables, if they depend on them. 2. Solution of differential equations: 2.1. Reduce the derivative if the equation is not reduced to the canonical (standard) form. for example: or 2.2. Separate variables, for example: or 2.4. Calculate the indefinite integrals on the left and right sides of the equation, for example: 2.3. If there are three variables in the equation, then make a change of variables, for example: and then separate the variables. Comment. Instead of evaluating indefinite integrals, one can evaluate definite integrals with a variable upper limit. The lower limits represent the initial values of the variables (initial conditions). Then there is no need to separately find the constant, which is automatically included in the solution, for example: Using the initial conditions, for example, t = 0, vx = vx0, determine the constant of integration: 2.5. Express the speed in terms of the time derivative of the coordinate, for example, and repeat steps 2.2 -2.4 Note. If the equation is reduced to a canonical form that has a standard solution, then this ready-made solution is used. The constants of integration are still found from the initial conditions. See, for example, oscillations (lecture 4, p. eight). Lecture 2 (continuation 2.2)

9 slide

Lecture 2 (continuation 2.3) Example 2 of solving the inverse problem: Force depends on time. A load of weight P begins to move along a smooth horizontal surface under the action of a force F, the magnitude of which is proportional to time (F = kt). Determine the distance traveled by the load in time t. 3. We compose the main equation of dynamics: 5. We lower the order of the derivative: 4. We project the main equation of dynamics onto the x-axis: or 7 6. We separate the variables: 7. We calculate the integrals from both parts of the equation: 9. We represent the projection of the velocity as the derivative of the coordinate with respect to time: 10. Calculate the integrals of both parts of the equation: 9. Separate the variables: 8. Determine the value of the constant C1 from the initial condition t = 0, vx = v0=0: As a result, we obtain the equation of motion (along the x axis), which gives the value of the distance traveled for time t: 1. We choose the reference system (Cartesian coordinates) so that the body has a positive coordinate: 2. We take the object of motion as a material point (the body moves forward), release it from the connection (reference plane) and replace it with the reaction (normal reaction of a smooth surface) : 11. Determine the value of the constant C2 from the initial condition t = 0, x = x0=0: Example 3 of solving the inverse problem: The force depends on the coordinate. A material point of mass m is thrown upwards from the Earth's surface with a speed v0. The force of gravity of the Earth is inversely proportional to the square of the distance from the point to the center of gravity (the center of the Earth). Determine the dependence of the speed on the distance y to the center of the Earth. 1. We choose the reference system (Cartesian coordinates) so that the body has a positive coordinate: 2. We compose the basic equation of dynamics: 3. We project the basic equation of dynamics onto the y axis: or The coefficient of proportionality can be found using the weight of a point on the surface of the Earth: R Hence the differential the equation looks like: or 4. Decrease the order of the derivative: 5. Change the variable: 6. Separate the variables: 7. Calculate the integrals of both sides of the equation: 8. Substitute the limits: As a result, we obtain an expression for the speed as a function of the y coordinate: Maximum height flight can be found by equating the speed to zero: The maximum flight altitude when the denominator turns to zero: From here, when setting the radius of the Earth and the acceleration of free fall, II cosmic speed is obtained:

10 slide

Lecture 2 (continuation 2.4) Example 2 of solving the inverse problem: Force depends on speed. A ship of mass m had a speed v0. The resistance of water to the movement of the vessel is proportional to the speed. Determine the time it takes the ship's speed to drop by half after turning off the engine, as well as the distance traveled by the ship to a complete stop. 8 1. We choose a reference system (Cartesian coordinates) so that the body has a positive coordinate: 2. We take the object of motion as a material point (the ship moves forward), free it from bonds (water) and replace it with a reaction (buoyant force - Archimedes force), and as well as the resistance to movement. 3. Add active force (gravity). 4. Compose the basic equation of dynamics: 5. Project the basic equation of dynamics onto the x-axis: or 6. Lower the order of the derivative: 7. Separate the variables: 8. Calculate the integrals of both sides of the equation: 9. Substitute the limits: An expression is obtained that relates the speed and time t, from which you can determine the time of movement: The time of movement, during which the speed will fall by half: It is interesting to note that when the speed approaches zero, the time of movement tends to infinity, i.e. final velocity cannot be zero. Why not "perpetual motion"? However, in this case, the distance traveled to the stop is a finite value. To determine the distance traveled, we turn to the expression obtained after lowering the order of the derivative and make a change of variable: After integrating and substituting the limits, we obtain: Distance traveled to the stop: ■ Motion of a point thrown at an angle to the horizon in a uniform gravity field without taking into account air resistance Eliminating time from the equations of motion, we obtain the trajectory equation: The flight time is determined by equating the y coordinate to zero: The flight range is determined by substituting the flight time:

11 slide

Lecture 3 Rectilinear oscillations of a material point - The oscillatory movement of a material point occurs under the condition: there is a restoring force that tends to return the point to the equilibrium position for any deviation from this position. 9 There is a restoring force, the equilibrium position is stable No restoring force, the equilibrium position is unstable No restoring force, the equilibrium position is indifferent It is always directed towards the equilibrium position, the value is directly proportional to the linear elongation (shortening) of the spring, equal to the deviation of the body from the equilibrium position: c is the spring stiffness coefficient, numerically equal to the force under which the spring changes its length by one, measured in N / m in the system SI. x y O Types of vibrations of a material point: 1. Free vibrations (without taking into account the resistance of the medium). 2. Free oscillations taking into account the resistance of the medium (damped oscillations). 3. Forced vibrations. 4. Forced oscillations taking into account the resistance of the medium. ■ Free oscillations - occur under the action of only a restoring force. Let's write down the basic law of dynamics: Let's choose a coordinate system centered at the equilibrium position (point O) and project the equation onto the x axis: Let's bring the resulting equation to the standard (canonical) form: of the equation obtained using the universal substitution: The roots of the characteristic equation are imaginary and equal: The general solution of the differential equation has the form: Point speed: Initial conditions: Define the constants: So, the equation of free vibrations has the form: The equation can be represented by a single-term expression: where a is the amplitude, - initial phase. The new constants a and - are related to the constants C1 and C2 by the relations: Let's define a and: The reason for the occurrence of free oscillations is the initial displacement x0 and/or the initial velocity v0.

12 slide

10 Lecture 3 (continuation 3.2) Damped oscillations of a material point - The oscillatory movement of a material point occurs in the presence of a restoring force and a force of resistance to movement. The dependence of the force of resistance to movement on displacement or speed is determined by the physical nature of the medium or connection that impedes movement. The simplest dependence is a linear dependence on speed (viscous resistance): - viscosity coefficient x y O from the values of the roots: 1. n< k – случай малого вязкого сопротивления: - корни комплексные, различные. или x = ae-nt x = -ae-nt Частота затухающих колебаний: Период: T* Декремент колебаний: ai ai+1 Логарифмический декремент колебаний: Затухание колебаний происходит очень быстро. Основное влияние силы вязкого сопротивления – уменьшение амплитуды колебаний с течением времени. 2. n >k - case of high viscous resistance: - real roots, different. or - these functions are aperiodic: 3. n = k: - roots are real, multiple. these functions are also aperiodic:

13 slide

Lecture 3 (continuation 3.3) Classification of solutions of free oscillations. Spring connections. equivalent hardness. y y 11 Diff. Equation Character. Equation Roots char. equation Solving differential equation Graph nk n=k

14 slide

Lecture 4 Forced vibrations of a material point - Along with the restoring force, a periodically changing force acts, called the perturbing force. The perturbing force can have a different nature. For example, in a particular case, the inertial effect of an unbalanced mass m1 of a rotating rotor causes harmonically changing force projections: The main equation of dynamics: The projection of the equation of dynamics on the axis: Let's bring the equation to the standard form: 12 The solution of this inhomogeneous differential equation consists of two parts x = x1 + x2: x1 is the general solution of the corresponding homogeneous equation and x2 is a particular solution of the inhomogeneous equation: We select the particular solution in the form of the right side: The resulting equality must be satisfied for any t . Then: or Thus, with the simultaneous action of restoring and disturbing forces, the material point performs a complex oscillatory motion, which is the result of the addition (superposition) of free (x1) and forced (x2) vibrations. If p< k (вынужденные колебания малой частоты), то фаза колебаний совпадает с фазой возмущающей силы: В итоге полное решение: или Общее решение: Постоянные С1 и С2, или a и определяются из начальных условий с использованием полного решения (!): Таким образом, частное решение: Если p >k (forced oscillations of high frequency), then the phase of the oscillations is opposite to the phase of the disturbing force:

15 slide

Lecture 4 (continuation 4.2) 13 Dynamic coefficient - the ratio of the amplitude of forced oscillations to the static deviation of a point under the action of a constant force H = const: The amplitude of forced oscillations: The static deviation can be found from the equilibrium equation: Here: From here: Thus, at p< k (малая частота вынужденных колебаний) коэффициент динамичности: При p >k (high frequency of forced oscillations) dynamic coefficient: Resonance - occurs when the frequency of forced oscillations coincides with the frequency of natural oscillations (p = k). This most often occurs when starting and stopping the rotation of poorly balanced rotors mounted on elastic suspensions. The differential equation of oscillations with equal frequencies: A particular solution in the form of the right side cannot be taken, because a linearly dependent solution will be obtained (see the general solution). General solution: Substitute in the differential equation: Let's take a particular solution in the form and calculate the derivatives: Thus, the solution is obtained: or Forced oscillations at resonance have an amplitude that increases indefinitely in proportion to time. Influence of resistance to motion during forced vibrations. The differential equation in the presence of viscous resistance has the form: The general solution is selected from the table (Lecture 3, p. 11) depending on the ratio of n and k (see). We take a particular solution in the form and calculate the derivatives: Substitute in the differential equation: Equating the coefficients for identical trigonometric functions, we obtain a system of equations: Raising both equations to a power and adding them, we obtain the amplitude of the forced oscillations: By dividing the second equation by the first, we obtain the phase shift of the forced oscillations: Thus , the equation of motion for forced oscillations, taking into account the resistance to motion, for example, for n< k (малое сопротивление): Вынужденные колебания при сопротивлении движению не затухают. Частота и период вынужденных колебаний равны частоте и периоду изменения возмущающей силы. Коэффициент динамичности при резонансе имеет конечную величину и зависит от соотношения n и к.

16 slide

Lecture 5 Relative motion of a material point - Let's assume that the moving (non-inertial) coordinate system Oxyz moves according to some law relative to the fixed (inertial) coordinate system O1x1y1z1. The motion of a material point M (x, y, z) relative to the mobile system Oxyz is relative, relative to the motionless system O1x1y1z1 is absolute. The motion of the mobile system Oxyz relative to the fixed system O1x1y1z1 is a portable motion. 14 z x1 y1 z1 O1 x y M x y z O Basic equation of dynamics: Absolute acceleration of a point: Substitute the absolute acceleration of a point into the basic equation of dynamics: Let’s transfer the terms with translational and Coriolis acceleration to the right side: The transferred terms have the dimension of forces and are considered as the corresponding inertial forces, equal: Then the relative motion of a point can be considered as absolute, if we add the translational and Coriolis forces of inertia to the acting forces: In projections onto the axes of the moving coordinate system, we have: rotation is uniform, then εe = 0: 2. Translational curvilinear motion: If the motion is rectilinear, then = : If the motion is rectilinear and uniform, then the moving system is inertial and the relative motion can be considered as absolute: No mechanical phenomena can detect a rectilinear uniform motion (principle of relativity of classical mechanics). Influence of the Earth's rotation on the equilibrium of bodies - Let's assume that the body is in equilibrium on the Earth's surface at an arbitrary latitude φ (parallels). The Earth rotates around its axis from west to east with an angular velocity: The radius of the Earth is about 6370 km. S R is the total reaction of a non-smooth surface. G - force of attraction of the Earth to the center. Ф - centrifugal force of inertia. Relative equilibrium condition: The resultant of the forces of attraction and inertia is the force of gravity (weight): The magnitude of the force of gravity (weight) on the surface of the Earth is P = mg. The centrifugal force of inertia is a small fraction of the force of gravity: The deviation of the force of gravity from the direction of the force of attraction is also small: Thus, the influence of the Earth's rotation on the balance of bodies is extremely small and is not taken into account in practical calculations. The maximum value of the inertial force (at φ = 0 - at the equator) is only 0.00343 of the value of gravity

17 slide

Lecture 5 (continuation 5.2) 15 Influence of the Earth's rotation on the motion of bodies in the Earth's gravitational field - Suppose a body falls to the Earth from a certain height H above the Earth's surface at latitude φ . Let's choose a moving frame of reference, rigidly connected with the Earth, directing the x, y axes tangentially to the parallel and to the meridian: Equation of relative motion: Here, the smallness of the centrifugal force of inertia compared to the force of gravity is taken into account. Thus, the force of gravity is identified with the force of gravity. In addition, we assume that gravity is directed perpendicular to the Earth's surface due to the smallness of its deflection, as discussed above. The Coriolis acceleration is equal to and directed parallel to the y-axis to the west. The Coriolis inertia force is directed in the opposite direction. We project the equation of relative motion on the axis: The solution of the first equation gives: Initial conditions: The solution of the third equation gives: Initial conditions: The third equation takes the form: Initial conditions: Its solution gives: The resulting solution shows that the body deviates to the east when it falls. Let us calculate the value of this deviation, for example, when falling from a height of 100 m. We find the fall time from the solution of the second equation: Thus, the influence of the Earth's rotation on the movement of bodies is extremely small for practical heights and speeds and is not taken into account in technical calculations. The solution of the second equation also implies the existence of a velocity along the y-axis, which should also cause and does cause the corresponding acceleration and Coriolis inertia force. The influence of this speed and the inertia force associated with it on the change in motion will be even less than the considered Coriolis inertia force associated with the vertical speed.

18 slide

Lecture 6 Dynamics of a mechanical system. A system of material points or a mechanical system - A set of material points or those material points united by general laws of interaction (the position or movement of each of the points or a body depends on the position and movement of all the others) The system of free points - the movement of which is not limited by any connections (for example, a planetary system , in which the planets are considered as material points). A system of non-free points or a non-free mechanical system - the movement of material points or bodies is limited by the constraints imposed on the system (for example, a mechanism, a machine, etc.). 16 Forces acting on the system. In addition to the previously existing classification of forces (active and reactive forces), a new classification of forces is introduced: 1. External forces (e) - acting on points and bodies of the system from points or bodies that are not part of this system. 2. Internal forces (i) - forces of interaction between material points or bodies included in the given system. The same force can be both external and internal force. It all depends on which mechanical system is considered. For example: In the system of the Sun, Earth and Moon, all gravitational forces between them are internal. When considering the Earth and Moon system, the gravitational forces applied from the side of the Sun are external: C Z L Based on the law of action and reaction, each internal force Fk corresponds to another internal force Fk’, equal in absolute value and opposite in direction. Two remarkable properties of internal forces follow from this: The main vector of all internal forces of the system is equal to zero: The main moment of all internal forces of the system relative to any center is equal to zero: Or in projections onto the coordinate axes: Note. Although these equations are similar to equilibrium equations, they are not, since internal forces are applied to various points or bodies of the system and can cause these points (bodies) to move relative to each other. It follows from these equations that internal forces do not affect the motion of a system considered as a whole. The center of mass of the system of material points. To describe the motion of the system as a whole, a geometric point is introduced, called the center of mass, the radius vector of which is determined by the expression, where M is the mass of the entire system: Or in projections onto the coordinate axes: The formulas for the center of mass are similar to those for the center of gravity. However, the concept of the center of mass is more general, since it is not related to the forces of gravity or the forces of gravity.

19 slide

Lecture 6 (continuation 6.2) 17 Theorem on the motion of the center of mass of the system - Consider a system of n material points. We divide the forces applied to each point into external and internal ones and replace them with the corresponding resultants Fke and Fki. Let's write down for each point the basic equation of dynamics: or Let's sum these equations over all points: On the left side of the equation, we will introduce the masses under the sign of the derivative and replace the sum of the derivatives with the derivative of the sum: From the definition of the center of mass: Substitute into the resulting equation: we obtain or: The product of the mass of the system and the acceleration of its center mass is equal to the main vector of external forces. In projections on the coordinate axes: The center of mass of the system moves as a material point with a mass equal to the mass of the entire system, to which all external forces acting on the system are applied. Consequences from the theorem on the motion of the center of mass of the system (conservation laws): 1. If in the time interval the main vector of the external forces of the system is zero, Re = 0, then the speed of the center of mass is constant, vC = const (the center of mass moves uniformly rectilinearly - the law of conservation of motion center of mass). 2. If in the time interval the projection of the main vector of the external forces of the system on the x axis is equal to zero, Rxe = 0, then the velocity of the center of mass along the x axis is constant, vCx = const (the center of mass moves uniformly along the axis). Similar statements are true for the y and z axes. Example: Two people of masses m1 and m2 are in a boat of mass m3. At the initial moment of time, the boat with people was at rest. Determine the displacement of the boat if a person of mass m2 moved to the bow of the boat at a distance a. 3. If in the time interval the main vector of external forces of the system is equal to zero, Re = 0, and at the initial moment the velocity of the center of mass is zero, vC = 0, then the radius vector of the center of mass remains constant, rC = const (the center of mass is at rest is the law of conservation of the position of the center of mass). 4. If in the time interval the projection of the main vector of the external forces of the system onto the x axis is equal to zero, Rxe = 0, and at the initial moment the velocity of the center of mass along this axis is zero, vCx = 0, then the coordinate of the center of mass along the x axis remains constant, xC = const (the center of mass does not move along this axis). Similar statements are true for the y and z axes. 1. The object of movement (a boat with people): 2. We discard connections (water): 3. We replace the connection with a reaction: 4. Add active forces: 5. Write down the theorem about the center of mass: Project onto the x-axis: O Determine how far you need to transfer to a person of mass m1, so that the boat stays in place: The boat will move a distance l in the opposite direction.

20 slide

Lecture 7 The impulse of force is a measure of mechanical interaction that characterizes the transfer of mechanical movement from the forces acting on a point over a given period of time: 18 In projections onto coordinate axes: In the case of a constant force: In projections onto coordinate axes: to the point of forces in the same time interval: Multiply by dt: Integrate over a given time interval: The amount of movement of the point is a measure of mechanical movement, determined by a vector equal to the product of the mass of the point and the vector of its velocity: Theorem on the change in the amount of movement of the system – Consider the system n material points. We divide the forces applied to each point into external and internal ones and replace them with the corresponding resultants Fke and Fki. Let's write down for each point the basic equation of dynamics: or Quantity of motion of a system of material points - the geometric sum of the quantities of motion of material points: By definition of the center of mass: The vector of the momentum of the system is equal to the product of the mass of the entire system and the velocity vector of the center of mass of the system. Then: In projections onto the coordinate axes: The time derivative of the momentum vector of the system is equal to the main vector of the system's external forces. Let's sum these equations over all points: On the left side of the equation, we introduce the masses under the sign of the derivative and replace the sum of the derivatives with the derivative of the sum: From the definition of the momentum of the system: In projections onto the coordinate axes:

21 slide

Euler's theorem - Application of the theorem on the change in the momentum of a system to the movement of a continuous medium (water). 1. We select as the object of movement the volume of water located in the curvilinear channel of the turbine: 2. We discard the connections and replace their action with reactions (Rpov - the resultant of surface forces) 3. Add active forces (Rb - the resultant of body forces): 4. Write down the theorem about change in the momentum of the system: The amount of movement of water at times t0 and t1 will be represented as the sums: Change in the momentum of water in the time interval : Change in the momentum of water over an infinitesimal time interval dt: , where F1 F2 Taking the product of density, cross-sectional area and velocity per second mass, we obtain: Substituting the differential of the momentum of the system into the change theorem, we obtain: Consequences from the theorem on the change in the momentum of the system (conservation laws): 1. If in the time interval the main vector of the external forces of the system is equal to zero, Re = 0, then the quantity vector motion is constant, Q = const is the law of conservation of momentum of the system). 2. If in the time interval the projection of the main vector of the external forces of the system onto the x axis is equal to zero, Rxe = 0, then the projection of the momentum of the system onto the x axis is constant, Qx = const. Similar statements are true for the y and z axes. Lecture 7 (continuation of 7.2) Example: A grenade of mass M, flying at a speed v, exploded into two parts. The speed of one of the fragments of mass m1 increased in the direction of motion to the value v1. Determine the speed of the second fragment. 1. The object of movement (grenade): 2. The object is a free system, there are no connections and their reactions. 3. Add active forces: 4. Write down the theorem on the change in momentum: Project onto the axis: β Divide the variables and integrate: The right integral is almost zero, because explosion time t

22 slide

Lecture 7 (continuation 7.3) 20 The angular momentum of a point or the kinetic moment of motion relative to a certain center is a measure of mechanical motion, determined by a vector equal to the vector product of the radius vector of a material point and the vector of its momentum: The kinetic moment of a system of material points relative to a certain center is geometric the sum of the moments of the momentum of all material points relative to the same center: In projections on the axis: In projections on the axis: Theorem on the change in the moment of momentum of the system - Let's consider a system of n material points. We divide the forces applied to each point into external and internal ones and replace them with the corresponding resultants Fke and Fki. Let's write down for each point the basic equation of dynamics: or Let's sum these equations for all points: Let's replace the sum of derivatives by the derivative of the sum: The expression in brackets is the moment of momentum of the system. From here: We vectorially multiply each of the equalities by the radius vector on the left: Let's see if it is possible to take the sign of the derivative outside the vector product: Thus, we got: center. In projections on the coordinate axes: The derivative of the moment of momentum of the system relative to some axis in time is equal to the main moment of the external forces of the system relative to the same axis.

23 slide

Lecture 8 21 ■ Consequences from the theorem on the change in the angular momentum of the system (conservation laws): 1. If in the time interval the vector of the main moment of the external forces of the system relative to a certain center is equal to zero, MOe = 0, then the vector of the angular momentum of the system relative to the same center is constant, KO = const is the law of conservation of momentum of the system). 2. If in the time interval the main moment of the external forces of the system relative to the x axis is equal to zero, Mxe = 0, then the angular momentum of the system relative to the x axis is constant, Kx = const. Similar statements are true for the y and z axes. 2. Moment of inertia of a rigid body about an axis: The moment of inertia of a material point about an axis is equal to the product of the mass of the point and the square of the distance of the point to the axis. The moment of inertia of a rigid body about an axis is equal to the sum of the products of the mass of each point and the square of the distance of this point from the axis. ■ Elements of the theory of moments of inertia - With the rotational motion of a rigid body, the measure of inertia (resistance to change in motion) is the moment of inertia about the axis of rotation. Consider the basic concepts of the definition and methods for calculating the moments of inertia. 1. Moment of inertia of a material point about the axis: In the transition from a discrete small mass to an infinitely small mass of a point, the limit of such a sum is determined by the integral: axial moment of inertia of a rigid body. In addition to the axial moment of inertia of a rigid body, there are other types of moments of inertia: the centrifugal moment of inertia of a rigid body. polar moment of inertia of a rigid body. 3. The theorem about the moments of inertia of a rigid body about parallel axes - the formula for the transition to parallel axes: Moment of inertia about the reference axis Static moments of inertia about the reference axes Body mass moments are zero:

24 slide

Lecture 8 (continuation 8.2) 22 Moment of inertia of a uniform rod of constant section about the axis: x z L Select the elementary volume dV = Adx at a distance x: x dx Elementary mass: To calculate the moment of inertia about the central axis (passing through the center of gravity), it is enough to change the location of the axis and set the integration limits (-L/2, L/2). Here we demonstrate the formula for the transition to parallel axes: zС 5. The moment of inertia of a homogeneous solid cylinder relative to the axis of symmetry: H dr r Let us single out the elementary volume dV = 2πrdrH (thin cylinder of radius r): Elementary mass: Here we use the cylinder volume formula V=πR2H. To calculate the moment of inertia of a hollow (thick) cylinder, it is enough to set the integration limits from R1 to R2 (R2> R1): 6. The moment of inertia of a thin cylinder about the axis of symmetry (t

25 slide

Lecture 8 (continuation 8.3) 23 ■ Differential equation of rotation of a rigid body about an axis: Let's write a theorem about changing the angular momentum of a rigid body rotating around a fixed axis: The momentum of a rotating rigid body is: The moment of external forces about the axis of rotation is equal to the torque (reactions and force do not create gravity moments): We substitute the kinetic moment and the torque into the theorem Example: Two people of the same weight G1 = G2 hang on a rope thrown over a solid block with a weight G3 = G1/4. At some point, one of them began to climb the rope with a relative speed u. Determine the lifting speed of each person. 1. Select the object of motion (block with people): 2. Discard the connections (supporting device of the block): 3. Replace the connection with reactions (bearing): 4. Add active forces (gravity): 5. Write down the theorem about changing the kinetic moment of the system with respect to axis of rotation of the block: R Since the moment of external forces is equal to zero, the kinetic moment must remain constant: At the initial moment of time t = 0, there was equilibrium and Kz0 = 0. After the beginning of the movement of one person relative to the rope, the whole system began to move, but the kinetic moment of the system must remain equal to zero: Kz = 0. The angular momentum of the system is the sum of the kinetic moments of both people and the block: Here v2 is the speed of the second person, equal to the speed of the cable. end to a fixed axis of rotation. Or: In the case of small oscillations sinφ φ: Period of oscillation: Moment of inertia of the bar:

26 slide

Lecture 8 (continuation 8.4 - additional material) 24 ■ Elementary theory of the gyroscope: A gyroscope is a rigid body rotating around the axis of material symmetry, one of the points of which is fixed. A free gyroscope is fixed in such a way that its center of mass remains stationary, and the axis of rotation passes through the center of mass and can take any position in space, i.e. the axis of rotation changes its position like the axis of the body's own rotation during spherical motion. The main assumption of the approximate (elementary) theory of the gyroscope is that the momentum vector (kinetic moment) of the rotor is considered to be directed along its own axis of rotation. Thus, despite the fact that in the general case the rotor participates in three rotations, only the angular velocity of its own rotation ω = dφ/dt is taken into account. The reason for this is that in modern technology the gyroscope rotor rotates with an angular velocity of the order of 5000-8000 rad / s (about 50000-80000 rpm), while the other two angular velocities associated with precession and nutation of their own axis of rotation tens of thousands of times less than this speed. The main property of a free gyroscope is that the rotor axis keeps the same direction in space with respect to the inertial (stellar) reference system (demonstrated by the Foucault pendulum, which keeps the swing plane unchanged with respect to the stars, 1852). This follows from the law of conservation of the kinetic moment relative to the center of mass of the rotor, provided that the friction in the bearings of the rotor suspension axes, the outer and inner frame is neglected: Force action on the axis of a free gyroscope. In the case of a force applied to the rotor axis, the moment of external forces relative to the center of mass is not equal to zero: ω ω С force, and towards the vector of the moment of this force, i.e. will rotate not about the x-axis (internal suspension), but about the y-axis (external suspension). Upon termination of the force, the rotor axis will remain in the same position, corresponding to the last time of the force, because from this point in time, the moment of external forces again becomes equal to zero. In the case of a short-term action of force (impact), the axis of the gyroscope practically does not change its position. Thus, the rapid rotation of the rotor gives the gyroscope the ability to counteract random influences that seek to change the position of the axis of rotation of the rotor, and with a constant action of the force, it maintains the position of the plane perpendicular to the acting force in which the axis of the rotor lies. These properties are used in the operation of inertial navigation systems.

As part of any curriculum, the study of physics begins with mechanics. Not from theoretical, not from applied and not computational, but from good old classical mechanics. This mechanics is also called Newtonian mechanics. According to legend, the scientist was walking in the garden, saw an apple fall, and it was this phenomenon that prompted him to discover the law of universal gravitation. 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 basics, 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 is of Greek origin and translates as "the art of building machines". But before building machines, we still have a long way to go, so let's follow in the footsteps of our ancestors, and we will study the movement of stones thrown at an angle to the horizon, and apples falling on heads from a height h.

Why does the study of physics begin with mechanics? Because it is completely natural, not to start it from 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 from something else, no matter how much they wanted to. 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 a person standing on the side of the road at a certain speed, and rests relative to his neighbor in a seat nearby, and moves at some other speed relative to a passenger in a car that overtakes 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 reference body relative to which cars, planes, people, animals move.

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

In order to move further, we need the notion of “ 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 sniffed an ideal gas, but they do exist! They are just 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, studies exactly how the body moves. In other words, this section deals with the quantitative characteristics of movement. Find speed, path - typical tasks of kinematics

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 equilibrium of bodies under the action of forces, that is, it answers the question: why does it not 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 scope of applicability. In general, the laws of classical mechanics are valid for the world familiar to us in terms of size (macroworld). They cease to work in the case of the world of particles, when classical mechanics is replaced by quantum mechanics. Also, classical mechanics is inapplicable 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, this is a special case when the dimensions of the body are large, and the speed is small.

Generally speaking, quantum and relativistic effects never disappear, they also take place during the usual motion of macroscopic bodies at a speed much lower than the speed of light. Another thing is that the action 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 the physical foundations of mechanics in future articles. For a better understanding of the mechanics, you can always refer to our authors, which individually shed light on the dark spot of the most difficult task.

View: this article has been read 32852 times

Pdf Select language... Russian Ukrainian English

Short review

Full material is downloaded above, after selecting the language

Statics
- Basic concepts of statics
- Force types
- Axioms of statics
- Connections and their reactions
- Converging force system
  - Methods for determining the resultant system of converging forces
  - Equilibrium conditions for a system of converging forces
- Moment of force about the center as a vector
  - Algebraic value of the moment of force
  - Properties of the moment of force about the center (point)
- Theory of pairs of forces
  - Addition of two parallel forces in the same direction
  - Addition of two parallel forces directed in opposite directions
  - Couples of forces
  - Couple of forces theorems
  - Conditions for the equilibrium of a system of pairs of forces
- Lever arm
- Arbitrary plane system of forces
  - Cases of reducing a flat system of forces to a simpler form
  - Analytical equilibrium conditions
- Center of Parallel Forces. Center of gravity
  - Center of Parallel Forces
  - The center of gravity of a rigid body and its coordinates
  - Center of gravity of volume, planes and lines
  - Methods for determining the position of the center of gravity
Basics of Strength Racsets
- Problems and methods of resistance of materials
- Load classification
- Classification of structural elements
- Rod deformations
- Main hypotheses and principles
- Internal forces. Section method
- Voltage
- Tension and compression
- Mechanical characteristics of the material
- Permissible stresses
- Material hardness
- Plots of longitudinal forces and stresses
- Shift
- Geometric characteristics of sections
- Torsion
- bend
  - Differential dependencies in bending
  - Flexural strength
  - normal stresses. Strength calculation
  - Shear stresses in bending
  - Bending stiffness
- Elements of the general theory of stress state
- Strength theories
- Bending with twist
Kinematics
- Point kinematics
  - Point trajectory
  - Methods for specifying the movement of a point
  - Point speed
  - point acceleration
- Kinematics of a rigid body
  - Translational motion of a rigid body
  - Rotational motion of a rigid body
  - Kinematics of gear mechanisms
  - Plane-parallel motion of a rigid body
- Complex point movement
Dynamics
- Basic laws of dynamics
- Point dynamics
  - Differential equations of a free material point
  - Two problems of point dynamics
- Rigid Body Dynamics
  - Classification of forces acting on a mechanical system
  - Differential equations of motion of a mechanical system
- General theorems of dynamics
  - Theorem on the motion of the center of mass of a mechanical system
  - Theorem on the change in momentum
  - Theorem on the change in angular momentum
  - Kinetic energy change theorem
Forces acting in machines
- Forces in engagement of a spur gear
- Friction in mechanisms and machines
  - Sliding friction
  - rolling friction
- Efficiency
Machine parts
- Mechanical transmissions
  - Types of mechanical gears
  - Basic and derived parameters of mechanical gears
  - gears
  - Gearboxes with flexible links
- Shafts
  - Purpose and classification
  - Design calculation
  - Check calculation of shafts
- Bearings
  - Plain bearings
  - Rolling bearings
- Connection of machine parts
  - Types of detachable and permanent connections
  - Keyed connections
Standardization of norms, interchangeability
- Tolerances and landings
- Unified system of tolerances and landings (ESDP)
- Deviation of shape and location

Format: pdf

Size: 4MV

Russian language

An example of the calculation of a spur gear
An example of the calculation of a spur gear. The choice of material, the calculation of allowable stresses, the calculation of contact and bending strength were carried out.

An example of solving the problem of beam bending
In the example, diagrams of transverse forces and bending moments are plotted, a dangerous section is found, and an I-beam is selected. In the problem, the construction of diagrams using differential dependencies was analyzed, a comparative analysis of various beam cross sections was carried out.

An example of solving the problem of shaft torsion
The task is to test the strength of a steel shaft for a given diameter, material and allowable stresses. During the solution, diagrams of torques, shear stresses and twist angles are built. Self weight of the shaft is not taken into account

An example of solving the problem of tension-compression of a rod
The task is to test the strength of a steel rod at given allowable stresses. During the solution, plots of longitudinal forces, normal stresses and displacements are built. Self weight of the bar is not taken into account

Application of the kinetic energy conservation theorem
An example of solving the problem of applying the theorem on the conservation of kinetic energy of a mechanical system

Determination of the speed and acceleration of a point according to the given equations of motion
An example of solving the problem of determining the speed and acceleration of a point according to the given equations of motion

Determination of velocities and accelerations of points of a rigid body during plane-parallel motion
An example of solving the problem of determining the velocities and accelerations of points of a rigid body during plane-parallel motion

Determination of Forces in Flat Truss Bars
An example of solving the problem of determining the forces in the bars of a flat truss by the Ritter method and the knot cutting method

state autonomous institution

Kaliningrad region

professional educational organization

College of Service and Tourism

Course of lectures with examples of practical tasks

"Fundamentals of Theoretical Mechanics"

by disciplineTechnical mechanics

for students3 course

specialties20.02.04 Fire safety

Kaliningrad

APPROVE

Deputy Director for SD GAU KO VEO KSTN.N. Myasnikov

APPROVED

Methodological Council of GAU KO VET KST

CONSIDERED

At a meeting of the PCC

Editorial team:

Kolganova A.A., methodologist

Falaleeva A.B., teacher of Russian language and literature

Tsvetaeva L.V., Chairman of the PCCgeneral mathematical and natural science disciplines

Compiled by:

Nezvanova I.V. Lecturer GAU KO VET KST

Content

Dynamics: basic concepts and axioms

Bibliography

Statics: basic concepts and axioms.

Theoretical information

Statics - a section of theoretical mechanics, which considers the properties of forces applied to the points of a rigid body, and the conditions for their equilibrium. Main goals:

1. Transformation of systems of forces into equivalent systems of forces.

2. Determining the conditions for the equilibrium of systems of forces acting on a rigid body.

material point called the simplest model of a material body

any shape, the dimensions of which are small enough and which can be taken as a geometric point having a certain mass. A mechanical system is any set of material points. An absolutely rigid body is a mechanical system, the distances between the points of which do not change under any interactions.

Strength is a measure of the mechanical interaction of material bodies with each other. Force is a vector quantity, since it is determined by three elements:

numerical value;

direction;

application point (A).

The unit of force is Newton (N).

Figure 1.1

A system of forces is a set of forces acting on a body.

A balanced (equal to zero) system of forces is a system that, being applied to a body, does not change its state.

The system of forces acting on the body can be replaced by one resultant acting as a system of forces.

Axioms of statics.

Axiom 1: If a balanced system of forces is applied to the body, then it moves uniformly and rectilinearly or is at rest (the law of inertia).

Axiom 2: An absolutely rigid body is in equilibrium under the action of two forces if and only if these forces are equal in absolute value, act in one straight line and are directed in opposite directions. Figure 1.2

Axiom 3: The mechanical state of the body will not be disturbed if a balanced system of forces is added to or subtracted from the system of forces acting on it.

Axiom 4: The resultant of the two forces applied to the body is equal to their geometric sum, that is, it is expressed in absolute value and direction by the diagonal of the parallelogram built on these forces as on the sides.

Figure 1.3.

Axiom 5: The forces with which two bodies act on each other are always equal in absolute value and directed along one straight line in opposite directions.

Figure 1.4.

Types of bonds and their reactions

connections are called any restrictions that prevent the movement of the body in space. The body, seeking under the action of the applied forces to move, which is prevented by the connection, will act on it with a certain force called force of pressure on the connection . According to the law of equality of action and reaction, the connection will act on the body with the same modulus, but oppositely directed force.
The force with which this connection acts on the body, preventing one or another movement, is called the reaction force (reaction) of the bond .
One of the fundamental principles of mechanics is liberation principle : any non-free body can be considered as free, if we discard the bonds and replace their action with the reactions of the bonds.

The bond reaction is directed in the opposite direction to where the bond does not allow the body to move. The main types of bonds and their reactions are shown in Table 1.1.

Table 1.1

Types of bonds and their reactions

Communication name

Symbol

Smooth surface (support) - surface (support), friction on which the given body can be neglected.
With free support, the reactionis directed perpendicular to the tangent through the pointBUT body contact1 with support surface2 .

Thread (flexible, inextensible). The connection, made in the form of an inextensible thread, does not allow the body to move away from the point of suspension. Therefore, the reaction of the thread is directed along the thread to the point of its suspension.

weightless rod – a rod, the weight of which can be neglected in comparison with the perceived load.
The reaction of a weightless hinged rectilinear rod is directed along the axis of the rod.

Movable hinge, articulated movable support. The reaction is directed along the normal to the supporting surface.

Rigid closure. In the plane of the rigid embedment there will be two components of the reaction, and moment of a pair of forces, which prevents the beam from turning1 relative to the pointBUT .
A rigid attachment in space takes away all six degrees of freedom from body 1 - three displacements along the coordinate axes and three rotations about these axes.
There will be three components in the spatial rigid embedment, , and three moments of pairs of forces.

Converging force system

A system of converging forces called a system of forces whose lines of action intersect at one point. Two forces converging at one point, according to the third axiom of statics, can be replaced by one force -resultant .
The main vector of the system of forces - a value equal to the geometric sum of the forces of the system.

The resultant of a plane system of converging forces can be definedgraphically and analytically.

Addition of a system of forces . The addition of a flat system of converging forces is carried out either by successive addition of forces with the construction of an intermediate resultant (Fig. 1.5), or by constructing a force polygon (Fig. 1.6).

Figure 1.5Figure 1.6

Projection of Force on the Axis - an algebraic quantity equal to the product of the modulus of force and the cosine of the angle between the force and the positive direction of the axis.
ProjectionFx(fig.1.7) forces per axle Xpositive if α is acute, negative if α is obtuse. If strengthis perpendicular to the axis, then its projection onto the axis is zero.

Figure 1.7

Projection of Force on a Plane Ohu– vector , concluded between the projections of the beginning and end of the forceto this plane. Those. the projection of the force onto the plane is a vector quantity, characterized not only by a numerical value, but also by the direction in the planeOhu (Fig. 1.8).

Figure 1.8

Then the projection module to the plane Ohu will be equal to:

Fxy = F cosα,

where α is the angle between the direction of the force and its projection.
Analytical way of specifying forces . For the analytical method of setting the forceit is necessary to choose a system of coordinate axesOhz, in relation to which the direction of force in space will be determined.
A vector depicting strength, can be constructed if the modulus of this force and the angles α, β, γ that the force forms with the coordinate axes are known. DotBUT application of force set separately by its coordinatesX, at, z. You can set the force by its projectionsfx, fy, fzon the coordinate axes. The modulus of force in this case is determined by the formula:

and direction cosines:

, .

Analytical way of adding forces : the projection of the sum vector onto some axis is equal to the algebraic sum of the projections of the terms of the vectors onto the same axis, i.e., if:

then , , .
Knowing Rx, Ry, Rz, we can define the module

and direction cosines:

, , .

Figure 1.9

For the equilibrium of a system of converging forces, it is necessary and sufficient that the resultant of these forces be equal to zero.
1) Geometric equilibrium condition for a converging system of forces : for the equilibrium of a system of converging forces, it is necessary and sufficient that the force polygon constructed from these forces

was closed (the end of the vector of the last term

force must coincide with the beginning of the vector of the first term of the force). Then the main vector of the system of forces will be equal to zero ()
2) Analytical equilibrium conditions . The module of the main vector of the system of forces is determined by the formula. =0. Because the , then the root expression can be equal to zero only if each term simultaneously vanishes, i.e.

Rx= 0, Ry= 0, R z = 0.

Therefore, for the equilibrium of the spatial system of converging forces, it is necessary and sufficient that the sums of the projections of these forces on each of the three coordinates of the axes be equal to zero:

For the equilibrium of a flat system of converging forces, it is necessary and sufficient that the sum of the projections of forces on each of the two coordinate axes be equal to zero:

Addition of two parallel forces in the same direction.

Figure 1.9

Two parallel forces directed in the same direction are reduced to one resultant force parallel to them and directed in the same direction. The magnitude of the resultant is equal to the sum of the magnitudes of these forces, and the point of its application C divides the distance between the lines of action of the forces internally into parts inversely proportional to the magnitudes of these forces, that is

B A C

R=F 1 +F 2

The addition of two unequal parallel forces directed in opposite directions.

Two unequal antiparallel forces are reduced to one resultant force parallel to them and directed towards the greater force. The magnitude of the resultant is equal to the difference between the magnitudes of these forces, and the point of its application, C, divides the distance between the lines of action of the forces externally into parts inversely proportional to the magnitudes of these forces, that is

Pair of forces and moment of force about a point.

Moment of force relative to the point O is called, taken with the appropriate sign, the product of the magnitude of the force by the distance h from the point O to the line of action of the force . This product is taken with a plus sign if the force tends to rotate the body counterclockwise, and with the - sign, if the force tends to rotate the body in a clockwise direction, that is . The length of the perpendicular h is calledshoulder of strength point O. The effect of the action of force i.e. the angular acceleration of the body is greater, the greater the magnitude of the moment of force.

Figure 1.11

A couple of forces A system is called a system consisting of two parallel forces of equal magnitude, directed in opposite directions. The distance h between the lines of action of forces is calledshoulder couple . Moment of a pair of forces m(F,F") is the product of the value of one of the forces that make up the pair and the arm of the pair, taken with the appropriate sign.

It is written as follows: m(F, F")= ± F × h, where the product is taken with a plus sign if the pair of forces tends to rotate the body counterclockwise and with a minus sign if the pair of forces tends to rotate the body clockwise.

The theorem on the sum of the moments of forces of a pair.

The sum of the moments of forces of the pair (F,F") with respect to any point 0 taken in the plane of action of the pair does not depend on the choice of this point and is equal to the moment of the pair.

Theorem on equivalent pairs. Consequences.

Theorem. Two pairs whose moments are equal to each other are equivalent, i.e. (F, F") ~ (P, P")

Consequence 1 . A pair of forces can be transferred to any place in the plane of its action, as well as rotated to any angle and change the arm and magnitude of the forces of the pair, while maintaining the moment of the pair.

Consequence 2. A pair of forces does not have a resultant and cannot be balanced by one force lying in the plane of the pair.

Figure 1.12

Addition and equilibrium condition for a system of pairs on a plane.

1. Theorem on the addition of pairs lying in the same plane. A system of pairs, arbitrarily located in the same plane, can be replaced by one pair, the moment of which is equal to the sum of the moments of these pairs.

2. Theorem on the equilibrium of a system of pairs on a plane.

In order for an absolutely rigid body to be at rest under the action of a system of pairs, arbitrarily located in the same plane, it is necessary and sufficient that the sum of the moments of all pairs be equal to zero, that is

Center of gravity

Gravity - the resultant of the forces of attraction to the Earth, distributed over the entire volume of the body.

Center of gravity of the body - this is such a point, invariably associated with this body, through which the line of action of the force of gravity of a given body passes at any position of the body in space.

Methods for finding the center of gravity

1. Symmetry method:

1.1. If a homogeneous body has a plane of symmetry, then the center of gravity lies in this plane

1.2. If a homogeneous body has an axis of symmetry, then the center of gravity lies on this axis. The center of gravity of a homogeneous body of revolution lies on the axis of revolution.

1.3 If a homogeneous body has two axes of symmetry, then the center of gravity is at the point of their intersection.

2. Partitioning method: The body is divided into the smallest number of parts, the forces of gravity and the position of the centers of gravity of which are known.

3. Method of negative masses: When determining the center of gravity of a body with free cavities, the partitioning method should be used, but the mass of free cavities should be considered negative.

Coordinates of the center of gravity of a flat figure:

The positions of the centers of gravity of simple geometric figures can be calculated using well-known formulas. (Figure 1.13)

Note: The center of gravity of the symmetry of the figure is on the axis of symmetry.

The center of gravity of the rod is at the middle of the height.

1.2. Examples of solving practical problems

Example 1: A weight is suspended on a rod and is in equilibrium. Determine the forces in the bar. (Figure 1.2.1)

Solution:

The forces that arise in the fastening rods are equal in magnitude to the forces with which the rods support the load. (5th axiom)

We determine the possible directions of the reactions of the bonds "rigid rods".

Efforts are directed along the rods.

Figure 1.2.1.

Let us free point A from bonds, replacing the action of bonds with their reactions. (Figure 1.2.2)

Let's start the construction with a known force by drawing a vectorFon some scale.

From the end of the vectorFdraw lines parallel to reactionsR 1 andR 2 .

Figure 1.2.2

Intersecting, the lines create a triangle. (Figure 1.2.3.). Knowing the scale of the constructions and measuring the length of the sides of the triangle, it is possible to determine the magnitude of the reactions in the rods.

For more accurate calculations, you can use geometric relationships, in particular, the sine theorem: the ratio of the side of the triangle to the sine of the opposite angle is a constant value

For this case:

Figure 1.2.3

Comment: If the direction of the vector (coupling reaction) on a given scheme and in the triangle of forces did not match, then the reaction on the scheme should be directed in the opposite direction.

Example 2: Determine the magnitude and direction of the resultant flat system of converging forces in an analytical way.

Solution:

Figure 1.2.4

1. We determine the projections of all forces of the system on Ox (Figure 1.2.4)

Algebraically adding the projections, we obtain the projection of the resultant on the Ox axis.

The sign indicates that the resultant is directed to the left.

2. We determine the projections of all forces on the Oy axis:

Algebraically adding the projections, we obtain the projection of the resultant onto the Oy axis.

The sign indicates that the resultant is directed downward.

3. Determine the modulus of the resultant by the magnitudes of the projections:

4. Determine the value of the angle of the resultant with the axis Ox:

and the value of the angle with the y-axis:

Example 3: Calculate the sum of the moments of forces relative to the point O (Figure 1.2.6).

OA= AB= ATD=DE=CB=2m

Figure 1.2.6

Solution:

1. The moment of force relative to a point is numerically equal to the product of the module and the arm of the force.

2. The moment of force is equal to zero if the line of action of the force passes through a point.

Example 4: Determine the position of the center of gravity of the figure shown in Figure 1.2.7

Solution:

We break the figure into three:

1-rectangle

BUT 1 =10*20=200cm 2

2-triangle

BUT 2 =1/2*10*15=75cm 2

3-lap

BUT 3 =3,14*3 2 =28.3cm 2

Figure 1 CG: x 1 =10cm, y 1 =5cm

Figure 2 CG: x 2 =20+1/3*15=25cm, u 2 =1/3*10=3.3cm

Figure 3 CG: x 3 =10cm, y 3 =5cm

It is defined similarly for With =4.5cm

Kinematics: basic concepts.

Basic kinematic parameters

Trajectory - the line that a material point outlines when moving in space. The trajectory can be a straight line and a curve, a flat and a spatial line.

Trajectory equation for plane motion: y =f ( x)

Distance traveled. The path is measured along the path in the direction of travel. Designation -S, units of measurement - meters.

Point motion equation is an equation that determines the position of a moving point as a function of time.

Figure 2.1

The position of a point at each moment of time can be determined by the distance traveled along the trajectory from some fixed point, considered as the origin (Figure 2.1). This kind of movement is callednatural . Thus, the equation of motion can be represented as S = f (t).

Figure 2.2

The position of a point can also be determined if its coordinates are known as a function of time (Figure 2.2). Then, in the case of motion on a plane, two equations must be given:

In the case of spatial motion, a third coordinate is also addedz= f 3 ( t)

This kind of movement is calledcoordinate .

Travel speed is a vector quantity that characterizes at the moment the speed and direction of movement along the trajectory.

Speed is a vector directed at any moment tangentially to the trajectory towards the direction of motion (Figure 2.3).

Figure 2.3

If a point covers equal distances in equal intervals of time, then the movement is calleduniform .

Average speed on the way ΔSdefined:

where∆S- distance traveled in time Δt; Δ t- time interval.

If a point travels unequal paths in equal intervals of time, then the movement is calleduneven . In this case, the speed is a variable and depends on timev= f( t)

The current speed is defined as

point acceleration - a vector quantity characterizing the rate of change of speed in magnitude and direction.

The speed of a point when moving from point M1 to point Mg changes in magnitude and direction. The average value of acceleration for this period of time

Current acceleration:

Usually, for convenience, two mutually perpendicular acceleration components are considered: normal and tangential (Figure 2.4)

Normal acceleration a n , characterizes the change in speed by

direction and is defined as

Normal acceleration is always directed perpendicular to the velocity towards the center of the arc.

Figure 2.4

Tangential acceleration a t , characterizes the change in velocity in magnitude and is always directed tangentially to the trajectory; during acceleration, its direction coincides with the direction of the velocity, and during deceleration, it is directed opposite to the direction of the velocity vector.

The full acceleration value is defined as:

Analysis of types and kinematic parameters of movements

Uniform movement - This is a movement at a constant speed:

For rectilinear uniform motion:

For curvilinear uniform motion:

Law of Uniform Motion :

Equal-variable motion – is a motion with a constant tangential acceleration:

For rectilinear uniform motion

For curvilinear uniform motion:

Law of uniform motion:

Kinematic graphs

Kinematic graphs - these are graphs of changes in path, speed and acceleration depending on time.

Uniform movement (Figure 2.5)

Figure 2.5

Equal-variable movement (figure 2.6)

Figure 2.6

The simplest motions of a rigid body

Forward motion called the movement of a rigid body, in which any straight line on the body during movement remains parallel to its initial position (Figure 2.7)

Figure 2.7

In translational motion, all points of the body move in the same way: the speeds and accelerations are the same at every moment.

Atrotary motion all points of the body describe circles around a common fixed axis.

The fixed axis around which all points of the body rotate is calledaxis of rotation.

To describe the rotational motion of a body around a fixed axis, onlycorner options. (Figure 2.8)

φ is the angle of rotation of the body;

ω – angular velocity, determines the change in the angle of rotation per unit time;

The change in angular velocity with time is determined by the angular acceleration:

2.2. Examples of solving practical problems

Example 1: The equation of motion of a point is given. Determine the speed of the point at the end of the third second of movement and the average speed for the first three seconds.

Solution:

1. Equation of speed

2. Speed at the end of the third second (t=3 c)

3. Average speed

Example 2: According to the given law of motion, determine the type of motion, the initial speed and tangential acceleration of the point, the time to stop.

Solution:

1. Type of movement: equally variable ()
2. When comparing the equations, it is obvious that

- the initial path traveled before the start of the countdown 10m;

- initial speed 20m/s

- constant tangential acceleration

- the acceleration is negative, therefore, the movement is slow, the acceleration is directed in the direction opposite to the speed of movement.

3. You can determine the time at which the speed of the point will be equal to zero.

3. Dynamics: basic concepts and axioms

Dynamics - a section of theoretical mechanics in which a connection is established between the movement of bodies and the forces acting on them.

In dynamics, two types of problems are solved:

determine the motion parameters according to the given forces;

determine the forces acting on the body, according to the given kinematic parameters of motion.

Undermaterial point imply a certain body that has a certain mass (i.e., contains a certain amount of matter), but does not have linear dimensions (an infinitesimal volume of space).
Isolated a material point is considered, which is not affected by other material points. In the real world, isolated material points, as well as isolated bodies, do not exist, this concept is conditional.

With translational motion, all points of the body move in the same way, so the body can be taken as a material point.

If the dimensions of the body are small compared to the trajectory, it can also be considered as a material point, while the point coincides with the center of gravity of the body.

During the rotational motion of the body, the points may not move in the same way, in this case, some provisions of the dynamics can be applied only to individual points, and the material object can be considered as a set of material points.

Therefore, the dynamics is divided into the dynamics of a point and the dynamics of a material system.

Axioms of dynamics

First axiom ( principle of inertia): in any isolated material point is in a state of rest or uniform and rectilinear motion until the applied forces take it out of this state.

This state is called the stateinertia. Remove the point from this state, i.e. give it some acceleration, maybe an external force.

Every body (point) hasinertia. The measure of inertia is the mass of the body.

Mass calledthe amount of matter in a body in classical mechanics, it is considered a constant value. The unit of mass is the kilogram (kg).

Second axiom (Newton's second law is the basic law of dynamics)

F=ma

wheret - point mass, kg;a - point acceleration, m/s 2 .

The acceleration imparted to a material point by a force is proportional to the magnitude of the force and coincides with the direction of the force.

Gravity acts on all bodies on Earth, it imparts to the body the acceleration of free fall, directed towards the center of the Earth:

G=mg

whereg- 9.81 m/s², free fall acceleration.

Third axiom (Newton's third law): withForces of interaction of two bodies are equal in magnitude and directed along the same straight line in different directions.

When interacting, the accelerations are inversely proportional to the masses.

Fourth axiom (law of independence of action of forces): toEach force in the system of forces acts as it would act alone.

The acceleration imparted to the point by the system of forces is equal to the geometric sum of the accelerations imparted to the point by each force separately (Figure 3.1):

Figure 3.1

The concept of friction. Types of friction.

Friction- resistance arising from the movement of one rough body on the surface of another. Sliding friction results in sliding friction, and rolling friction results in rocking friction.

Sliding friction

Figure 3.2.

The reason is the mechanical engagement of the protrusions. The force of resistance to movement during sliding is called the force of sliding friction (Figure 3.2)

Laws of sliding friction:

1. The force of sliding friction is directly proportional to the force of normal pressure:

whereR- force of normal pressure, directed perpendicular to the supporting surface;f- coefficient of sliding friction.

Figure 3.3.

In the case of a body moving along an inclined plane (Figure 3.3)

rolling friction

The rolling resistance is related to the mutual deformation of the ground and the wheel and is much less than the sliding friction.

For uniform rolling of the wheel, it is necessary to apply forceF dv (Figure 3.4)

The wheel rolling condition is that the moving moment must not be less than the moment of resistance:

Figure 3.4.

Example 1: Example 2: To two material points of massm 1 =2kg andm 2 = 5 kg equal forces are applied. Compare the values faster.

Solution:

According to the third axiom, the acceleration dynamics are inversely proportional to the masses:

Example 3: Determine the work of gravity when moving a load from point A to point C along an inclined plane (Figure 3. 7). The force of gravity of the body is 1500N. AB=6m, BC=4m. Example 3: Determine the work of the cutting force in 3 minutes. The rotation speed of the workpiece is 120 rpm, the diameter of the workpiece is 40mm, the cutting force is 1kN. (Figure 3.8)

Solution:

1. Working with rotary motion:

2. Angular speed 120 rpm

Figure 3.8.

3. The number of revolutions for a given time isz\u003d 120 * 3 \u003d 360 rev.

Rotation angle during this time φ=2πz\u003d 2 * 3.14 * 360 \u003d 2261 rad

4. Work for 3 turns:W\u003d 1 * 0.02 * 2261 \u003d 45.2 kJ

Bibliography

Olofinskaya, V.P. "Technical Mechanics", Moscow "Forum" 2011

Erdedi A.A. Erdedi N.A. Theoretical mechanics. Strength of materials.- R-n-D; Phoenix, 2010