The vector sum of all forces acting on a body. The main vector is the vector sum of all forces applied to the body

A) circle.

C) parabola.

D) the trajectory can be any.

E) straight.

2. If the bodies are separated by airless space, then heat transfer between them is possible

A) thermal conductivity and convection.

B) radiation.

C) thermal conductivity.

D) convection and radiation.

E) convection.

3. Electrons and neutrons have electrical charges

A) electron – negative, neutron – positive.

B) electron and neutron – negative.

C) electron – positive, neutron – negative.

D) electron and neutron – positive.

E) electron – negative, neutron – has no charge.

4. The current required to perform work equal to 250 J with a light bulb rated at 4V and for 3 minutes is equal to

5. As a result of a spontaneous transformation, the nucleus of a helium atom flew out of the atomic nucleus as a result of the following radioactive decay

A) gamma radiation.

B) two-proton decay.

C) alpha decay.

D) proton decay.

E) beta decay.

6. A point on the celestial sphere, which is designated by the same sign as the constellation Cancer, is a point

A) parade of planets

B) vernal equinox

C) autumnal equinox

D) summer solstice

E) winter solstice

7. The movement of a truck is described by the equations x1= - 270 + 12t, and the movement of a pedestrian along the side of the same highway by the equation x2= - 1.5t. The meeting time is

8. If a body is thrown upward at a speed of 9 m/s, then it will reach its maximum height in (g = 10 m/s2)

9. Under the influence of a constant force equal to 4 N, a body with a mass of 8 kg will move

A) uniformly accelerated with an acceleration of 0.5 m/s2

B) uniformly accelerated with an acceleration of 2 m/s2

C) uniformly accelerated with an acceleration of 32 m/s2

D) uniformly at a speed of 0.5 m/s

E) uniformly at a speed of 2 m/s

10. The power of the trolleybus traction motor is 86 kW. The work that the engine can do in 2 hours is

A) 619200 kJ.

C) 14400 kJ.

E) 17200 kJ.

11. Potential energy of an elastically deformed body when the deformation increases by 4 times

A) will not change.

B) will decrease by 4 times.

C) will increase 16 times.

D) will increase by 4 times.

E) will decrease by 16 times.

12. Balls with masses m1 = 5 g and m2 = 25 g move towards each other at speeds υ1 = 8 m/s and υ2 = 4 m/s. After an inelastic impact, the speed of the ball m1 is equal (the direction of the coordinate axis coincides with the direction of motion of the first body)

13. With mechanical vibrations

A) only potential energy is constant

B) both potential energy and kinetic energy are constant

C) only kinetic energy is constant

D) only the total mechanical energy is constant

E) energy is constant in the first half of the period

14. If tin is at the melting point, then melting 4 kg will require an amount of heat equal to (J/kg)

15. An electric field of intensity 0.2 N/C acts on a charge of 2 C with a force

16. Establish the correct sequence of electromagnetic waves as the frequency increases

1) radio waves, 2) visible light, 3) x-rays, 4) infrared radiation, 5) ultraviolet radiation

A) 4, 1, 5, 2, 3

B) 5, 4, 1, 2, 3

C) 3, 4, 5, 1, 2

D) 2, 1, 5, 3, 4

E) 1, 4, 2, 5, 3

17. A student cuts sheet metal by applying a force of 40 N to the handles of the scissors. The distance from the axis of the scissors to the point of application of the force is 35 cm, and the distance from the axis of the scissors to the sheet metal is 2.5 cm. The force required to cut the sheet metal

18. The area of ​​the small piston of a hydraulic press is 4 cm2, and the area of ​​the large one is 0.01 m2. The pressure force on the large piston is greater than the pressure force on the small piston in

B) 0.0025 times

E) 0.04 times

19. A gas, expanding at a constant pressure of 200 Pa, did 1000 J of work. If the gas initially occupied a volume of 1.5 m, then the new volume of gas is equal to

20. The distance from the object to the image is 3 times greater than the distance from the object to the lens. This is a lens...

A) biconcave

B) flat

C) collecting

D) scattering

E) flat-concave

The mechanical action of bodies on each other is always their interaction.

If body 1 acts on body 2, then body 2 necessarily acts on body 1.

For example,the driving wheels of an electric locomotive (Fig. 2.3) are acted upon by static friction forces from the rails, directed towards the movement of the electric locomotive. The sum of these forces is the traction force of the electric locomotive. In turn, the drive wheels act on the rails by static friction forces directed in the opposite direction.

A quantitative description of mechanical interaction was given by Newton in his third law of dynamics.

For material points this law is formulated So:

Two material points act on each other with forces equal in magnitude and directed oppositely along a straight line connecting these points(Fig.2.4):
.

The third law is not always true.

Performed strictly

in case of contact interactions,

during the interaction of bodies at rest at some distance from each other.

Let us move from the dynamics of an individual material point to the dynamics of a mechanical system consisting of material points.

For -of that material point of the system, according to Newton’s second law (2.5), we have:

. (2.6)

Here And - mass and speed -that material point, - the sum of all forces acting on it.

The forces acting on a mechanical system are divided into external and internal. External forces act on points of a mechanical system from other, external bodies.

Inner forces act between points of the system itself.

Then force in expression (2.6) can be represented as the sum of external and internal forces:

, (2.7)

Where
– the resultant of all external forces acting on -that point of the system; - internal force acting on this point from the side th.

Let's substitute expression (2.7) into (2.6):

, (2.8)

summing the left and right sides of equations (2.8), written for all material points of the system, we obtain

. (2.9)

According to Newton's third law, the interaction forces -that and -points of the system are equal in magnitude and opposite in direction
.

Therefore, the sum of all internal forces in equation (2.9) is equal to zero:

. (2.10)

The vector sum of all external forces acting on the system is called the main vector of external forces

. (2.11)

Reversing the operations of summation and differentiation in expression (2.9) and taking into account the results (2.10) and (2.11), as well as the definition of the momentum of the mechanical system (2.3), we obtain

- basic equation for the dynamics of translational motion of a rigid body.

This equation expresses law of change of momentum of a mechanical system: the time derivative of the momentum of a mechanical system is equal to the main vector of external forces acting on the system.

2.6. Center of mass and the law of its motion.

Center of mass(inertia) of a mechanical system is called dot , whose radius vector is equal to the ratio of the sum of the products of the masses of all material points of the system by their radius vectors to the mass of the entire system:

(2.12)

Where And - mass and radius vector -that material point, -the total number of these points,
–total mass of the system.

If the radius vectors are drawn from the center of mass , That
.

Thus, the center of mass is a geometric point , for which the sum of the products of the masses of all material points forming a mechanical system by their radius vectors drawn from this point is equal to zero.

In the case of continuous distribution of mass in the system (in the case of an extended body), the radius vector of the center of mass of the system is:

,

Where r– radius vector of a small element of the system, the mass of which is equal todm, integration is carried out over all elements of the system, i.e. throughout the entire mass m.

Differentiating formula (2.12) with respect to time, we obtain

expression for center of mass velocity:

Center of mass speed of a mechanical system is equal to the ratio of the momentum of this system to its mass.

Then impulse of the systemis equal to the product of its mass and the speed of the center of mass:

.

Substituting this expression into the basic equation of the dynamics of translational motion of a rigid body, we have:

(2.13)

- the center of mass of a mechanical system moves as a material point, the mass of which is equal to the mass of the entire system and which is acted upon by a force equal to the main vector of external forces applied to the system.

Equation (2.13) shows that to change the speed of the center of mass of the system, it is necessary that an external force act on the system. Internal forces of interaction between parts of the system can cause changes in the speeds of these parts, but cannot affect the total momentum of the system and the speed of its center of mass.

If the mechanical system is closed, then
and the speed of the center of mass does not change over time.

Thus, center of mass of a closed system either at rest or moving at a constant speed relative to an inertial reference frame. This means that a reference system can be associated with the center of mass, and this system will be inertial.

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

Definition 1

The vector sum of all forces simultaneously acting on a body is the force resultant, which is determined by the rule of vector addition of forces:

R → = F 1 → + F 2 → + F 3 → + . . . + F n → = ∑ i = 1 n F i → .

The resultant force acts on a body in the same way as the sum of all forces acting on it.

To add 2 forces use rule parallelogram(picture 1).

Picture 1 . Addition of 2 forces according to the parallelogram rule

Let us derive the formula for the modulus of the resultant force using the cosine theorem:

R → = F 1 → 2 + F 2 → 2 + 2 F 1 → 2 F 2 → 2 cos α

Definition 3

If it is necessary to add more than 2 forces, use polygon rule: from the end
The 1st force must draw a vector equal and parallel to the 2nd force; from the end of the 2nd force it is necessary to draw a vector equal and parallel to the 3rd force, etc.

Figure 2. Addition of forces using the polygon rule

The final 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 force. Figure 2 clearly illustrates an example of finding the resultant forces from 4 forces: F 1 →, F 2 →, F 3 →, F 4 →. Moreover, the summed vectors do not necessarily have to be in the same plane.

The result of the force acting on a material point will depend only on its modulus and direction. A solid body has certain dimensions. Therefore, forces with the same magnitudes and directions cause different movements of a rigid body depending on the point of application.

Definition 4

Line of action of force called a straight line passing through the force vector.

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 (Figure 3 ). A point will be in equilibrium if the vector sum of all forces acting on it is equal to 0: ∑ i = 1 n F i → = 0 → . In this case, the sum of the projections of these forces onto any coordinate axis is also equal to 0.

Decomposition of forces into two components- this is the replacement of one force by 2, applied at the same point and producing the same effect on the body as this one force. The decomposition of forces is carried out, like addition, by the parallelogram rule.

The problem of decomposing one force (the modulus and direction of which are given) into 2, applied at one point and acting at an angle to each other, has a unique solution in the following cases when the following are known:

• directions of 2 component forces;
• module and direction of one of the component forces;
• modules of 2 component forces.

It is necessary to decompose the force F into 2 components located in the same plane with F and directed along straight lines a and b (Figure 4 ). Then it is enough to draw 2 straight lines from the end of the vector F, parallel to straight lines a and b. The segment F A and the segment F B represent the required forces.

Figure 4. Decomposition of the force vector in directions

Example 2

The second version of this problem is to find one of the projections of the force vector using the given force vectors and the 2nd projection (Figure 5 a).

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

In the second version of the problem, it is necessary to construct a parallelogram along the diagonal and one of the sides, as in planimetry. Figure 5 b shows such a parallelogram and indicates the desired component F 2 → force F → .

So, the 2nd solution: add to the force a force equal to - F 1 → (Figure 5 c). As a result, we obtain the desired force F →.

Example 3

Three forces F 1 → = 1 N; F 2 → = 2 N; F 3 → = 3 N are applied to one point, are in the same plane (Figure 6 a) and make angles with the horizontal α = 0 °; β = 60°; γ = 30° respectively. It is necessary to find the resultant force.

Solution

Figure 6. Finding the resultant force from given vectors

Let's draw mutually perpendicular axes O X and O Y so that the O X axis coincides with the horizontal along which the force F 1 → is directed. Let's make a projection of these forces onto the coordinate axes (Figure 6 b). The projections F 2 y and F 2 x are negative. The sum of the projections of forces onto the coordinate axis O X is equal to the projection onto this axis of the resultant: F 1 + F 2 cos β - F 3 cos γ = F x = 4 - 3 3 2 ≈ - 0.6 N.

Similarly, for projections onto the O Y axis: - F 2 sin β + F 3 sin γ = F y = 3 - 2 3 2 ≈ - 0.2 N.

We determine the modulus of the resultant using the Pythagorean theorem:

F = F x 2 + F y 2 = 0.36 + 0.04 ≈ 0.64 N.

We find the direction of the resultant using the angle between the resultant and the axis (Figure 6 c):

t g φ = F y F x = 3 - 2 3 4 - 3 3 ≈ 0.4.

Example 4

A force F = 1 kN is applied at point B of the bracket and is directed vertically downward (Figure 7 a). It is necessary to find the components of this force in the directions of the bracket rods. All necessary data is shown in the figure.

Solution

Figure 7. Finding the components of force F in the directions of the bracket rods

Given:

F = 1 k N = 1000 N

Let the rods be screwed to the wall at points A and C. Figure 7 b shows the decomposition of the force F → into components along the directions A B and B C. From here it is clear that

F 1 → = F t g β ≈ 577 N;

F 2 → = F cos β ≈ 1155 N.

Answer: F 1 → = 557 N; F 2 → = 1155 N.

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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 - 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; $\left|(\overrightarrow(F))_2\right|=1155\ Н$

According to Newton's first law, in inertial frames of reference, a body can change its speed only if other bodies act on it. The mutual action of bodies on each other is expressed quantitatively using such a physical quantity as force (). A force can change the speed of a body, both in magnitude and in direction. Force is a vector quantity; it has a modulus (magnitude) and a direction. The direction of the resultant force determines the direction of the acceleration vector of the body on which the force in question acts.

The basic law by which the direction and magnitude of the resultant force is determined is Newton’s second law:

where m is the mass of the body on which the force acts; - the acceleration that the force imparts to the body in question. The essence of Newton's second law is that the forces that act on a body determine the change in the speed of the body, and not just its speed. It must be remembered that Newton's second law works for inertial frames of reference.

If several forces act on a body, then their combined action is characterized by the resultant force. Let us assume that several forces act on the body simultaneously, and the body moves with an acceleration equal to the vector sum of the accelerations that would appear under the influence of each of the forces separately. The forces acting on the body and applied to one point must be added according to the rule of vector addition. The vector sum of all forces acting on a body at one moment in time is called the resultant force ():

When several forces act on a body, Newton's second law is written as:

The resultant of all forces acting on the body can be equal to zero if there is mutual compensation of the forces applied to the body. In this case, the body moves at a constant speed or is at rest.

When depicting forces acting on a body in a drawing, in the case of uniformly accelerated movement of the body, the resultant force directed along the acceleration should be depicted longer than the oppositely directed force (sum of forces). In the case of uniform motion (or rest), the magnitude of the vectors of forces directed in opposite directions is the same.

To find the resultant force, you should depict in the drawing all the forces that must be taken into account in the problem acting on the body. Forces should be added according to the rules of vector addition.

Examples of solving problems on the topic “Resultant force”

EXAMPLE 1

EXAMPLE 2