Let us consider the movement of a material point (Fig. 46) in an inertial reference system under the influence of forces caused by the interaction of points with other points and bodies (i.e., arising as a result of the interaction of material objects).

Note that when moving in a non-inertial reference system, relative movements are partially determined by the movement of the reference system itself.

The equations of motion are compiled based on Newton's laws.

Treatise “Mathematical principles of natural philosophy”:

1687 – year of origin theoretical mechanics.

Newton's laws are idealized laws of nature, but for practice this is acceptable within very wide limits.

Let's introduce measures of movement.

Quantity of movement– equal to the product of mass m by the point velocity vector:

where m = const > 0 is a measure of the inertia of matter.

Moment of momentum relative to the origin (Fig. 47):

.

Kinetic energy of a material point:

Later we will show that in a number of cases the movement of a point is more clearly described through or T.

When formulating Newton's laws, we denote:

The force of interaction between points and;

The total force applied to a point M interacting with many points.

Newton's first law: a material point remains in a state of rest or uniform rectilinear motion relative to an inertial reference system until the forces acting on it change this state.

That is, an isolated point is either at rest or moving rectilinearly and uniformly. The reason for the change in movement is outside the point itself.

Newton's second law: the time derivative of the momentum of a material point is geometrically equal to the force applied to the point. Or, with constant mass, the product of the mass of a point and its absolute acceleration is geometrically equal to the force applied to the material point, i.e.

or if m = const.

The connection between the kinematic quantity – acceleration and the dynamic quantity – force through the coefficient of proportionality – mass.

Newton's third law: any two material points interact with each other with forces directed along a straight line connecting these points, equal in magnitude and oppositely directed (Fig. 48).

Let's consider the influence of point M1 with other points (Fig. 49).

For we have acceleration:

The principle of independent action of forces: the acceleration caused by a force is determined only by that force and does not depend on other forces.

Consequence:

; denoting

The geometric sum of accelerations caused by the forces of interaction of point M1 with other points is proportional to the geometric sum of interaction forces – parallelogram rule for adding forces.

What does strength depend on? ?

1) from the coordinates of the point at a given time;

2) from the prehistory of movement (aging);

3) from the environment (temperature);

4) air resistance.

Idealization: forces depend only on the coordinates of the point, on the first derivatives and explicitly on time:

In practice, this is acceptable.

The development of physics has led to a change in some outdated concepts and to the clarification of the boundaries of the region within which Newton’s mechanics is valid: his concept of absolute space has now been replaced by the concept of an inertial frame of reference; it has been established that Newtonian mechanics - classical mechanics - is not applicable if the relative velocities of points are comparable to the speed of light [this is the field of relativistic or Einsteinian mechanics]; Classical mechanics is also inapplicable to the study of microworld phenomena [this is the field of quantum mechanics]. But they are based on classical mechanics. In other areas => classical mechanics gives fairly accurate results.

Security questions:

1. What is called dynamics?

2. List the measures of motion of a material point

3. Formulate Newton's laws.

4. What are the limits of the scope of application of Newton's classical mechanics?

Lecture 16. Differential equations of motion of a point

Let us consider the motion of a free material point in an inertial reference system in Cartesian coordinates. From Newton's 2nd law:

, ,

Moreover, Fx, Fy, Fz – can depend on coordinates, first derivatives, time: .

If the law of motion is known (for example from kinematics):

then => Fx(t), Fy(t), Fz(t). This the first (direct) point dynamics problem.

If the force is known, then to study the motion it is necessary to integrate differential equations - this is the second (inverse) point dynamics problem.

Forms of differential equations of motion

1) Newton’s 2nd law – for momentum.

2) Multiply by (vectorally):

or -angular momentum equation.

[Why? - on one's own. Take into account].

The time derivative of the moment of momentum is geometrically equal to the moment of force.

Detailed entry (coordinate):

3) Multiply scalarly by elementary displacements:

.

- kinetic energy equation.

The differential of the kinetic energy of a point is equal to the elementary work of the sum of forces applied to the point on the actual displacement.

About first integrals(conservation laws).

From differential equations: a function of coordinates, their time derivatives, which is constant by virtue of the equations (that is, its time derivative is zero) => is called the first integral.

We get the following conditions.

If - first integral, then

1) If Fx = 0, then , - integral of momentum ( law of conservation of momentum).

2) If (that is, the projection of the moment of force on the z axis),

,

Integral of angular momentum ( law of conservation of angular momentum).

3) Let us obtain the energy integral.

.

Let the right-hand side be the total differential of some scalar function – force field potential .

To be a total differential:

1) - that is, the field stationary(does not depend on t).

2) with conditions from higher mathematics:

; ;

Otherwise: if and, then and the equation for kinetic energy will be in total differentials:

.

Integrating:

.

Let's introduce potential energy:

.

Then: - energy integral ( law of conservation of mechanical energy).

If the force field is potential and stationary, then the sum of the kinetic and potential energies of a free material point is equal to a constant.

E0 – mechanical energy; is found from the initial conditions.

Energy is conserved, that is, conserved => the field is called conservative.

Let us show that the work of the conservative field forces does not depend on the type of trajectory, but is equal to the difference in the values ​​of the function P at the end and beginning of the movement (Fig. 51).

,

Q.E.D.

.

The work of conservative field forces on a closed displacement is zero (Fig. 52).

Security questions:

1. Formulate the direct and inverse problems of dynamics.

2. Write the equation for the angular momentum of a point.

3. What is called the feather integral of a differential equation?

4. Which force field is called conservative?

Lecture 17. Particular types of force fields

1) Strength depends only from time to time– the field is homogeneous, but not stationary.

.

;

.

Likewise for y and z.

2) Force projections depend only on the corresponding coordinates.

.

Multiplying by dx and integrating:

.

Differentiate again to check:

; .

.

(the sign is taken from the initial conditions).

Separating variables:

.

3) The projection of force depends only from the velocity projection on the same axis.

.

Denoting:

.

Separating variables:

.

Thus, in each of the three special cases of force fields, given the force, mass and initial conditions, expressions for the speed and acceleration of the point are determined.

Security questions:

1. What is the essence of the method of separating variables when solving differential equations?

2. What is special about integrating the equation of motion of a point if the force depends only on the coordinate?

3. In what real problems does force depend on the speed of the point?

Lecture 18. Basics of Point System Dynamics

Let's consider the motion of n free material points relative to the inertial frame of reference (Fig. 53).

Point mass.

Weight of the entire system:

Let us call the center of mass of the system point C, the radius of which is the vector

,

Basic measures of motion of a system of material points:

1. The total momentum of the system (the geometric sum of the momentum of material points).

Where is the speed of the point.

Consider a system of points with constant masses => differentiating:

;

where is the speed of the center of mass.

So,

The amount of motion of a system of material points is equal to the amount of motion of the mass of the entire system concentrated at the center of mass.

2. Sum of angular momentum or angular momentum of the system:

.

is represented as a monomial only in the case of equal velocities of all points of the system.

3. Kinetic energy of the system:

It is also not always presented in a single-term form.

We divide forces into external and internal.

External forces act on the part of the masses outside the system.

Inner forces– interaction forces between points of the system.

Let's denote:

Total external force to a point

The total force of interaction between a point and other points in the system.

The division into internal and external forces is conditional.

Let us obtain some properties of internal forces.

Let's consider points and (Fig. 54).

From Newton's 3rd law:

Internal force per point:

.

Obviously:

.

So, the sum of internal forces and the sum of moments of internal forces are equal to zero relative to any point and any axis.

Let's consider the amount basic work internal forces.

Let , Where,

Distance between points.

Work on elementary actual displacements of interaction forces between two points:

[ - projection onto, including the sign].

Let us denote the sum of elementary works of internal forces:

(d – means “on elementary movements”)

Security questions:

1. What is called the center of mass of a system of material points?

2. Name the main measures of motion of a system of material points.

Physics. 7th grade

Topic: Interaction of bodies

Lesson 21. Addition of forces

Yudina N.A., physics teacher of the highest category, Central Educational Center No. 1409, finalist of the city competition “Teacher of the Year” (Moscow, 2008)

October 27, 2010

Addition of forces - resultant force, resultant force

Good afternoon.

Today is the twenty-first lesson.

Section "Interaction of bodies". And today we will get acquainted with the method of adding forces, when a body is acted upon not by one, but by several forces at once, a resultant force or a resultant force.

Let's take an example. We will hang two weights from the spring, the mass of each of which is 100 g. So, the total mass of the resulting body is 200 g.

This means that the force of gravity that acts on this resulting body is 2 N. Let's try to depict this force of gravity graphically to scale.

Drawing

The scale chosen is 1H - this is a single segment. Then the force of gravity acting on the body =.

Now we will try to attach another weight weighing 100 g.

As we can see, the spring has stretched. The dynamometer shows us a total force of 3N.

Let us again depict the force acting on the first two loads.

Then we add the force of gravity acting on the additional load, .

Please note that both forces are directed along the same straight line in the same direction. The resultant force, let’s find it, for this we need to add the modules of these forces R=F1+F2.

The direction of the resultant will be in the same direction where both forces were directed.

Now let’s turn to an example that will allow us to analyze the situation when forces are directed in different directions.

So, the two teams are in a tug of war. The total force of one team is =500 N. The total force of the second team is =700 N.

Scale: 100 N.

I chose the scale - a single segment corresponds to 100 N.

And then the figure clearly shows: 5 single segments - the force of the first team is 500 N; 7 unit segments - the force of the second command is 700 N. The figure shows that these two forces are directed in different directions along the same straight line. In order to find the resultant of these two forces, it is necessary to subtract the smaller force R = F2-F1 from the larger force in magnitude, and the direction of the resulting force will be in the direction of the larger force.

On the drawing we can indicate the name: – resultant or resultant force.

In the case when not one, but several forces act on a body at once, it is necessary to find their resultant.

It must also be remembered that if several forces act on a body, but, as in this case, these forces are equal in magnitude and opposite in direction, the force of gravity acting on these loads towards the ground, downward, and the elastic force acting upward are these the forces are equal in magnitude and opposite in direction.

In this case, the body will either be at rest, or it can move uniformly and rectilinearly.

Thank you. Goodbye.

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Lesson type: formation of new knowledge.

Lesson methods: research method.

Lesson objectives:

• Educational: show the connection between the material being studied and real life using examples; familiarize students with the concept of resultant force;
• Developmental: developing skills in working with instruments; improve group work skills;
• Educational: cultivate diligence, accuracy and clarity when answering, the ability to see physics around you.

Equipment: dynamometer (spring, demonstration), bodies of various masses, trolley, spring, ruler, multi-media projector. Self-work card.

Lesson progress

1. Goal setting

– What concept have we been studying for several lessons?

– Would you like to know more about power? What exactly?

2. Repetition

• Tell me what you know about strength?
• What significance does it have in life? What is it intended for?
• What forces exist in nature?

– Let’s show the effect of forces on a car. Not one, but several forces can act on a body.

– Give examples in which several forces act on a body.

3. Formation of new knowledge

Let's conduct an experiment:

We hang two weights (a) from the spring, one below the other, and note the length to which the spring stretches. Let's remove these weights and replace them with one weight (b), which stretches the spring to the same length. Let us conclude that there is a force that produces the same effect as several simultaneously acting forces, called resultant.

The designation of this force is R, units of measurement – 1 N.

Fill out the table.

4. Consolidation of the studied material

– Solving problems involving the resultant. ( In the presentation)

– Independent work to find various forces.

Independent work “Strength. Resultant"

5. Homework: paragraph 29, rep. to questions, ex. 11 (1, 2, 3 letters).

Strength. Addition of forces

Any changes in nature occur as a result of interaction between bodies. The ball lies on the ground and will not begin to move unless you push it with your foot; the spring will not stretch if you attach a weight to it, etc. When a body interacts with other bodies, the speed of its movement changes. In physics, they often do not indicate which body and how it acts on a given body, but say that “a force acts on the body.”

Force is a physical quantity that quantitatively characterizes the action of one body on another, as a result of which the body changes its speed. Force is a vector quantity. That is, in addition to the numerical value, the force has a direction. Force is designated by the letter F and in the International System is measured in newtons. 1 newton is the force that a body weighing 1 kg at rest produces in 1 second at a speed of 1 meter per second in the absence of friction. You can measure strength using a special device - a dynamometer.

Depending on the nature of the interaction in mechanics, three types of forces are distinguished:

• gravity,
• elastic force,
• friction force.

As a rule, not one, but several forces act on the body. In this case, the resultant of forces is considered. A resultant force is a force that acts in the same way as several forces simultaneously acting on a body. Using the results of the experiments, we can conclude: the resultant of forces directed along one straight line in one direction is directed in the same direction, and its value is equal to the sum of the values ​​of these forces. The resultant of two forces directed along one straight line in opposite directions is directed towards the greater force and is equal to the difference in the values ​​of these forces.

When several forces act simultaneously on one body, the body moves with acceleration, which is the vector sum of the accelerations that would arise under the action of each force separately. The forces acting on a body and applied to one point are added according to the rule of vector addition.

The vector sum of all forces simultaneously acting on a body is called the resultant force and is determined by the rule of vector addition of forces: $\overrightarrow(R)=(\overrightarrow(F))_1+(\overrightarrow(F))_2+(\overrightarrow(F)) _3+\dots +(\overrightarrow(F))_n=\sum^n_(i=1)((\overrightarrow(F))_i)$.

The resultant force has the same effect on a body as the sum of all forces applied to it.

To add two forces, the parallelogram rule is used (Fig. 1):

Figure 1. Addition of two forces according to the parallelogram rule

In this case, we find the modulus of the sum of two forces using the cosine theorem:

\[\left|\overrightarrow(R)\right|=\sqrt((\left|(\overrightarrow(F))_1\right|)^2+(\left|(\overrightarrow(F))_2\right |)^2+2(\left|(\overrightarrow(F))_1\right|)^2(\left|(\overrightarrow(F))_2\right|)^2(cos \alpha \ ))\ ]

If you need to add more than two forces applied at one point, then use the polygon rule: ~ from the end of the first force draw a vector equal and parallel to the second force; from the end of the second force is a vector equal and parallel to the third force, and so on.

Figure 2. Addition of forces according to the polygon rule

The closing vector drawn from the point of application of forces to the end of the last force is equal in magnitude and direction to the resultant. In Fig. 2 this rule is illustrated by the example of finding the resultant of four forces $(\overrightarrow(F))_1,\ (\overrightarrow(F))_2,(\overrightarrow(F))_3,(\overrightarrow(F) )_4$. Note that the vectors being added do not necessarily belong to the same plane.

The result of a force acting on a material point depends only on its modulus and direction. A solid body has certain dimensions. Therefore, forces of equal magnitude and direction cause different movements of a rigid body depending on the point of application. The straight line passing through the force vector is called the line of action of the force.

Figure 3. Addition of forces applied to different points of the body

If forces are applied to different points of the body and do not act parallel to each other, then the resultant is applied to the point of intersection of the lines of action of the forces (Fig. 3).

A point is in equilibrium if the vector sum of all forces acting on it is equal to zero: $\sum^n_(i=1)((\overrightarrow(F))_i)=\overrightarrow(0)$. In this case, the sum of the projections of these forces onto any coordinate axis is also zero.

The replacement of one force by two, applied at the same point and producing the same effect on the body as this one force, is called the decomposition of forces. The decomposition of forces is carried out, as is their addition, according to the parallelogram rule.

The problem of decomposing one force (the modulus and direction of which are known) into two, applied at one point and acting at an angle to each other, has a unique solution in the following cases, if known:

1. directions of both components of forces;
2. module and direction of one of the component forces;
3. modules of both components of forces.

Let, for example, we want to decompose the force $F$ into two components lying in the same plane with F and directed along straight lines a and b (Fig. 4). To do this, it is enough to draw two lines parallel to a and b from the end of the vector representing F. The segments $F_A$ and $F_B$ will depict the required forces.

Figure 4. Decomposition of the force vector by directions

Another version of this problem is to find one of the projections of the force vector given the force vectors and the second projection. (Fig. 5 a).

Figure 5. Finding the projection of the force vector using given vectors

The problem comes down to constructing a parallelogram along the diagonal and one of the sides, known from planimetry. In Fig. 5b such a parallelogram is constructed and the required component $(\overrightarrow(F))_2$ of the force $(\overrightarrow(F))$ is indicated.

The second solution is to add to the force a force equal to - $(\overrightarrow(F))_1$ (Fig. 5c). As a result, we obtain the desired force $(\overrightarrow(F))_2$.

Three forces~$(\overrightarrow(F))_1=1\ N;;\ (\overrightarrow(F))_2=2\ N;;\ (\overrightarrow(F))_3=3\ N$ applied to one point, lie in the same plane (Fig. 6 a) and make angles~ with the horizontal $\alpha =0()^\circ ;;\beta =60()^\circ ;;\gamma =30()^\ circ $respectively. Find the resultant of these forces.

Let us draw two mutually perpendicular axes OX and OY so that the OX axis coincides with the horizontal along which the force $(\overrightarrow(F))_1$ is directed. Let's project these forces onto the coordinate axes (Fig. 6 b). The projections $F_(2y)$ and $F_(2x)$ are negative. The sum of the projections of forces onto the OX axis is equal to the projection onto this axis of the resultant: $F_1+F_2(cos \beta \ )-F_3(cos \gamma \ )=F_x=\frac(4-3\sqrt(3))(2)\ approx -0.6\ H$. Similarly, for projections onto the OY axis: $-F_2(sin \beta \ )+F_3(sin \gamma =F_y=\ )\frac(3-2\sqrt(3))(2)\approx -0.2\ H$ . The modulus of the resultant is determined by the Pythagorean theorem: $F=\sqrt(F^2_x+F^2_y)=\sqrt(0.36+0.04)\approx 0.64\ Н$. The direction of the resultant is determined using the angle between the resultant and the axis (Fig. 6 c): $tg\varphi =\frac(F_y)(F_x)=\ \frac(3-2\sqrt(3))(4-3\sqrt (3))\approx 0.4$

The force $F = 1kH$ is applied at point B of the bracket and is directed vertically downwards (Fig. 7a). Find the components of this force in the directions of the bracket rods. The required data is shown in the figure.

F = 1 kN = 1000N

$(\mathbf \beta )$ = $30^(\circ)$

$(\overrightarrow(F))_1,\ (\overrightarrow(F))_2$ - ?

Let the rods be attached to the wall at points A and C. The decomposition of the force $(\overrightarrow(F))$ into components along the directions AB and BC is shown in Fig. 7b. This shows that $\left|(\overrightarrow(F))_1\right|=Ftg\beta \approx 577\ H;\ \ $

\[\left|(\overrightarrow(F))_2\right|=F(cos \beta \ )\approx 1155\ H. \]

Answer: $\left|(\overrightarrow(F))_1\right|$=577 N; $\left|(\overrightarrow(F))_2\right|=1155\ Н$