Let the body mass m for some small time interval Δ t force acted Under the influence of this force, the speed of the body changed by Therefore, during the time Δ t the body moves with acceleration

From the basic law of dynamics ( Newton's second law) follows:

The physical quantity equal to the product of the mass of the body and the speed of its movement is called body momentum(or amount of movement). The momentum of the body is a vector quantity. The SI unit of momentum is kilogram-meter per second (kg m/s).

The physical quantity equal to the product of the force and the time of its action is called momentum of force . The momentum of a force is also a vector quantity.

In new terms Newton's second law can be formulated as follows:

ANDthe change in the momentum of the body (momentum) is equal to the momentum of the force.

Denoting the momentum of the body by the letter Newton's second law can be written as

It was in this general form that Newton himself formulated the second law. The force in this expression is the resultant of all forces applied to the body. This vector equality can be written in projections onto the coordinate axes:

Thus, the change in the projection of the momentum of the body on any of the three mutually perpendicular axes is equal to the projection of the momentum of the force on the same axis. Consider as an example one-dimensional movement, i.e., the movement of the body along one of the coordinate axes (for example, the axis OY). Let the body fall freely with an initial velocity υ 0 under the action of gravity; the fall time is t. Let's direct the axis OY vertically down. The momentum of gravity F t = mg during t equals mgt. This momentum is equal to the change in momentum of the body

This simple result coincides with the kinematicformulafor the speed of uniformly accelerated motion. In this example, the force remained unchanged in absolute value over the entire time interval t. If the force changes in magnitude, then the average value of the force must be substituted into the expression for the impulse of the force F cf on the time interval of its action. Rice. 1.16.1 illustrates a method for determining the impulse of a time-dependent force.

Let us choose a small interval Δ on the time axis t, during which the force F (t) remains virtually unchanged. Impulse of force F (t) Δ t in time Δ t will be equal to the area of ​​the shaded bar. If the entire time axis on the interval from 0 to t split into small intervals Δ ti, and then sum the force impulses on all intervals Δ ti, then the total impulse of the force will be equal to the area formed by the step curve with the time axis. In the limit (Δ ti→ 0) this area is equal to the area bounded by the graph F (t) and axis t. This method for determining the momentum of a force from a graph F (t) is general and applicable to any laws of force change with time. Mathematically, the problem is reduced to integration functions F (t) on the interval .

The impulse of force, the graph of which is shown in fig. 1.16.1, on the interval from t 1 = 0 s to t 2 = 10 s is equal to:

In this simple example

In some cases, the average force F cp can be determined if the time of its action and the impulse imparted to the body are known. For example, a strong impact of a football player on a ball weighing 0.415 kg can give him a speed υ = 30 m/s. The impact time is approximately equal to 8·10 -3 s.

Pulse p acquired by the ball as a result of a stroke is:

Therefore, the average force F cf, with which the football player's foot acted on the ball during the kick, is:

This is a very big power. It is approximately equal to the weight of a body weighing 160 kg.

If the movement of the body during the action of the force occurred along a certain curvilinear trajectory, then the initial and final momenta of the body may differ not only in absolute value, but also in direction. In this case, to determine the change in momentum, it is convenient to use pulse diagram , which depicts the vectors and , as well as the vector constructed according to the parallelogram rule. As an example, in fig. 1.16.2 shows an impulse diagram for a ball bouncing off a rough wall. ball mass m hit the wall with a speed at an angle α to the normal (axis OX) and rebounded from it with a speed at an angle β. During contact with the wall, a certain force acted on the ball, the direction of which coincides with the direction of the vector

With a normal fall of a ball with a mass m on an elastic wall with a speed , after the rebound the ball will have a speed . Therefore, the change in the momentum of the ball during the rebound is

In projections on the axis OX this result can be written in the scalar form Δ px = –2mυ x. Axis OX directed away from the wall (as in Fig. 1.16.2), so υ x < 0 и Δpx> 0. Therefore, the module Δ p momentum change is related to the modulus υ of the ball speed by the relation Δ p = 2mυ.

His movements, i.e. value .

Pulse is a vector quantity coinciding in direction with the velocity vector.

The unit of momentum in the SI system: kg m/s .

The impulse of a system of bodies is equal to the vector sum of the impulses of all bodies included in the system:

Law of conservation of momentum

If additional external forces act on the system of interacting bodies, for example, then in this case the relation is valid, which is sometimes called the law of momentum change:

For a closed system (in the absence of external forces), the law of conservation of momentum is valid:

The action of the law of conservation of momentum can explain the phenomenon of recoil when shooting from a rifle or during artillery shooting. Also, the operation of the law of conservation of momentum underlies the principle of operation of all jet engines.

When solving physical problems, the law of conservation of momentum is used when knowledge of all the details of motion is not required, but the result of the interaction of bodies is important. Such problems, for example, are the problems of impact or collision of bodies. The law of conservation of momentum is used when considering the motion of bodies of variable mass, such as launch vehicles. Most of the mass of such a rocket is fuel. In the active phase of the flight, this fuel burns out, and the mass of the rocket rapidly decreases in this part of the trajectory. Also, the law of conservation of momentum is necessary in cases where the concept is inapplicable. It is difficult to imagine a situation where a motionless body acquires some speed instantly. In normal practice, bodies always accelerate and pick up speed gradually. However, during the movement of electrons and other subatomic particles, the change in their state occurs abruptly without staying in intermediate states. In such cases, the classical concept of "acceleration" cannot be applied.

Examples of problem solving

EXAMPLE 1

EXAMPLE 2

EXAMPLE 3

body momentum

The momentum of a body is a quantity equal to the product of the mass of the body and its speed.

It should be remembered that we are talking about a body that can be represented as a material point. The momentum of a body ($p$) is also called the momentum. The concept of momentum was introduced into physics by René Descartes (1596-1650). The term "impulse" appeared later (impulsus in Latin means "push"). Momentum is a vector quantity (like velocity) and is expressed by the formula:

$p↖(→)=mυ↖(→)$

The direction of the momentum vector always coincides with the direction of the velocity.

The unit of momentum in SI is the momentum of a body with a mass of $1$ kg moving at a speed of $1$ m/s, therefore, the unit of momentum is $1$ kg $·$ m/s.

If a constant force acts on a body (material point) during the time interval $∆t$, then the acceleration will also be constant:

$a↖(→)=((υ_2)↖(→)-(υ_1)↖(→))/(∆t)$

where, $(υ_1)↖(→)$ and $(υ_2)↖(→)$ are the initial and final velocities of the body. Substituting this value into the expression of Newton's second law, we get:

$(m((υ_2)↖(→)-(υ_1)↖(→)))/(∆t)=F↖(→)$

Opening the brackets and using the expression for the momentum of the body, we have:

$(p_2)↖(→)-(p_1)↖(→)=F↖(→)∆t$

Here $(p_2)↖(→)-(p_1)↖(→)=∆p↖(→)$ is the momentum change over time $∆t$. Then the previous equation becomes:

$∆p↖(→)=F↖(→)∆t$

The expression $∆p↖(→)=F↖(→)∆t$ is a mathematical representation of Newton's second law.

The product of a force and its duration is called momentum of force. That's why the change in the momentum of a point is equal to the change in the momentum of the force acting on it.

The expression $∆p↖(→)=F↖(→)∆t$ is called body motion equation. It should be noted that the same action - a change in the momentum of a point - can be obtained by a small force in a long period of time and by a large force in a small period of time.

Impulse of the system tel. Law of change of momentum

The impulse (momentum) of a mechanical system is a vector equal to the sum of the impulses of all material points of this system:

$(p_(syst))↖(→)=(p_1)↖(→)+(p_2)↖(→)+...$

The laws of change and conservation of momentum are a consequence of Newton's second and third laws.

Consider a system consisting of two bodies. The forces ($F_(12)$ and $F_(21)$ in the figure, with which the bodies of the system interact with each other, are called internal.

Let, in addition to internal forces, external forces $(F_1)↖(→)$ and $(F_2)↖(→)$ act on the system. For each body, the equation $∆p↖(→)=F↖(→)∆t$ can be written. Adding the left and right parts of these equations, we get:

$(∆p_1)↖(→)+(∆p_2)↖(→)=((F_(12))↖(→)+(F_(21))↖(→)+(F_1)↖(→)+ (F_2)↖(→))∆t$

According to Newton's third law $(F_(12))↖(→)=-(F_(21))↖(→)$.

Hence,

$(∆p_1)↖(→)+(∆p_2)↖(→)=((F_1)↖(→)+(F_2)↖(→))∆t$

On the left side is the geometric sum of the changes in the momentum of all the bodies of the system, equal to the change in the momentum of the system itself - $(∆p_(syst))↖(→)$. With this in mind, the equality $(∆p_1)↖(→)+(∆p_2) ↖(→)=((F_1)↖(→)+(F_2)↖(→))∆t$ can be written:

$(∆p_(sys))↖(→)=F↖(→)∆t$

where $F↖(→)$ is the sum of all external forces acting on the body. The result obtained means that only external forces can change the momentum of the system, and the change in the momentum of the system is directed in the same way as the total external force. This is the essence of the law of change in the momentum of a mechanical system.

Internal forces cannot change the total momentum of the system. They only change the impulses of the individual bodies of the system.

Law of conservation of momentum

From the equation $(∆p_(syst))↖(→)=F↖(→)∆t$ the momentum conservation law follows. If no external forces act on the system, then the right side of the equation $(∆p_(sys))↖(→)=F↖(→)∆t$ vanishes, which means that the total momentum of the system remains unchanged:

$(∆p_(sys))↖(→)=m_1(υ_1)↖(→)+m_2(υ_2)↖(→)=const$

A system on which no external forces act or the resultant of external forces is equal to zero is called closed.

The law of conservation of momentum states:

The total momentum of a closed system of bodies remains constant for any interaction of the bodies of the system with each other.

The result obtained is valid for a system containing an arbitrary number of bodies. If the sum of external forces is not equal to zero, but the sum of their projections on some direction is equal to zero, then the projection of the momentum of the system on this direction does not change. So, for example, a system of bodies on the surface of the Earth cannot be considered closed due to the force of gravity acting on all bodies, however, the sum of the projections of impulses on the horizontal direction can remain unchanged (in the absence of friction), since in this direction the force of gravity does not is valid.

Jet propulsion

Consider examples that confirm the validity of the law of conservation of momentum.

Let's take a children's rubber balloon, inflate it and let it go. We will see that when the air starts to come out of it in one direction, the balloon itself will fly in the other direction. The movement of the ball is an example of jet propulsion. It is explained by the law of conservation of momentum: the total momentum of the system "ball plus air in it" before the outflow of air is zero; it must remain equal to zero during the movement; therefore, the ball moves in the direction opposite to the direction of the outflow of the jet, and with such a speed that its momentum is equal in absolute value to the momentum of the air jet.

jet propulsion called the motion of a body that occurs when a part of it separates from it at some speed. Due to the law of conservation of momentum, the direction of motion of the body is opposite to the direction of motion of the separated part.

Rocket flights are based on the principle of jet propulsion. A modern space rocket is a very complex aircraft. The mass of the rocket is the sum of the mass of the working fluid (i.e., hot gases resulting from the combustion of fuel and ejected in the form of a jet stream) and the final, or, as they say, “dry” mass of the rocket, remaining after the ejection of the working fluid from the rocket.

When a reactive gas jet is ejected from a rocket at high speed, the rocket itself rushes in the opposite direction. According to the momentum conservation law, the momentum $m_(p)υ_p$ acquired by the rocket must be equal to the momentum $m_(gas) υ_(gas)$ of the ejected gases:

$m_(p)υ_p=m_(gas) υ_(gas)$

It follows that the speed of the rocket

$υ_p=((m_(gas))/(m_p)) υ_(gas)$

It can be seen from this formula that the greater the speed of the rocket, the greater the speed of the ejected gases and the ratio of the mass of the working fluid (i.e., the mass of fuel) to the final ("dry") mass of the rocket.

The formula $υ_p=((m_(gas))/(m_p))·υ_(gas)$ is approximate. It does not take into account that as the fuel burns, the mass of the flying rocket becomes smaller and smaller. The exact formula for the speed of a rocket was obtained in 1897 by K. E. Tsiolkovsky and bears his name.

Force work

The term "work" was introduced into physics in 1826 by the French scientist J. Poncelet. If in everyday life only human labor is called work, then in physics and, in particular, in mechanics, it is generally accepted that work is done by force. The physical quantity of work is usually denoted by the letter $A$.

Force work- this is a measure of the action of a force, depending on its module and direction, as well as on the displacement of the point of application of the force. For a constant force and rectilinear movement, the work is determined by the equality:

$A=F|∆r↖(→)|cosα$

where $F$ is the force acting on the body, $∆r↖(→)$ is the displacement, $α$ is the angle between the force and the displacement.

The work of the force is equal to the product of the modules of force and displacement and the cosine of the angle between them, i.e. the scalar product of the vectors $F↖(→)$ and $∆r↖(→)$.

Work is a scalar quantity. If $α 0$, and if $90°

When several forces act on a body, the total work (the sum of the work of all forces) is equal to the work of the resulting force.

The SI unit of work is joule($1$ J). $1$ J is the work done by a force of $1$ N on a path of $1$ m in the direction of this force. This unit is named after the English scientist J. Joule (1818-1889): $1$ J = $1$ N $·$ m. Kilojoules and millijoules are also often used: $1$ kJ $= 1,000$ J, $1$ mJ $= 0.001$ J.

The work of gravity

Let us consider a body sliding along an inclined plane with an inclination angle $α$ and a height $H$.

We express $∆x$ in terms of $H$ and $α$:

$∆x=(H)/(sinα)$

Considering that gravity $F_т=mg$ makes an angle ($90° - α$) with the direction of movement, using the formula $∆x=(H)/(sin)α$, we obtain an expression for the work of gravity $A_g$:

$A_g=mg cos(90°-α)(H)/(sinα)=mgH$

From this formula it can be seen that the work of gravity depends on the height and does not depend on the angle of inclination of the plane.

From this it follows that:

1. the work of gravity does not depend on the shape of the trajectory along which the body moves, but only on the initial and final position of the body;
2. when a body moves along a closed trajectory, the work of gravity is zero, i.e., gravity is a conservative force (conservative forces are forces that have this property).

The work of reaction forces, is zero because the reaction force ($N$) is directed perpendicular to the displacement $∆x$.

The work of the friction force

The friction force is directed opposite to the displacement $∆x$ and makes an angle $180°$ with it, so the work of the friction force is negative:

$A_(tr)=F_(tr)∆x cos180°=-F_(tr) ∆x$

Since $F_(tr)=μN, N=mg cosα, ∆x=l=(H)/(sinα),$ then

$A_(tr)=μmgHctgα$

The work of the elastic force

Let an external force $F↖(→)$ act on an unstretched spring of length $l_0$, stretching it by $∆l_0=x_0$. In position $x=x_0F_(control)=kx_0$. After the termination of the force $F↖(→)$ at the point $x_0$, the spring is compressed under the action of the force $F_(control)$.

Let us determine the work of the elastic force when the coordinate of the right end of the spring changes from $х_0$ to $х$. Since the elastic force in this area changes linearly, in Hooke's law, its average value in this area can be used:

$F_(ex.av.)=(kx_0+kx)/(2)=(k)/(2)(x_0+x)$

Then the work (taking into account the fact that the directions $(F_(exp.av.))↖(→)$ and $(∆x)↖(→)$ coincide) is equal to:

$A_(exerc)=(k)/(2)(x_0+x)(x_0-x)=(kx_0^2)/(2)-(kx^2)/(2)$

It can be shown that the form of the last formula does not depend on the angle between $(F_(exp.av.))↖(→)$ and $(∆x)↖(→)$. The work of the elastic forces depends only on the deformations of the spring in the initial and final states.

Thus, the elastic force, like gravity, is a conservative force.

Power of force

Power is a physical quantity measured by the ratio of work to the period of time during which it is produced.

In other words, power shows how much work is done per unit of time (in SI, for $1$ s).

Power is determined by the formula:

where $N$ is the power, $A$ is the work done in the time $∆t$.

Substituting $A=F|(∆r)↖(→)|cosα$ into the formula $N=(A)/(∆t)$ instead of the work $A$, we get:

$N=(F|(∆r)↖(→)|cosα)/(∆t)=Fυcosα$

The power is equal to the product of the modules of the force and velocity vectors and the cosine of the angle between these vectors.

Power in the SI system is measured in watts (W). One watt ($1$ W) is the power at which $1$ J of work is done in $1$ s: $1$ W $= 1$ J/s.

This unit is named after the English inventor J. Watt (Watt), who built the first steam engine. J. Watt himself (1736-1819) used a different unit of power - horsepower (hp), which he introduced in order to be able to compare the performance of a steam engine and a horse: $ 1 $ hp. $= 735.5$ Tue.

In technology, larger units of power are often used - kilowatts and megawatts: $1$ kW $= 1000$ W, $1$ MW $= 1000000$ W.

Kinetic energy. Law of change of kinetic energy

If a body or several interacting bodies (a system of bodies) can do work, then they say that they have energy.

The word "energy" (from the Greek. energia - action, activity) is often used in everyday life. So, for example, people who can quickly do work are called energetic, with great energy.

The energy possessed by a body due to motion is called kinetic energy.

As in the case of the definition of energy in general, we can say about kinetic energy that kinetic energy is the ability of a moving body to do work.

Let us find the kinetic energy of a body of mass $m$ moving with a speed of $υ$. Since kinetic energy is the energy due to motion, the zero state for it is the state in which the body is at rest. Having found the work necessary to communicate a given speed to the body, we will find its kinetic energy.

To do this, we calculate the work done on the displacement section $∆r↖(→)$ when the directions of the force vectors $F↖(→)$ and displacement $∆r↖(→)$ coincide. In this case, the work is

where $∆x=∆r$

For the movement of a point with acceleration $α=const$, the expression for movement has the form:

$∆x=υ_1t+(at^2)/(2),$

where $υ_1$ is the initial speed.

Substituting the expression for $∆x$ from $∆x=υ_1t+(at^2)/(2)$ into the equation $A=F ∆x$ and using Newton's second law $F=ma$, we get:

$A=ma(υ_1t+(at^2)/(2))=(mat)/(2)(2υ_1+at)$

Expressing the acceleration in terms of initial $υ_1$ and final $υ_2$ speeds $a=(υ_2-υ_1)/(t)$ and substituting into $A=ma(υ_1t+(at^2)/(2))=(mat)/ (2)(2υ_1+at)$ we have:

$A=(m(υ_2-υ_1))/(2) (2υ_1+υ_2-υ_1)$

$A=(mυ_2^2)/(2)-(mυ_1^2)/(2)$

Now equating the initial velocity to zero: $υ_1=0$, we obtain an expression for kinetic energy:

$E_K=(mυ)/(2)=(p^2)/(2m)$

Thus, a moving body has kinetic energy. This energy is equal to the work that must be done to increase the speed of the body from zero to $υ$.

It follows from $E_K=(mυ)/(2)=(p^2)/(2m)$ that the work done by a force to move a body from one position to another is equal to the change in kinetic energy:

$A=E_(K_2)-E_(K_1)=∆E_K$

The equality $A=E_(K_2)-E_(K_1)=∆E_K$ expresses theorem on the change in kinetic energy.

Change in the kinetic energy of the body(material point) for a certain period of time is equal to the work done during this time by the force acting on the body.

Potential energy

Potential energy is the energy determined by the mutual arrangement of interacting bodies or parts of the same body.

Since energy is defined as the ability of a body to do work, potential energy is naturally defined as the work of a force that depends only on the relative position of the bodies. This is the work of gravity $A=mgh_1-mgh_2=mgH$ and the work of elasticity:

$A=(kx_0^2)/(2)-(kx^2)/(2)$

The potential energy of the body interacting with the Earth is called the value equal to the product of the mass $m$ of this body and the free fall acceleration $g$ and the height $h$ of the body above the Earth's surface:

The potential energy of an elastically deformed body is the value equal to half the product of the coefficient of elasticity (stiffness) $k$ of the body and the square of deformation $∆l$:

$E_p=(1)/(2)k∆l^2$

The work of conservative forces (gravity and elasticity), taking into account $E_p=mgh$ and $E_p=(1)/(2)k∆l^2$, is expressed as follows:

$A=E_(p_1)-E_(p_2)=-(E_(p_2)-E_(p_1))=-∆E_p$

This formula allows us to give a general definition of potential energy.

The potential energy of the system is a value that depends on the position of the bodies, the change of which during the transition of the system from the initial state to the final state is equal to the work of the internal conservative forces of the system, taken with the opposite sign.

The minus sign on the right side of the equation $A=E_(p_1)-E_(p_2)=-(E_(p_2)-E_(p_1))=-∆E_p$ means that when work is done by internal forces (for example, falling body to the ground under the action of gravity in the "stone-Earth" system), the energy of the system decreases. Work and change in potential energy in a system always have opposite signs.

Since work determines only the change in potential energy, only the change in energy has physical meaning in mechanics. Therefore, the choice of the zero energy level is arbitrary and is determined solely by considerations of convenience, for example, the ease of writing the corresponding equations.

The law of change and conservation of mechanical energy

Total mechanical energy of the system the sum of its kinetic and potential energies is called:

It is determined by the position of the bodies (potential energy) and their speed (kinetic energy).

According to the kinetic energy theorem,

$E_k-E_(k_1)=A_p+A_(pr),$

where $А_р$ is the work of potential forces, $А_(pr)$ is the work of nonpotential forces.

In turn, the work of potential forces is equal to the difference in the potential energy of the body in the initial $E_(p_1)$ and final $E_p$ states. With this in mind, we get an expression for the law of change of mechanical energy:

$(E_k+E_p)-(E_(k_1)+E_(p_1))=A_(pr)$

where the left side of the equality is the change in the total mechanical energy, and the right side is the work of nonpotential forces.

So, law of change of mechanical energy reads:

The change in the mechanical energy of the system is equal to the work of all nonpotential forces.

A mechanical system in which only potential forces act is called conservative.

In a conservative system $A_(pr) = 0$. this implies law of conservation of mechanical energy:

In a closed conservative system, the total mechanical energy is conserved (does not change with time):

$E_k+E_p=E_(k_1)+E_(p_1)$

The law of conservation of mechanical energy is derived from the laws of Newtonian mechanics, which are applicable to a system of material points (or macroparticles).

However, the law of conservation of mechanical energy is also valid for a system of microparticles, where Newton's laws themselves no longer apply.

The law of conservation of mechanical energy is a consequence of the homogeneity of time.

Uniformity of time is that, under the same initial conditions, the course of physical processes does not depend on the moment at which these conditions are created.

The law of conservation of total mechanical energy means that when the kinetic energy in a conservative system changes, its potential energy must also change, so that their sum remains constant. This means the possibility of converting one type of energy into another.

In accordance with various forms of the motion of matter, various types of energy are considered: mechanical, internal (equal to the sum of the kinetic energy of the chaotic movement of molecules relative to the center of mass of the body and the potential energy of the interaction of molecules with each other), electromagnetic, chemical (which consists of the kinetic energy of the movement of electrons and electric the energy of their interaction with each other and with atomic nuclei), nuclear energy, etc. It can be seen from the foregoing that the division of energy into different types is rather arbitrary.

Natural phenomena are usually accompanied by the transformation of one type of energy into another. So, for example, the friction of parts of various mechanisms leads to the conversion of mechanical energy into heat, i.e., into internal energy. In heat engines, on the contrary, internal energy is converted into mechanical energy; in galvanic cells, chemical energy is converted into electrical energy, etc.

Currently, the concept of energy is one of the basic concepts of physics. This concept is inextricably linked with the idea of ​​the transformation of one form of movement into another.

Here is how the concept of energy is formulated in modern physics:

Energy is a general quantitative measure of the movement and interaction of all types of matter. Energy does not arise from nothing and does not disappear, it can only pass from one form to another. The concept of energy binds together all the phenomena of nature.

simple mechanisms. mechanism efficiency

Simple mechanisms are devices that change the magnitude or direction of the forces applied to the body.

They are used to move or lift large loads with little effort. These include the lever and its varieties - blocks (movable and fixed), the gate, the inclined plane and its varieties - the wedge, the screw, etc.

Lever arm. Lever rule

The lever is a rigid body capable of rotating around a fixed support.

The leverage rule says:

A lever is in equilibrium if the forces applied to it are inversely proportional to their arms:

$(F_2)/(F_1)=(l_1)/(l_2)$

From the formula $(F_2)/(F_1)=(l_1)/(l_2)$, applying the property of proportion to it (the product of the extreme terms of the proportion is equal to the product of its middle terms), we can obtain the following formula:

But $F_1l_1=M_1$ is the moment of force tending to turn the lever clockwise, and $F_2l_2=M_2$ is the moment of force tending to turn the lever counterclockwise. Thus, $M_1=M_2$, which was to be proved.

The lever began to be used by people in ancient times. With its help, it was possible to lift heavy stone slabs during the construction of the pyramids in ancient Egypt. Without leverage, this would not have been possible. Indeed, for example, for the construction of the pyramid of Cheops, which has a height of $147$ m, more than two million stone blocks were used, the smallest of which had a mass of $2.5$ tons!

Nowadays, levers are widely used both in production (for example, cranes) and in everyday life (scissors, wire cutters, scales).

Fixed block

The action of a fixed block is similar to the action of a lever with equal leverage: $l_1=l_2=r$. The applied force $F_1$ is equal to the load $F_2$, and the equilibrium condition is:

Fixed block used when you need to change the direction of a force without changing its magnitude.

Movable block

The movable block acts similarly to a lever, whose arms are: $l_2=(l_1)/(2)=r$. In this case, the equilibrium condition has the form:

where $F_1$ is the applied force, $F_2$ is the load. The use of a movable block gives a gain in strength twice.

Polyspast (block system)

An ordinary chain hoist consists of $n$ movable and $n$ fixed blocks. Applying it gives a gain in strength of $2n$ times:

$F_1=(F_2)/(2n)$

Power chain hoist consists of n movable and one fixed block. The use of a power chain hoist gives a gain in strength of $2^n$ times:

$F_1=(F_2)/(2^n)$

Screw

The screw is an inclined plane wound on the axis.

The condition for the balance of forces acting on the screw has the form:

$F_1=(F_2h)/(2πr)=F_2tgα, F_1=(F_2h)/(2πR)$

where $F_1$ is an external force applied to the screw and acting at a distance $R$ from its axis; $F_2$ is the force acting in the direction of the screw axis; $h$ - screw pitch; $r$ is the average thread radius; $α$ is the angle of the thread. $R$ is the length of the lever (wrench) that rotates the screw with the force $F_1$.

Efficiency

Coefficient of performance (COP) - the ratio of useful work to all the work expended.

Efficiency is often expressed as a percentage and denoted by the Greek letter $η$ ("this"):

$η=(A_p)/(A_3) 100%$

where $A_n$ is useful work, $A_3$ is all the work expended.

Useful work is always only a part of the total work that a person expends using this or that mechanism.

Part of the work done is spent on overcoming the forces of friction. Since $А_3 > А_п$, the efficiency is always less than $1$ (or $< 100%$).

Since each of the works in this equation can be expressed as the product of the corresponding force and the distance traveled, it can be rewritten as follows: $F_1s_1≈F_2s_2$.

From this it follows that, winning with the help of the mechanism in force, we lose the same number of times on the way, and vice versa. This law is called the golden rule of mechanics.

The golden rule of mechanics is an approximate law, since it does not take into account the work to overcome friction and gravity of the parts of the devices used. Nevertheless, it can be very useful when analyzing the operation of any simple mechanism.

So, for example, thanks to this rule, we can immediately say that the worker shown in the figure, with a double gain in the lifting force of $ 10 $ cm, will have to lower the opposite end of the lever by $ 20 $ cm.

Collision of bodies. Elastic and inelastic impacts

The laws of conservation of momentum and mechanical energy are used to solve the problem of the motion of bodies after a collision: the known momenta and energies before the collision are used to determine the values ​​of these quantities after the collision. Consider the cases of elastic and inelastic impacts.

An absolutely inelastic impact is called, after which the bodies form a single body moving at a certain speed. The problem of the speed of the latter is solved using the law of conservation of momentum for a system of bodies with masses $m_1$ and $m_2$ (if we are talking about two bodies) before and after the impact:

$m_1(υ_1)↖(→)+m_2(υ_2)↖(→)=(m_1+m_2)υ↖(→)$

Obviously, the kinetic energy of bodies is not conserved during an inelastic impact (for example, at $(υ_1)↖(→)=-(υ_2)↖(→)$ and $m_1=m_2$ it becomes equal to zero after the impact).

An absolutely elastic impact is called, in which not only the sum of impulses is preserved, but also the sum of the kinetic energies of the colliding bodies.

For an absolutely elastic impact, the equations

$m_1(υ_1)↖(→)+m_2(υ_2)↖(→)=m_1(υ"_1)↖(→)+m_2(υ"_2)↖(→);$

$(m_(1)υ_1^2)/(2)+(m_(2)υ_2^2)/(2)=(m_1(υ"_1)^2)/(2)+(m_2(υ"_2 )^2)/(2)$

where $m_1, m_2$ are the masses of the balls, $υ_1, υ_2$ are the velocities of the balls before the impact, $υ"_1, υ"_2$ are the velocities of the balls after the impact.

Let's do some simple transformations with formulas. According to Newton's second law, the force can be found: F=m*a. The acceleration is found as follows: a=v⁄t . Thus we get: F= m*v/t.

Determination of body momentum: formula

It turns out that the force is characterized by a change in the product of mass and speed in time. If we denote this product by a certain value, then we will get a change in this value over time as a characteristic of the force. This quantity is called the momentum of the body. The momentum of the body is expressed by the formula:

where p is the momentum of the body, m is the mass, v is the velocity.

Momentum is a vector quantity, and its direction always coincides with the direction of velocity. The unit of momentum is kilogram per meter per second (1 kg*m/s).

What is the momentum of the body: how to understand?

Let's try in a simple way, "on the fingers" to figure out what the momentum of the body is. If the body is at rest, then its momentum is zero. Logically. If the speed of the body changes, then the body has a certain momentum, which characterizes the magnitude of the force applied to it.

If there is no impact on the body, but it moves at a certain speed, that is, it has a certain momentum, then its momentum means what effect this body can have when interacting with another body.

The momentum formula includes the mass of the body and its speed. That is, the greater the mass and / or speed of the body, the greater the impact it can have. This is clear from life experience.

To move a body of small mass, a small force is needed. The greater the mass of the body, the more effort will have to be applied. The same applies to the speed that is reported to the body. In the case of the impact of the body itself on another, the momentum also shows the amount with which the body is able to act on other bodies. This value directly depends on the speed and mass of the original body.

Impulse in the interaction of bodies

Another question arises: what will happen to the momentum of the body when it interacts with another body? The mass of a body cannot change if it remains intact, but the speed can easily change. In this case, the speed of the body will change depending on its mass.

In fact, it is clear that when bodies with very different masses collide, their speed will change in different ways. If a soccer ball flying at high speed crashes into a person who is not ready for this, for example, a spectator, then the spectator may fall, that is, it will acquire some small speed, but it will definitely not fly like a ball.

And all because the mass of the spectator is much greater than the mass of the ball. But at the same time, the total momentum of these two bodies will remain unchanged.

Law of conservation of momentum: formula

This is the law of conservation of momentum: when two bodies interact, their total momentum remains unchanged. The law of conservation of momentum is valid only in a closed system, that is, in a system in which there is no influence of external forces or their total action is zero.

In reality, a system of bodies is almost always influenced by a third party, but the general impulse, like energy, does not disappear into nowhere and does not arise from nowhere, it is distributed among all participants in the interaction.

A 22-caliber bullet has a mass of only 2 g. If someone throws such a bullet, he can easily catch it even without gloves. If you try to catch such a bullet that has flown out of the muzzle at a speed of 300 m / s, then even gloves will not help here.

If a toy cart is rolling towards you, you can stop it with your toe. If a truck is rolling towards you, you should keep your feet out of the way.

Let's consider a problem that demonstrates the connection between the momentum of a force and a change in the momentum of a body.

Example. The mass of the ball is 400 g, the speed acquired by the ball after the impact is 30 m/s. The force with which the foot acted on the ball was 1500 N, and the impact time was 8 ms. Find the momentum of the force and the change in the momentum of the body for the ball.

Change in body momentum

Example. Estimate the average force from the side of the floor acting on the ball during impact.

1) During the impact, two forces act on the ball: support reaction force, gravity.

The reaction force changes during the impact time, so it is possible to find the average floor reaction force.

2) Change in momentum body shown in the picture

3) From Newton's second law

The main thing to remember

1) Formulas of body impulse, force impulse;
2) The direction of the momentum vector;
3) Find the change in body momentum

General derivation of Newton's second law

F(t) chart. variable force

The force impulse is numerically equal to the area of ​​the figure under the graph F(t).

If the force is not constant in time, for example, it increases linearly F=kt, then the momentum of this force is equal to the area of ​​the triangle. You can replace this force with such a constant force that will change the momentum of the body by the same amount in the same period of time.

Average resultant force

LAW OF CONSERVATION OF MOMENTUM

Online testing

Closed system of bodies

This is a system of bodies that interact only with each other. There are no external forces of interaction.

In the real world, such a system cannot exist, there is no way to remove any external interaction. A closed system of bodies is a physical model, just like a material point is a model. This is a model of a system of bodies that allegedly interact only with each other, external forces are not taken into account, they are neglected.

Law of conservation of momentum

In a closed system of bodies vector the sum of the momenta of the bodies does not change when the bodies interact. If the momentum of one body has increased, then this means that at that moment the momentum of some other body (or several bodies) has decreased by exactly the same amount.

Let's consider such an example. Girl and boy are skating. A closed system of bodies - a girl and a boy (we neglect friction and other external forces). The girl stands still, her momentum is zero, since the speed is zero (see the body momentum formula). After the boy, moving at some speed, collides with the girl, she will also begin to move. Now her body has momentum. The numerical value of the momentum of the girl is exactly the same as the momentum of the boy decreased after the collision.

One body of mass 20 kg is moving at a speed , the second body of mass 4 kg is moving in the same direction with a speed of . What is the momentum of each body. What is the momentum of the system?

Impulse of the body system is the vector sum of the impulses of all bodies in the system. In our example, this is the sum of two vectors (since two bodies are considered), which are directed in the same direction, therefore

Now let's calculate the momentum of the system of bodies from the previous example if the second body moves in the opposite direction.

Since the bodies move in opposite directions, we get the vector sum of the multidirectional impulses. More on the sum of vectors.

The main thing to remember

1) What is a closed system of bodies;
2) Law of conservation of momentum and its application