Solution.
To solve the problem, let us consider the physical system "body - gravitational field of the Earth". The body will be considered a material point, and the Earth's gravitational field - homogeneous. The selected physical system is not closed, since during body movement interacts with air.
If we do not take into account the buoyant force acting on the body from the air side, then the change in the total mechanical energy of the system is equal to the work of the air resistance force, i.e.∆ E = A c.
We choose the zero level of potential energy on the surface of the Earth. The only external force in relation to the "body - Earth" system is the air resistance force directed vertically upward. Initial energy of the system E 1, final E 2.
Resistance force work A.
Because the angle between the drag force and the displacement is 180 °, then the cosine is -1, so A = - F c h. Let us equate A.
The considered open physical system can also be described by a theorem on the change in the kinetic energy of a system of interacting objects, according to which the change in the kinetic energy of the system is equal to the work done by external and internal forces during its transition from the initial state to the final state. If you do not take into account the buoyancy force acting on the body from the air side, and the internal one - the force of gravity. Hence∆ E к = A 1 + A 2, where A 1 = mgh - work of gravity, A 2 = F c hcos 180 ° = - F c h - work of the resistance force;∆ E = E 2 - E 1.
Resistance forces are forces that prevent a vehicle from moving. These forces are directed against his movement.
When driving uphill, characterized by the height H p, the projection length V NS on the horizontal plane and the elevation angle of the road α, the following resistance forces act on the car (Figure 3.12): rolling resistance force R To , equal to the sum of the rolling resistance forces of the front (P K |) and rear (P K2) wheels, the resistance to lifting R NS , air resistance force D and resistance force to acceleration R AND . The rolling and rolling resistance forces are related to the characteristics of the road. The sum of these forces is called the road resistance force. R D .
Rice. 3.13. Energy loss for internal friction in the tire:
a - point corresponding to the maximum values of load and deflection of the tire
Rolling resistance force
The appearance of the rolling resistance force during movement is due to energy losses due to internal friction in tires, surface friction of tires on the road and rutting (on deformable roads). Energy losses due to internal friction in a tire can be judged from Fig. 3.13, which shows the relationship between the vertical load on the wheel and the deformation of the tire - its deflection f NS .
When the wheel moves on an uneven surface, the tire deforms under the action of an alternating load. Line α O, which corresponds to the increase in the load deforming the tire, does not coincide with the line aO, corresponding to the removal of the load. The area of the area enclosed between the indicated curves characterizes the energy losses for internal friction between individual parts of the tire (tread, carcass, cord layers, etc.).
The frictional energy loss in the tire is called hysteresis, and the line ОαО - hysteresis loop.
Friction losses in a tire are irreversible, since during deformation it heats up and heat is released from it, which is dissipated into the environment. The energy expended on deformation of the tire is not fully recovered during the subsequent restoration of its shape.
Rolling resistance force R To reaches the highest value when driving on a horizontal road. In this case
where G - vehicle weight, N; f is the rolling resistance coefficient.
When driving uphill and downhill, the rolling resistance force decreases compared to R To on a horizontal road, and the more significant the steeper they are. For this case of motion, the rolling resistance force
where α is the ascent angle, °.
Knowing the force of rolling resistance, you can determine the power, kW,
spent on overcoming this resistance:
where v is the vehicle speed, m / s 2
For a horizontal road cos0 ° = 1 and
Z 
rolling resistance force dependence R To
and power N K from vehicle speed v
are shown in Fig. 3.14
Rolling resistance coefficient
The rolling resistance coefficient significantly affects the energy loss when the vehicle is moving. It depends on many design and operational
Fig 3.15. Rolling resistance coefficient versus
Travel speeds (a), air pressure in the tire (b) and the moment transmitted through the wheel (c)
factors and is determined experimentally. Its average values for various roads at normal tire pressure are 0.01 ... 0.1. Consider the influence of various factors on the rolling resistance coefficient.
Travel speed... When the speed changes in the range of 0 ... 50 km / h, the rolling resistance coefficient changes slightly and can be considered constant in the specified speed range.
With an increase in the speed of movement outside the specified interval, the rolling resistance coefficient significantly increases (Fig. 3.15, a) due to increased energy losses in the tire for friction.
The rolling resistance coefficient depending on the speed of movement can be approximately calculated by the formula
where 
-
vehicle speed, km / h.
The type and condition of the road surface. On paved roads, rolling resistance is mainly due to the deformation of the tire.
As the number of road irregularities increases, the rolling resistance coefficient increases.
On deformable roads, the rolling resistance coefficient is determined by the deformations of the tire and the road. In this case, it depends not only on the type of tire, but also on the depth of the resulting track and the condition of the soil.
The rolling resistance values at the recommended air pressure and tire load levels and average travel speeds on various roads are shown below:
Asphalt and cement concrete highway:
v good condition..................................... 0,007...0,015
in a satisfactory condition ............... 0.015 ... 0.02
Gravel road in good condition .... 0.02 ... 0.025
Cobblestone road in good condition ...... 0.025 ... 0.03
Dirt road, dry, rolled .............. 0.025 ... 0.03
Sand................................................. ................... 0.1 ... 0.3
Icy road, ice ............................... 0.015 ... 0.03
Rolled snow road ............................. 0.03 ... 0.05
Bus type. The rolling resistance coefficient largely depends on the tread pattern, wear, carcass design and the quality of the tire material. Tread wear, fewer cords and improved material quality result in a drop in rolling resistance due to reduced tire energy loss.
Tire pressure... On roads with a hard surface, with a decrease in air pressure in the tire, the rolling resistance coefficient increases (Fig. 3.15, b). On deformable roads, when the tire pressure drops, the track depth decreases, but the internal friction losses in the tire increase. Therefore, for each type of road, a certain air pressure in the tire is recommended, at which the rolling resistance coefficient has a minimum value.
... With an increase in the vertical load on the wheel, the rolling resistance coefficient increases significantly on deformable roads and insignificantly on paved roads.The moment transmitted through the wheel... When the moment is transmitted through the wheel, the rolling resistance coefficient increases (Fig. 3.15, v) due to slippage losses of the tire in the place of its contact with the road. For driven wheels, the rolling resistance coefficient is 10 ... 15% higher than for driven wheels.
The rolling resistance coefficient has a significant effect on fuel consumption and therefore on a vehicle's fuel economy. Studies have shown that even small reductions in this ratio provide measurable fuel savings. Therefore, it is no coincidence that designers and researchers are striving to create such tires, when using which the rolling resistance coefficient will be negligible, but this is a very difficult problem.
When any object moves on the surface or in the air, forces arise that prevent this. They are called forces of resistance or friction. In this article, we will show you how to find the strength of resistance and consider the factors that influence it.
To determine the resistance force, it is necessary to use Newton's third law. This value is numerically equal to the force that must be applied to force the object to move evenly on a flat horizontal surface. This can be done with a dynamometer. The resistance force is calculated by the formula F = μ * m * g. According to this formula, the sought value is directly proportional to body weight. It is worth considering that for the correct calculation it is necessary to choose μ - a coefficient that depends on the material from which the support is made. The material of the object is also taken into account. This coefficient is selected according to the table. For the calculation, the constant g is used, which is equal to 9.8 m / s2. How to calculate the resistance if the body does not move in a straight line, but along an inclined plane? To do this, the cos of the angle must be entered into the original formula. It is on the angle of inclination that the friction and resistance of the surface of bodies to motion depend. The formula for determining the friction on an inclined plane will look like this: F = μ * m * g * cos (α). If the body is moving at a height, then the air friction force acts on it, which depends on the speed of the object. The required value can be calculated by the formula F = v * α. Where v is the speed of movement of the object, and α is the coefficient of resistance of the medium. This formula is only suitable for bodies that move at low speeds. To determine the drag force of jet aircraft and other high-speed units, another one is used - F = v2 * β. To calculate the friction force of high-speed bodies, the square of the speed and the coefficient β are used, which is calculated for each object separately. When an object moves in a gas or liquid, when calculating the friction force, it is necessary to take into account the density of the medium, as well as the mass and volume of the body. Resistance to movement significantly reduces the speed of trains and vehicles. Moreover, moving objects are affected by two types of forces - permanent and temporary. The total frictional force is represented by the sum of two values. To reduce drag and increase machine speed, designers and engineers have come up with a variety of materials with a sliding surface that pushes air off. That is why the front of high-speed trains is streamlined. Fish move very quickly in the water thanks to a streamlined body covered with mucus, which reduces friction. The resistance force does not always have a negative effect on the movement of cars. To get the car out of the mud, it is necessary to pour sand or gravel under the wheels. Due to the increased friction, the car copes well with swampy soil and mud.Resistance to movement in the air is used during skydiving. As a result of the resulting friction between the canopy and the air, the speed of the parachutist is reduced, which makes it possible to practice parachuting without prejudice to life.
3.5. The laws of conservation and change of energy
3.5.1. Law of change full mechanical energy
The change in the total mechanical energy of a system of bodies occurs when the work is done by forces acting both between the bodies of the system and from the side of external bodies.
The change in mechanical energy ∆E of a system of bodies is determined the law of variation of the total mechanical energy:
∆E = E 2 - E 1 = A ext + A tr (res),
where E 1 is the total mechanical energy of the initial state of the system; E 2 - total mechanical energy of the final state of the system; A extern - work performed on the bodies of the system by external forces; A tr (res) is the work performed by the friction (resistance) forces acting inside the system.
Example 30. At a certain height, a body at rest has a potential energy equal to 56 J. By the time it falls to the Earth, the body has a kinetic energy equal to 44 J. Determine the work of the air resistance forces.
Solution. The figure shows two positions of the body: at a certain height (first) and by the time it falls to the Earth (second). The zero level of potential energy is chosen at the surface of the Earth.
The total mechanical energy of a body relative to the Earth's surface is determined by the sum of potential and kinetic energy:
- at some height
E 1 = W p 1 + W k 1;
- by the time of falling to Earth
E 2 = W p 2 + W k 2,
where W p 1 = 56 J - potential energy of the body at a certain height; W k 1 = 0 - kinetic energy of a body at rest at a certain height; W p 2 = 0 J is the potential energy of the body by the time it falls to the Earth; W k 2 = 44 J is the kinetic energy of the body by the time it falls to the Earth.
We find the work of the air resistance forces from the law of change in the total mechanical energy of a body:
where E 1 = W p 1 - total mechanical energy of the body at a certain height; E 2 = W k 2 is the total mechanical energy of the body by the time it falls to the Earth; A ext = 0 - work of external forces (there are no external forces); A res is the work of the air resistance forces.
The sought work of the air resistance forces is thus determined by the expression
A res = W k 2 - W p 1.
Let's make the calculation:
A res = 44 - 56 = −12 J.
The work of the air resistance forces is negative.
Example 31. Two springs with stiffness factors of 1.0 kN / m and 2.0 kN / m are connected in parallel. What work needs to be done to stretch the spring system by 20 cm?
Solution. The figure shows two springs with different stiffness factors connected in parallel.
The external force F → stretching the springs depends on the amount of deformation of the composite spring, therefore, the calculation of the work of the specified force according to the formula for calculating the work of a constant force is invalid.
To calculate the work, we will use the law of change in the total mechanical energy of the system:
E 2 - E 1 = A ext + A res,
where E 1 is the total mechanical energy of the composite spring in the undeformed state; E 2 - total mechanical energy of the deformed spring; A ext - the work of an external force (the required value); A res = 0 - work of resistance forces.
The total mechanical energy of a compound spring is the potential energy of its deformation:
- for an undeformed spring
E 1 = W p 1 = 0,
- for extended spring
E 2 = W p 2 = k total (Δ l) 2 2,
where k total is the general coefficient of stiffness of the composite spring; ∆l is the amount of tension of the spring.
The total stiffness coefficient of two springs connected in parallel is the sum
k total = k 1 + k 2,
where k 1 - coefficient of stiffness of the first spring; k 2 - coefficient of stiffness of the second spring.
We find the work of the external force from the law of change in the total mechanical energy of the body:
A ext = E 2 - E 1,
substituting in this expression the formulas that determine E 1 and E 2, as well as the expression for the total coefficient of stiffness of the composite spring:
A ext = k total (Δ l) 2 2 - 0 = (k 1 + k 2) (Δ l) 2 2.
Let's do the calculation:
A ext = (1.0 + 2.0) ⋅ 10 3 ⋅ (20 ⋅ 10 - 2) 2 2 = 60 J.
Example 32. A bullet weighing 10.0 g, flying at a speed of 800 m / s, hits a wall. The modulus of the force of resistance to the movement of the bullet in the wall is constant and amounts to 8.00 kN. Determine how far the bullet will go into the wall.
Solution. The figure shows two positions of the bullet: when it approaches the wall (first) and to the moment the bullet stops (gets stuck) in the wall (second).
The total mechanical energy of a bullet is the kinetic energy of its movement:
- when the bullet approaches the wall
E 1 = W k 1 = m v 1 2 2;
- by the time the bullet stops (gets stuck) in the wall
E 2 = W k 2 = m v 2 2 2,
where W k 1 - kinetic energy of the bullet when approaching the wall; W k 2 - kinetic energy of the bullet at the moment of its stopping (getting stuck) in the wall; m is the mass of the bullet; v 1 - bullet velocity module when approaching the wall; v 2 = 0 - the magnitude of the bullet velocity by the time it stops (gets stuck) in the wall.
The distance the bullet will penetrate into the wall can be found from the law of change in the total mechanical energy of the bullet:
E 2 - E 1 = A ext + A res,
where E 1 = m v 1 2 2 is the total mechanical energy of the bullet when approaching the wall; E 2 = 0 is the total mechanical energy of the bullet at the moment of its stopping (getting stuck) in the wall; A ext = 0 - work of external forces (there are no external forces); A res is the work of resistance forces.
The work of the resistance forces is determined by the product:
A res = F res l cos α,
where F res is the modulus of the force of resistance to the movement of the bullet; l is the distance the bullet will penetrate into the wall; α = 180 ° - the angle between the directions of the resistance force and the direction of movement of the bullet.
Thus, the law of change in the total mechanical energy of a bullet is explicitly as follows:
- m v 1 2 2 = F res l cos 180 °.
The sought distance is determined by the ratio
l = - m v 1 2 2 F res cos 180 ° = m v 1 2 2 F res
l = 10.0 ⋅ 10 - 3 ⋅ 800 2 2 ⋅ 8.00 ⋅ 10 3 = 0.40 m = 400 mm.
This is a creative assignment for a computer science master class for schoolchildren at FEFU.
The goal of the assignment is to find out how the trajectory of the body will change if the air resistance is taken into account. It is also necessary to answer the question of whether the flight range will still reach its maximum value at a throw angle of 45 °, given air resistance.
In the "Analytical Research" section, the theory is presented. This section can be skipped, but it should be mostly understandable for you, because b O Most of this you went through in school.
The section "Numerical study" contains a description of the algorithm that must be implemented on a computer. The algorithm is simple and concise, so everyone should be fine.
Analytical research
Let's introduce a rectangular coordinate system as shown in the figure. At the initial moment of time, a body with mass m is at the origin. The gravitational acceleration vector is directed vertically downward and has coordinates (0, - g).is the vector of the initial velocity. Let's expand this vector in the basis:
... Here, where is the modulus of the velocity vector, is the throwing angle. Let's write Newton's second law:.
Acceleration at each moment of time is the (instantaneous) rate of change of speed, that is, the derivative of the speed with respect to time:.
Therefore, Newton's 2nd law can be rewritten as follows:
, where is the resultant of all forces acting on the body.
Since the body is affected by the force of gravity and the force of air resistance, then 
.
We will consider three cases:
1) The force of air resistance is 0:.
2) The force of air resistance is oppositely directed with the velocity vector, and its value is proportional to the velocity: 
.
3) The force of air resistance is oppositely directed with the velocity vector, and its value is proportional to the square of the velocity: 
.
First, consider the 1st case.
In this case 
, or . 
It follows that 
(uniformly accelerated motion).
Because ( r is the radius vector), then 
.
From here 
.
This formula is nothing more than the familiar formula for the law of motion of a body with uniformly accelerated motion.
Since then 
.
Considering that and 
, we obtain scalar equalities from the last vector equality:
Let's analyze the resulting formulas.
Find flight time body. Equating y to zero, we get
It follows from this formula that the maximum flight range is achieved at.
Now we will find body trajectory equation... To do this, we express t across x
And substitute the resulting expression for t into equality for y.
The resulting function y(x) is a quadratic function, its graph is a parabola, the branches of which are directed downward.
The movement of a body thrown at an angle to the horizon (excluding air resistance) is described in this video.
Now consider the second case: .
The second law takes the form 
,
from here 
.
Let's write this equality in scalar form:
We got two linear differential equations.
The first equation has a solution
This can be verified by substituting this function into the equation for v x and in the initial condition 
.
Here e = 2.718281828459 ... is Euler's number.
The second equation has a solution
Because 
,
, then in the presence of air resistance, the movement of the body tends to be uniform, in contrast to case 1, when the speed increases indefinitely.
The following video says that the skydiver first moves at an accelerated rate, and then begins to move evenly (even before the parachute is deployed).
Let's find expressions for x and y.
Because x(0) = 0, y(0) = 0, then
It remains for us to consider case 3, when
.Newton's second law has the form
, or 
.In scalar form, this equation has the form:
it system of nonlinear differential equations. This system cannot be solved explicitly, therefore it is necessary to apply numerical modeling.
Numerical research
In the previous section, we saw that in the first two cases, the law of motion of the body can be obtained explicitly. However, in the third case, it is necessary to solve the problem numerically. With the help of numerical methods, we will get only an approximate solution, but a small accuracy will be fine for us. (The number π or the square root of 2, by the way, cannot be written absolutely exactly, therefore, when calculating, some finite number of digits are taken, and this is quite enough.)We will consider the second case, when the force of air resistance is determined by the formula ... Note that for k= 0 we get the first case.
Body speed 
obeys the following equations:
Acceleration components are written on the left-hand sides of these equations 
.
Recall that acceleration is the (instantaneous) rate of change in speed, that is, the derivative of the speed with respect to time.
The velocity components are written on the right-hand sides of the equations. Thus, these equations show how the rate of change of speed is related to speed.
Let's try to find solutions to these equations using numerical methods. To do this, we introduce on the time axis grid: choose a number and consider the moments of time of the form:.
Our task is to approximately calculate the values 
at the nodes of the grid.
Replace the acceleration in the equations ( instant speed speed change) by average speed changes in speed, considering the movement of the body over a period of time:
Now let's substitute the obtained approximations into our equations.
The resulting formulas allow us to calculate the values of the functions at the next grid point, if the values of these functions at the previous grid point are known.
Using the described method, we can obtain a table of approximate values of the velocity components.
How to find the law of motion of a body, i.e. table of approximate coordinates x(t), y(t)? Likewise!
We have
The vx [j] value is equal to the function value, for other arrays it is the same.
Now it remains to write a loop, inside which we will calculate vx through the already calculated value vx [j], and the same with the rest of the arrays. The cycle will be on j from 1 to N.
Do not forget to initialize the initial values vx, vy, x, y with the formulas, x 0 = 0, y 0 = 0.
In Pascal and C for calculating the sine and cosine, there are functions sin (x), cos (x). Note that these functions take an argument in radians.
You need to build a graph of body movement when k= 0 and k> 0 and compare the resulting graphs. Graphs can be built in Excel.
Note that the calculation formulas are so simple that you can use only Excel for calculations and not even use a programming language.
However, in the future, you will need to solve a problem in CATS, in which you need to calculate the time and range of flight of a body, where you cannot do without a programming language.
Please note that you can to test your program and check your graphs by comparing the calculation results for k= 0 with the exact formulas given in the Analytical Study section.
Experiment with your program. Make sure that if there is no air resistance ( k= 0) the maximum flight range at a fixed initial speed is achieved at an angle of 45 °.
And taking into account the air resistance? At what angle is the maximum flight range achieved?
The figure shows the trajectories of the body at v 0 = 10 m / s, α = 45 °, g= 9.8 m / s 2, m= 1 kg, k= 0 and 1, obtained using numerical simulation at Δ t = 0,01.
You can familiarize yourself with the wonderful work of 10-graders from Troitsk, presented at the conference "Start to Science" in 2011. The work is devoted to modeling the motion of a tennis ball thrown at an angle to the horizon (taking into account air resistance). Both numerical modeling and field experiment are used.
Thus, this creative task allows you to get acquainted with the methods of mathematical and numerical modeling, which are actively used in practice, but are little studied at school. For example, these methods were used in the implementation of atomic and space projects in the USSR in the middle of the 20th century.
