So far, we have considered the comparison when two (or more) forces act on the body, the vector sum of which is equal to zero. In this case, the body can either be at rest or move uniformly. If the body is at rest, then the total work of all forces applied to it is zero. Equal to zero and the work of each individual force. If the body moves uniformly, then the total work of all forces is still zero. But each force separately, if it is not perpendicular to the direction of motion, does a certain work - positive or negative.

Let us now consider the case when the resultant of all forces applied to the body is not equal to zero or when only one force acts on the body. In this case, as follows from Newton's second law, the body will move with acceleration. The speed of the body will change, and the work done by the forces in this case is not zero, it can be positive or negative. It can be expected that there is some connection between the change in the speed of the body and the work done by the forces applied to the body. Let's try to install it. Imagine, for simplicity of reasoning, that the body moves along a straight line and the resultant of the forces applied to it is constant in absolute value; and directed along the same line. Let's designate this resultant force as and the projection of displacement onto the direction of the force as Let's direct the coordinate axis along the direction of the force. Then, as shown in § 75, the work done is equal to Let us direct the coordinate axis along the displacement of the body. Then, as was shown in § 75, the work A done by the resultant is: If the directions of the force and displacement coincide, then it is positive and the work is positive. If the resultant is directed opposite to the direction of motion of the body, then its work is negative. The force imparts acceleration to the body. According to Newton's second law. On the other hand, in the second chapter we found that in a rectilinear uniformly accelerated motion

Hence it follows that

Here - the initial speed of the body, i.e. its speed at the beginning of the movement - its speed at the end of this section.

We have obtained a formula that relates the work done by a force to the change in speed (more precisely, the square of the speed) of a body caused by this force.

Half of the product of the mass of a body and the square of its speed has a special name - the kinetic energy of the body, and formula (1) is often called the kinetic energy theorem.

The work of the force is equal to the change in the kinetic energy of the body.

It can be shown that formula (1), derived by us for a force that is constant in magnitude and directed along the movement, is also valid in cases where the force changes and its direction does not coincide with the direction of movement.

Formula (1) is remarkable in many respects.

First, it follows from it that the work of the force acting on the body depends only on the initial and final values ​​of the body's velocity and does not depend on the speed with which it moved at other points.

Secondly, from formula (1) it can be seen that its right side can be both positive and negative, depending on whether the speed of the body increases or decreases. If the speed of the body increases, then the right side of formula (1) is positive, therefore, the work So it should be, because to increase the speed of the body (according to absolute value) the force acting on it must be directed in the same direction as the displacement. On the contrary, when the speed of the body decreases, the right side of formula (1) takes negative meaning(the force is in the opposite direction of the displacement).

If the velocity of the body at the initial point is zero, the expression for work takes the form:

Formula (2) allows you to calculate the work that needs to be done in order to tell a body at rest a speed equal to

The opposite is obvious: to stop a body moving at a speed, it is necessary to do work

very reminiscent of the formula obtained in the previous chapter (see § 59), which establishes between the impulse of a force and a change in the momentum of a body

Indeed, the left side of formula (3) differs from the left side of formula (1) in that in it the force is multiplied not by the displacement performed by the body, but by the duration of the force. On the right side of formula (3) is the product of the body mass and its speed (momentum) instead of half the product of the body mass and the square of its speed, which appears on the right side of formula (1). Both of these formulas are a consequence of Newton's laws (from which they were derived), and the quantities are characteristics of motion.

But there is also a fundamental difference between formulas (1) and (3): formula O) establishes a connection between scalar quantities, while formula (3) is a vector formula.

Task I. What work must be done so that a train moving at a speed increases its speed Mass of the train. What force must be applied to the train if this speed increase is to occur over a 2 km section? The movement is considered to be uniformly accelerated.

Solution. Work A can be found by the formula

Substituting the data given in the problem here, we get:

But by definition, therefore,

Task 2, What height will a body thrown up with an initial velocity reach?

Solution. The body will rise up until its velocity is zero. Only the force of gravity acts on the body where is the mass of the body and is the acceleration of free fall (we neglect the force of air resistance and the Archimedean force).

Applying the formula

We have already obtained this expression earlier (see p. 60) in a more complicated way.

Exercise 48

1. How is the work of force related to the kinetic energy of the body?

2 How does the kinetic energy of a body change if the force applied to it does positive work?

3. How does the kinetic energy of a body change if the force applied to it does negative work.

4. The body moves uniformly along a circle with a radius of 0.5 m, having a kinetic energy of 10 J. What is the force acting on the body? How is it directed? What is the work done by this force?

5. A force of 40 N is applied to a body at rest with a mass of 3 kg. After that, the body passes along a smooth horizontal plane without friction for 3 m. Then the force decreases to 20 n, and the body travels another 3 m. Find the kinetic energy of the body at the end point of its movement.

6. What work must be done to stop a train weighing 1,000 tons moving at a speed of 108 km/h?

7. A body with a mass of 5 kg, moving at a speed of 6 m / s, is subjected to a force of 8 n, directed in the direction opposite to the movement. As a result, the speed of the body decreases to 2 m/sec. What is the magnitude and sign of the work done by the force? What is the distance traveled by the body?

8. A force of 4 N begins to act on a body that was originally at rest, directed at an angle of 60 ° to the horizon. A body moves on a smooth horizontal surface without friction. Calculate the work done by the force if the body traveled a distance of 1 m.

9. What is the kinetic energy theorem?

Newton's laws are a mathematical abstraction. In reality, the cause of the movement or rest of bodies, as well as their deformation, are several forces at once. Therefore, an important addition to the laws of mechanics will be the introduction of the concept of the resultant force and its application.

About the reasons for the changes

Classical mechanics is divided into two sections - kinematics, which describes the trajectory of motion of bodies with the help of equations, and dynamics, which deals with the reasons for changing the position of objects or the objects themselves.

The reason for the changes is a certain force, which is a measure of the action on the body of other bodies or force fields (for example, an electromagnetic field or gravity). For example, the force of elasticity causes the deformation of the body, the force of gravity - the fall of bodies to the Earth.

Force is a vector quantity, that is, its action is directed. Force modulus in general case is proportional to a certain coefficient (for the deformation of a spring, this is its stiffness), as well as to the parameters of the action (mass, charge).

For example, in the case of the Coulomb force, this is the magnitude of both charges, taken modulo, the squared distance between the charges and the coefficient k, in the SI system, defined by the expression: $k = (1 \over 4 \pi \epsilon)$, where $\epsilon$ is the dielectric constant.

Addition of forces

In the case when n forces act on the body, they speak of the resultant force, and the formula for Newton's second law takes the form:

$m\vec a = \sum\limits_(i=1)^n \vec F_i$.

Rice. 1. The resultant of forces.

Since F is a vector quantity, the sum of forces is called geometric (or vector). Such addition is performed according to the rule of a triangle or parallelogram, or by components. Let's explain each method with an example. To do this, we write the formula for the resultant force in a general form:

$F = \sum\limits_(i=1)^n \vec F_i$

And the force $F_i$ can be represented as:

$F = (F_(xi), F_(yi), F_(zi))$

Then the sum of the two forces will be the new vector $F_(ab) = (F_(xb) + F_(xa), F_(yb) + F_(ya), F_(zb) + F_(za))$.

Rice. 2. Componentwise addition of vectors.

The absolute value of the resultant can be calculated as follows:

$F = \sqrt((F_(xb) + F_(xa))^2 + (F_(yb) + F_(ya))^2 + (F_(zb) + F_(za))^2)$

Now let's give a strict definition: the resultant force is the vector sum of all the forces that affect the body.

Let's analyze the rules of a triangle and a parallelogram. Graphically it looks like this:

Rice. 3. Rule of triangle and parallelogram.

Outwardly, they seem different, but when it comes to calculations, they come down to finding the third side of a triangle (or, which is the same thing, the diagonal of a parallelogram) using the cosine theorem.

If there are more than two forces, it is sometimes more convenient to use the polygon rule. At its core, this is still the same triangle, only repeated in one figure a number of times. If, as a result, the contour turned out to be closed, the total action of forces is equal to zero and the body is at rest.

Tasks

• A box placed in the center of a Cartesian rectangular coordinate system is subject to two forces: $F_1 = (5, 0)$ and $F_2 = (3, 3)$. Calculate the resultant in two ways: according to the triangle rule and using component-by-component addition of vectors.

Solution

The resultant force will be the vector sum of $F_1$ and $F_2$.

Therefore, we write:

$\vec F = \vec F_1 + \vec F_2 = (5+3, 0+3) = (8, 3)$
The absolute value of the resultant force:

$F = \sqrt(8^2 + 3^2) = \sqrt(64 + 9) = 8.5 N$

Now we get the same value using the triangle rule. To do this, we first find the absolute values ​​of $F_1$ and $F_2$, as well as the angle between them.

$F_1 = \sqrt(5^2 + 0^2) = 5 N$

$F_2 = \sqrt(3^2 + 3^2) = 4.2 N$

The angle between them is 45˚, since the first force is parallel to the Ox axis, and the second divides the first coordinate plane in half, that is, it is the bisector of a right angle.

Now, having placed the vectors according to the triangle rule, we calculate the resultant using the cosine theorem:

$F = \sqrt(F_1^2 + F_2^2 - 2F_1F_2 cos135) = \sqrt(F_1^2 + F_2^2 + 2F_1F_2 sin45) = \sqrt(25 + 18 + 2 \cdot 5 \cdot 4,2 \ cdot sin45) = 8.5 N$

• Three forces act on the machine: $F_1 = (-5, 0)$, $F_2 = (-2, 0)$, $F_1 = (7,0)$. What is their resultant?

Solution

It is enough to add the x components of the vectors:

$F = -5 - 2 + 7 = 0$

What have we learned?

During the lesson, the concept of the resultant of forces was introduced and various methods for its calculation were considered, as well as Newton's second law was introduced for the general case when the number of forces is unlimited.

Topic quiz

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Statics is a branch of mechanics that studies the conditions of equilibrium of bodies.

It follows from Newton's second law that if the geometric sum of all external forces applied to a body is zero, then the body is at rest or performs a uniform rectilinear motion. In this case, it is customary to say that the forces applied to the body balance each other. When calculating resultant all forces acting on a body can be applied to center of gravity .

For a non-rotating body to be in equilibrium, it is necessary that the resultant of all forces applied to the body be equal to zero.

On fig. 1.14.1 an example of equilibrium is given solid body under the influence of three forces. Intersection point O lines of action of forces and does not coincide with the point of application of gravity (center of mass C), but at equilibrium these points are necessarily on the same vertical. When calculating the resultant, all forces are reduced to one point.

If the body can rotate about some axis, then for its equilibrium it is not enough to equal zero the resultant of all forces.

The rotating action of a force depends not only on its magnitude, but also on the distance between the line of action of the force and the axis of rotation.

The length of the perpendicular drawn from the axis of rotation to the line of action of the force is called shoulder of strength.

The product of the modulus of force per shoulder d called moment of force M. The moments of those forces that tend to rotate the body counterclockwise are considered positive (Fig. 1.14.2).

moment rule : a body with a fixed axis of rotation is in equilibrium if algebraic sum moments of all forces applied to the body about this axis is equal to zero:

In the International System of Units (SI), moments of forces are measured in Hnewton— meters (N∙m) .

In the general case, when a body can move translationally and rotate, both conditions must be met for equilibrium: the resultant force must be equal to zero and the sum of all moments of forces must be equal to zero.

Wheel rolling on a horizontal surface - example indifferent balance(Fig. 1.14.3). If the wheel is stopped at any point, it will be in equilibrium. Along with indifferent equilibrium in mechanics, states are distinguished sustainable And unstable balance.

A state of equilibrium is called stable if, with small deviations of the body from this state, forces or moments of forces arise that tend to return the body to an equilibrium state.

With a small deviation of the body from the state of unstable equilibrium, forces or moments of forces arise that tend to remove the body from the equilibrium position.

A ball lying on a flat horizontal surface is in a state of indifferent equilibrium. A ball located at the top of a spherical ledge is an example of an unstable equilibrium. Finally, the ball at the bottom of the spherical cavity is in a state of stable equilibrium (Fig. 1.14.4).

For a body with a fixed axis of rotation, all three types of equilibrium are possible. Indifferent equilibrium occurs when the axis of rotation passes through the center of mass. In stable and unstable equilibrium, the center of mass is on a vertical line passing through the axis of rotation. In this case, if the center of mass is below the axis of rotation, the state of equilibrium is stable. If the center of mass is located above the axis, the equilibrium state is unstable (Fig. 1.14.5).

A special case is the equilibrium of a body on a support. In this case, the elastic force of the support is not applied to one point, but is distributed over the base of the body. A body is in equilibrium if a vertical line drawn through the center of mass of the body passes through footprint, i.e., inside the contour formed by lines connecting the support points. If this line does not cross the area of ​​support, then the body overturns. An interesting example of the equilibrium of a body on a support is the leaning tower in the Italian city of Pisa (Fig. 1.14.6), which, according to legend, was used by Galileo when studying the laws of free fall of bodies. The tower has the shape of a cylinder with a height of 55 m and a radius of 7 m. The top of the tower deviates from the vertical by 4.5 m.

A vertical line drawn through the center of mass of the tower intersects the base approximately 2.3 m from its center. Thus, the tower is in a state of equilibrium. The balance will be disturbed and the tower will fall when the deviation of its top from the vertical reaches 14 m. Apparently, this will not happen very soon.

IN inertial systems reference, a change in the speed of a body is possible only under the action of another body on it. The action of one body on another is expressed in terms of physical quantity like force(). The impact of one body on another can cause a change in the speed of the body, both in magnitude and in direction. Therefore, the force is a vector and is determined not only by the magnitude (modulus), but also by the direction. The direction of the force determines the direction of the acceleration vector of the body affected by the force in question.

The magnitude and direction of force is determined by 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 meaning of Newton's second law lies in the fact that the forces that act on the body determine how the speed of the body changes, and not just its speed. Note that Newton's second law is valid only in inertial frames of reference.

If several forces act simultaneously on the body, then the body moves with an acceleration that is equal to the vector sum of the accelerations that would appear under the influence of each of the bodies separately. The forces acting on the body and applied to its one point should be added in accordance with the rule of vector addition.

DEFINITION

The vector sum of all forces acting on the body at the same time is called resultant force ():

If several forces act on the body, then Newton's second law is written as:

The resultant of all forces acting on the body can be equal to zero if there is a 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 the forces acting on the body, in the drawing, in the case of a uniformly accelerated movement of the body, the resultant force directed along the acceleration should be depicted longer than the oppositely directed force (the sum of forces). When uniform motion(or rest) the dyne of force vectors directed in opposite directions is the same.

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

Examples of problem solving

EXAMPLE 1

EXAMPLE 2

Igor Babin (St. Petersburg) 14.05.2012 17:33

in the condition it is written that you need to find the weight of the body.

and in solving the modulus of gravity.

How can weight be measured in Newtons.

In the condition error (

Alexey (St. Petersburg)

Good afternoon!

You are confusing the concepts of mass and weight. The weight of the body is the force (and therefore the weight is measured in Newtons), with which the body presses on the support or stretches the suspension. As follows from the definition, this force is applied not even to the body, but to the support. Weightlessness is a state when the body loses not mass, but weight, that is, the body ceases to put pressure on other bodies.

I agree, some liberties were allowed in the decision in the definitions, now it has been corrected.

Yuri Shoitov (Kursk) 26.06.2012 21:20

The concept of "body weight" was introduced in educational physics extremely unsuccessful. If in the everyday concept weight means mass, then in school physics, as you correctly noted, the weight of the body is the force (and therefore the weight is measured in Newtons), with which the body presses on the support or stretches the suspension. notice, that we are talking about one support and about one thread. If there are several supports or threads, the concept of weight disappears.

I give an example. Let a body be suspended on a thread in a liquid. It stretches the thread and presses on the liquid with a force equal to minus the force of Archimedes. Why, speaking of the weight of a body in a fluid, do we not add up these forces, as you do in your decision?

I registered on your site, but did not notice what has changed in our communication. Please excuse my stupidity, but I, being an old man, do not navigate the site freely enough.

Alexey (St. Petersburg)

Good afternoon!

Indeed, the concept of body weight is very vague when the body has several supports. Usually, the weight in this case is defined as the sum of interactions with all supports. In this case, the impact on gaseous and liquid media, as a rule, is excluded. This just falls under the example you described, with a weight suspended in the water.

Here, a children's problem immediately comes to mind: "What weighs more: a kilogram of down or a kilogram of lead?" If we solve this problem honestly, then we must undoubtedly take into account the power of Archimedes. And by weight, most likely, we will understand what the scales will show us, that is, the force with which fluff and lead press, say, on the scales. That is, here the force of interaction with air is, as it were, excluded from the concept of weight.

On the other hand, if we assume that we have pumped out all the air and put on the scales the body to which the rope is tied. Then the force of gravity will be balanced by the sum of the reaction force of the support and the force of the thread tension. If we understand weight as the force of action on the supports that prevent falling, then the weight here will be equal to this sum of the tension force of the thread and the pressure force on the scale pan, that is, it will coincide in magnitude with the force of gravity. Again the question arises: why is the thread better or worse than the Archimedes force?

In general, one can agree here that the concept of weight makes sense only in empty space, where there is only one support and a body. How to be here, this is a matter of terminology, which, unfortunately, everyone here has their own, since this is not such an important question :) And if the force of Archimedes in the air in all ordinary cases can be neglected, which means that it will especially affect the amount of weight cannot, then for a body in a liquid this is already critical.

To be completely honest, the division of forces into types is very arbitrary. Imagine a box being dragged along a horizontal surface. It is usually said that two forces act on the box from the side of the surface: the reaction force of the support, directed vertically, and the friction force, directed horizontally. But these are two forces acting between the same bodies, why don't we just draw one force, which is their vector sum (this, by the way, is sometimes done). It's probably a matter of convenience :)

So I'm a bit confused as to what to do with this particular task. The easiest way, probably, is to reformulate it and ask a question about the magnitude of gravity.

Don't worry, it's all right. When registering, you must provide an e-mail. If you now go to the site under your account, then when you try to leave a comment in the "Your e-mail" window, the same address should immediately appear. After that, the system will automatically sign your messages.