Definition

The concept of "traction force" often found in physics problems involving ideas about mechanical power or vehicle movement. Generally speaking, this is a hypothetical force that is introduced for convenience in solving problems.

Let's clarify this idea. Consider the movement of a bus. The traction force (let's denote it as $(\overline(F))_t$) in this case is the static friction force, which acts on the lower points of the wheels from the side of the highway surface. To implement the movement of the bus on the road, wheels vehicle rotates the motor so that the friction force is directed in the direction of movement (Fig. 1). In this case, we define the traction force as the friction force that arises between the drive wheels and the surface on which the wheels roll. If there is no friction force (the wheel is on the ice), then the bus does not move because the wheels slip. The friction that appears between the wheels and the road surface creates forward movement.

Since the traction force depends on the friction force, then to increase the value of $F_t\ $ the friction should be increased. Friction increases with increasing friction coefficient and (or) with increasing normal pressure force, which depends on body weight.

The question arises about the need to introduce some kind of traction force instead of using the usual friction force. When separating the traction force and the movement resistance force from the external forces that act on our bus, the equations of motion have a universal form, and using the traction force, the useful mechanical power ($N$) is simply expressed:

where $\overline(v)$ is the speed of the body (in our case, a bus).

Note that the traction force does not have a clearly defined formula, like, for example, the gravitational force or the Archimedes force and other forces. It is often calculated using Newton's second law and considering all the forces that act on the body.

Thrust force

I.V. was the first to obtain the equations of motion of bodies of variable mass and the formula for calculating the reactive force. Meshchersky in 1897. The reactive force formula is the basis for calculating the thrust force of rocket and turbo-rocket engines of all systems.

Let the rocket move with speed $\overline(v)$ relative to the Earth. Along with it, part of the fuel moves at the same speed, which burns out in the next second. During combustion, the combustion products of this part of the fuel receive an additional speed $\overline(u)$ relative to the rocket. Relative to the Earth, they have a speed of $\overline(v)-\overline(u)$. At the same time, the rocket itself increases speed. After ejection, combustion products do not interact with the rocket. Therefore, the rocket system plus fuel combustion products is considered as a system of two bodies that interact during combustion according to the laws of inelastic impact. Let a rocket jet engine emit a mass of $\mu$ fuel combustion products every second. Using the law of conservation of momentum and Newton’s second law, it is obtained that the modulus of the reactive thrust of the engine ($R$) of the rocket is equal to:

Formula (2) shows that the reactive force that acts on a body of variable mass is proportional to the mass of separated particles per unit time and the speed of movement of these particles relative to the body.

Examples of problems with solutions

Example 1

Exercise. The traction force acting on a body located on an inclined plane (Fig. 2) is directed upward along this plane (Fig. 2). What is its value if the mass of the body is $m$, the angle of inclination of the plane is $\alpha ,\ $the acceleration of the body is $a$? The coefficient of friction of the body on the plane is equal to $\mu $. A body moves at a constant speed uphill.

Solution. Let's write down Newton's second law for forces acting on a body, taking into account that the body moves uniformly:

Let us write the projections of equation (1.1) on the X and Y axes:

\[\left\( \begin(array)(c) X:\ -mg(\sin \alpha +\ )F-F_(tr)=0\left(1.2\right);;\ \\ Y:\ N-mg(\cos \alpha =0\left(1.3\right).\ ) \end(array) \right.\]

The friction force is related to the normal pressure force as:

Let us express $N$ from (1.3), use expression (1.4), and obtain from (1.2) the thrust force:

\[-mg(\sin \alpha +\ )F-\mu mg(\cos \alpha \ )=0\to F=\mu mg(\cos \alpha \ )+mg(\sin \alpha .\ ) \]

Answer.$F=mg(\mu (\cos \alpha \ )+(\sin \alpha).\ )$

Example 2

Exercise. A rocket with a mass (at the initial moment of time) equal to $M,$ was launched vertically upward. The relative emission rate of combustion products is $u$, the fuel consumption is $\mu$. What will be the acceleration of the rocket after time $t$ after launch, if air resistance is not taken into account and the gravity field is assumed to be uniform.

Solution. Let's make a drawing.

Two forces will act on the rocket (from the conditions of the problem): gravity and reactive thrust. Let's write down the equation of rocket motion:

In the projection onto the Y axis, we write equation (2.1) as:

The thrust force can be found as:

Taking into account equality (2.3), we transform the equation to the form:

\[\mu u-mg=ma\to a=\frac(\mu u-mg)(m)\left(2.4\right).\]

The mass of the rocket at time $t$ is equal to:

Substituting (2.5) into (2.4) we have:

Answer.$a=\frac(\mu u)(M-\mu t)-g.$

Jet thrust is usually considered as the reaction force of particles being separated. The point of application is considered to be the center of the outflow - the center of the engine nozzle cut, and the direction is opposite to the velocity vector of the outflow of combustion products (or the working fluid, in the case of a non-chemical engine). That is, jet thrust:

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✪ Conservation of momentum: jet propulsion

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✪ Is it true that...? #4-Jet thrust?!

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Jet propulsion in nature

Proof

M p ⋅ Δ v → Δ t = − Δ m t Δ t ⋅ u → (\displaystyle m_(p)\cdot (\frac (\Delta (\vec (v)))(\Delta t))=-(\ frac (\Delta m_(t))(\Delta t))\cdot (\vec (u)))

F → p = m p ⋅ a → = − u → ⋅ Δ m t Δ t (\displaystyle (\vec (F))_(p)=m_(p)\cdot (\vec (a))=-(\vec (u))\cdot (\frac (\Delta m_(t))(\Delta t)))

Meshchersky equation

If on a rocket, except reactive force F → p (\displaystyle (\vec (F))_(p)), external force acts F → (\displaystyle (\vec (F))), then the motion dynamics equation will take the form:

M p ⋅ Δ v → Δ t = F → + F → p ⇔ (\displaystyle m_(p)\cdot (\frac (\Delta (\vec (v)))(\Delta t))=(\vec ( F))+(\vec (F))_(p)\Leftrightarrow ) m p ⋅ Δ v → Δ t = F → + (− u → ⋅ Δ m t Δ t) (\displaystyle m_(p)\cdot (\frac (\Delta (\vec (v)))(\Delta t)) =(\vec (F))+(-(\vec (u))\cdot (\frac (\Delta m_(t))(\Delta t))))

Meshchersky's formula is a generalization

Any problem in mechanics can be solved using Newton's laws. However, the application of the law of conservation of momentum in many cases greatly simplifies the solution. The law of conservation of momentum is of great importance for the study of jet propulsion.

What kind of movement is called reactive?

Jet motion is understood as the movement of a body that occurs when some part of it separates at a certain speed relative to the body, for example, when combustion products flow out of the nozzle of a jet aircraft. In this case, a so-called reactive force appears, imparting acceleration to the body.

Observing jet motion is very simple. Inflate a child's rubber ball and release it. The ball will quickly rise upward (Fig. 5.4). The movement, however, will be short-lived. The reactive force acts only as long as the outflow of air continues.

Rice. 5.4

The main feature of the reactive force is that it occurs without any interaction with external bodies. There is only interaction between the rocket and the stream of matter flowing out of it.

The force that imparts acceleration to a car or pedestrian on the ground, a steamship on the water or a propeller-driven airplane in the air arises only due to the interaction of these bodies with the ground, water or air.

When the fuel combustion products flow out, due to the pressure in the combustion chamber, they acquire a certain speed relative to the rocket and, therefore, a certain momentum. Therefore, in accordance with the law of conservation of momentum, the rocket itself receives an impulse of the same magnitude, but directed in the opposite direction.

The mass of the rocket decreases over time. A rocket in flight is a body of variable mass. To calculate its motion, it is convenient to apply the law of conservation of momentum.

Meshchersky equation

Let us derive the equation of motion of the rocket and find an expression for the reactive force. We will assume that the speed of the gases flowing out of the rocket relative to the rocket is constant and equal to . External forces do not act on the rocket: it is in outer space far from stars and planets.

Let at some point in time the speed of the rocket relative to the inertial system associated with the stars be equal to (Fig. 5.5, a), and the mass of the rocket be equal to M. After a short time interval Δt, the mass of the rocket will become equal

where μ is fuel consumption(1).

Rice. 5.5

During this period of time, the speed of the rocket will change by Δ and become equal to 1 = + Δ. The speed of gas outflow relative to the selected inertial reference frame is equal to + (Fig. 5.5,b), since before the start of combustion the fuel had the same speed as the rocket.

Let us write down the law of conservation of momentum for the rocket-gas system:

Opening the brackets, we get:

The term μΔtΔ can be neglected in comparison with the others, since it contains the product of two small quantities (this quantity is said to be of the second order of smallness). After bringing similar terms we will have:

This is one of Meshchersky’s equations (2) for the motion of a body of variable mass, obtained by him in 1897.

If we introduce the notation p = -μ, then equation (5.4.1) will coincide in form with Newton’s second law. However, the body mass M is not constant here, but decreases with time due to the loss of matter.

The value p = -μ is called reactive force. It appears as a result of the outflow of gases from the rocket, is applied to the rocket and is directed opposite to the speed of the gases relative to the rocket. Reactive force is determined only by the speed of gas flow relative to the rocket and fuel consumption. It is important that it does not depend on the details of the engine design. It is only important that the engine ensures the outflow of gases from the rocket at a speed with fuel consumption μ. The reactive force of space rockets reaches 1000 kN.

If external forces act on a rocket, then its movement is determined by the reactive force and the sum of external forces. In this case, equation (5.4.1) will be written as follows:

The principle of jet propulsion is based on the fact that flowing from jet engine gases receive impulse. The rocket acquires the same magnitude of impulse.

Self-test questions

(1) Fuel consumption is the ratio of the mass of burned fuel to the time of its combustion.

(2) Meshchersky I. V. (1859-1935) - Professor of the St. Petersburg Polytechnic Institute. His works on the mechanics of bodies of variable mass steel theoretical basis rocket technology.

The main goal in the transmission of electricity is to increase the efficiency of networks. Therefore, it is necessary to reduce losses. The main cause of losses is reactive power, the compensation of which significantly improves the quality of electricity.

Reactive power causes unnecessary heating of wires, electrical substations are overloaded. Transformer power and cable sections are forced to be overestimated, the network voltage decreases.

The concept of reactive power

To find out what reactive power is, it is necessary to define other possible types power. When an active load (resistor) exists in the circuit, only active power is consumed, which is completely spent on energy conversion. This means that we can formulate what active power is - that at which the current makes effective work.

On DC Only active power is consumed, calculated according to the formula:

Measured in watts (W).

In electrical circuits with alternating current in the presence of active and reactive loads, the power indicator is summed from two components: active and reactive power.

Capacitive (capacitors). Characterized by a phase advance of current compared to voltage;
Inductive (coils). Characterized by a phase lag of the current relative to the voltage.

If we consider a circuit with alternating current and a connected active load (heaters, kettles, filament light bulbs), the current and voltage will be in phase, and the total power taken at a certain time cutoff is calculated by multiplying the voltage and current readings.

However, when the circuit contains reactive components, the voltage and current readings will not be in phase, but will differ by a certain amount determined by the offset angle "φ". Taking advantage in simple language, it is said that a reactive load returns as much energy to the circuit as it consumes. As a result, it turns out that for the active power consumption, the indicator will be zero. At the same time, a reactive current flows through the circuit, doing no effective work. Consequently, reactive power is consumed.

Reactive power is the part of the energy that allows you to set the electromagnetic fields required by the equipment alternating current.

Reactive power is calculated using the formula:

Q = U x I x sin φ.

The unit of measure for reactive power is VAr (reactive voltampere).

Expression for active power:

P = U x I x cos φ.

The relationship of active, reactive and apparent power for a sinusoidal variable current is represented geometrically by the three sides of a right-angled triangle, called the power triangle. AC electrical circuits consume two types of energy: active power and reactive power. In addition, the value of active power is never negative, while reactive power can be either positive (with an inductive load) or negative (with a capacitive load).

Important! It can be seen from the power triangle that it is always beneficial to reduce the reactive component in order to increase the efficiency of the system.

Apparent power is not found as an algebraic sum of active and reactive power values, it is a vector sum of P and Q. Its quantitative value is calculated by taking the square root of the sum of squares of power indicators: active and reactive. The apparent power can be measured in VA (voltampere) or its derivatives: kVA, mVA.

In order for the apparent power to be calculated, the phase difference between the sinusoidal values U and I must be known.

Power factor

Using a geometrically represented vector picture, you can find the ratio of the sides of the triangle corresponding to the useful and total power, which will be equal to the cosine phi or power coefficient:

This coefficient determines the efficiency of the network.

The number of watts consumed is the same as the number of volts consumed at a power factor of 1 or 100%.

Important! The greater the cos φ, or the smaller the shift angle of the sinusoidal values of current and voltage, the closer the total power is to the active value.

If, for example, there is a coil for which:

P = 80 W;
Q = 130 VAr;
then S = 152.6 BA as root mean square;
cos φ = P/S = 0.52 or 52%

We can say that the coil requires 130 VAr of total power to do 80 W of useful work.

Correction cos φ

To correct cos φ, the fact is used that with a capacitive and inductive load, the reactive energy vectors are in antiphase. Since most loads are inductive, by connecting a capacitor, you can increase cos φ.

Main consumers of reactive energy:

Transformers. They are windings that have inductive coupling and transform currents and voltages through magnetic fields. These devices are the main element of electrical networks that transmit electricity. Losses especially increase when working at Idling and at low load. Transformers are widely used in production and in everyday life;
Induction furnaces, in which metals are melted by creating eddy currents in them;
Asynchronous motors. The largest consumer of reactive energy. The torque in them is created by means of an alternating magnetic field stator;
Electric power converters, such as power rectifiers used to power the contact network of railway transport and others.

Capacitor banks are connected at electrical substations to control the voltage within specified levels. The load varies throughout the day with morning and evening peaks, as well as throughout the week, decreasing on weekends, which changes the voltage readings. By connecting and disconnecting capacitors its level is varied. This is done manually and using automation.

How and where cos φ is measured

Reactive power is checked by changing cos φ with a special device - a phase meter. Its scale is graduated in quantitative values of cos φ from zero to one in the inductive and capacitive sectors. Fully compensate Negative influence inductance will not succeed, but it is possible to get closer to the desired value - 0.95 in the inductive zone.

Phase meters are used when working with installations that can affect the operating mode of the electrical network through regulation of cos φ.

Since financial calculations for consumed energy also take into account its reactive component, factories install automatic compensators on capacitors, the capacity of which can vary. Networks typically use static capacitors;
When regulating cos φ in synchronous generators by changing the exciting current, it is necessary to monitor it visually in manual operating modes;
Synchronous compensators, which are synchronous motors operating without load, supply energy to the network in overexcitation mode, which compensates for the inductive component. To regulate the exciting current, observe the readings of cos φ using a phase meter.

Power factor correction is one of the most effective investments for reducing energy costs. At the same time, the quality of the energy received improves.

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Traction force can be determined through useful power, and vehicle speed (v):

For a car going up a hill that has a slope, the mass of the car m traction force (F T) will enter into the equation:

where a is the acceleration with which the car is moving.

Traction force units

The basic unit of force in the SI system is: =N

In GHS: =din

Traction formula

If a body has acceleration when moving, then, in addition to all others, it is necessarily acted upon by a certain force, which is the traction force at the moment in time under consideration. In fact, if a body moves in a straight line and at a constant speed, then the traction force also acts, since the body must overcome resistance forces. Typically, the traction force is found by considering the forces acting on the body, finding the resultant and applying Newton's second law. There is no strictly defined formula for traction force.

It should not be assumed that the traction force of, for example, a vehicle acts from the engine, since internal forces cannot change the speed of the system as a whole, which would be in conflict with the law of conservation of momentum. However, it should be noted that in order to obtain the required direction from the static friction force, the motor rotates the wheels, the wheels “cling to the road” and a traction force is generated. Theoretically, it would be possible not to use the concept of “thrust force”, but to talk about the static friction force or the air reaction force. But it is more convenient to divide the external forces that act on transport into two parts, with some forces called traction forces, and others - resistance forces. This is done so that the equations of motion do not lose their universal form and the useful mechanical power (P) has a simple expression:

Examples of problem solving

Example

Exercise. A car with a mass of 1 ton, when moving on a horizontal surface, is subject to a friction force equal to = 0.1 of the force of gravity. What will be the traction force if the car moves with an acceleration of 2 m/s?

Solution. Let's make a drawing.