Statics- this is the section theoretical mechanics, in which the conditions of equilibrium of material bodies under the influence of forces are studied.

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

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

The 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.

Curing principle

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

Material point statics

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 equal to zero.

Geometric interpretation... If the beginning of the second vector is placed at the end of the first vector, and the beginning of the third is placed at the end of the second vector, and then this process is continued, then the end of the last, n -th vector will be aligned with the beginning of the first vector. That is, we get a closed geometric figure, the lengths of the sides of which are equal to the moduli 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 axis are equal to zero:

If you choose any direction given by some vector, then the sum of the projections of the force vectors on this direction is equal to zero:
.
Let us multiply equation (1) scalarly by a vector:
.
Here is the scalar product of 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 relative to a point

Determination of the 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 the force F by the shoulder OH.

Let the vectors and be located in the plane of the drawing. According to the property of the 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 moment vector is directed at us. Absolute torque value:
.
Since, then
(3) .

Using geometry, you 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 drop the perpendicular OH to this 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 with respect to the center O equals force per shoulder this force relative to the selected center O.

When calculating the moment, it is often convenient to decompose the force into two components:
,
where . The force passes through point O. Therefore, its moment is zero. Then
.
Absolute torque value:
.

Moment components in a rectangular coordinate system

If we choose a rectangular coordinate system Oxyz centered 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, from 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 will not change with this movement.

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

The same applies to forces whose continuation lines intersect at one point.

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

A couple of forces

A couple of forces- these are two forces, equal in absolute value 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 included in the pair is equal to 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 balance, the nature of the forces included in the pair is irrelevant. A pair of forces is used to indicate that a moment of forces is acting on the body, which has a certain value.

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 force relative to a selected point, but only need to know the moment of force relative to the selected axis.

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

The properties of the moment of force about the axis

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

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

Calculation of the moment of force about the axis

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

Let's build a rectangular coordinate system. Let the Oz axis coincide with O′O ′ ′. From point A we drop the perpendicular OH to O′O ′ ′. Draw the Ox axis through points O and A. 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 crosses the O′O ′ ′ axis. Therefore, its moment is zero. The force is parallel to the O'O '' axis. Therefore, its moment is also zero. By 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.

Equilibrium conditions for a rigid body

In equilibrium, the vector sum of all forces acting on the body is zero and the vector sum of the moments of these forces relative to an arbitrary stationary center is 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 outside it. Usually the center O is chosen to make the calculations simpler.

Equilibrium conditions can be formulated in another way.

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

Sometimes these conditions are more convenient. There are times when, by choosing the axes, you can make the calculations simpler.

Body center of gravity

Let's consider one of the most important forces - the force of gravity. Here forces are not applied at certain points of the body, but are continuously distributed over its volume. For each part of the body with an infinitely small volume Δ V, the force of gravity acts. Here ρ is the density of the substance of the body, is the acceleration of gravity.

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

Let's find the sum of the forces of gravity formed by all parts of the body:
,
where is the body weight. Thus, the sum of the gravity forces of individual infinitesimal parts of the body can be replaced by one vector of the gravity of the whole body:
.

Let us find the sum of the moments of gravity, relative to the chosen center O in an arbitrary way:

.
Here we have introduced point C which is called center of gravity body. 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 balance, the sum of the forces of gravity 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 respective reference books. If the body has an axis or plane of symmetry, then the center of gravity is located on this axis or plane. So, the centers of gravity of a sphere, circle or circle are in the centers of the circles of these figures. Centers of gravity rectangular parallelepiped, rectangle or square are also located in their centers - at the intersection points 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 the resultant force of the quantity applied at the center of gravity of the plot. 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 with a resultant. The value of the resultant is equal to the area of ​​the diagram:
.
The application point is at the center of gravity of the plot. The center of gravity of a triangle with height h is 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 from 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 largest value is equal to:
,
where f is the coefficient of friction. The friction coefficient is dimensionless.

Rolling friction... Let the rounded body roll or can 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, which prevents the body from moving. Greatest value the moment of friction is equal to:
,
where δ is the rolling friction coefficient. It has the dimension of length.

References:
S. M. Targ, A short course in theoretical mechanics, " graduate School", 2010.

Point kinematics.

1. The subject of theoretical mechanics. Basic abstractions.

Theoretical mechanicsis a science in which the general laws of mechanical motion and mechanical interaction of material bodies are studied

Mechanical movementis called the movement of a body in relation to another body, which occurs in space and time.

Mechanical interaction such an interaction of material bodies is called, which changes the nature of their mechanical movement.

Statics - This is a branch of theoretical mechanics, in which methods of transforming systems of forces into equivalent systems are studied and conditions for the equilibrium of forces applied to a solid are established.

Kinematics - this is a branch of theoretical mechanics that studies the movement of material bodies in space from a geometric point of view, regardless of the forces acting on them.

Dynamics - this is a section of mechanics, which studies the movement of material bodies in space, depending on the forces acting on them.

Objects of study in theoretical mechanics:

material point,

system of material points,

Absolutely solid.

Absolute space and absolute time are independent of one another. Absolute space - three-dimensional, homogeneous, stationary Euclidean space. Absolute time - flows from the past to the future continuously, it is homogeneous, the same at all points in space and does not depend on the movement of matter.

2. The subject of kinematics.

Kinematics - this is the section of mechanics in which the geometric properties the movement of bodies without taking into account their inertia (i.e. mass) and the forces acting on them

To determine the position of a moving body (or point) with the body, in relation to which the movement of the given body is being studied, some coordinate system is rigidly connected, which, together with the body, forms frame of reference.

The main task of kinematics is, knowing the law of motion of a given body (point), to determine all the kinematic quantities that characterize its motion (speed and acceleration).

3. Methods for specifying the movement of a point

· Natural way

It should be known:

Point motion trajectory;

Start and direction of counting;

The law of motion of a point along a given trajectory in the form (1.1)

· Coordinate way

Equations (1.2) are the equations of motion of the point M.

The equation of the trajectory of point M can be obtained by excluding the time parameter « t » from equations (1.2)

· Vector way

(1.3) The relationship between the coordinate and vector methods of specifying the movement of a point (1.4)

The relationship between coordinate and natural ways point movement assignments

Determine the trajectory of a point, excluding time from equations (1.2);

-- find the law of motion of a point along a trajectory (use the expression for the differential of the arc)

After integration, we obtain the law of motion of a point along a given trajectory:

The relationship between the coordinate and vector methods of specifying the motion of a point is determined by the equation (1.4)

4. Determination of the speed of a point in the vector method of specifying the movement.

Let at the moment of timetthe position of the point is determined by the radius vector, and at the moment of timet 1 - radius vector, then for a period of time the point will move.

(1.5) average point speed, the vector is directed as well as the vector

Point speed at a given time

To get the speed of a point at a given moment in time, it is necessary to make the passage to the limit

(1.6)

(1.7)

The velocity vector of a point at a given time is equal to the first derivative of the radius vector in time and is directed tangentially to the trajectory at a given point.

(unit¾ m / s, km / h)

Average acceleration vector has the same direction as the vectorΔ v , that is, directed towards the concavity of the trajectory.

Acceleration vector of a point at a given time is equal to the first derivative of the velocity vector or the second derivative of the radius vector of the point with respect to time.

(unit of measure -)

How is the vector positioned in relation to the path of the point?

In straight-line motion, the vector is directed along the straight line along which the point is moving. If the trajectory of a point is a flat curve, then the acceleration vector, as well as the vector cp, lies in the plane of this curve and is directed towards its concavity. If the trajectory is not a plane curve, then the vector cp will be directed towards the concavity of the trajectory and will lie in the plane passing through the tangent to the trajectory at the pointM and a straight line parallel to the tangent at an adjacent pointM 1 . V limit when pointM 1 strives for M this plane occupies the position of the so-called contacting plane. Therefore, in the general case, the acceleration vector lies in the contacting plane and is directed towards the concavity of the curve.

Theoretical mechanics- this is a section of mechanics, which sets out the basic laws of mechanical motion and mechanical interaction of material bodies.

Theoretical mechanics is the science in which the movements of bodies over time (mechanical movements) are studied. It serves as the basis for other branches of mechanics (theory of elasticity, resistance of materials, theory of plasticity, theory of mechanisms and machines, hydro-aerodynamics) and many technical disciplines.

Mechanical movement Is change over time mutual position in the space of material bodies.

Mechanical interaction- this is such an interaction as a result of which mechanical movement changes or the relative position of body parts changes.

Rigid body statics

Statics- this is a section of theoretical mechanics, which deals with the problems of equilibrium of rigid bodies and the transformation of one system of forces into another, equivalent to it.

Basic concepts and laws of statics
• Absolutely solid(solid, body) is a material body, the distance between any points in which does not change.
• Material point Is a body whose dimensions, according to the conditions of the problem, can be neglected.
• Free body Is a body, the movement of which is not subject to any restrictions.
• Unfree (bound) body Is a body with restrictions imposed on its movement.
• Connections- these are bodies that prevent the movement of the object under consideration (body or system of bodies).
• Communication reaction Is a force that characterizes the effect of a bond on a rigid body. If we consider the force with which a rigid body acts on a bond as an action, then the bond reaction is a reaction. In this case, the force - the action is applied to the bond, and the bond reaction is applied to the solid.
• Mechanical system Is a set of interconnected bodies or material points.
• Solid can be considered as a mechanical system, the position and distance between the points of which do not change.
• Force Is a vector quantity that characterizes the mechanical action of one material body on another.
  Force as a vector is characterized by the point of application, direction of action and absolute value. The unit of measure for the modulus of force is Newton.
• Force action line Is a straight line along which the force vector is directed.
• Concentrated power- force applied at one point.
• Distributed forces (distributed load)- these are the forces acting on all points of the volume, surface or length of the body.
  The distributed load is set by the force acting on a unit of volume (surface, length).
  The dimension of the distributed load is N / m 3 (N / m 2, N / m).
• External force Is a force acting from a body that does not belong to the considered mechanical system.
• Inner strength Is a force acting on a material point of a mechanical system from another material point belonging to the system under consideration.
• Force system Is a set of forces acting on a mechanical system.
• Flat system of forces Is a system of forces, the lines of action of which lie in the same plane.
• Spatial system of forces Is a system of forces, the lines of action of which do not lie in the same plane.
• System of converging forces Is a system of forces whose lines of action intersect at one point.
• Arbitrary system of forces Is a system of forces, the lines of action of which do not intersect at one point.
• Equivalent systems of forces- these are systems of forces, the replacement of which one with another does not change the mechanical state of the body.
  Accepted designation:.
• Equilibrium- this is a state in which the body under the action of forces remains stationary or moves uniformly in a straight line.
• Balanced system of forces Is a system of forces that, when applied to a free solid, does not change its mechanical state (does not unbalance).
  .
• Resultant force Is a force, the action of which on the body is equivalent to the action of the system of forces.
  .
• Moment of power Is a value that characterizes the rotational ability of a force.
• A couple of forces Is a system of two parallel, equal in magnitude, oppositely directed forces.
  Accepted designation:.
  Under the action of a pair of forces, the body will rotate.
• Axis force projection Is a segment enclosed between perpendiculars drawn from the beginning and end of the force vector to this axis.
  The projection is positive if the direction of the line segment coincides with the positive direction of the axis.
• Force projection onto plane Is a vector on a plane, enclosed between perpendiculars drawn from the beginning and end of the force vector to this plane.
• Law 1 (law of inertia). An isolated material point is at rest or moves evenly and rectilinearly.
  The uniform and rectilinear motion of a material point is motion by inertia. Under the state of equilibrium of a material point and solid understand not only the state of rest, but also the motion by inertia. For a solid, there are different kinds inertial motion, for example, uniform rotation of a rigid body around a fixed axis.
• Law 2. A solid body is in equilibrium under the action of two forces only if these forces are equal in magnitude and directed in opposite directions along the common line of action.
  These two forces are called balancing forces.
  In general, forces are called balancing if the rigid body to which these forces are applied is at rest.
• Law 3. Without disturbing the state (the word "state" here means a state of motion or rest) of a rigid body, one can add and drop counterbalancing forces.
  Consequence. Without violating the state of a rigid body, force can be transferred along its line of action to any point in the body.
  Two systems of forces are called equivalent if one of them can be replaced by another without violating the state of a rigid body.
• Law 4. The resultant of two forces applied at one point, applied at the same point, is equal in magnitude to the diagonal of the parallelogram built on these forces, and is directed along this
  diagonals.
  The modulus of the resultant is equal to:
  
• Law 5 (the law of equality of action and reaction)... The forces with which two bodies act on each other are equal in magnitude and directed in opposite directions along one straight line.
  It should be borne in mind that action- force applied to the body B, and counteraction- force applied to the body A are not balanced, since they are attached to different bodies.
• Law 6 (law of hardening)... The equilibrium of a non-solid body is not disturbed when it solidifies.
  It should not be forgotten that the conditions of equilibrium, which are necessary and sufficient for a solid, are necessary, but not sufficient for the corresponding non-solid.
• Law 7 (the law of release from ties). A non-free rigid body can be considered as free if it is mentally freed from bonds, replacing the action of bonds with the corresponding reactions of bonds.

Connections and their reactions
• Smooth surface constrains movement along the normal to the support surface. The reaction is directed perpendicular to the surface.
• Articulated movable support constrains the movement of the body along the normal to the reference plane. The reaction is directed along the normal to the support surface.
• Articulated fixed support counteracts any movement in a plane perpendicular to the axis of rotation.
• Articulated weightless rod counteracts the movement of the body along the line of the bar. The reaction will be directed along the line of the bar.
• Blind termination counteracts any movement and rotation in the plane. Its action can be replaced by a force represented in the form of two components and a pair of forces with a moment.

Kinematics

Kinematics- a section of theoretical mechanics, which examines the general geometric properties of mechanical motion, as a process that occurs in space and time. Moving objects are considered as geometric points or geometric bodies.

Basic concepts of kinematics
• The law of motion of a point (body) Is the dependence of the position of a point (body) in space on time.
• Point trajectory- this is geometric place positions of a point in space during its movement.
• Point (body) speed- This is a characteristic of the change in time of the position of a point (body) in space.
• Point (body) acceleration- This is a characteristic of the change in time of the speed of a point (body).

Determination of kinematic characteristics of a point
• Point trajectory
  In the vector frame of reference, the trajectory is described by the expression:.
  In the coordinate system of reference, the trajectory is determined according to the law of motion of a point and is described by the expressions z = f (x, y)- in space, or y = f (x)- in the plane.
  In the natural frame of reference, the trajectory is set in advance.
• Determining the speed of a point in a vector coordinate system
  When specifying the movement of a point in a vector coordinate system, the ratio of the movement to the time interval is called the average value of the speed in this time interval:.
  Taking the time interval as an infinitely small value, the speed value is obtained at a given time (instantaneous speed value): .
  The average velocity vector is directed along the vector in the direction of the point's movement, the instantaneous velocity vector is directed tangentially to the trajectory in the direction of the point's movement.
  Output: the speed of a point is a vector quantity equal to the derivative of the law of motion with respect to time.
  Derivative property: the derivative of any quantity with respect to time determines the rate of change of this quantity.
• Determining the speed of a point in a coordinate system
  Point coordinates change rates:
  .
  The modulus of the full speed of a point with a rectangular coordinate system will be equal to:
  .
  The direction of the velocity vector is determined by the cosines of the direction angles:
  ,
  where are the angles between the velocity vector and the coordinate axes.
• Determining the speed of a point in the natural frame of reference
  The speed of a point in the natural frame of reference is determined as a derivative of the law of motion of a point:.
  According to the previous conclusions, the velocity vector is directed tangentially to the trajectory in the direction of movement of the point and in the axes is determined by only one projection.

Rigid body kinematics
• In the kinematics of solids, two main tasks are solved:
  1) the task of movement and the determination of the kinematic characteristics of the body as a whole;
  2) determination of the kinematic characteristics of the points of the body.
• The translational motion of a rigid body
  Translational movement is a movement in which a straight line drawn through two points of the body remains parallel to its original position.
  Theorem: during translational motion, all points of the body move along the same trajectories and at each moment of time have the same velocity and acceleration in magnitude and direction.
  Output: the translational movement of a rigid body is determined by the movement of any of its points, and therefore, the task and study of its movement is reduced to the kinematics of the point.
• Rotational movement of a rigid body around a fixed axis
  The rotational movement of a rigid body around a fixed axis is the movement of a rigid body in which two points belonging to the body remain motionless during the entire time of movement.
  The position of the body is determined by the angle of rotation. The angle unit is radians. (Radian is the central angle of a circle whose arc length is equal to the radius, the total angle of the circle contains 2π radians.)
  Law rotary motion bodies around a fixed axis.
  Angular velocity and angular acceleration we define the body by the method of differentiation:
  - angular velocity, rad / s;
  - angular acceleration, rad / s².
  If you cut the body with a plane perpendicular to the axis, select the point on the axis of rotation WITH and an arbitrary point M then point M will describe around the point WITH circle radius R... During dt an elementary rotation through an angle occurs, while the point M will move along the trajectory at a distance .
  Linear speed module:
  .
  Point acceleration M with a known trajectory, it is determined by its components:
  ,
  where .
  As a result, we get the formulas
  tangential acceleration: ;
  normal acceleration: .

Dynamics

Dynamics- This is a section of theoretical mechanics in which the mechanical movements of material bodies are studied, depending on the reasons that cause them.

Basic concepts of dynamics
• Inertia Is the property of material bodies to maintain a state of rest or uniform straight motion until external forces change this state.
• Weight Is a quantitative measure of body inertia. The unit of measure for mass is kilogram (kg).
• Material point Is a body with a mass, the dimensions of which are neglected when solving this problem.
• Center of gravity of the mechanical system- geometric point, the coordinates of which are determined by the formulas:
  
  where m k, x k, y k, z k- mass and coordinates k-th point of the mechanical system, m Is the mass of the system.
  In a homogeneous gravity field, the position of the center of mass coincides with the position of the center of gravity.
• Moment of inertia of a material body about the axis Is a quantitative measure of rotational inertia.
  The moment of inertia of a material point about the axis is equal to the product of the point's mass by the square of the point's distance from the axis:
  .
  The moment of inertia of the system (body) about the axis is equal to the arithmetic sum of the moments of inertia of all points:
  
• The force of inertia of a material point Is a vector quantity equal in magnitude to the product of the point mass by the acceleration modulus and directed opposite to the acceleration vector:
• The force of inertia of a material body Is a vector quantity equal in modulus to the product of the body mass by the modulus of acceleration of the center of mass of the body and directed opposite to the vector of acceleration of the center of mass:,
  where is the acceleration of the center of mass of the body.
• Elementary Force Impulse Is a vector quantity equal to the product of the force vector by an infinitely small time interval dt:
  .
  The total impulse of force for Δt is equal to the integral of elementary impulses:
  .
• Elementary work of strength Is a scalar dA equal to the scalar proi

Within any training course the study of physics begins with mechanics. Not with theoretical, not with applied and not computational, but with 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 pushed him to the discovery of the law of universal gravitation. Of course, the law has always existed, and Newton only gave it a form that people understand, 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 has Greek origin and translates as "the art of building machines." But before the construction of machines, we are still like the Moon, so we will 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 of 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 from the foundations of mechanics. Placed within the framework of time and space, people, in fact, could not start from something else, with all their desire. Moving bodies are the first thing we turn our attention to.

What is movement?

Mechanical movement 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 on the seat next to him, and moves at a different 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 frame of reference - 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 of our measurements in a geocentric frame of reference associated with the Earth. Earth is a reference body, relative to which cars, airplanes, people, 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 constructs a mathematical description of motion and finds connections between physical quantities characterizing it.

In order to move further, we need the concept “ material point ”. They say physics - exact science, but physicists know how many approximations and assumptions have to be made to agree on this very accuracy. Nobody has ever seen a material point or smelled ideal gas, but they are! It's just much easier to live with them.

Material point is a body, the size and shape of which 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 the body moves. In other words, this section deals with quantitative characteristics movement. Find speed, path - typical tasks kinematics

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

Statics studies the balance of bodies under the action of forces, that is, answers the question: why does it not fall at all?

The limits of applicability of classical mechanics.

Classic 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 for the world we are accustomed to in terms of size (macrocosm). They stop working in the case of the particle world, when the classical one 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. You can learn more about it from our article.

Generally speaking, quantum and relativistic effects never go anywhere; they also take place during the ordinary motion of macroscopic bodies with a speed much less 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. Thus, classical mechanics will never lose their fundamental importance.

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

Force. The system of forces. Equilibrium of an absolutely rigid body

In mechanics, force is understood as a measure of the mechanical interaction of material bodies, as a result of which interacting bodies can impart acceleration to each other or deform (change their shape). Force is a vector quantity. It is characterized by a numerical value, or modulus, point of application, and direction. The point of application of the force and its direction determine the line of action of the force. The figure shows how the force is applied to point A. Segment AB = modulus of force F. Line LM is called the line of action of the force. In sist. SI force meas. in newtons (N). There is also 1MN = 10 6 N, 1 kN = 10 3 N. There are 2 ways to set the force: by direct description and vector (through the projection on the coordinate axes). F = F x i + F y j + F z k, where F x, F y, F z are the projections of the force on the coordinate axes, and i, j, k are unit unit vectors. Absolutely solid body-body in which the distance m-du 2 its points stop. unchanged regardless of the effect of forces on him.

The combination of several forces (F 1, F 2, ..., F n) is called a system of forces. If, without violating the state of the body, one system of forces (F 1, F 2, ..., F n) can be replaced by another system (P 1, P 2, ..., P n) and vice versa, then such systems of forces are called equivalent. This is symbolically denoted as follows: (F 1, F 2, ..., F n) ~ (P 1, P 2, ..., P n). However, this does not mean that if two systems of forces have the same effect on the body, they will be equivalent. Equivalent systems cause the same system state. When the system of forces (F 1, F 2, ..., F n) is equivalent to one force R, then R is called. resultant. The resulting force can replace the action of all these forces. But not every system of forces has a resultant. In the inertial coordinate system, the law of inertia is fulfilled. This means, in particular, that the body, which is at rest at the initial moment, will remain in this state if no forces act on it. If an absolutely rigid body remains at rest under the action of a system of forces (F 1, F 2, ..., F n) on it, then this system is called balanced, or a system of forces equivalent to zero: (F 1, F 2,. .., F n) ~ 0. In this case, the body is said to be in equilibrium. In mathematics, two vectors are considered equal if they are parallel, directed in the same direction and equal in absolute value. For the equivalence of the two forces, this is not enough, and the relation F ~ P still does not follow from the equality F = P. Two forces are equivalent if they are equal in vector and applied to one point of the body.

Statics axioms and their consequences

The body, under the action of force, acquires acceleration and cannot be at rest. The first axiom sets the conditions under which the system of forces will be balanced.

Axiom 1. Two forces applied to an absolutely rigid body will be balanced (equivalent to zero) if and only if they are equal in absolute value, act along one straight line and are directed in opposite directions... This means that if an absolutely rigid body is at rest under the action of two forces, then these forces are equal in magnitude, act in one straight line and are directed in opposite directions. Conversely, if two forces of equal magnitude act on an absolutely rigid body along one straight line in opposite directions and the body was at rest at the initial moment, then the state of rest of the body will remain.

In fig. 1.4 shows the balanced forces F 1, F 2 and P 1, P 2, satisfying the relations: (F 1, F 2) ~ 0, (P 1, P 2) ~ 0. When solving some problems of statics, one has to consider the forces applied to the ends of rigid rods, the weight of which can be neglected, and it is known that the rods are in equilibrium. From the formulated axiom, the forces acting on such a rod are directed along a straight line passing through the ends of the rod, opposite in direction and equal to each other in modulus (Fig. 1.5, a). The same is the case when the axis of the rod is curved (Fig. 1.5, b).

Axiom 2. Without violating the state of an absolutely rigid body, forces can be applied or discarded to it if and only if they constitute a balanced system, in particular, if this system consists of two forces, equal in magnitude, acting in one straight line and directed in opposite directions. This axiom implies a consequence: without violating the state of the body, the point of application of the force can be transferred along the line of its action. Indeed, let the force F A be applied to point A (Fig. 1.6, a). We apply at point B on the line of action of the force FA two balanced forces FB and F "B, assuming that FB = FA (Fig. 1.6, b). Then, according to axiom 2, we will have FA ~ FA, FB, F` B). So as the forces F А and FB also form a balanced system of forces (axiom 1), then according to axiom 2 they can be discarded (Fig. 1.6, c). Thus, FA ~ FA, FB, F` B) ~ FB, or FA ~ FB, which proves the corollary. This corollary shows that the force applied to an absolutely rigid body is a sliding vector. Both axioms and the proved corollary cannot be applied to deformable bodies, in particular, the transfer of the point of application of the force along the line of its action changes the stress deformed body condition.

Axiom 3.Without changing the state of the body, two forces applied to one of its points can be replaced by one resultant force applied at the same point and equal to them geometric sum(axiom of the parallelogram of forces). This axiom establishes two circumstances: 1) two forces F 1 and F 2 (Fig. 1.7), applied to one point, have a resultant, that is, they are equivalent to one force (F 1, F 2) ~ R; 2) the axiom completely determines the modulus, application point and direction of the resultant force R = F 1 + F 2. (1.5) In other words, the resultant R can be constructed as a parallelogram diagonal with sides coinciding with F 1 and F 2. The modulus of the resultant is determined by the equality R = (F 1 2 + F 2 2 + 2F l F 2 cosa) 1/2, where a is the angle between these vectors F 1 and F 2. The third axiom is applicable to any body. The second and third axioms of statics make it possible to pass from one system of forces to another system equivalent to it. In particular, they make it possible to decompose any force R into two, three, etc. components, that is, to pass to another system of forces for which the force R is the resultant. By setting, for example, two directions that lie with R in the same plane, you can build a parallelogram in which the diagonal represents the force R. Then the forces directed along the sides of the parallelogram will form a system for which the force R will be the resultant (Fig. 1.7). A similar construction can be carried out in space. To do this, it is enough to draw three straight lines from the point of application of the force R that do not lie in the same plane, and build on them a parallelepiped with a diagonal representing the force R, and with edges directed along these straight lines (Fig. 1.8).

Axiom 4 (Newton's 3rd Law). The forces of interaction of two bodies are equal in magnitude and are directed along one straight line in opposite directions. Note that the forces of interaction between two bodies do not constitute a system of balanced forces, since they are applied to different bodies. If body I acts on body II with force P, and body II acts on body I with force F (Fig. 1.9), then these forces are equal in magnitude (F = P) and are directed along one straight line in opposite directions, i.e. . F = –Р. If we denote by F the force with which the Sun attracts the Earth, then the Earth attracts the Sun with the same modulus, but oppositely directed force - F. When the body moves along the plane, a friction force T will be applied to it, directed in the direction opposite to the motion. This is the force with which the fixed plane acts on the body. On the basis of the fourth axiom, the body acts on the plane with the same force, but its direction will be opposite to the force T.

In fig. 1.10 shows a body moving to the right; the friction force T is applied to the moving body, and the force T "= –T - to the plane. Consider still the resting system, shown in Fig. 1.11, a. It consists of an engine A mounted on a foundation B, which in turn is located on the base C. The forces of gravity F 1 and F 2 act on the engine and the foundation, respectively.The forces also act: F 3 - the force of action of body A on body B (it is equal to the weight of body A); F`z - force of the reverse action of body B on body A ; F 4 is the force of action of bodies A and B on the base C (it is equal to the total weight of bodies A and B); F` 4 is the force of the reverse action of the base C on body B. These forces are shown in Fig. 1.11, b, c, d .According to the axiom 4 F 3 = –F` 3, F 4 = –F` 4, and these forces of interaction are determined by the given forces F 1 and F 2. To find the forces of interaction, it is necessary to proceed from the axiom 1. Due to the rest of the body A (Fig. 1.11,6) should be F s = –F 1, which means that F 3 = F 1. In the same way, from the equilibrium condition of the body B (Fig. 1.11, c) it follows F` 4 = - (F 2 + F 3) , i.e., F` 4 = - (F 1 + F 2) and F 4 = F 1 + F 2.

Axiom 5. The balance of a deformable body will not be violated if its points are rigidly connected and the body is considered absolutely rigid. This axiom is used when it comes to the equilibrium of bodies that cannot be considered rigid. External forces applied to such bodies must satisfy the equilibrium conditions of a rigid body, but for non-rigid bodies these conditions are only necessary, but not sufficient. For example, for the balance of an absolutely solid weightless rod, it is necessary and sufficient that the forces F and F "applied to the ends of the rod act along a straight line connecting its ends, be equal in magnitude and directed in different directions. The same conditions are also necessary for the equilibrium of a segment of a weightless thread , but for a thread they are insufficient - it is necessary to additionally require that the forces acting on the thread be tensile (Fig. 1.12, b), while for a rod they can also be compressive (Fig. 1.12, a).

Consider the case of equivalence to zero of three non-parallel forces applied to a rigid body (Fig. 1.13, a). The three non-parallel forces theorem. If, under the action of three forces, the body is in equilibrium and the lines of action of the two forces intersect, then all the forces lie in the same plane, and their lines of action intersect at one point. Let a system of three forces F 1, F 3 and F 3 act on the body, and the lines of action of the forces F 1 and F 2 intersect at point A (Fig. 1.13, a). According to the corollary from Axiom 2, the forces F 1 and F 2 can be transferred to point A (Fig. 1.13, b), and according to Axiom 3, they can be replaced by one force R, and (Fig. 1.13, c) R = F 1 + F 2 ... Thus, the considered system of forces is reduced to two forces R and F 3 (Fig. 1.13, c). According to the conditions of the theorem, the body is in equilibrium, therefore, according to axiom 1, the forces R and F 3 must have a common line of action, but then the lines of action of all three forces must intersect at one point.

Active forces and reactions of bonds

The body is called free if its movements are not limited by anything. A body whose movements are limited by other bodies is called unfree, and the bodies limiting the movements of this body are connections... At the points of contact, forces of interaction arise between the given body and the bonds. The forces with which the connections act on a given body are called bond reactions.

The principle of release : any non-free body can be regarded as free if the action of bonds is replaced by their reactions applied to the given body. In statics, it is possible to fully determine the reactions of bonds using the conditions or equations of equilibrium of the body, which will be established later, but their directions in many cases can be determined from considering the properties of the bonds. As a simple example, Fig. 1.14, and a body is represented, the point M of which is connected to a fixed point O by means of a rod, the weight of which can be neglected; the ends of the rod have hinges allowing free rotation. In this case, the OM rod serves as the connection for the body; constraint on the freedom of movement of point M is expressed in the fact that it is forced to be at a constant distance from point O. ... Thus, the direction of the reaction of the rod coincides with the straight line OM (Fig. 1.14, b). Likewise, the reaction force of a flexible, inextensible thread should be directed along the thread. In fig. 1.15 shows a body hanging on two strands and the reactions of strands R 1 and R 2. Forces acting on a non-free body are divided into two categories. One category is formed by forces that do not depend on connections, and the other - by the reactions of connections. In this case, the reactions of connections are passive - they arise because the forces of the first category act on the body. Forces that do not depend on connections are called active, and the reactions of connections are called passive forces. In fig. 1.16, and the top shows two active forces F 1 and F 2 of equal modulus, stretching the bar AB, the bottom shows the reactions R 1 and R 2 of the stretched bar. In fig. 1.16, b, the top shows the active forces F 1 and F 2, compressing the bar, the bottom shows the reactions R 1 and R 2 of the compressed bar.

Link properties

1. If a rigid body rests on a perfectly smooth (frictionless) surface, then the point of contact of the body with the surface can freely slide along the surface, but cannot move in the direction along the normal to the surface. The reaction of a perfectly smooth surface is directed along a common normal to the contacting surfaces (Fig. 1.17, a). If a solid body has a smooth surface and rests on a tip (Fig. 1.17, b), then the reaction is directed along the normal to the surface of the body itself. If a solid body is abuts the tip against the corner (Figure 1.17, c), then the connection prevents the tip from moving both horizontally and vertically. Accordingly, the reaction R angle can be represented by two components - horizontal R x and vertical R y, the values ​​and directions of which are ultimately determined by the given forces.

2. A spherical joint is the device shown in fig. 1.18, a, which makes the point O of the body under consideration fixed. If the spherical contact surface is ideally smooth, then the reaction of the spherical hinge is normal to this surface. The reaction passes through the center of the hinge O; the direction of the reaction can be any and is determined in each case.

It is also impossible to determine in advance the direction of reaction of the thrust bearing shown in Fig. 1.18, b. 3. Cylindrical hinge-fixed support (Fig. 1.19, a). The reaction of such a support passes through its axis, and the direction of the reaction can be any (in a plane perpendicular to the axis of the support). 4. A cylindrical hinge-movable support (Fig. 1.19, b) prevents the movement of the fixed point of the body along the perpendicular to plane I-I; accordingly, the reaction of such a support also has the direction of this perpendicular.

In mechanical systems formed by joining several solids, there are internal connections with external connections (supports). In these cases, the system is sometimes mentally dismembered and the discarded not only external, but also internal connections are replaced with appropriate reactions. The forces of interaction between the individual points of a given body are called internal, and the forces acting on a given body and caused by other bodies are called external.

Basic tasks of statics

1. The problem of reducing a system of forces: how to replace a given system of forces with another, the simplest, equivalent to it?

2. The problem of equilibrium: what conditions must a system of forces applied to a given body (or material point) satisfy in order for it to be a balanced system?

The second problem is often posed in cases where equilibrium is known to exist, for example, when it is known in advance that the body is in equilibrium, which is provided by the constraints imposed on the body. In this case, the conditions of equilibrium establish a relationship between all the forces applied to the body. With the help of these conditions, it is possible to determine the support reactions. It should be borne in mind that the determination of the reactions of bonds (external and internal) is necessary for the subsequent calculation of the strength of the structure.

In a more general case, when a system of bodies that can move relative to each other is considered, one of the main problems of statics is the problem of determining possible equilibrium positions.

Reduction of the system of converging forces to the resultant

Forces are called converging if the lines of action of all the forces that make up the system intersect at one point. Let's prove the theorem: The system of converging forces is equivalent to one force (resultant), which is equal to the sum of all these forces and passes through the point of intersection of their lines of action. Let a system of converging forces F 1, F 2, F 3, ..., F n, applied to an absolutely rigid body, be given (Fig. 2.1, a). We transfer the points of application of forces along the lines of their action to the point of intersection of these lines (21, b). We got a system of forces, attached to one point. It is equivalent to the given one. Add F 1 and F 2, we get their resultant: R 2 = F 1 + F 2. Add R 2 to F 3: R 3 = R 2 + F 3 = F 1 + F 2 + F 3. Add F 1 + F 2 + F 3 +… + F n = R n = R = åF i. Ch.t.d. Instead of parallelograms, you can build a power polygon. Let the system consist of 4 forces (Figure 2.2.). From the end of the vector F 1 we postpone the vector F 2. The vector connecting the start O and the end of the vector F 2 will be the vector R 2. Next, we postpone the vector F 3 by placing its beginning at the end of the vector F 2. Then we get a vector R 8 going from point O to the end of the vector F 3. Add the vector F 4 in the same way; in this case, we obtain that the vector going from the beginning of the first vector F 1 to the end of the vector F 4 is the resultant R. Such a spatial polygon is called a force polygon. If the end of the last force does not coincide with the beginning of the first force, then the power polygon is called open... If a geometer is right to find the resultant use, then this method is called geometric.

They use the analytical method more to determine the resultant. The projection of the sum of vectors on a certain axis is equal to the sum of projections on the same axis of the terms of the vectors, we obtain R x = åF kx = F 1x + F 2x +… + F nx; R y = åF ky = F 1y + F 2y + ... + F ny; R z = åF kz = F 1z + F 2z + ... + F nz; where F kx, F ky, F kz are the projections of the force F k on the axis, and R x, R y, R z are the projections of the resultant on the same axes. The projections of the resultant system of converging forces on the coordinate axes are equal to the algebraic sums of the projections of these forces on the corresponding axes. The modulus of the resultant R is equal to: R = (R x 2 + R y 2 + R z 2) 1/2. The direction cosines are: cos (x, R) = R x / R, cos (y, R) = R y / R, cos (z, R) = R z / R. If the forces are located in the area, then everything is the same, there is no Z axis.

Equilibrium conditions for a system of converging forces

(F 1, F 2, ..., F n) ~ R => for the equilibrium of a body under the action of a system of converging forces, it is necessary and sufficient that their resultant be equal to zero: R = 0. Therefore, in the force polygon of a balanced system converging forces, the end of the last force must coincide with the beginning of the first force; in this case, the force polygon is said to be closed (Fig. 2.3). This condition is used in the graphical solution of problems for plane systems of forces. The vector equality R = 0 is equivalent to three scalar equalities: R x = åF kx = F 1x + F 2x +… + F nx = 0; R y = åF ky = F 1y + F 2y + ... + F ny = 0; R z = åF kz = F 1z + F 2z + ... + F nz = 0; where F kx, F ky, F kz are the projections of the force F k on the axis, and R x, R y, R z are the projections of the resultant on the same axes. That is, for the equilibrium of the converging system of forces, it is necessary and sufficient that the algebraic sums of the projections of all the forces of the given system on each of the coordinate axes be equal to zero. For a plane system of forces, the condition associated with the Z axis disappears. The equilibrium conditions make it possible to control whether a given system of forces is in equilibrium.

Addition of two parallel forces

1) Let parallel and equally directed forces F 1 and F 2 be applied to points A and B of the body and you need to find their resultant (Fig. 3.1). We apply to points A and B equal in magnitude and oppositely directed forces Q 1 and Q 2 (their modulus can be any); such an addition can be done on the basis of axiom 2. Then at points A and B we get two forces R 1 and R 2: R 1 ~ (F 1, Q 1) and R 2 ~ (F 2, Q 2). The lines of action of these forces intersect at some point O. Let's transfer the forces R 1 and R 2 to point O and decompose each into components: R 1 ~ (F 1 ', Q 2') and R 2 ~ (F 2 ', Q 2' ). It can be seen from the construction that Q 1 ’= Q 1 and Q 2’ = Q 2, therefore, Q 1 ’= –Q 2’ and these two forces, according to axiom 2, can be discarded. In addition, F 1 '= F 1, F 2' = F 2. Forces F 1 'and F 2' act in one straight line, and they can be replaced by one force R = F 1 + F 2, which will be the desired resultant. The modulus of the resultant is equal to R = F 1 + F 2. The line of action of the resultant is parallel to the lines of action F 1 and F 2. From the similarity of the triangles Oac 1 and OAC, as well as Obc 2 and OBC, we obtain the ratio: F 1 / F 2 = BC / AC. This relation determines the point of application of the resultant R. The system of two parallel forces directed in one direction has a resultant parallel to these forces, and its modulus is equal to the sum modules of these forces.

2) Let two parallel forces act on the body, directed in different directions and not equal in magnitude. Given: F 1, F 2; F 1> F 2.

Using the formulas R = F 1 + F 2 and F 1 / F 2 = BC / AC, the force F 1 can be decomposed into two components, F "2 and R, directed towards the force F 1. Let's do it so that the force F" 2 turned out to be applied to point B, and put F "2 = –F 2. Thus, (F l, F 2) ~ (R, F "2, F 2)... Forces F 2, F 2 ' can be discarded as being equivalent to zero (axiom 2), therefore, (F 1, F 2) ~ R, that is, the force R is the resultant. Let us define the force R satisfying such an expansion of the force F 1. Formulas R = F 1 + F 2 and F 1 / F 2 = BC / AC give R + F 2 '= F 1, R / F 2 = AB / AC (*). this implies R = F 1 –F 2 '= F 1 + F 2, and since the forces F t and F 2 are directed in different directions, then R = F 1 –F 2. Substituting this expression into the second formula (*), we get after simple transformations F 1 / F 2 = BC / AC. the ratio determines the point of application of the resultant R. Two oppositely directed parallel forces not equal in magnitude have a resultant parallel to these forces, and its modulus is equal to the difference in the moduli of these forces.

3) Let two parallels act on the body, equal in magnitude, but opposite in force. This system is called a pair of forces and is indicated by the symbol (F 1, F 2)... Suppose that the modulus F 2 gradually increases, approaching the value of the modulus F 1. Then the difference in modules will tend to zero, and the system of forces (F 1, F 2) - to a pair. In this case, | R | Þ0, and the line of its action is to move away from the lines of action of these forces. A pair of forces is an unbalanced system that cannot be replaced by a single force. A pair of forces has no resultant.

Moment of force about a point and axis. Moment of a pair of forces

The moment of force relative to a point (center) is a vector numerically equal to the product of the modulus of the force by the shoulder, that is, the shortest distance from the specified point to the line of action of the force. It is directed perpendicular to the plane passing through the selected point and the line of action of the force. If the moment of force is on the clock of the hand, then the moment is negative, and if against, then it is positive. If O is the point relative to the moment of force F, then the moment of force is denoted by the symbol M o (F). If the point of application of the force F is determined by the radius vector r relative to O, then the relation M o (F) = r x F. (3.6) That is. the moment of force is equal to the vector product of the vector r by the vector F. The modulus of the vector product is M o (F) = rF sin a = Fh, (3.7) where h is the shoulder of the force. The vector Mo (F) is directed perpendicular to the plane passing through the vectors r and F, and counterclockwise. Thus, formula (3.6) completely determines the modulus and direction of the moment of force F. Formula (3.7) can be written in the form M O (F) = 2S, (3.8) where S is the area of ​​the triangle ОАВ. Let x, y, z be the coordinates of the point of application of the force, a F x, F y, F z - the projection of the force on the coordinate axes. If t. About tries. at the origin, then the moment of force:

This means that the projections of the moment of force on the coordinate axes are determined by the f-mi: M ox (F) = yF z –zF y, M oy (F) = zF x –xF z, M oz (F) = xF y –yF x (3.10 ).

Let us introduce the concept of the projection of a force onto a plane. Let the force F and some space be given. Let us drop perpendiculars from the beginning and end of the force vector to this plane (Fig. 3.5). The projection of a force onto a plane is a vector, the beginning and end of which coincide with the projection of the beginning and the projection of the end of the force onto this plane. The projection of the force F onto the area xOy will be F xy. Moment of force F xy rel. m. О (if z = 0, F z = 0) will be M o (F xy) = (xF y –yF x) k. This moment is directed along the z-axis, and its projection onto the z-axis exactly coincides with the projection onto the same axis of the moment of force F relative to the point O.Te, M Oz (F) = M Oz (F xy) = xF y –yF x. (3.11). The same result can be obtained by projecting the force F onto any other plane parallel to the xOy plane. In this case, the point of intersection of the axis with the plane will be different (denote O 1). However, all values ​​x, y, F x, F y included in the right-hand side of equality (3.11) will remain unchanged: M Oz (F) = M Olz (F xy). The projection of a moment of force about a point onto an axis passing through that point does not depend on the choice of a point on the axis. Instead of M Oz (F), we write M z (F). This projection of the moment is called the moment of force about the z-axis. Before calculations, the force F is projected onto the plane, perpendicular to the axis. М z (F) = М z (F xy) = ± F xy h (3.12). h- shoulder. If clockwise, then +, against -. To calculate mom. forces you need to: 1) select an arbitrary point on the axis and build a plane perpendicular to the axis; 2) project a force on this plane; 3) determine the shoulder of the projection of the force h. The moment of force about the axis is equal to the product of the modulus of the projection of the force on its shoulder, taken with the corresponding sign. From (3.12) it follows that the moment of force relative to the axis is zero: 1) when the projection of the force onto the plane perpendicular to the axis is zero, that is, when the force and the axis are parallel; 2) when the shoulder of the projection h is equal to zero, i.e. when the line of action of the force intersects the axis. Or: the moment of force about the axis is zero if and only if the line of action of the force and the axis are in the same plane.

Let us introduce the concept of the moment of a pair. Let us find what is the sum of the moments of forces that make up a pair, relative to an arbitrary point. Let O be an arbitrary point in space (Fig. 3.8), and F and F "are the forces that make up a pair. Then M o (F) = OAxF, M o (F") = OBxF ", whence M o (F) + M o (F ") = ОАxF + OBxF", but since F "= - F, then M 0 (F) + M 0 (F") = OAxF – ОBхF = (ОА– OB) xF. Taking into account the equality ОА –ОВ = VA, we finally find: M 0 (F) + M 0 (F ") = BAхF. That is, the sum of the moments of forces that make up a pair does not depend on the position of the point relative to which the moments are taken. The vector product BAxF is called the moment of the pair. The moment of the pair is denoted by the symbol M (F, F "), and M (F, F") = BAxF = ABxF ", or, M = BAxF = ABxF". (3.13). The moment of the pair is a vector perpendicular to the plane of the pair, equal in magnitude to the product of the modulus of one of the forces of the pair on the shoulder of the pair (that is, the shortest distance between the lines of action of the forces that make up the pair) and directed in the direction from which the “rotation” of the pair is visible going counterclockwise. If h is the shoulder of the pair, then M (F, F ") = hF. In order for the pair of forces to balance the system, it is necessary: ​​that the moment of the pair = 0, or the shoulder = 0.

Pair theorems

Theorem 1.Two pairs lying in the same plane can be replaced by one pair lying in the same plane, with a moment equal to the sum of the moments of these two pairs ... For docking, consider two pairs (F 1, F` 1) and (F 2, F` 2) (Fig. 3.9) and transfer the points of application of all forces along the lines of their action to points A and B, respectively. Adding forces according to axiom 3, we get R = F 1 + F 2 and R "= F` 1 + F` 2, but F" 1 = –F 1 and F` 2 = –F 2. Therefore, R = –R ", that is, the forces R and R" form a pair. The moment of this pair: M = M (R, R ") = BAxR = BAx (F 1 + F 2) = BAxF 1 + BAxF 2. (3.14). When the forces that make up a pair are transferred along the lines of their action, neither the shoulder nor the direction of rotation of the pair does not change, therefore, the moment of the pair does not change either. Hence, BAxF 1 = M (F 1, F "1) = M 1, BAxF 2 = M (F 2, f` 2) = M 2, and the formula (З.14) will take the form M = M 1 + M 2, (3.15) p.t.d. Let's make two remarks. 1. The lines of action of the forces that make up the pair may turn out to be parallel. The theorem remains valid in this case as well. 2. After addition, it can turn out that M (R, R ") = 0; based on Remark 1, it follows from this that the set of two pairs (F 1, F` 1, F 2, F` 2) ~ 0.

Theorem 2.Two pairs having equal moments are equivalent. Let a pair (F 1, F` 1) act on a body in plane I with moment M 1. Let us show that this pair can be replaced by another pair (F 2, F` 2) located in plane II, if only its moment М 2 is equal to М 1. Note that planes I and II must be parallel, in particular, they can coincide. Indeed, from the parallelism of the moments M 1, and M 2 it follows that the planes of action of the pairs, perpendicular to the moments, are also parallel. Let us introduce into consideration a new pair (F 3, F` 3) and apply it together with the pair (F 2, F` 2) to the body, placing both pairs in plane II. To do this, according to axiom 2, you need to select a pair (F 3, F` 3) with a moment M 3 so that the applied system of forces (F 2, F` 2, F 3, F` 3) is balanced. We put F 3 = –F` 1 and F` 3 = –F 1 and we match the points of application of these forces with the projections A 1 and B 1 of points A and B onto plane II (see Fig. 3.10). In accordance with the construction, we will have: M 3 ​​= –M 1 or, taking into account that M 1 = M 2, M 2 + M 3 = 0, we get (F 2, F` 2, F 3, F` 3) ~ 0. Thus, the pairs (F 2, F` 2) and (F 3, F` 3) are mutually balanced and their attachment to the body does not violate its state (axiom 2), so (F 1, F` 1) ~ (F 1, F` 1, F 2, F` 2, F 3, F` 3). (3.16). On the other hand, the forces F 1 and F 3, as well as F` 1 and F` 3 can be added according to the rule of addition of parallel forces directed in one direction. They are equal in absolute value, so their resultants R and R "must be applied at the intersection of the diagonals of the rectangle ABB 1 A 1, in addition, they are equal in absolute value and directed in opposite directions. This means that they constitute a system equivalent to zero. So , (F 1, F` 1, F 3, F` 3) ~ (R, R ") ~ 0. Now we can write (F 1, F` 1, F 2, F` 2, F 3, F` 3) ~ (F 2, F` 2). (3.17). Comparing relations (3.16) and (3.17), we obtain (F 1, F` 1) ~ (F 2, F` 2), etc. It follows from this theorem that a pair of forces can be moved and rotated in the plane of its action, transferred to a parallel plane; in a pair, you can simultaneously change the forces and the shoulder, keeping only the direction of rotation of the pair and the modulus of its moment (F 1 h 1 = F 2 h 2).

Theorem 3. Two pairs lying in intersecting planes are equivalent to one pair, the moment of which is equal to the sum of the moments of the two given pairs. Let the pairs (F 1, F` 1) and (F 2, F` 2) be located in intersecting planes I and II, respectively. Using the corollary of Theorem 2, we bring both pairs to the arm AB (Fig. 3.11), located on the line of intersection of planes I and II. Let us denote the transformed pairs by (Q 1, Q` 1) and (Q 2, Q` 2). In this case, the equalities must be satisfied: M 1 = M (Q 1, Q` 1) = M (F 1, F` 1) and M 2 = M (Q 2, Q` 2) = M (F 2, F` 2 ). Let us add the forces applied at points A and B, respectively, according to axiom 3. Then we get R = Q 1 + Q 2 and R "= Q` 1 + Q` 2. Taking into account that Q` 1 = –Q 1 and Q` 2 = –Q 2, we get: R = –R". Thus, we have proved that a system of two pairs is equivalent to one pair (R, R "). Find the moment M of this pair. M (R, R") = BAxR, but R = Q 1 + Q 2 and M (R , R ") = BAx (Q 1 + Q 2) = BAxQ 1 + BAxQ 2 = M (Q 1, Q` 1) + M (Q 2, Q` 2) = M (F 1, F" 1) + M (F 2, F` 2), or M = M 1 + M 2, that is, the theorem is proved.

Conclusion: the moment of a pair is a free vector and completely determines the action of a pair on an absolutely rigid body. For deformable bodies, the theory of pairs is inapplicable.

Reducing a system of pairs to its simplest form. Equilibrium of a system of pairs

Let a system of n pairs (F 1, F 1`), (F 2, F` 2) ..., (F n, F` n) be given, arbitrarily located in space, the moments of which are equal to M 1, M 2. .., M n. The first two pairs can be replaced by one pair (R 1, R` 1) with the moment M * 2: M * 2 = M 1 + M 2. We add the resulting pair (R 1, R` 1) with the pair (F 3, F` 3), then we get a new pair (R 2, R` 2) with the moment M * 3: M * 3 = M * 2 + M 3 = M 1 + M 2 + M 3. Continuing further the sequential addition of the moments of the pairs, we obtain the last resulting pair (R, R ") with the moment M = M 1 + M 2 + ... + M n = åM k. (3.18). The system of pairs is reduced to one pair, the moment of which is equal to the sum of the moments of all pairs.Now it is easy to solve the second problem of statics, that is, to find the equilibrium conditions for the body on which the system of pairs acts.In order for the system of pairs to be equivalent to zero, that is, to be reduced to two balanced forces, it is necessary and it is sufficient that the moment of the resulting pair is equal to 0. Then from formula (3.18) we obtain the following equilibrium condition in vector form: М 1 + М 2 + М 3 + ... + М n = 0. (3.19).

In projections onto the coordinate axes, equation (3.19) gives three scalar equations. Equilibrium condition (3.19) is simplified when all pairs lie in the same plane. In this case, all moments are perpendicular to this plane, and therefore equation (3.19) is sufficient to project only on one axis, for example, an axis perpendicular to the plane of pairs. Let it be the z-axis (Figure 3.12). Then from equation (3.19) we get: М 1Z + М 2Z + ... + М nZ = 0. It is clear that M Z = M, if the rotation of the pair is seen from the positive direction of the z axis counterclockwise, and M Z = –M in the opposite direction of rotation. Both of these cases are shown in Fig. 3.12.

Parallel Force Transfer Lemma

Let us prove the lemma:A force applied at any point of a rigid body is equivalent to the same force applied at any other point of this body, and a pair of forces, the moment of which is equal to the moment of this force relative to the new point of application. Let a force F be applied at point A of a rigid body (Fig. 4.1). We now apply at point B of the body a system of two forces F "and F²-, equivalent to zero, and we choose F" = F (hence, F "= - F). Then the force F ~ (F, F", F "), since (F ", F") ~ 0. But, on the other hand, the system of forces (F, F ", F") is equivalent to the force F "and a pair of forces (F, F"); therefore, the force F is equivalent to the force F "and pair of forces (F, F "). The moment of the pair (F, F") is equal to M = M (F, F ") = BAxF, i.e. equal to the moment of force F relative to point B M = MB (F). Thus , the lemma on parallel force transfer is proved.

Basic theorem of statics

Let an arbitrary system of forces (F 1, F 2, ..., F n) be given. The sum of these forces F = åF k is called the main vector of the system of forces. The sum of the moments of forces relative to any pole is called the main moment of the considered system of forces relative to this pole.

The main theorem of statics (Poinsot's theorem ):Any spatial system of forces in the general case can be replaced by an equivalent system consisting of one force applied at some point of the body (reference center) and equal to the main vector of this system of forces, and one pair of forces, the moment of which is equal to the main moment of all forces relative to the selected center of reference. Let О be the center of reference taken as the origin of coordinates, r 1, r 2, r 3, ..., rn are the corresponding radius vectors of the points of application of the forces F 1, F 2, F 3, ..., F n that make up this system forces (Fig. 4.2, a). Let us transfer the forces F 1, F a, F 3, ..., F n to the point O. Let us add these forces as converging; we get one force: F о = F 1 + F 2 +… + F n = åF k, which is equal to the main vector (Fig. 4.2, b). But with the sequential transfer of forces F 1, F 2, ..., F n to point O, we get each time the corresponding pair of forces (F 1, F "1), (F 2, F" 2), ..., ( F n, F "n). The moments of these pairs are respectively equal to the moments of these forces relative to the point O: M 1 = M (F 1, F” 1) = r 1 x F 1 = M o (F 1), M 2 = M (F 2, F "2) = r 2 x F 2 = M about (F 2), ..., M p = M (F n, F" n) = rnx F n = M about (F n). Based on the rule of reducing the system of pairs to the simplest form, all the indicated pairs can be replaced with one pair. Its moment is equal to the sum of the moments of all forces of the system relative to point O, that is, it is equal to the main moment, since according to formulas (3.18) and (4.1) we have (Fig. 4.2, c) M 0 = M 1 + M 2 + .. . + М n = М о (F 1) + М о (F 2) + ... + М о (F n) == åМ о (F k) = år kx F k. The system of forces, arbitrarily located in space, can be replaced at an arbitrarily chosen center of reference by the force F o = åF k (4.2) and a pair of forces with the moment M 0 = åM 0 (F k) = år k x F k. (4.3). In technique, it is often easier to set not the strength or the pair, but their moments. For example, the characteristic of an electric motor does not include the force with which the stator acts on the rotor, but the torque.

Equilibrium conditions for the spatial system of forces

Theorem.For the balance of the spatial system of forces, it is necessary and sufficient that the main vector and main point of this system were zero. Adequacy: when F o = 0, the system of converging forces applied at the center of reduction O is equivalent to zero, and when Mo = 0, the system of pairs of forces is equivalent to zero. Therefore, the original system of forces is equivalent to zero. Need: Let the given system of forces be equivalent to zero. Bringing the system to two forces, we note that the system of forces Q and P (Fig. 4.4) must be equivalent to zero, therefore, these two forces must have a common line of action and the equality Q = –Р must be fulfilled. But this can be if the line of action of the force P passes through the point O, that is, if h = 0. This means that the main moment is zero (M o = 0). Because Q + P = 0, a Q = F o + P ", then F o + P" + P = 0, and, consequently, F o = 0. Necessary and sufficient conditions are equal to the spatial system of forces they are of the form: F o = 0 , M o = 0 (4.15),

or, in projections onto the coordinate axes, Fox = åF kx = F 1x + F 2x +… + F nx = 0; F Oy = åF ky = F 1y + F 2y + ... + F ny = 0; F oz = åF kz = F 1z + F 2z +… + F nz = 0 (4.16). M Ox = åM Ox (F k) = M Ox (F 1) + M ox (F 2) + ... + M Ox (F n) = 0, M Oy = åM Oy (F k) = M oy ( F 1) + M oy (F 2) + ... + M oy (F n) = 0, M oz = åM Oz (F k) = M Oz (F 1) + M oz (F 2) + ... + M oz (F n) = 0. (4.17)

That. when solving problems, having 6 levels, you can find 6 unknowns. Note: a pair of forces cannot be reduced to a resultant. Special cases: 1) Equilibrium of the spatial system of parallel forces. Let the Z axis be parallel to the lines of force action (Figure 4.6), then the projections of forces on x and y are 0 (F kx = 0 and F ky = 0), and only F oz remains. As for the moments, only M ox and M oy remain, and M oz is absent. 2) Equilibrium of a plane system of forces. There remain ur-I F ox, F oy and the moment M oz (Figure 4.7). 3) Equilibrium of a plane system of parallel forces. (fig. 4.8). Only 2 ur-I remain: F oy and M oz. When drawing up ur-th equilibrium, any point can be selected for the center of the ghost.

Reducing a flat system of forces to its simplest form

Consider a system of forces (F 1, F 2, ..., F n) located in one plane. Let us combine the coordinate system Oxy with the plane of location of forces and, choosing its origin as the center of reference, we reduce the system of forces under consideration to one force F 0 = åF k, (5.1) equal to the main vector, and to a pair of forces, the moment of which is equal to the main moment M 0 = åM 0 (F k), (5.2) where M o (F k) is the moment of force F k relative to the center of reference O. Since the forces are located in one plate, the force F o also lies in this plane. The moment of the M o pair is directed perpendicular to this plane, since the pair itself is divided into the action of the forces in question. Thus, for a plane system of forces, the main vector and the main moment are always perpendicular to each other (Fig. 5.1). The moment is fully characterized by the algebraic value M z, equal to the product of the pair's shoulder by the value of one of the forces that make up the pair, taken with a plus sign, if the pair's "rotation" occurs counterclockwise, and with a minus sign, if it occurs clockwise arrows. Let, for example, be given two pairs, (F 1, F` 1) and (F 2, F` 2) (Fig. 5.2); then, according to this definition, we have M z (F 1, F` 1) = h 1 F 1, MZ (F 2, F "2) = - h 2 F 2. The moment of force relative to a point is an algebraic quantity equal to the projection of the moment vector forces relative to this point on an axis perpendicular to the plane, i.e. equal to the product of the force modulus per shoulder, taken with the corresponding sign.For the cases shown in Fig.5.3, a and b, respectively, will be M oz (F 1) = hF 1 , M oz (F 2) = - hF 2 (5.4). The index z in formulas (5.3) and (5.4) is kept in order to indicate the algebraic nature of the moments. The moduli of the moment of a pair and the moment of force are denoted as follows: М (F , F ") = | М z (F, F`) |, М о (F) = | М Оz (F) |. We get M oz = åM oz (F z). For the analytical determination of the main vector, the following formulas are used: F ox = åF kx = F 1x + F 2x +… + F nx, F oy = åF ky = F 1y, + F 2y +… + F ny, F o = (F 2 ox + F 2 oy) 1/2 = ([åF kx] 2 + [åF ky] 2) 1/2 (5.8); cos (x, F o) = F ox / F o, cos (y, F o) = F Oy / F o. (5.9). And the main moment is M Оz = åM Oz (F k) = å (x k F ky –y k F kx), (5.10) where x k, y k are the coordinates of the point of application of the force F k.

Let us prove that if the principal vector of a plane system of forces is not equal to zero, then the given system of forces is equivalent to one force, that is, it is reduced to the resultant. Let Fo ≠ 0, MOz ≠ 0 (Fig. 5.4, a). The arc arrow in Fig. 5.4, ​​but symbolically depicts a couple with the moment MOz. A pair of forces, the moment of which is equal to the main moment, we represent in the form of two forces F1 and F`1, equal in magnitude to the main vector Fo, that is, F1 = F`1 = Fo. In this case, we will apply one of the forces (F`1) that make up the pair to the center of reduction and direct it in the direction opposite to the direction of the force Fo (Fig. 5.4, b). Then the system of forces Fo and F`1 is equivalent to zero and can be rejected. Therefore, the given system of forces is equivalent to a single force F1 applied to the point 01; this force is the resultant. The resultant will be denoted by the letter R, i.e. F1 = R. Obviously, the distance h from the previous center of reduction O to the line of action of the resultant can be found from the condition | MOz | = hF1 = hFo, i.e. h = | MOz | / Fo. The distance h must be postponed from the point O so that the moment of the pair of forces (F1, F`1) coincides with the main moment MOz (Fig. 5.4, b). As a result of reducing the system of forces to a given center, the following cases can occur: (1) Fo ≠ 0, MOz ≠ 0. In this case, the system of forces can be reduced to one force (resultant), as shown in Fig. 5.4, ​​c. (2) Fo ≠ 0, MOz = 0. In this case, the system of forces is reduced to one force (resultant) passing through the given center of reference. (3) Fo = 0, MOz ≠ 0. In this case, the system of forces is equivalent to one pair of forces. (4) Fo = 0, MOz = 0. In this case, the considered system of forces is equivalent to zero, that is, the forces that make up the system are mutually balanced.

Varignon's theorem

Varignon's theorem. If the considered flat system of forces is reduced to a resultant, then the moment of this resultant relative to any point is equal to the algebraic sum of the moments of all forces of the given system relative to that same point. Suppose that the system of forces is reduced to the resultant R passing through the point O. Let us now take another point O 1 as the center of reduction. The main moment (5.5) relative to this point is equal to the sum of the moments of all forces: M O1Z = åM o1z (F k) (5.11). On the other hand, we have M O1Z = M Olz (R), (5.12) since the principal moment for the center of reduction O is equal to zero (M Oz = 0). Comparing relations (5.11) and (5.12), we obtain M O1z (R) = åM OlZ (F k); (5.13) h.t.d. Using Varignon's theorem, one can find the equation of the line of action of the resultant. Let the resultant R 1 be applied at some point O 1 with coordinates x and y (Fig. 5.5) and the principal vector F o and the principal moment M Oya are known at the reference center at the origin. Since R 1 = F o, the components of the resultant along the x and y axes are equal to R lx = F Ox = F Ox i and R ly = F Oy = F oy j. According to Varignon's theorem, the moment of the resultant relative to the origin is equal to the principal moment at the reference center at the origin, that is, Moz = M Oz (R 1) = xF Oy –yF Ox. (5.14). The values ​​of M Oz, F Ox and F oy do not change when the point of application of the resultant is transferred along its line of action, therefore, the coordinates x and y in equation (5.14) can be viewed as the current coordinates of the line of action of the resultant. Thus, equation (5.14) is the equation of the line of action of the resultant. For F ox ≠ 0, it can be rewritten as y = (F oy / F ox) x– (M oz / F ox).

Equilibrium conditions for a plane system of forces

A necessary and sufficient condition for the equilibrium of the system of forces is the equality to zero of the main vector and the main moment. For a plane system of forces, these conditions get the form F o = åF k = 0, M Oz = åM oz (F k) = 0, (5.15), where O is an arbitrary point in the plane of action of the forces. We get: F ox = åF kx = F 1x + F 2x + ... + F nx = 0, P ox = åF ky = F 1y + F 2y + ... + F ny = 0, М Оz = åM Oz (F k) = M oz (F 1) + M oz (F 2) + ... + M oz (F n) = 0, i.e. for the equilibrium of a plane system of forces, it is necessary and sufficient that the algebraic sums of the projections of all forces on two coordinate axes and the algebraic sum of the moments of all forces relative to an arbitrary point are equal to zero. The second form of the equilibrium equation is the equality to zero of the algebraic sums of the moments of all forces with respect to any three points that do not lie on one straight line; åM Az (F k) = 0, åM Bz (F k) = 0, åM Cz (F k) = 0, (5.17), where A, B and C are the indicated points. The need to satisfy these equalities follows from conditions (5.15). Let us prove their sufficiency. Suppose that all equalities (5.17) are satisfied. Equality to zero of the main moment at the center of reference at point A is possible, either if the system is reduced to the resultant (R ≠ 0) and its line of action passes through point A, or R = 0; similarly, the equality to zero of the principal moment with respect to points B and C means that either R ≠ 0 and the resultant passes through both points, or R = 0. But the resultant cannot pass through all these three points A, B and C (by condition, they do not lie on one straight line). Consequently, equalities (5.17) are possible only for R = 0, that is, the system of forces is in equilibrium. Note that if points A, B and C lie on one straight line, then the fulfillment of conditions (5.17) will not be a sufficient condition for equilibrium, - in this case, the system can be reduced to a resultant, the line of action of which passes through these points.

The third form of equilibrium equations for a plane system of forces

The third form of equilibrium equations for a plane system of forces is the equality to zero of the algebraic sums of the moments of all forces of the system relative to any two points and the equality to zero of the algebraic sum of the projections of all forces of the system onto an axis not perpendicular to the straight line passing through two selected points; åМ Аz (F k) = 0, åМ Bz (F k) = 0, åF kx = 0 (5.18) (the x-axis is not perpendicular to the segment А В). The necessity of these equalities for the balance of forces follows directly from conditions (5.15). Let us make sure that the fulfillment of these conditions is sufficient for the balance of forces. From the first two equalities, as in the previous case, it follows that if the system of forces has a resultant, then its line of action passes through points A and B (Fig. 5.7). Then the projection of the resultant on the x-axis, which is not perpendicular to the segment AB, will be nonzero. But this possibility is excluded by the third equation (5.18) since R x = åF hx). Consequently, the resultant must be equal to zero and the system is in equilibrium. If the x axis is perpendicular to the segment AB, then equations (5.18) will not be sufficient equilibrium conditions, since in this case the system can have a resultant, the line of action of which passes through points A and B. Thus, the system of equilibrium equations can contain one an equation of moments and two equations of projections, or two equations of moments and one equation of projections, or three equations of moments. Let the lines of action of all forces be parallel to the y-axis (Fig. 4.8). Then the equilibrium equations for the considered system of parallel forces will be åF ky = 0, åM Oz (F k) = 0. (5.19). åM Az (F k) = 0, åM Bz (F k) = 0, (5.20) and points A and B should not lie on a straight line parallel to the y-axis. The system of forces acting on a rigid body can consist of both concentrated (isolated) forces and distributed forces. Distinguish between forces distributed along the line, along the surface and over the volume of the body.

Body balance in the presence of sliding friction

If two bodies I and II (Fig. 6.1) interact with each other, touching at point A, then always the reaction RA, acting, for example, from the side of body II and applied to body I, can be decomposed into two components: NA, directed along the common normal to the surface of the contacting bodies at point A, and T A, lying in the tangent plane. The component N A is called the normal reaction, the force T A is called the sliding friction force - it prevents the sliding of body I over body II. In accordance with axiom 4 (Newton's third law), body II from the side of body I is acted upon by an equal in magnitude and oppositely directed reaction force. Its component perpendicular to the tangent plane is called the normal pressure force. Friction force T A = 0 if the contacting surfaces are perfectly smooth. In real conditions, the surfaces are rough and in many cases the friction force cannot be neglected. The maximum friction force is approximately proportional to the normal pressure, i.e. T max = fN. (6.3) - Amonton-Coulomb law. The coefficient f is called the coefficient of sliding friction. Its value does not depend on the area of ​​the contacting surfaces, but depends on the material and the degree of roughness of the contacting surfaces. The friction force can be calculated by f-le T = fN only if there is a critical case. In other cases, the friction force should be determined from ur-th equals. The figure shows the reaction R (here active forces tend to move the body to the right). The angle j between the limiting reaction R and the normal to the surface is called the angle of friction. tgj = T max / N = f.

The locus of all possible directions of the limiting reaction R forms a conical surface - a friction cone (Figure 6.6, b). If the coefficient of friction f is the same in all directions, then the friction cone will be circular. In cases where the friction coefficient f depends on the direction of the possible motion of the body, the friction cone will not be circular. If the resultant of active forces. is inside the friction cone, then by increasing its modulus, the balance of the body cannot be disturbed; for the body to start moving, it is necessary (and sufficient) that the resultant of active forces F be outside the friction cone. Consider the friction of flexible bodies (Figure 6.8). Euler's formula helps to find the smallest force P capable of balancing the force Q. P = Qe -fj *. You can also find a force P capable of overcoming the frictional resistance together with the force Q. In this case, only the sign of f will change in the Euler formula: P = Qe fj *.

Body balance in the presence of rolling friction

Consider a cylinder (roller) resting on a horizontal plane when a horizontal active force S acts on it; besides it, the force of gravity P acts, as well as the normal reaction N and the force of friction T (Fig. 6.10, a). With a sufficiently small modulus of force S, the cylinder remains at rest. But this fact cannot be explained if one is satisfied with the introduction of the forces shown in Fig. 6.10, a. According to this scheme, equilibrium is impossible, since the main moment of all forces acting on the cylinder М Сz = –Sr is nonzero, and one of the equilibrium conditions is not fulfilled. The reason for this discrepancy lies in the fact that we imagine this body as absolutely solid and assume that the cylinder touches the surface along the generatrix. To eliminate the noted discrepancy between theory and experiment, it is necessary to abandon the hypothesis of an absolutely rigid body and take into account that in reality the cylinder and the plane near point C are deformed and there is a certain contact area of ​​finite width. As a result, in its right side, the cylinder is pressed more strongly than in the left, and full reaction R is applied to the right of point C (see point C 1 in Fig. 6.10, b). The resulting scheme of acting forces is statically satisfactory, since the moment of the pair (S, T) can be balanced by the moment of the pair (N, P). Unlike the first scheme (Fig. 6.10, a), a pair of forces is applied to the cylinder with the moment М T = Nh. (6.11). This moment is called the rolling frictional moment. h = Sr /, where h is the distance from C to C 1. (6.13). With an increase in the modulus of active force S, the distance h increases. But this distance is related to the contact surface area and, therefore, cannot increase indefinitely. This means that a state will come when an increase in the force S will lead to an imbalance. Let us denote the maximum possible value of h by the letter d. The d value is proportional to the radius of the cylinder and is different for different materials. Therefore, if equilibrium takes place, then the condition is fulfilled: h<=d.(6.14). d называется коэффициентом трения качения; она имеет размерность длины. Условие (6.14) можно также записать в виде М т <=dN, или, учитывая (6.12), S<=(d/r)N.(6.15). Очевидно, что максимальный момент трения качения M T max =dN пропорционален силе нормального давления.

Center of Parallel Forces

The conditions for reducing the system of parallel forces to the resultant are reduced to one inequality F ≠ 0. What happens to the resultant R when the lines of action of these parallel forces are simultaneously rotated by the same angle, if the points of application of these forces remain unchanged and the rotations of the lines of action of the forces occur around the parallel axes. Under these conditions, the resultant of a given system of forces is also simultaneously rotated by the same angle, and the rotation occurs around some fixed point, which is called the center of parallel forces. Let us proceed to the proof of this statement. Suppose that for the considered system of parallel forces F 1, F 2, ..., F n, the principal vector is not equal to zero, therefore, this system of forces is reduced to the resultant. Let point О 1 be any point of the line of action of this resultant. Let now r is the radius vector of the point 0 1 relative to the selected pole O, and r k is the radius vector of the point of application of the force F k (Fig. 8.1). According to Varignon's theorem, the sum of the moments of all forces of the system with respect to the point 0 1 is equal to zero: å (r k –r) xF k = 0, that is, år k xF k –årxF k = år k xF k –råF k = 0. We introduce a unit vector e, then any force F k can be represented in the form F k = F * ke (where F * k = F h, if the directions of the force F h and the vector e coincide, and F * k = –F h, if F k and e are directed opposite to each other); åF k = eåF * k. We get: år k xF * k e – rxeåF * k = 0, whence [år k F * k –råF * k] xe = 0. The last equality is satisfied for any direction of forces (that is, the direction of the unit vector e) only under the condition that the first factor is zero: år k F * k –råF * k = 0. This ravine has a unique solution with respect to the radius vector r, which determines the point of application of the resultant, which does not change its position when the lines of action of the forces are rotated. This point is the center of parallel forces. Denoting the radius vector of the center of parallel forces through rc: rc = (år k F * k) / (åF * k) = (r 1 F * 1 + r 2 F * 2 + ... + rn F * n) / (F * 1 + F * 2 + ... + F * n). Let x c, y c, z c - coordinates of the center of parallel forces, a x k, y k, z k - coordinates of the point of application of an arbitrary force F k; then the coordinates of the center of parallel forces can be found from the formulas:

xc = (xk F * k) / (F * k) = (x 1 F * 1 + x 2 F * 2 +… + xn F * n) / (F * 1 + F * 2 +… + F * n ), yc = (yk F * k) / (F * k) =

= (y 1 F * 1 + y 2 F * 2 +… + y n F * n) / (F * 1 + F * 2 +… + F * n), z c =

= (z k F * k) / (åF * k) = (z 1 F * 1 + z 2 F * 2 +… + z n F * n) / (F * 1 + F * 2 +… + F * n)

Expressions x k F * k, y k F * k, z k F * k are called static moments of a given system of forces, respectively, relative to the coordinate planes yOz, xOz, xOy. If the origin of coordinates is chosen at the center of parallel forces, then x c = y c = z c = 0, and the static moments of the given system of forces are equal to zero.

The center of gravity

A body of arbitrary shape, located in a gravity field, can be divided by sections parallel to the coordinate planes into elementary volumes (Fig. 8.2). If we neglect the size of the body in comparison with the radius of the Earth, then the forces of gravity acting on each elementary volume can be considered parallel to each other. We denote by DV k the volume of an elementary parallelepiped centered at the point M k (see Fig. 8.2), and the force of gravity acting on this element, by DP k. Then the average specific gravity of a volume element is called the ratio DP k / DV k. By contracting the parallelepiped to the point М k, we obtain the specific gravity at a given point of the body as the limit of the average specific gravity g (x k, y k, z k) = lim DVk®0 (8.10). Thus, the specific gravity is a function of coordinates, i.e. g = g (x, y, z). We will assume that along with the geometric characteristics of the body, the specific gravity at each point of the body is also given. Let's return to the division of the body into elementary volumes. If we exclude the volumes of those elements that border on the surface of the body, then you can get a stepped body, consisting of a set of parallelepipeds. We apply to the center of each parallelepiped the force of gravity DP k = g k DV k, where g h is the specific gravity at the point of the body coinciding with the center of the parallelepiped. For a system of n parallel gravity forces formed in this way, one can find the center of parallel forces r (n) = (år k DP k) / (åDP k) = (r 1 DP 1 + r 2 DP 2 +… + rn DP n) / (DP 1 + DP 2 +… + DP n). This formula determines the position of some point C n. The center of gravity is a point that is the limiting point for the points C n at п®µ.