Rotary motion is a type of mechanical motion. During the rotational motion of an absolutely rigid body, its points describe circles located in parallel planes. The centers of all circles lie in this case on one straight line, perpendicular to the planes of the circles and called the axis of rotation. The axis of rotation can be located inside the body and outside it. The axis of rotation in a given reference system can be either movable or fixed. For example, in the reference frame associated with the Earth, the axis of rotation of the generator rotor at the power plant is fixed.

Kinetic characteristics:

The rotation of a rigid body as a whole is characterized by an angle, measured in angular degrees or radians, angular velocity (measured in rad / s) and angular acceleration (unit - rad / s²).

With uniform rotation (T revolutions per second):

Frequency of rotation - the number of revolutions of the body per unit time.-

The period of rotation is the time of one complete revolution. The rotation period T and its frequency are related by the relation.

Linear speed of a point located at a distance R from the axis of rotation

Angular velocity of body rotation

Moment of force (synonyms: torque, torque, torque, torque) is a vector physical quantity equal to the vector product of the radius vector (drawn from the axis of rotation to the point of application of the force - by definition) by the vector of this force. Characterizes the rotational action of force on a rigid body.

The moment of force is measured in Newton meters. 1 Nm - the moment of force that produces a force of 1 N on a lever 1 m long. The force is applied to the end of the lever and is directed perpendicular to it.

The angular momentum (kinetic momentum, angular momentum, orbital momentum, angular momentum) characterizes the amount of rotational motion. A quantity that depends on how much mass is rotating, how it is distributed about the axis of rotation, and how fast the rotation occurs. The angular momentum of a closed system is conserved

The law of conservation of angular momentum (the law of conservation of angular momentum) is one of the fundamental laws of conservation. It is expressed mathematically in terms of the vector sum of all angular momenta about the chosen axis for a closed system of bodies and remains constant until external forces act on the system. In accordance with this, the angular momentum of a closed system in any coordinate system does not change with time.

The law of conservation of angular momentum is a manifestation of the isotropy of space with respect to rotation.

16. Equation of the dynamics of rotational motion. Moment of inertia.

The basic equation of the dynamics of the rotational motion of a material point is the angular acceleration of a point during its rotation around a fixed axis, which is proportional to the torque and inversely proportional to the moment of inertia.

M = E*J or E = M/J

Comparing the obtained expression with Newton's second law with the translational law, we see that the moment of inertia J is a measure of the body's inertia in rotational motion. Like mass, the quantity is additive.

Moment of inertia - scalar (in general case- tensor) physical quantity, a measure of inertia in rotational motion around an axis, just as the mass of a body is a measure of its inertia in forward movement. It is characterized by the distribution of masses in the body: the moment of inertia is equal to the sum of the products of elementary masses and the square of their distances to the base set (point, line or plane).

SI unit: kg m². Designation: I or J.

There are several moments of inertia - depending on the manifold, from which the distance of points is measured.

Moment of inertia properties:

1. The moment of inertia of the system is equal to the sum of the moment of inertia of its parts.

2. The moment of inertia of a body is a quantity immanently inherent in this body.

The moment of inertia of a rigid body is a veline that characterizes the distribution of mass in the body and is a measure of the inertia of the body during rotational motion.

Moment of inertia formula:

Steiner's theorem:

The moment of inertia of the body about any axis is equal to the moment of inertia about a parallel axis passing through the center of inertia, added to the value m*(R*R), where R is the distance between the axes.

moment of inertia mechanical system relative to a fixed axis (“axial moment of inertia”) is the value Ja, equal to the sum of the products of the masses of all n material points of the system and the squares of their distances to the axis:

The axial moment of inertia of the body Ja is a measure of the inertia of the body in rotational motion around the axis, just as the mass of the body is a measure of its inertia in translational motion.

The central moment of inertia (or the moment of inertia about point O) is the quantity

.

Moment of force about the axis is the moment of the projection of a force onto a plane perpendicular to the axis, relative to the point of intersection of the axis with this plane

The moment about an axis is positive if the force tends to rotate a plane perpendicular to the axis counterclockwise when viewed towards the axis.

The moment of force about the axis is 0 in two cases:

    If the force is parallel to the axis

    If the force crosses the axis

If the line of action and the axis lie in the same plane, then the moment of force about the axis is 0.

27. The relationship between the moment of force about an axis and the vector moment of force about a point.

Mz(F)=Mo(F)*cosαThe moment of force, relative to the axis, is equal to the projection of the vector of the moment of forces, relative to the point of the axis, on this axis.

28. The main theorem of statics about bringing the system of forces to a given center (Poinsot's theorem). Principal vector and principal moment of the system of forces.

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 (center of reduction) 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 referral center.

The main vector of the force system called vector R equal to the vector sum of these forces:

R = F 1 + F 2 + ... + F n= F i .

For a flat system of forces, its main vector lies in the plane of action of these forces.

The main moment of the system of forces about the center O is called the vector L O , equal to the sum of the vector moments of these forces relative to the point O:

L O= M O( F 1) + M O( F 2) + ... + M O( F n) = M O( F i).

Vector R does not depend on the choice of the center O, and the vector L O when changing the position of the center O can generally change.

Poinsot's theorem: An arbitrary spatial system of forces can be replaced by one force with the main vector of the system of forces and a pair of forces with the main moment without disturbing the state of the rigid body. The main vector is the geometric sum of all forces acting on a rigid body and is located in the plane of action of the forces. The main vector is considered through its projections on the coordinate axes.

To bring forces to a given center applied at some point of a rigid body, it is necessary: ​​1) to transfer the force to itself in parallel to a given center without changing the force modulus; 2) in a given center, apply a pair of forces, the vector moment of which is equal to the vector moment of the transferred force of the relative new center, this pair is called an attached pair.

Dependence of the main moment on the choice of the center of reduction. The principal moment with respect to the new reduction center is equal to the geometric sum of the principal moment with respect to the old reduction center and the cross product of the radius-vector connecting new center cast with the old, on the main vector.

29 Special cases of reducing the spatial system of forces

Values ​​of principal vector and principal moment

Cast result

Force system is reduced to a pair of forces, the moment of which is equal to the main moment (the main moment of the system of forces does not depend on the choice of the center of reduction O).

The system of forces is reduced to a resultant equal to passing through the center O.

The system of forces is reduced to a resultant equal to the main vector and parallel to it and separated from it at a distance. The position of the line of action of the resultant must be such that the direction of its moment relative to the center of reduction O coincides with the direction relative to the center O.

, and the vectors are not perpendicular

The system of forces is reduced to a dynamo (power screw) - a combination of a force and a pair of forces lying in a plane perpendicular to this force.

The system of forces applied to a rigid body is balanced.

30. Reduction to dynamism. In mechanics, a dynamo is such a set of forces and a pair of forces () acting on a rigid body, in which the force is perpendicular to the plane of action of the pair of forces. Using the vector moment of a couple of forces, one can also define a dynamo as a combination of a force and a couple whose force is parallel to the vector moment of a couple of forces.

Central helical axis equation Suppose that in the center of reduction, taken as the origin of coordinates, the main vector with projections on the coordinate axes and the main moment with projections are obtained. When the system of forces is reduced to the center of reduction O 1 (Fig. 30), a dynamo is obtained with the main vector and the main moment , Vectors and as forming a linam. are parallel and therefore can differ only by a scalar factor k 0. We have, since .The principal moments and , satisfy the relation

When solving problems of moving objects, in some cases their spatial dimensions are neglected, introducing the concept of a material point. For another type of problem, in which bodies at rest or rotating bodies are considered, it is important to know their parameters and application points. external forces. In this case we are talking about the moment of forces about the axis of rotation. Let's consider this issue in the article.

The concept of the moment of force

Before bringing about a fixed axis of rotation, it is necessary to clarify what phenomenon will be discussed. Below is a figure that shows a wrench of length d, a force F is applied to its end. It is easy to imagine that the result of its action will be the rotation of the wrench counterclockwise and unscrewing the nut.

According to the definition, the moment of force about the axis of rotation is the product of the shoulder (d in this case) and the force (F), that is, the following expression can be written: M = d * F. It should immediately be noted that the above formula is written in scalar form, that is, it allows you to calculate the absolute value of the moment M. As can be seen from the formula, the unit of measurement of the considered quantity is newtons per meter (N * m).

- vector quantity

As discussed above, the moment M is actually a vector. To clarify this statement, consider another figure.

Here we see a lever of length L, which is fixed on the axis (shown by the arrow). A force F is applied to its end at an angle Φ. It is not difficult to imagine that this force will cause the lever to rise. The formula for the moment in vector form in this case will be written as follows: M¯ = L¯*F¯, here the line over the symbol means that the quantity in question is a vector. It should be clarified that L¯ is directed from to the point of application of the force F¯.

The above expression is a vector product. Its resulting vector (M¯) will be perpendicular to the plane formed by L¯ and F¯. To determine the direction of the moment M¯, there are several rules ( right hand, gimlet). In order not to memorize them and not get confused in the order of multiplication of the vectors L¯ and F¯ (the direction of M¯ depends on it), you should remember one simple thing: the moment of force will be directed in such a way that if you look from the end of its vector, then the acting force F ¯ will rotate the lever counterclockwise. This direction of the moment is conditionally taken as positive. If the system rotates clockwise, then the resulting moment of forces has a negative value.

Thus, in the considered case with the lever L, the value of M¯ is directed upwards (from the figure to the reader).

In scalar form, the formula for the moment is written as: M = L*F*sin(180-Φ) or M = L*F*sin(Φ) (sin(180-Φ) = sin(Φ)). According to the definition of the sine, we can write the equality: M = d*F, where d = L*sin(Φ) (see the figure and the corresponding right triangle). The last formula is similar to the one given in the previous paragraph.

The above calculations demonstrate how to work with vector and scalar quantities of moments of forces in order to avoid errors.

The physical meaning of M¯

Since the two cases considered in the previous paragraphs are associated with rotational motion, one can guess what meaning the moment of force carries. If the force acting on a material point is a measure of the increase in the speed of the linear displacement of the latter, then the moment of force is a measure of its rotational ability in relation to the system under consideration.

Let's bring good example. Any person opens the door by holding its handle. It can also be done by pushing the door in the area of ​​the handle. Why doesn't anyone open it by pushing in the hinge area? Very simple: the closer the force is applied to the hinges, the more difficult it is to open the door, and vice versa. The derivation of the previous sentence follows from the formula for the moment (M = d*F), which shows that when M = const, the quantities d and F are inversely related.

Moment of force - additive quantity

In all the cases considered above, there was only one acting force. When solving real problems, the situation is much more complicated. Usually systems that rotate or are in equilibrium are subject to several torsional forces, each of which creates its own moment. In this case, the solution of problems is reduced to finding the total moment of forces relative to the axis of rotation.

The total moment is found by the usual sum of the individual moments for each force, however, remember to use the correct sign for each of them.

Problem solution example

To consolidate the acquired knowledge, it is proposed to solve the following problem: it is necessary to calculate the total moment of force for the system shown in the figure below.

We see that three forces (F1, F2, F3) act on a lever 7 m long, and they have different points applications relative to the axis of rotation. Since the direction of forces is perpendicular to the lever, there is no need to use a vector expression for the moment of torsion. It is possible to calculate the total moment M using a scalar formula and remembering to set the desired sign. Since the forces F1 and F3 tend to turn the lever counterclockwise, and F2 - clockwise, the moment of rotation for the first will be positive, and for the second - negative. We have: M \u003d F1 * 7-F2 * 5 + F3 * 3 \u003d 140-50 + 75 \u003d 165 N * m. That is, the total moment is positive and directed upwards (at the reader).

Most best definition Torque is the tendency of a force to rotate an object around an axis, fulcrum, or pivot point. Torque can be calculated using force and moment arm (perpendicular distance from the axis to the line of action of the force), or using moment of inertia and angular acceleration.

Steps

Using force and leverage

  1. Determine the forces acting on the body and the corresponding moments. If the force is not perpendicular to the moment arm under consideration (i.e. it acts at an angle), then you may need to find its components using trigonometric functions such as sine or cosine.

    • The force component considered will depend on the perpendicular force equivalent.
    • Imagine a horizontal rod, to which a force of 10 N must be applied at an angle of 30° above the horizontal plane in order to rotate it around the center.
    • Since you need to use a force that is not perpendicular to the moment arm, you need the vertical component of the force to rotate the rod.
    • Therefore, the y-component must be considered, or F = 10sin30°N should be used.

  2. Use the moment equation, τ = Fr, and simply replace the variables with the given or received data.

    • A simple example: Imagine a 30 kg child sitting on one end of a seesaw. The length of one side of the swing is 1.5 m.
    • Because the pivot of the swing is in the center, you don't need to multiply the length.
    • You need to determine the force exerted by the child using mass and acceleration.
    • Since the mass is given, you need to multiply it by the gravitational acceleration, g, which is 9.81 m/s 2 . Hence:
    • Now you have all the necessary data to use the moment equation:

  3. Use the signs (plus or minus) to show the direction of the moment. If the force rotates the body clockwise, then the moment is negative. If the force rotates the body counterclockwise, then the moment is positive.

    • In the case of multiple applied forces, simply add up all the moments in the body.
    • Because each force tends to cause a different direction of rotation, it is important to use the rotation sign to keep track of the direction of each force.
    • For example, two forces were applied to the rim of a wheel having a diameter of 0.050 m, F 1 = 10.0 N, directed clockwise, and F 2 = 9.0 N, directed counterclockwise.
    • Since the given body is a circle, the fixed axis is its center. You need to divide the diameter to get the radius. The size of the radius will serve as the shoulder of the moment. Therefore, the radius is 0.025 m.
    • For clarity, we can solve separate equations for each of the moments arising from the corresponding force.
    • For force 1, the action is directed clockwise, therefore, the moment it creates is negative:
    • For force 2, the action is directed counterclockwise, therefore, the moment it creates is positive:
    • Now we can add up all the moments to get the resulting torque:

    Using moment of inertia and angular acceleration

    1. To begin solving the problem, understand how the moment of inertia of a body works. The moment of inertia of a body is the body's resistance to rotational motion. The moment of inertia depends on both the mass and the nature of its distribution.

      • To understand this clearly, imagine two cylinders of the same diameter but different masses.
      • Imagine that you need to rotate both cylinders around their central axis.
      • Obviously, a cylinder with more mass will be harder to turn than another cylinder because it is "heavier".
      • Now imagine two cylinders of different diameters but the same mass. To look cylindrical and have different masses, but at the same time have different diameters, the shape, or mass distribution, of both cylinders must be different.
      • A larger diameter cylinder will look like a flat, rounded plate, while a smaller one will look like a solid tube of fabric.
      • A cylinder with a larger diameter will be harder to turn because you need to apply more force to overcome the longer moment arm.

    2. Select the equation you will use to calculate the moment of inertia. There are several equations that can be used for this.

      • The first equation is the simplest: the summation of the masses and moment arms of all particles.
      • This equation is used for material points, or particles. An ideal particle is a body that has mass but does not occupy space.
      • In other words, the only significant characteristic of this body is the mass; you don't need to know its size, shape, or structure.
      • The idea of ​​a material particle is widely used in physics to simplify calculations and use ideal and theoretical schemes.
      • Now imagine an object like a hollow cylinder or a solid uniform sphere. These objects have a clear and defined shape, size and structure.
      • Therefore, you cannot consider them as a material point.
      • Fortunately, formulas that apply to some common objects can be used:

    3. Find the moment of inertia. To start calculating the torque, you need to find the moment of inertia. Use the following example as a guide:

      • Two small “weights” weighing 5.0 kg and 7.0 kg are mounted at a distance of 4.0 m from each other on a light rod (whose mass can be neglected). The axis of rotation is in the middle of the rod. The rod spins up from rest to an angular velocity of 30.0 rad/s in 3.00 s. Calculate the generated torque.
      • Since the axis of rotation is in the middle of the rod, the moment arm of both weights is equal to half of its length, i.e. 2.0 m
      • Since the shape, size and structure of the “weights” is not specified, we can assume that the weights are material particles.
      • The moment of inertia can be calculated as follows:

    4. Find the angular acceleration, α. To calculate the angular acceleration, you can use the formula α= at/r.

      • The first formula, α= at/r, can be used if the tangential acceleration and radius are given.
      • Tangential acceleration is an acceleration directed tangentially to the direction of motion.
      • Imagine an object moving along a curved path. The tangential acceleration is simply its linear acceleration at any point along the way.
      • In the case of the second formula, it is easiest to illustrate it by relating it to concepts from kinematics: displacement, linear velocity and linear acceleration.
      • Displacement is the distance traveled by an object (SI unit - meters, m); linear speed is a measure of the change in displacement per unit of time (SI unit - m / s); linear acceleration is an indicator of the change in linear speed per unit of time (SI unit - m / s 2).
      • Now let's look at the analogues of these quantities during rotational motion: angular displacement, θ - the angle of rotation of a certain point or segment (SI unit - rad); angular velocity, ω - change in angular displacement per unit of time (SI unit - rad/s); and angular acceleration, α - change in angular velocity per unit time (SI unit - rad / s 2).
      • Returning to our example, we were given data for angular momentum and time. Since the rotation started from rest, the initial angular velocity is 0. We can use the equation to find:

    5. Use the equation, τ = Iα, to find the torque. Just replace the variables with the answers from the previous steps.

      • You may notice that the unit "rad" doesn't fit in with our units of measurement, as it's considered a dimensionless quantity.
      • This means that you can ignore it and continue with your calculations.
      • For unit analysis, we can express angular acceleration in s -2 .
    • In the first method, if the body is a circle and its axis of rotation is in the center, then it is not necessary to calculate the components of the force (provided that the force is not applied obliquely), since the force lies on the tangent to the circle, i.e. perpendicular to the moment arm.
    • If you find it difficult to imagine how the rotation occurs, then take a pen and try to recreate the problem. For a more accurate reproduction, do not forget to copy the position of the axis of rotation and the direction of the applied force.

Moment of a pair of forces

The moment of force relative to some point (center) is a vector numerically equal to the product of the force modulus and the arm, i.e. the shortest distance from the specified point to the line of action of the force, and directed perpendicular to the plane passing through the chosen point and the line of action of the force in the direction from which the "rotation" performed by the force around the point appears to be counterclockwise. The moment of force characterizes its rotational action.

If ABOUT- the point relative to which the moment of force is located F, then the moment of force is denoted by the symbol M o (F). Let us show that if the point of application of the force F determined by the radius vector r, then the relation

M o (F)=r×F. (3.6)

According to this ratio the moment of force is equal to the vector product of the vector r to the vector F.

Indeed, the modulus of the cross product is

M o ( F)=RF sin= Fh, (3.7)

Where h- arm of strength. Note also that the vector M o (F) directed perpendicular to the plane passing through the vectors r And F, in the direction from which the shortest turn of the vector r to the direction of the vector F appears to be counter-clockwise. Thus, formula (3.6) completely determines the modulus and direction of the moment of force F.

Sometimes it is useful to write formula (3.7) in the form

M o ( F)=2S, (3.8)

Where S- area of ​​a triangle OAB.

Let x, y, z are the coordinates of the force application point, and Fx, Fy, Fz are the force projections on the coordinate axes. Then if the point ABOUT located at the origin, the moment of force is expressed as follows:

It follows that the projections of the moment of force on the coordinate axes are determined by the formulas:

M Ox(F)=yF z -zF y,

M Oy(F)=zF x -xF z ,

M Oy(F)=xF y -yF x. (3.10)

Let us now introduce the concept of the projection of a force onto a plane.

May strength be given F and some plane. Let us drop perpendiculars to this plane from the beginning and end of the force vector.

The projection of force on a plane called vector , the beginning and end of which coincide with the projection of the beginning and the projection of the end of the force on this plane.

If we take the plane as the considered plane hoy, then the projection of the force F on this plane there will be a vector Fhu.



Moment of power Fhu relative to the point ABOUT(points of intersection of the axis z with plane hoy) can be calculated by formula (3.9) if we take z=0, Fz=0. Get

MO(Fhu)=(xF y -yF x)k.

Thus, the moment is directed along the axis z, and its projection onto the axis z exactly coincides with the projection onto the same axis of the moment of force F relative to the point ABOUT. In other words,

M Oz(F)=M Oz(Fhu)= xF y -yF x. (3.11)

Obviously, the same result can be obtained by projecting the force F to any other plane parallel to hoy. In this case, the point of intersection of the axis z with the plane will be different (we denote new point crossing through ABOUT 1). However, all the quantities on the right-hand side of equality (3.11) X, at, F x, F remain unchanged, and therefore we can write

M Oz(F)=M O 1 z ( Fhu).

In other words, the projection of the moment of force about a point on the axis passing through this point does not depend on the choice of a point on the axis . Therefore, in what follows, instead of the symbol M Oz(F) we will use the symbol Mz(F). This moment projection is called moment of force about the axis z. The calculation of the moment of a force about an axis is often more conveniently done by force projection. F onto a plane perpendicular to the axis, and calculating the quantity Mz(Fhu).

In accordance with formula (3.7) and taking into account the sign of the projection, we obtain:

Mz(F)=Mz(Fhu)=± F xy h*. (3.12)

Here h*- arm of strength Fhu relative to the point ABOUT. If the observer sees from the side of the positive direction of the z-axis, that the force Fhu tends to rotate the body around an axis z counterclockwise, then the "+" sign is taken, and otherwise - the "-" sign.

Formula (3.12) makes it possible to formulate the following rule for calculating the moment of force about the axis. For this you need:

select an arbitrary point on the axis and construct a plane perpendicular to the axis;

project a force onto this plane;

Determine the projection arm of the force h*.

The moment of force about the axis is equal to the product of the module of the force projection on its shoulder, taken with the appropriate sign (see the above rule).

From formula (3.12) it follows that the moment of force about the axis is zero in two cases:

· when the projection of the force on a plane perpendicular to the axis is equal to zero, i.e. when force and axis are parallel ;

when shoulder projection h* equals zero, i.e. when the line of action crosses the axis .

Both of these cases can be combined into one: 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 .

Task 3.1. Calculate relative to a point ABOUT moment of power F applied to the point A and a diagonally directed face of a cube with side A.

When solving such problems, it is advisable to first calculate the moments of force F relative to the coordinate axes x, y, z. Point coordinates A application of force F will

Force projections F on the coordinate axes:

Substituting these values ​​into equalities (3.10), we find

The same expressions for the moments of force F relative to the coordinate axes can be obtained using formula (3.12). To do this, we design a force F on a plane perpendicular to the axis X And at. It's obvious that . Applying the above rule, we get, as expected, the same expressions:

, , .

The modulus of the moment is determined by the equality

.

Let us now introduce the concept of the moment of a pair. Let us first find what the sum of the moments of the forces that make up the pair is, relative to an arbitrary point. Let ABOUT is an arbitrary point in space, and F And F"- forces that make up a couple.

Then M o (F)= OA × F, M o (F") = OV × F",

M o (F) + M o (F ") = OA × F+ OV × F",

but since F= -F", That

M o (F) + M o (F ") = OA × F- OV × F=(OA-OV)× F.

Taking into account the equality OA-OV=VA , we finally find:

M o (F) + M o (F ") = VA × F.

Hence, the sum of the moments of the forces that make up the pair does not depend on the position of the point relative to which the moments are taken .

vector product VA × F and called pair moment . The moment of the pair is denoted by the symbol M(F, F"), and

M(F, F")=VA × F= AB × F",

or, in short,

M=VA × F= AB × F". (3.13)

Considering the right side of this equality, we notice that the moment of a pair is a vector perpendicular to the plane of the pair, equal in absolute value to the product of the modulus of one of the forces of the pair and the arm of the pair (i.e., 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 seen to occur anti-clockwise . If h is the shoulder of the pair, then M(F, F")=h×F.

It can be seen from the definition itself that the moment of a pair of forces is a free vector, the line of action of which is not defined (additional justification for this remark follows from Theorems 2 and 3 of this chapter).

In order for a pair of forces to form a balanced system (a system of forces equivalent to zero), it is necessary and sufficient that the moment of the pair be equal to zero. Indeed, if the moment of the pair is zero, M=h×F, then either F=0, i.e. no strength, or the shoulder of a couple h equals zero. But in this case, the forces of the couple will act in one straight line; since they are equal in absolute value and directed to opposite sides, then on the basis of axiom 1 they will constitute a balanced system. Conversely, if two forces F1 And F2, which make up a pair, are balanced, then, based on the same axiom 1, they act along one straight line. But in this case, the leverage of the pair h equals zero and therefore M=h×F=0.

Pair theorems

Let us prove three theorems by which equivalent transformations of pairs become possible. In all considerations, it should be remembered that they refer to pairs acting on any one solid body.

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 the given two pairs.

To prove this theorem, consider two pairs ( F1,F" 1) And ( F2,F" 2) and transfer the points of application of all forces along the lines of their action to the points A And IN respectively. Adding the forces according to axiom 3, we get

R=F1+F2 And R"=F" 1+F" 2,

But F1=-F" 1 And F2=-F" 2.

Hence, R=-R", i.e. strength R And R" form a couple. Let's find the moment of this pair using formula (3.13):

M=M(R, R")=VA× R= VA× (F1+F2)=VA× F1+VA× F2. (3.14)

When the forces that make up the pair are transferred along the lines of their action, neither the arm nor the direction of rotation of the pairs change, therefore, the moment of the pair does not change either. Means,

VA × F 1 \u003d M(F1,F" 1)=M 1, VA× F 2 \u003d M(F2,F" 2)=M 2

and formula (3.14) takes the form

M \u003d M 1 + M 2, (3.15)

which proves the validity of the above theorem.

Let us make two remarks on this theorem.

1. The lines of action of the forces that make up the pairs may turn out to be parallel. The theorem remains valid in this case as well, but to prove it, one should use the rule of addition of parallel forces.

2. After addition, it may turn out that M(R, R")=0; Based on the remark made earlier, this implies that the set of two pairs ( F1,F" 1, F2,F" 2)=0.

Theorem 2. Two pairs having geometrically equal moments are equivalent.

Let on the body in the plane I a couple ( F1,F" 1) with moment M 1. Let us show that this pair can be replaced by another one with the pair ( F2,F" 2) located in the plane II, if only its moment M 2 equals M 1(according to the definition (see 1.1) this will mean that the pairs ( F1,F" 1) And ( F2,F" 2) are equivalent). First of all, we note that the planes I And II must be parallel, in particular they may coincide. Indeed, from the parallelism of the moments M 1 And M 2(in our case M 1=M 2) it follows that the planes of action of the pairs, perpendicular to the moments, are also parallel.

We introduce into consideration a new pair (F3,F" 3) and apply it together with the pair ( F2,F" 2) to the body, placing both pairs in the plane II. To do this, according to Axiom 2, we need to choose a pair ( F3,F" 3) with moment M 3 so that the applied system of forces ( F2,F" 2, F3,F" 3) was balanced. This can be done, for example, as follows: we set F3=-F" 1 And F" 3 =-F1 and let us combine the points of application of these forces with the projections A 1 and IN 1 points A And IN to the plane II. According to the construction, we will have: M 3 \u003d -M 1 or considering that M 1 = M 2,

M 2 + M 3 = 0.

Taking into account the second remark to the previous theorem, we obtain ( F2,F" 2, F3,F" 3)=0. So the pairs ( F2,F" 2) And ( F3,F" 3) are mutually balanced and their attachment to the body does not violate its state (axiom 2), so that

(F1,F" 1)= (F1,F" 1, F2,F" 2, F3,F" 3). (3.16)

On the other hand, forces F1 And F3, and F" 1 And F" 3 can be added according to the rule of addition of parallel forces directed in one direction. Modulo, all these forces are equal to each other, so their resultant R And R" must be applied at the intersection point 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,

(F1,F" 1, F3,F" 3)=(R, R")=0.

Now we can write

(F1,F" 1, F2,F" 2, F3,F" 3)=(F3,F" 3). (3.17)

Comparing relations (3.16) and (3.17), we obtain ( F1,F" 1)=(F2,F" 2), which was to be proved.

It follows from this theorem that a pair of forces can be moved in the plane of its action, transferred to a parallel plane; finally, in a pair, you can change the forces and the shoulder at the same time, retaining only the direction of rotation of the pair and the modulus of its momentum ( F 1 h 1 =F 2 h 2).

In what follows, we will make extensive use of such equivalent transformations of a pair.

Theorem 3. Two pairs lying in intersecting planes are equivalent to one pair whose moment is equal to the sum of the moments of the two given pairs.

Let couples ( F1,F" 1) And ( F2,F" 2) are located in intersecting planes I And II respectively. Using the corollary of Theorem 2, we reduce both pairs to the shoulder AB located on the line of intersection of the planes I And II. Denote the transformed pairs by ( Q1,Q" 1) And ( Q2,Q" 2). In this case, the equalities

M 1 =M(Q1,Q" 1)=M(F1,F" 1) And M 2 =M(Q2,Q" 2)=M(F2,F" 2).

Let us add according to axiom 3 the forces applied at the points A And IN respectively. Then we get R \u003d Q 1 + Q 2 And R"= Q" 1 +Q" 2. Given that Q" 1 \u003d -Q 1 And Q" 2 \u003d -Q 2, we get R=-R". Thus, we have proved that the system of two pairs is equivalent to one pair ( R,R").

Let's find a moment M this couple. Based on formula (3.13), we have

M(R,R")=VA× (Q1+Q2)=VA× Q1+ VA× Q2=

=M(Q1,Q" 1)+M(Q2,Q" 2)=M(F1,F" 1)+M(F2,F" 2)

M \u003d M 1 + M 2,

those. the theorem is proven.

Note that the result obtained is also valid for pairs lying in parallel planes. By Theorem 2, such pairs can be reduced to a single plane, and by Theorem 1, they can be replaced by a single pair whose moment is equal to the sum of the moments of the component pairs.

The pair theorems proved above lead to an important conclusion: the moment of the pair is a free vector and completely determines the action of the pair on an absolutely rigid body . Indeed, we have already proved that if two pairs have the same moments (and therefore lie in the same plane or in parallel planes), then they are equivalent to each other (Theorem 2). On the other hand, two pairs lying in intersecting planes cannot be equivalent, because this would mean that one of them and the pair opposite to the other are equivalent to zero, which is impossible, since the sum of the moments of such pairs is different from zero.

Thus, the introduced concept of the moment of a couple is extremely useful, since it fully reflects the mechanical action of a couple on a body. In this sense, we can say that the moment exhaustively represents the action of a pair on a rigid body.

For deformable bodies, the above theory of pairs is not applicable. Two opposite pairs, acting, for example, on the ends of the rod, are equivalent to zero from the point of view of the statics of a rigid body. Meanwhile, their action on the deformable rod causes its torsion, and the more, the greater the modules of the moments.

Let's move on to solving the first and second problems of statics, when only pairs of forces act on the body.