The work of gravity depends only on the change in height and is equal to the product of the gravity modulus and the vertical displacement of the point (Fig. 15.6):

Where Δh- height change. When lowering, the work is positive, when rising, it is negative.

Work done by resultant force

Under the influence of a system of forces, a point with mass T moves from position M 1 to position M 2(Fig. 15.7).

In the case of motion under the action of a system of forces, the resultant work theorem is used.

The work of the resultant on a certain displacement is equal to the algebraic sum of the work of the system of forces on the same displacement.

Examples of problem solving

Example 1. A body weighing 200 kg is lifted along an inclined plane (Fig. 15.8).

Determine the work done when moving 10 m at a constant speed. Coefficient of friction between a body and a plane f = 0,15.

Solution

1. With a uniform rise, the driving force is equal to the sum of the forces of resistance to movement. We plot the forces acting on the body on the diagram:

1. We use the resultant work theorem:

1. We substitute the input quantities and determine the lifting work:

Example 2. Determine the work done by gravity when moving a load from a point A exactly WITH along an inclined plane (Fig. 15.9). The force of gravity of the body is 1500 N. AB = 6 m, BC = 4 m.

Solution

1. The work of gravity depends only on changes in the height of the load. Change in height when moving from point A to C:

2. Work of gravity:

Example 3. Determine the work done by the cutting force in 3 minutes. The rotation speed of the part is 120 rpm, the diameter of the workpiece is 40 mm, the cutting force is 1 kN (Fig. 15.10).

Solution

1. Rotary work

where F res is the cutting force.

2. Angular speed 120 rpm.

3. The number of revolutions for a given time is z = 120 3 = 360 rpm.

Angle of rotation during this time

4. Work in 3 minutes Wp= 1 0.02 2261 = 45.2 kJ.

Example 4. Body mass m= 50 kg is moved along the floor using a horizontal force Q to a distance S= 6 m. Determine the work done by the friction force if the coefficient of friction between the surface of the body and the floor f= 0.3 (Fig. 1.63).

Solution

According to the Ammonton-Coulomb law, the friction force

The friction force is directed in the direction opposite to the movement, so the work done by this force is negative:

Example 5. Determine the tension of the belt drive branches (Fig. 1.65), if the power transmitted by the shaft is N=20 kW, shaft speed n = 150 rpm

Solution

The torque transmitted by the shaft is

Example 6. Wheel radius R= 0.3 m rolls without sliding on a horizontal rail (Fig. 1.66). Find the work of rolling friction when the center of the wheel moves a distance S= 30 m, if the vertical load on the wheel axle is P = 100 kN. The rolling friction coefficient of a wheel on a rail is equal to k= 0.005 cm.

Solution

Rolling friction occurs due to deformations of the wheel and rail in the area of ​​their contact. Normal reaction N moves forward in the direction of movement and forms a vertical pressure force R on the wheel axle a pair whose shoulder is equal to the rolling friction coefficient k, and the moment

This pair tends to turn the wheel in the direction opposite to its rotation. Therefore, the work of rolling friction will be negative and will be determined as the product of the constant friction moment and the angle of rotation of the wheel φ , i.e.

The path traveled by a wheel can be defined as the product of its rotation angle and radius

Entering a value φ into the work expression and substituting numerical values, we get

Test questions and assignments

1. What forces are called driving forces?

2. What forces are called resistance forces?

3. Write down formulas for determining work during translational and rotational motions.

4. What force is called circumferential? What is torque?

5. State the resultant work theorem.

Let's calculate the work done by gravity when a body moves along different trajectories.

Let us assume that a body of mass was raised to a height above the surface of the Earth. Let us determine the work that gravity will do in the case when this body freely falls vertically to the surface of the Earth.

In the direction of motion, a constant force will act on the body. Under the influence of this force, the body will travel a distance k. By definition, the work done by this force will be equal to

Let us give the body the opportunity to move under the influence of gravity along an inclined plane (Fig. 5.12) at an angle a to the horizon (there is no friction). A force will act on the body along the inclined plane. This force is constant throughout the movement. The distance traveled by a body along an inclined plane can be expressed in terms of the height k at which the body was initially located. From the triangle it is clear that

Knowing the force and the distance I traveled by the body under the influence of this force, we can calculate the work A performed by gravity during such movement:

This means that when moving along an inclined plane, the work done by gravity does not depend on the angle of inclination of the plane and is still equal to the product of gravity and the difference in heights at which the starting and ending points of the movement are located:

Let us now give the body the opportunity to descend from a height along some curvilinear trajectory (Fig. 5.13). Let's calculate the work done by gravity during this movement of the body.

As we know, any trajectory with the required accuracy can always be represented as a sequence of small linear movements. For example, a section can be represented by a straight line segment.

Each such section will be an inclined plane of short length. As has just been proven, the work of gravity in such a section does not depend on the angle of inclination and will be equal to the product of gravity and the difference in heights of points A and B:

This is true for all sections of the curved trajectory. Therefore, the total work done by gravity when moving along any arbitrary curved path will always be equal to the product of gravity and the difference in heights of the starting and ending points of movement.

This result is of extraordinary importance and can be expressed as follows: the work of gravity does not depend on the shape of the trajectory along which the body moves, and is determined only by the initial and final positions of the body.

The same result can be expressed in another, even more general way. Let us assume that a body of mass descended from point A to point C along a certain curvilinear trajectory (Fig. 5.14). Then it was raised from point C to point A along the trajectory. The force of gravity performed work during all these movements. On the section, it did some positive work, proportional to the difference in the heights of the points. On the section (when rising with the help of extraneous forces), gravity did negative work. The amount of this work is also proportional to the difference in the heights of the points

The independence of the work of gravity from the shape of the trajectory can now be formulated as follows: the work of gravity on any closed trajectory is always equal to zero.

This remarkable property of gravity makes it possible to significantly simplify the solution of problems associated with calculating the work of this force. Many other forces also have this property, for example, the forces of universal gravitation (a special case of which is the force of gravity), the forces of elasticity, the forces of the electric field created by stationary charges, etc.

All forces whose work on a closed trajectory is zero are called conservative forces. A remarkable property of such forces is that the work expended against them is then completely returned when the body is freed from the bonds that hold it.

Work of gravity. Problem solving

The purpose of the lesson: determine the formula for the work of gravity; determine that the work done by gravity does not depend on the trajectory of the body; develop practical problem solving skills.

During the classes.

1. Organizational moment. Greeting students, checking absentees, setting lesson goals.

2.Checking homework.

3. Studying new material. In the previous lesson, we defined a formula for determining work. What formula determines the work done by a constant force? (A=FScosα)

What is A andS?

Now let's apply this formula to gravity. But first, let's remember what the force of gravity is? (F= mg)

Consider case a) the body falls vertically down. As you and I know, gravity is always directed straight down. In order to determine the directionSyou need to remember the definition. (Displacement is a vector connecting the starting and ending points. It is directed from beginning to end)

That. for determining , Since the direction of movement and gravity coincide, thenα =0 and the work done by gravity is:

Consider case b) the body moves vertically upward. Because the direction of gravity and movement are opposite, thenα =0 and the work done by gravity is .

That. Thus, if you compare two formulas modulo, they will be the same.

Consider case c) the body moves along an inclined plane. The work of force is equal to the scalar product of the force vector and the vector of displacement of the body performed under the influence of a given force, that is, the work of gravity in this case will be equal to, Where – the angle between the vectors of gravity and displacement. The figure shows that the movement () represents the hypotenuse of a right triangle, and the altitudeh– leg. According to the property of a right triangle:

.Hence

That. what can be concluded?(that the work done by gravity does not depend on the trajectory of movement.)

Let's consider the last example, when the trajectory movement will be a closed line. Who will say what the work will be equal to and why? (A=0, because the displacement is 0)

Let's note!: the work done by gravity when a body moves along a closed trajectory is zero.

4. Fixing the material.

Task 1. A hunter shoots from a cliff at an angle of 40° to the horizon. During the time the bullet fell, the work done by gravity was 5 J. If the bullet entered the ground at a distance of 250 m from the rock, then what is its mass?

Task 2. While on Neptune, the body moved as shown in the figure. During this movement, the work done by gravity was 840 J. If the mass of this body is 5 kg, then what is the acceleration of gravity on Neptune?

5. Homework.

Work of gravity. Gravity R material point with mass T near the Earth's surface can be considered a constant equal to mg

directed vertically downwards.

Job A strength R on moving from a point M 0 to the point M

Where h = z 0 - z x - height of lowering the point.

The work done by gravity is equal to the product of this force by the height of descent (work is positive) or the height of rise (work is negative). The work done by gravity does not depend on the shape of the trajectory between points M 0 and M|, and if these points coincide, then the work of gravity is zero (the case of a closed path). It is also equal to zero if the points M 0 And M lie in the same horizontal plane.

Work of linear elastic force. The linear elastic force (or linear restoring force) is the force acting according to Hooke’s law (Fig. 63):

F = - Withr,

Where r- the distance from the point of static equilibrium, where the force is zero, to the point in question M; With- constant coefficient - stiffness coefficient.

A=--().

Using this formula, the work of the linear elastic force is calculated. If the point M 0 coincides with the point of static equilibrium ABOUT, so then r 0 =0 and for the work of force on displacement from a point ABOUT to the point M we have

Magnitude r- the shortest distance between the point in question and the point of static equilibrium. Let us denote it by λ and call it deformation. Then

The work of the linear elastic force on displacement from a state of static equilibrium is always negative and equal to half the product of the stiffness coefficient and the square of the deformation. The work of the linear elastic force does not depend on the shape of the movement and the work of any closed movement is zero. It is also equal to zero if the points Mo And M lie on the same sphere described from the point of static equilibrium.

Work of variable force during curvilinear motion.

Work of force on a curved section

Let us consider the general case of finding the work of a variable force, the point of application of which moves along a curvilinear trajectory. Let the point M of application of a variable force F move along an arbitrary continuous curve. Let us denote by the vector of infinitesimal displacement of point M. This vector is directed tangentially to the curve in the same direction as the velocity vector.

Elementary work of a variable force F on an infinitesimal displacement

ds is the scalar product of the vectors F and ds:

Where A- angle between vectors F and ds

That is, the elementary work of force is equal to the product of the magnitudes of the force vectors and infinitesimal displacement, multiplied by the cosine of the angle between these vectors.

Let us decompose the force vector F into two components: - directed tangentially to the trajectory - and - directed along the normal. Line of action of force

is perpendicular to the tangent to the trajectory along which the point moves, and its work is zero. Then:

dA= Ftds.

In order to calculate the work of a variable force F on the final section of the curve from A to b, you should calculate the integral of the elementary work:

Potential and kinetic energy.

Potential energy P mateerial point in considerationmy force field point M is called work, which forces carry outla acting on a material point when it moves from the pointMto the starting pointM 0 , i.e.

P = Umm 0

P = =-U=- U

The constant C 0 is the same for all points of the field, depending on which point of the field is chosen as the initial one. It is obvious that potential energy can only be introduced for a potential force field, in which the work does not depend on the form of movement between points M And M 0 . A non-potential force field has no potential energy, and there is no force function for it.

dA = dU= -dП; A = U - U 0 = P 0 - P

From the above formulas it follows that P is determined up to an arbitrary constant, which depends on the choice of the starting point, but this arbitrary constant does not affect the forces calculated through potential energy and the work of these forces. Considering this:

P= - U+ const or P =- U.

The potential energy at any point in the field, up to an arbitrary constant, can be determined as the value of the force function at the same point, taken with a minus sign.

Kinetic energy system is called a scalar quantity T, equal to the sum of the kinetic energies of all points of the system:

Kinetic energy is a characteristic of both translational and rotational motions of a system. Kinetic energy is a scalar quantity and, moreover, essentially positive. Therefore, it does not depend on the directions of movement of parts of the system and does not characterize changes in these directions.

Let us also note the following important circumstance. Internal forces act on parts of the system in mutually opposite directions. Changes in kinetic energy are affected by the action of both external and internal forces

Uniform motion of a point.

Uniform motion of a point- movement, with which it touches. acceleration ω t of a point (in the case of rectilinear motion, the total acceleration ω )constantly. The law of uniform motion of a point and the law of change in its speed υ with this movement are given by the equalities:

where s is the distance of the point measured along the arc of the trajectory from the reference point selected on the trajectory, t- time, s 0 - value of s at the beginning. moment of time t = = 0. - start. point speed. When the signs υ And ω identical, uniformly alternating motion. is accelerated, and when different - slowed down.

Upon admission in uniformly alternating motion of a rigid body, everything said applies to each point of the body; with uniform rotation around a fixed angular axis. the acceleration e of the body is constant, and the law of rotation and the law of change in angle. the velocities ω of the body are given by the equalities

where φ is the angle of rotation of the body, φ 0 is the value of φ at the beginning. moment of time t= 0, ω 0 - start. ang. body speed. When the signs of ω and ε coincide, the rotation is accelerated, and when they do not coincide, it is slowed down.

Work done by a constant force during linear motion.

Let us define the work for the case when the acting force is constant in magnitude and direction, and the point of its application moves along a straight path. Let's consider a material point C, to which a force is applied that is constant in value and direction (Fig. 134, a).

Over a certain period of time t, point C moved to position C1 along a straight path at a distance s.

The work W of a constant force during rectilinear motion of the point of its application is equal to the product of the force modulus F by the distance s and by the cosine of the angle between the direction of the force and the direction of movement, i.e.

The angle α between the direction of force and the direction of movement can vary from 0 to 180°. At α< 90° работа положительна, при α >90° is negative, at α = 90° the work is zero.

If a force makes an acute angle with the direction of movement, it is called a driving force; the work done by the force is always positive. If the angle between the directions of force and displacement is obtuse, the force resists the movement, does negative work and is called the drag force. Examples of resistance forces include cutting forces, friction, air resistance and others, which are always directed in the direction opposite to movement.

When α = 0°, i.e. when the direction of the force coincides with the direction of the velocity, then W = F s, since cos 0° = 1. The product F cos α is the projection of the force onto the direction of motion of the material point. Consequently, the work of a force can be defined as the product of the displacement s and the projection of the force on the direction of motion of the point.

33. Inertia forces of a rigid body

In classical mechanics, concepts of forces and their properties are based on Newton's laws and are inextricably linked with the concept of a mineral reference frame.

Indeed, a physical quantity called force is introduced into consideration by Newton’s second law, while the law itself is formulated only for inertial reference systems. Accordingly, the concept of force initially turns out to be defined only for such reference systems.

The equation of Newton's second law, which connects the acceleration and mass of a material point with the force acting on it, is written in the form

It directly follows from the equation that the acceleration of bodies is caused only by forces, and vice versa: the action of uncompensated forces on a body necessarily causes its acceleration.

Newton's third law complements and develops what was said about forces in the second law.

force is a measure of the mechanical action of other bodies on a given material body

in accordance with Newton's third law, forces can only exist in pairs, and the nature of the forces in each such pair is the same.

any force acting on a body has its source in the form of another body. In other words, forces necessarily represent the result interactions tel.

No other forces in mechanics are introduced or used. The possibility of the existence of forces that arise independently, without interacting bodies, is not allowed by mechanics.

Although the names of Euler and d’Alembertian inertia forces contain the word force, these physical quantities are not forces in the sense accepted in mechanics.

34. The concept of plane-parallel motion of a rigid body

The motion of a rigid body is called plane-parallel if all points of the body move in planes parallel to some fixed plane (the main plane). Let some body V perform plane motion, π is the main plane. From the definition of plane-parallel motion and the properties of an absolutely rigid body it follows that any segment of straight line AB perpendicular to the plane π will undergo translational motion. That is, the trajectories, velocities and accelerations of all points on the segment AB will be the same. Thus, the movement of each point of the section s parallel to the plane π determines the movement of all points of the body V lying on a segment perpendicular to the section at a given point. Examples of plane-parallel motion are: rolling of a wheel along a straight segment, since all its points move in planes parallel to a plane perpendicular to the axis of the wheel; A special case of such motion is the rotation of a rigid body around a fixed axis; in fact, all points of a rotating body move in planes parallel to some perpendicular to the axis of rotation of the fixed plane.

35. Inertia forces during rectilinear and curvilinear motion of a material point

The force with which a point resists a change in motion is called the force of inertia of a material point. The inertial force is directed opposite to the acceleration of the point and is equal to the mass multiplied by the acceleration.

When moving in a straight line the direction of acceleration coincides with the trajectory. The inertial force is directed in the direction opposite to the acceleration, and its numerical value is determined by the formula:

During accelerated motion, the directions of acceleration and velocity coincide and the force of inertia is directed in the direction opposite to the movement. During slow motion, when acceleration is directed in the direction opposite to speed, the force of inertia acts in the direction of movement.

Atcurvilinear and unevenmovement acceleration can be decomposed into normal an and tangent at components. Similarly, the inertial force of a point also consists of two components: normal and tangential.

Normal the component of the inertial force is equal to the product of the mass of the point and the normal acceleration and is directed opposite to this acceleration:

Tangent the component of the inertial force is equal to the product of the mass of the point and the tangential acceleration and is directed opposite to this acceleration:

Obviously, the total inertia force of a point M equal to the geometric sum of the normal and tangential components, i.e.

Considering that the tangent and normal components are mutually perpendicular, the total inertial force is:

36. Theorems on the addition of velocities and accelerations of a point in complex motion

Velocity addition theorem:

In mechanics, the absolute speed of a point is equal to the vector sum of its relative and transfer speeds:

The speed of movement of a body relative to a fixed frame of reference is equal to the vector sum of the speed of this body relative to a moving frame of reference and the speed (relative to a fixed frame) of the point of the moving frame of reference in which the body is located.

in complex motion, the absolute speed of a point is equal to the geometric sum of the portable and relative speeds. The magnitude of the absolute speed is determined by where α – angle between vectors And .

Theorem on the addition of accelerations ( CORIOLIS THEOREM)

acor = aper + aot + acor

The formula expresses the following Coriolis theorem on the addition of acceleration

rhenium:1 in complex motion, the acceleration of a point is equal to geometric

the sum of three accelerations: relative, portable and rotary, or

Coriolis

acor = 2(ω × vot)

37.D'Alembert's principle

d'Alembert's principle for a material point: at each moment of movement of a material point, active forces, reaction reactions and the force of inertia form a balanced system of forces.

D'Alembert principle- in mechanics: one of the basic principles of dynamics, according to which, if inertial forces are added to the given forces acting on the points of a mechanical system and the reactions of the superimposed connections, then a balanced system of forces is obtained.

According to this principle, for each i-th point of the system the equality is true

where is the active force acting on this point, is the reaction of the connection imposed on the point, is the inertial force, numerically equal to the product of the mass of the point and its acceleration and directed opposite to this acceleration ().

In fact, we are talking about transferring the term ma from right to left in Newton’s second law() separately for each of the material points under consideration and calling this term the D’Alembert force of inertia.

D'Alembert's principle allows one to apply simpler statics methods to solving problems of dynamics, which is why it is widely used in engineering practice, the so-called. kinetostatic method. It is especially convenient to use it to determine the reactions of bonds in cases where the law of the occurring motion is known or found from solving the corresponding equations.

Let's calculate the work of gravity m g, performed when moving a material point (body) with mass m from position 1 to position 2. Using formula (4.2) we obtain,

From the drawing it is clear that dScos=dh; then the expression for A 12 can be transformed as follows:

The resulting expression for A 12 shows that, regardless of the type of trajectory, the work of moving a material point (body) in a gravitational field depends only on its initial and final height:

4.1.2. Work done by the force of universal gravitation

Let us calculate the work done by the force of universal gravitation on the part of a body of mass M when a body of mass m moves from a position characterized by the radius vector r 1 to position with radius vector r 2 (see Fig. 4.5).

The gravitational field is central, since the gravitational force acts along the line connecting the material point m (or the center of mass of this body) with the center O of the gravitational field. By definition of work (4.2) we have:

,

where the force F is determined by law (2.12).

From the figure it is clear that dScos=dr, therefore dA=F(r)dr, and for A 12 we have:

The resulting expression does not contain information about the trajectory of the body, and it can be argued that the work of the central force depends only on the initial and final distances r 1 and r 2 of the moving point to the force center.

4.1.3. Work of elastic force

The derivation of the formula for the work of the elastic force is carried out similarly to the derivation for the force of universal gravity. This work is equal to

Here r 1 and r 2 are the magnitude of the absolute deformation of the body in the initial and final states. These deformations represent the coordinates of the point of application of an external (deforming) force, provided that the origin of coordinates corresponds to the undeformed state of the body. As in the previously considered cases, the work of the force turns out to be independent of the shape of the trajectory of the point of application of the force, and is determined only by its initial and final positions.

Chapter 5. Energy

Energy is the ability of a body (system) to do work.

Energy serves as a universal quantitative measure of the movement and interaction of all types of matter. There are two types of mechanical energy: potential and kinetic.

5.1. Potential energy

Let only conservative and gyroscopic forces, independent of time, act on a material point or mechanical system. In other words, a material point is in a stationary field of forces. Let us arbitrarily assume that any state of the system is zero. Considering other states, let us call the potential energy of a system in some other state a value U equal to the work of conservative forces performed when transferring the system from this state to zero.

Potential energy systems in a certain state are called a scalar quantity U, equal to the work of conservative forces performed when transferring the system from this state to a state conventionally taken as zero.

Since the work of conservative forces does not depend on the trajectory of a material point, its potential energy depends only on the initial state of the system. This means that the potential energy of a system is determined by its state. The ability to arbitrarily choose a zero state (zero level of potential energy) means that the potential energy of the system is not determined uniquely, but to an accuracy of an arbitrary constant C, depending on the choice made. Indeed, if we conditionally take the state represented by point O as the zero state (see Fig. 5.1), then the potential energy U M of a system located in the state represented by point M is equal to the work A MO performed by field forces during the transition from state M to O's condition

If we take O I as the starting point, then the potential energy of point M will be equal to the work
on moving from M to O I. Due to the conservatism of the field forces, the work along the trajectory of the MO is equal to the work along the trajectory of the MO I O:

A MO =
+
.

Note that the work
a completely definite value, depending only on the choice of points O and O I. Thus, when the position of the initial point O changes, the potential energy changes by a constant amount:

.

From the above it follows that the potential energy at position O is zero. However, it can be considered not equal to zero, but to some arbitrary value. Then, when the system transitions from state M to zero, it is necessary to talk not about the potential energy of state M, but about the difference in potential energies in states M and O. Arbitrariness in the choice of constant C does not affect either theoretical conclusions, or, especially, the course of physical processes. It is not the absolute value of the potential energy U that is significant, but its change -
, which does not contain an arbitrary constant C.

Let the system move from state M to state N. The work A MN done by conservative forces can be expressed in terms of the potential energies of states M and N.

Let (see Fig. 5.2) this transition be made through point O, along the trajectory MON. Then A MN =A MON =A MO +A ON . By definition of potential energy, we can write: U M =A MO +C, U N = A NO +C, where C is the same constant. We have:

The difference between the potential energies of the initial and final states U M -U N represents its decrease (the decrease is equal to the increment taken with the opposite sign). The resulting relationship plays an important role: it allows us to state that:

the work of conservative forces acting on the bodies of a mechanical system is equal to the decrease in the potential energy of the system:

The specific form of the function U, which determines the magnitude of potential energy, depends on the nature of the acting forces, or on the nature of the force field. In sections 4.1.1 – 4.1.3, expressions for the work of conservative forces of various natures are obtained. Comparing relations (4.11), (4.12) and (4.13) with relation (5.1) it is easy to come to the conclusion that potential energy:

in the gravity field is determined by the relation

in the elastic force field is determined by the relation

.

The determination of potential energy in the field of universal gravity has a peculiarity. Relation (4.12) was obtained by directly calculating the work done by the force of universal gravity:

As a rule, bodies are considered equal to zero. This is justified by the fact that at an infinitely large distance (r 2 =) the gravitational force becomes zero and there is no interaction energy, i.e. U  =0. From formula (4.17) it follows that A 1  =-U=U  -U 1 .

So, for the potential energy in the gravitational field we have the relation

Notes

1. When deriving relation (4.12), the possible movement of the center of gravity was not taken into account. It can be shown that the resulting relationship remains valid when taking into account the motion of the gravitating center. The amount of work depends only on the relative movement of the gravitating bodies and does not depend on the absolute movements of each of them.

2. The potential energy of the system in the most general case is the sum

,

Where
– external potential energy of the system associated with the action of external conservative forces on it. This component of potential energy is always additive. Internal potential energy of the system
, caused by the action of internal conservative forces, must take into account the interaction of all parts of the system, and, in the general case, is not an additive quantity. The condition of additivity of the total potential energy is fulfilled only in the case of weak interaction between parts of the system, when it can be neglected.