If on a body of mass m for a certain period of time Δ t force F → acts, then the body speed changes ∆ v → = v 2 → - v 1 → . We find that during the time Δ t the body continues to move with acceleration:
a → = ∆ v → ∆ t = v 2 → - v 1 → ∆ t .
Based on the fundamental law of dynamics, that is, Newton's second law, we have:
F → = m a → = m v 2 → - v 1 → ∆ t or F → ∆ t = m v 2 → - m v 1 → = m ∆ v → = ∆ m v → .
Definition 1Body impulse, or momentum is a physical quantity equal to the product of the mass of a body and the speed of its movement.
The momentum of a body is considered a vector quantity, which is measured in kilogram-meter per second (kg m/s).
Definition 2
Impulse force is a physical quantity equal to the product of a force and the time of its action.
Momentum is classified as a vector quantity. There is another formulation of the definition.
Definition 3
The change in the momentum of the body is equal to the impulse of the force.
When denoting momentum p → Newton's second law is written as:
F → ∆ t = ∆ p → .
This type allows us to formulate Newton's second law. Force F → is the resultant of all forces acting on the body. Equality is written as projections onto coordinate axes type:
F x Δ t = Δ p x ; F y Δ t = Δ p y ; F z Δ t = Δ p z .
Figure 1. 16. 1. Body impulse model.
The change in the projection of the body's momentum onto any of the three mutually perpendicular axes is equal to the projection of the force impulse onto the same axis.
Definition 4
One-dimensional movement– this is the movement of a body along one of the coordinate axes.
Example 1
Using an example, consider the free fall of a body with an initial speed v 0 under the influence of gravity over a period of time t. When the O Y axis is directed vertically downward, the gravity impulse F t = mg, acting during time t, is equal to m g t. Such an impulse is equal to the change in the momentum of the body:
F t t = m g t = Δ p = m (v – v 0), whence v = v 0 + g t.
The entry coincides with the kinematic formula for determining speed uniformly accelerated motion. The magnitude of the force does not change over the entire interval t. When it is variable in magnitude, then the momentum formula requires substituting the average value of force F with p from the time interval t. Figure 1. 16. 2 shows how the impulse of a force that depends on time is determined.
Figure 1. 16. 2. Calculation of force impulse from the graph of dependence F (t)
It is necessary to select the interval Δ t on the time axis; it is clear that the force F(t) practically unchanged. Force impulse F (t) Δ t over a period of time Δ t will be equal to the area of the shaded figure. When dividing the time axis into intervals by Δ t i in the interval from 0 to t, add up the impulses of all acting forces from these intervals Δ t i , then the total impulse of force will be equal to the area of formation using the step and time axes.
By applying the limit (Δ t i → 0), you can find the area that will be limited by the graph F(t) and t axis. Using the definition of force impulse from a graph is applicable to any laws where there are changing forces and time. This decision leads to integration of function F(t) from the interval [ 0 ; t ] .
Figure 1. 16. 2 shows a force impulse located in the interval from t 1 = 0 s to t 2 = 10.
From the formula we find that F c p (t 2 - t 1) = 1 2 F m a x (t 2 - t 1) = 100 N s = 100 k g m / s.
That is, from the example we can see F with p = 1 2 F m a x = 10 N.
There are cases when determining the average force F c p is possible with known time and data on the reported impulse. With a strong impact on a ball with a mass of 0.415 kg g, a speed of v = 30 m/s can be reported. Approximate time impact is the value 8 · 10 – 3 s.
Then the momentum formula takes the form:
p = m v = 12.5 k g m/s.
To determine average strength F with p during the impact, it is necessary F with p = p ∆ t = 1.56 10 3 N.
We received very great value, which is equal to a body weighing 160 kg.
When the movement occurs curvilinear trajectory, That initial value p 1 → and final
p 2 → can be different in magnitude and direction. To determine the momentum ∆ p →, a momentum diagram is used, where there are vectors p 1 → and p 2 →, and ∆ p → = p 2 → - p 1 → is constructed according to the parallelogram rule.
Example 2
As an example, see Figure 1. 16. 2, where a diagram of the impulses of a ball bouncing off a wall is drawn. When serving, a ball with mass m with a speed v 1 → hits the surface at an angle α to the normal and rebounds with a speed v 2 → with an angle β. When hitting the wall, the ball was subjected to the action of a force F →, directed in the same way as the vector ∆ p →.
Figure 1. 16. 3. Rebounding of a ball from a rough wall and impulse diagram.
If a ball with mass m falls normally onto an elastic surface with a speed v 1 → = v → , then upon rebound it will change to v 2 → = - v → . This means that over a certain period of time the impulse will change and will be equal to ∆ p → = - 2 m v → . Using projections onto O X, the result will be written as Δ p x = – 2 m v x. From the drawing 1 . 16 . 3 it is clear that the O X axis is directed from the wall, then v x follows< 0 и Δ p x >0 . From the formula we find that the module Δ p is associated with the velocity module, which takes the form Δ p = 2 m v .
If you notice an error in the text, please highlight it and press Ctrl+Enter
Body impulse
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're talking about about a body that can be represented as a material point. The momentum of the 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 speed) and is expressed by the formula:
$p↖(→)=mυ↖(→)$
The direction of the momentum vector always coincides with the direction of the velocity.
The SI unit of impulse is the impulse of a body with a mass of $1$ kg moving at a speed of $1$ m/s; therefore, the unit of impulse is $1$ kg $·$ m/s.
If a constant force acts on a body (material point) during a period of time $∆t$, then the acceleration will also be constant:
$a↖(→)=((υ_2)↖(→)-(υ_1)↖(→))/(∆t)$
where, $(υ_1)↖(→)$ and $(υ_2)↖(→)$ are the initial and final speed bodies. 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 change in momentum over time $∆t$. Then the previous equation will take the form:
$∆p↖(→)=F↖(→)∆t$
The expression $∆p↖(→)=F↖(→)∆t$ is mathematical notation Newton's second law.
The product of a force and the duration of its action is called impulse 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 equation of body motion. It should be noted that the same action - a change in the momentum of a point - can be achieved by a small force in big gap time and great strength in a short period of time.
Impulse of the system tel. Law of Momentum Change
Impulse (amount of movement) mechanical system is called a vector equal to the sum of the momenta of all material points 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.
Let us 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, act on the system external forces$(F_1)↖(→)$ and $(F_2)↖(→)$. For each body we can write the equation $∆p↖(→)=F↖(→)∆t$. Adding the left and right sides 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 geometric sum changes in the impulses of all bodies of the system, equal to the change in the impulse of the system itself - $(∆p_(syst))↖(→)$. Taking this into account, the equality $(∆p_1)↖(→)+(∆p_2)↖(→)=( (F_1)↖(→)+(F_2)↖(→))∆t$ can be written:
$(∆p_(syst))↖(→)=F↖(→)∆t$
where $F↖(→)$ is the sum of all external forces acting on the body. The result obtained means that the momentum of the system can only be changed by external forces, 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 momentum of a mechanical system.
Internal forces cannot change the total momentum of the system. They only change the impulses of individual bodies of the system.
Law of conservation of momentum
From the equation $(∆p_(syst))↖(→)=F↖(→)∆t$ the law of conservation of momentum follows. If no external forces act on the system, then right side equation $(∆p_(syst))↖(→)=F↖(→)∆t$ becomes zero, which means the total momentum of the system remains unchanged:
$(∆p_(syst))↖(→)=m_1(υ_1)↖(→)+m_2(υ_2)↖(→)=const$
A system on which no external forces act or the resultant of external forces is 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 obtained result is valid for a system containing arbitrary number tel. If the sum of external forces is not equal to zero, but the sum of their projections to some direction is equal to zero, then the projection of the system’s momentum to 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 works.
Jet propulsion
Let us consider examples that confirm the validity of the law of conservation of momentum.
Let's take a children's rubber ball, inflate it and release it. We will see that when the air begins to leave it in one direction, the ball itself will fly in the other. The movement of the ball is an example jet propulsion. It is explained by the law of conservation of momentum: the total momentum of the “ball plus air in it” system before the air flows out is zero; it must remain equal to zero during movement; so the ball moves to the side, opposite direction outflow of the jet, and at such a speed that its momentum in absolute value equal to impulse air jet.
Jet motion call the movement of a body that occurs when some part of it is separated from it at any speed. Due to the law of conservation of momentum, the direction of movement of the body is opposite to the direction of movement 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 consists of the mass of the working fluid (i.e., hot gases formed as a result of fuel combustion and emitted in the form of a jet stream) and the final, or, as they say, “dry” mass of the rocket remaining after the working fluid is ejected from the rocket.
When a jet gas stream high speed is thrown out of the rocket, the rocket itself rushes into the opposite side. According to the law of conservation of momentum, 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
$υ_p=((m_(gas))/(m_p))·υ_(gas)$
From this formula it is clear that the greater the speed of the rocket, the greater the speed of the emitted gases and the ratio of the mass of the working fluid (i.e., the mass of the 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 less and less. The exact formula for rocket speed was obtained in 1897 by K. E. Tsiolkovsky and bears his name.
Work of force
The term “work” was introduced into physics in 1826 by the French scientist J. Poncelet. If in everyday life If only human labor is called work, then in physics and, in particular, in mechanics it is generally accepted that work is performed by force. The physical quantity of work is usually denoted by the letter $A$.
Work of force is a measure of the action of a force, depending on its magnitude and direction, as well as on the displacement of the point of application of the force. For constant force And linear movement 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 force is equal to the product of the moduli of force and displacement and the cosine of the angle between them, i.e. scalar product 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 unit of work in SI is joule($1$ J). $1$ J is the work done by a force of $1$ N along a path of $1$ m in the direction of action 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.
Work of gravity
Let us consider a body sliding along an inclined plane with an angle of inclination $α$ and a height $H$.
Let us express $∆x$ in terms of $H$ and $α$:
$∆x=(H)/(sinα)$
Considering that the force of 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 is clear that the work done by gravity depends on the height and does not depend on the angle of inclination of the plane.
It follows that:
- 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;
- when a body moves along a closed trajectory, the work done by gravity is zero, i.e., gravity is a conservative force (forces that have this property are called conservative).
Work of reaction forces, is equal to zero, since the reaction force ($N$) is directed perpendicular to the displacement $∆x$.
Work of friction force
The friction force is directed opposite to the displacement $∆x$ and makes an angle of $180°$ with it, therefore 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
$A_(tr)=μmgHctgα$
Work of 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 force $F↖(→)$ ceases to act at point $x_0$, the spring is compressed under the action of force $F_(control)$.
Let us determine the work of the elastic force when the coordinate of the right end of the spring changes from $x_0$ to $x$. Since the elastic force in this area changes linearly, Hooke’s law can use its average value in this area:
$F_(control av.)=(kx_0+kx)/(2)=(k)/(2)(x_0+x)$
Then the work (taking into account the fact that the directions $(F_(control av.))↖(→)$ and $(∆x)↖(→)$ coincide) is equal to:
$A_(control)=(k)/(2)(x_0+x)(x_0-x)=(kx_0^2)/(2)-(kx^2)/(2)$
It can be shown that the form of the last formula does not depend on the angle between $(F_(control av.))↖(→)$ and $(∆x)↖(→)$. The work of 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 power
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 - per $1$ s).
Power is determined by the formula:
where $N$ is power, $A$ is work done during time $∆t$.
Substituting into the formula $N=(A)/(∆t)$ instead of the work $A$ its expression $A=F|(∆r)↖(→)|cosα$, we obtain:
$N=(F|(∆r)↖(→)|cosα)/(∆t)=Fυcosα$
Power is equal to the product of the magnitudes 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 for $1$ s: $1$ W $= 1$ J/s.
This unit is named in part English inventor J. Watt (Watt), who built the first steam engine. J. Watt himself (1736-1819) used another 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. $= 735.5$ W.
In technology, larger power units are often used - kilowatt and megawatt: $1$ 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 are said to have energy.
The word “energy” (from the Greek energia - action, activity) is often used in everyday life. For example, people who can do work quickly are called energetic, having 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 $υ$. Since kinetic energy is energy due to motion, zero state for it is the state in which the body is at rest. Having found the work necessary to impart a given speed to a body, we will find its kinetic energy.
To do this, let us calculate the work in the area of displacement $∆r↖(→)$ when the directions of the force vectors $F↖(→)$ and displacement $∆r↖(→)$ coincide. In this case the work is equal
where $∆x=∆r$
For the motion of a point with acceleration $α=const$, the expression for displacement has the form:
$∆x=υ_1t+(at^2)/(2),$
where $υ_1$ is the initial speed.
Substituting the expression for $∆x$ into the equation $A=F·∆x$ from $∆x=υ_1t+(at^2)/(2)$ and using Newton’s second law $F=ma$, we obtain:
$A=ma(υ_1t+(at^2)/(2))=(mat)/(2)(2υ_1+at)$
Expressing the acceleration through the initial $υ_1$ and final $υ_2$ velocities $a=(υ_2-υ_1)/(t)$ and substituting in $A=ma(υ_1t+(at^2)/(2))=(mat)/ (2)(2υ_1+at)$ 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 speed 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 the value $υ$.
From $E_K=(mυ)/(2)=(p^2)/(2m)$ it follows 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 body kinetic energy(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 relative position 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 done by a force, depending only on relative position tel. This is the work of gravity $A=mgh_1-mgh_2=mgH$ and the work of elasticity:
$A=(kx_0^2)/(2)-(kx^2)/(2)$
Potential energy of the body interacting with the Earth is called the quantity equal to the product mass $m$ of this body for acceleration free fall$g$ and to the height $h$ of the body above the Earth’s surface:
The potential energy of an elastically deformed body is the quantity equal to half product of the elasticity (stiffness) coefficient $k$ of the body by the square of the deformation $∆l$:
$E_p=(1)/(2)k∆l^2$
Job 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 you to give general definition potential energy.
The potential energy of a system is a quantity that depends on the position of the bodies, the change in 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 performed by internal forces (for example, a fall bodies on the ground under the influence of gravity in the “rock-Earth” system), the energy of the system decreases. Work and changes in potential energy in a system always have opposite signs.
Since work only determines the change in potential energy, then physical meaning in mechanics has only a change in energy. Therefore the choice zero level energy 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 bodies (potential energy) and their speed (kinetic energy).
According to the kinetic energy theorem,
$E_k-E_(k_1)=A_p+A_(pr),$
where $A_p$ is the work of potential forces, $A_(pr)$ is the work of non-potential 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. Taking this into account, we obtain an expression for law of change mechanical energy:
$(E_k+E_p)-(E_(k_1)+E_(p_1))=A_(pr)$
Where left side equality is the change in total mechanical energy, and the right one is the work of non-potential 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 non-potential forces.
A mechanical system in which only potential forces, is called conservative.
In a conservative system $A_(pr) = 0$. It follows 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 Newton's laws of 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 uniformity of time.
Uniformity of time is that for the same initial conditions leakage physical processes does not depend on at what point in time 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.
According to various forms the movements of matter are considered various types energy: mechanical, internal ( equal to the amount kinetic energy of the chaotic movement of molecules relative to the center of mass of the body and the potential energy of interaction of molecules with each other), electromagnetic, chemical (which consists of the kinetic energy of the movement of electrons and electrical energy their interactions with each other and with atomic nuclei), nuclear, etc. From the above it is clear that the division of energy into different types Quite conditional.
Natural phenomena are usually accompanied by the transformation of one type of energy into another. For example, friction of parts of various mechanisms leads to the conversion of mechanical energy into heat, i.e. internal energy. In heat engines, on the contrary, the transformation occurs internal energy to mechanical; V 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's how in modern physics the concept of energy is formulated:
Energy is a general quantitative measure of movement and interaction of all types of matter. Energy does not appear from nothing and does not disappear, it can only move from one form to another. The concept of energy links together all natural phenomena.
Simple mechanisms. Mechanism efficiency
Simple mechanisms are devices that change the magnitude or direction of forces applied to a body.
They are used to move or lift large loads with little effort. These include the lever and its varieties - blocks (movable and fixed), gates, inclined plane and its varieties - wedge, screw, etc.
Lever. Leverage rule
The lever is solid, capable of rotating around a fixed support.
The rule of leverage 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 a 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 trying to turn the lever counterclockwise. Thus, $M_1=M_2$, which is what needed to be proven.
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 pyramids in Ancient Egypt. Without leverage this would not be possible. After all, for example, for the construction of the Cheops pyramid, which has a height of $147$ m, more than two million stone blocks were used, the smallest of which weighed $2.5$ 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 arms: $l_1=l_2=r$. 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 moving block acts similarly to a lever, the arms of which 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 moving block gives a double gain in strength.
Pulley hoist (block system)
A conventional chain hoist consists of $n$ moving and $n$ fixed blocks. Using 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 pulley gives a gain in strength of $2^n$ times:
$F_1=(F_2)/(2^n)$
Screw
A screw is an inclined plane wound around an axis.
The equilibrium condition for the forces acting on the propeller has the form:
$F_1=(F_2h)/(2πr)=F_2tgα, F_1=(F_2h)/(2πR)$
where $F_1$ is the external force applied to the propeller and acting at a distance $R$ from its axis; $F_2$ is the force acting in the direction of the propeller axis; $h$ — propeller pitch; $r$ — average radius threads; $α$ is the angle of inclination of the thread. $R$ is the length of the lever (wrench) rotating the screw with a force of $F_1$.
Efficiency
Coefficient useful action(efficiency) - the ratio of useful work to all expended work.
The efficiency factor is often expressed as a percentage and denoted Greek letter$η$ (“this”):
$η=(A_п)/(A_3)·100%$
where $A_n$ is useful work, $A_3$ is all expended work.
Useful work is always only a part full work which a person spends using one or another mechanism.
Part perfect work is spent to overcome friction forces. Since $A_3 > A_n$, the efficiency is always less than $1$ (or $< 100%$).
Since each of the works in this equality can be expressed as a product of the corresponding force and the distance traveled, it can be rewritten as follows: $F_1s_1≈F_2s_2$.
It follows that, winning with the help of a mechanism in force, we lose the same number of times along 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 of overcoming friction and gravity of the parts of the devices used. Nevertheless, it can be very useful in 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 force of lifting the load by $10$ 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: from the known impulses and energies before the collision, the values of these quantities are determined after the collision. Let us consider the cases of elastic and inelastic impacts.
An impact is called absolutely inelastic, 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 of a system of bodies with masses $m_1$ and $m_2$ (if we are talking about two bodies) before and after the impact:
$m_1(υ_1)↖(→)+m_2(υ_2)↖(→)=(m_1+m_2)υ↖(→)$
It is obvious that the kinetic energy of bodies during an inelastic impact is not conserved (for example, with $(υ_1)↖(→)=-(υ_2)↖(→)$ and $m_1=m_2$ it becomes equal to zero after the impact).
An impact is called absolutely elastic, in which not only the sum of impulses is preserved, but also the sum kinetic energies hitting bodies.
For absolutely elastic impact the equations are valid
$m_1(υ_1)↖(→)+m_2(υ_2)↖(→)=m_1(υ"_1)↖(→)+m_2(υ"_2)↖(→);$
$(m_(1)υ_1^2)/(2)+(m_(2)υ_2^2)/(2)=(m_1(υ"_1)^2)/(2)+(m_2(υ"_2 )^2)/(2)$
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.
Force impulse and body impulse
As has been shown, Newton's second law can be written as
Ft=mv-mv o =p-p o =D p.
The vector quantity Ft, equal to the product of the force and the time of its action, is called impulse of force. The vector quantity p=mv, equal to the product of the mass of a body and its speed, is called body impulse.
In SI, the unit of impulse is taken to be the impulse of a body weighing 1 kg moving at a speed of 1 m/s, i.e. The unit of impulse is the kilogrammeter per second (1 kg m/s).
The change in the momentum of the body D p over time t is equal to the impulse of the force Ft acting on the body during this time.
The concept of impulse is one of fundamental concepts physics. The momentum of a body is one of the quantities capable of maintaining its value unchanged under certain conditions.(but in modulus and in direction).
Conservation of total momentum of a closed-loop system
Closed system call a group of bodies that do not interact with any other bodies that are not part of this group. The forces of interaction between bodies included in a closed system are called internal. (Internal forces are usually denoted by the letter f).
Let's consider the interaction of bodies inside a closed system. Let two balls of the same diameter, made of different substances(i.e. having different masses), roll on a perfectly smooth horizontal surface and collide with each other. During an impact, which we will consider central and absolutely elastic, the velocities and impulses of the balls change. Let the mass of the first ball m 1, its speed before the impact V 1, and after the impact V 1 "; the mass of the second ball m 2, its speed before the impact v 2, after the impact v 2". According to Newton's third law, the interaction forces between the balls are equal in magnitude and opposite in direction, i.e. f 1 = -f 2 .
According to Newton’s second law, the change in the impulses of the balls as a result of their collision is equal to the impulses of the interaction forces between them, i.e.
m 1 v 1 "-m 1 v 1 =f 1 t (3.1)
m 2 v 2 "-m 2 v 2 =f 2 t (3.2)
where t is the interaction time of the balls.
Adding expressions (3.1) and (3.2) term by term, we find that
m 1 v 1 "-m 1 v 1 +m 2 v 2 "-m 2 v 2 =0.
Hence,
m 1 v 1 "+m 2 v 2 "=m 1 v 1 +m 2 v 2
or else
p 1 "+p 2 "=p 1 +p 2 . (3.3)
Let us denote p 1 "+p 2 "=p" and p 1 +p 2 =p.
The vector sum of the momenta of all bodies included in the system is called full impulse of this system. From (3.3) it is clear that p"=p, i.e. p"-p=D p=0, therefore,
p=p 1 +p 2 =const.
Formula (3.4) expresses law of conservation of momentum in closed system
, which is formulated as follows: the total momentum of a closed system of bodies remains constant during any interactions of the bodies of this system with each other.
In other words, internal forces cannot change the total momentum of the system either in magnitude or in direction.
Change in total momentum of an open-loop system
A group of bodies that interact not only with each other, but also with bodies that are not part of this group is called open system. The forces with which bodies not included in this system act on the bodies of a given system are called external (usually external forces are denoted by the letter F).
Let us consider the interaction of two bodies in open system. Changes in the impulses of these bodies occur both under the influence of internal forces and under the influence of external forces.
According to Newton's second law, the changes in the momenta of the bodies in question for the first and second bodies are
D р 1 =f 1 t+F 1 t (3.5)
D р 2 =f 2 t+F 2 t (3.6)
where t is the time of action of external and internal forces.
Adding expressions (3.5) and (3.6) term by term, we find that
D (p 1 +p 2)=(f 1 +f 2)t +(F 1 +F 2)t (3.7)
In this formula, p=p 1 +p 2 is the total momentum of the system, f 1 +f 2 =0 (since according to Newton’s third law (f 1 = -f 2), F 1 +F 2 =F is the resultant of all external forces , acting on the bodies of this system. Taking into account the above, formula (3.7) takes the form
D р=Ft. (3.8)
From (3.8) it is clear that the total momentum of the system changes only under the influence of external forces. If the system is closed, i.e. F=0, then D р=0 and, therefore, р=const. Thus, formula (3.4) is a special case of formula (3.8), which shows under what conditions the total momentum of the system is conserved and under what conditions it changes.
Jet propulsion.
The significance of Tsiolkovsky’s work for astronautics
The movement of a body resulting from the separation of part of its mass from it at a certain speed is called reactive.
All types of motion, except reactive, are impossible without the presence of forces external to a given system, i.e., without the interaction of the bodies of a given system with environment, A to achieve jet propulsion, no interaction of the body with the environment is required. Initially the system is at rest, i.e. its total momentum is zero. When part of its mass begins to be ejected from the system at a certain speed, then (since the total momentum of a closed system, according to the law of conservation of momentum, must remain unchanged) the system receives a speed directed in the opposite direction. Indeed, since m 1 v 1 +m 2 v 2 =0, then m 1 v 1 =-m 2 v 2, i.e.
v 2 = -v 1 m 1 / m 2 .
From this formula it follows that the speed v 2 obtained by a system with mass m 2 depends on the ejected mass m 1 and the speed v 1 of its ejection.
A heat engine in which the traction force arising from the reaction of a jet of escaping hot gases is applied directly to its body is called a reactive engine. Unlike others vehicles device with jet engine can move in outer space.
The founder of the theory space flights is the outstanding Russian scientist Tsiolkovsky (1857 - 1935). He gave general basics theory of jet propulsion, developed the basic principles and schemes of jet aircraft, proved the necessity of using a multi-stage rocket for interplanetary flights. Tsiolkovsky's ideas were successfully implemented in the USSR during the construction of artificial Earth satellites and spacecraft.
Founder practical astronautics is Soviet scientist academician Korolev (1906 - 1966). Under his leadership, the world's first artificial satellite Earth, the first human flight into space took place in human history. The first cosmonaut on Earth was soviet man Yu.A. Gagarin (1934 - 1968).
Questions for self-control:
- How is Newton's second law written in impulse form?
- What is called a force impulse? body impulse?
- What system of bodies is called closed?
- What forces are called internal?
- Using the example of the interaction of two bodies in a closed system, show how the law of conservation of momentum is established. How is it formulated?
- What is the total momentum of a system?
- Can internal forces change the total momentum of a system?
- What system of bodies is called unclosed?
- What forces are called external?
- Establish a formula showing under what conditions the total momentum of the system changes and under what conditions it is conserved.
- What kind of movement is called reactive?
- Can it occur without interaction of a moving body with the environment?
- What law is jet propulsion based on?
- What is the significance of Tsiolkovsky’s work for astronautics?
They change because interaction forces act on each of the bodies, but the sum of the impulses remains constant. This is called law of conservation of momentum.
Newton's second law is expressed by the formula. It can be written in another way, if we remember that acceleration is equal to the rate of change in the speed of a body. For uniformly accelerated motion, the formula will look like:
If we substitute this expression into the formula, we get:
,
This formula can be rewritten as:
The right-hand side of this equality records the change in the product of a body’s mass and its speed. The product of body mass and speed is a physical quantity called body impulse or amount of body movement.
Body impulse is called the product of a body's mass and its speed. This vector quantity. The direction of the momentum vector coincides with the direction of the velocity vector.
In other words, a body of mass m, moving with speed has momentum. The SI unit of impulse is the impulse of a body weighing 1 kg moving at a speed of 1 m/s (kg m/s). When two bodies interact with each other, if the first acts on the second body with a force, then, according to Newton’s third law, the second acts on the first with a force. Let us denote the masses of these two bodies by m 1 and m 2, and their speeds relative to any reference system through and. After a while t as a result of the interaction of bodies, their velocities will change and become equal and . Substituting these values into the formula, we get:
,
,
Hence,
Let us change the signs of both sides of the equality to their opposites and write them in the form
On the left side of the equation is the sum of the initial impulses of two bodies, on the right side is the sum of the impulses of the same bodies over time t. The amounts are equal. So, despite that. that the impulse of each body changes during interaction, the total impulse (the sum of the impulses of both bodies) remains unchanged.
Valid also when several bodies interact. However, it is important that these bodies interact only with each other and are not affected by forces from other bodies not included in the system (or that external forces are balanced). A group of bodies that does not interact with other bodies is called closed system valid only for closed systems.
Having studied Newton's laws, we see that with their help it is possible to solve the basic problems of mechanics if we know all the forces acting on the body. There are situations in which it is difficult or even impossible to determine these values. Let's consider several such situations.When two billiard balls or cars collide, we can state that current forces, that this is their nature, elastic forces act here. However, we will not be able to accurately determine either their modules or their directions, especially since these forces have an extremely short duration of action.When rockets and jets move, we also have little to say about the forces driving specified bodies into motion.In such cases, methods are used that allow one to avoid solving the equations of motion and immediately use the consequences of these equations. At the same time, new physical quantities. Let's consider one of these quantities, called the momentum of the body
An arrow fired from a bow. The longer the contact of the string with the arrow continues (∆t), the greater the change in the arrow's momentum (∆), and therefore, the higher its final speed.
Two colliding balls. While the balls are in contact, they act on each other with forces equal in magnitude, as Newton’s third law teaches us. This means that the changes in their momenta must also be equal in magnitude, even if the masses of the balls are not equal.
After analyzing the formulas, two important conclusions can be drawn:
1. Identical forces acting for the same period of time cause the same changes in momentum different bodies, regardless of the mass of the latter.
2. The same change in the momentum of a body can be achieved either by acting with a small force over a long period of time, or by acting briefly with a large force on the same body.
According to Newton's second law, we can write:
∆t = ∆ = ∆ / ∆t
The ratio of the change in the momentum of a body to the period of time during which this change occurred is equal to the sum of the forces acting on the body.
Having analyzed this equation, we see that Newton's second law allows us to expand the class of problems to be solved and include problems in which the mass of bodies changes over time.
If we try to solve problems with variable mass bodies using the usual formulation of Newton's second law:
then attempting such a solution would lead to an error.
An example of this is the already mentioned jet plane or space rocket, which burn fuel while moving, and the products of this combustion are released into the surrounding space. Naturally, the mass of an aircraft or rocket decreases as fuel is consumed.
Despite the fact that Newton’s second law in the form “the resultant force is equal to the product of the mass of a body and its acceleration” allows us to solve a fairly wide class of problems, there are cases of motion of bodies that cannot be completely described by this equation. In such cases, it is necessary to apply another formulation of the second law, connecting the change in the momentum of the body with the impulse of the resultant force. In addition, there are a number of problems in which solving the equations of motion is mathematically extremely difficult or even impossible. In such cases, it is useful for us to use the concept of momentum.
Using the law of conservation of momentum and the relationship between the momentum of a force and the momentum of a body, we can derive Newton's second and third laws.
Newton's second law is derived from the relationship between the impulse of a force and the momentum of a body.
The impulse of force is equal to the change in the momentum of the body:
Having made the appropriate transfers, we obtain the dependence of force on acceleration, because acceleration is defined as the ratio of the change in speed to the time during which this change occurred:
Substituting the values into our formula, we obtain the formula for Newton’s second law:
To derive Newton's third law, we need the law of conservation of momentum.
Vectors emphasize the vector nature of speed, that is, the fact that speed can change in direction. After transformations we get:
Since the period of time in a closed system was a constant value for both bodies, we can write:
We have obtained Newton's third law: two bodies interact with each other with forces equal in magnitude and opposite in direction. The vectors of these forces are directed towards each other, respectively, the modules of these forces are equal in value.
References
- Tikhomirova S.A., Yavorsky B.M. Physics ( basic level) - M.: Mnemosyne, 2012.
- Gendenshtein L.E., Dick Yu.I. Physics 10th grade. - M.: Mnemosyne, 2014.
- Kikoin I.K., Kikoin A.K. Physics - 9, Moscow, Education, 1990.
Homework
- Define the impulse of a body, the impulse of force.
- How are the impulse of a body related to the impulse of force?
- What conclusions can be drawn from the formulas for body impulse and force impulse?
- Internet portal Questions-physics.ru ().
- Internet portal Frutmrut.ru ().
- Internet portal Fizmat.by ().
