If the resultant force is zero. Conditions for the equilibrium of bodies. Finding the resultant force

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 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 gives an example of the equilibrium of a rigid body under the action 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 the algebraic sum of the moments of all forces applied to the body about this axis is 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 forward 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 equilibrium(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 reference systems, a change in the speed of a body is possible only when another body acts on it. Quantitatively, the action of one body on another is expressed using such a physical quantity as 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). In the case of 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 into educational physics extremely unsuccessfully. 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. Note that we are talking about one support and 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 question 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 value 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.

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 the formula (1) is positive, therefore, the work It should be so because to increase the speed of the body (in absolute value), the force acting on it must be directed in the same direction as the movement. On the contrary, when the speed of the body decreases, the right side of formula (1) takes a negative value (the force is directed opposite to 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/s. 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?

Systematization of knowledge about the resultant of all forces applied to the body; about vector addition.

Lesson objectives (for teacher):

Educational:

• Clarify and expand knowledge about the resultant force and how to find it.
• To form the ability to apply the concept of the resultant force to the justification of the laws of motion (Newton's laws)
• Determine the level of mastering the topic;
• Continue to develop the skills of self-analysis of the situation and self-control.

Educational:

• To contribute to the formation of the worldview idea of ​​the cognizability of phenomena and properties of the surrounding world;
• Emphasize the importance of modulation in the cognizability of matter;
• Pay attention to the formation of universal human qualities:
  a) efficiency,
  b) independence;
  c) accuracy;
  d) discipline;
  e) responsible attitude to learning.

Developing:

Equipment and demonstrations.

1. Illustrations:
   sketch for the fable by I.A. Krylov "Swan, crayfish and pike",
   sketch of the painting by I. Repin “Barge haulers on the Volga”,
   to problem No. 108 “Turnip” - “Physicist's Task Book” by G. Oster.
2. Arrows colored on a polyethylene basis.
3. Copy paper.
4. Kodoscope and film with the solution of two problems of independent work.
5. Shatalov "Supporting notes".
6. Faraday's portrait.

Board layout:

“If you are in this
figure it out properly
you better be able to follow
following my train of thought
in what follows."
M. Faraday

During the classes

1. Organizational moment

Examination:

• absent;
• the presence of diaries, notebooks, pens, rulers, pencils;

Appearance rating.

2. Repetition

As we talk in class, we repeat:

• I Newton's law.
• Force is the cause of acceleration.
• Newton's second law.
• Addition of vectors to the rule of a triangle and a parallelogram.

3. Main material

Lesson problem.

“Once a Swan, Cancer and Pike
Carried with luggage, a cart came from
And together, three, all harnessed to it;
Out of the skin climb out
And the cart still does not move!
The luggage would have seemed easy for them:
Yes, the swan breaks into the clouds,
Cancer moves back
And Pike pulls into the water!
Who is guilty of them, who is right -
It is not for us to judge;
Yes, only things are still there!”

(I.A. Krylov)

The fable expresses a skeptical attitude towards Alexander I, it ridicules the turmoil in the State Council of 1816, the reforms and committees started by Alexander I were unable to budge the deeply bogged down cart of autocracy. In this, from a political point of view, Ivan Andreevich was right. But let's find out the physical aspect. Is Krylov right? To do this, it is necessary to become more familiar with the concept of the resultant of forces applied to the body.

A force equal to the geometric sum of all forces applied to the body (point) is called the resultant or resultant force.

Picture 1

How does this body behave? Either it is at rest or it moves in a straight line and uniformly, since it follows from Newton's I law that there are such frames of reference with respect to which a progressively moving body retains its speed constant if no other bodies act on it or the action of these bodies is compensated,

i.e. |F 1 | = |F 2 | (the definition of the resultant is introduced).

A force that produces the same effect on a body as several simultaneously acting forces is called the resultant of these forces.

Finding the resultant of several forces is the geometric addition of the acting forces; is carried out according to the rule of a triangle or parallelogram.

In figure 1 R=0, because .

To add two vectors, the beginning of the second is applied to the end of the first vector and the beginning of the first is connected to the end of the second (manipulation on a board with polyethylene-based arrows). This vector is the resultant of all forces applied to the body, i.e. R \u003d F 1 - F 2 \u003d 0

How can one formulate Newton's first law based on the definition of the resultant force? The well-known formulation of Newton's first law:

“If other bodies do not act on a given body or the actions of other bodies are compensated (balanced), then this body is either at rest or moves in a straight line and uniformly.”

New formulation of Newton's I law (give the formulation of Newton's I law for the record):

“If the resultant of the forces applied to the body is zero, then the body retains its state of rest or uniform rectilinear motion.”

How to proceed when finding the resultant, if the forces applied to the body are directed in one direction along one straight line?

Task #1 (solution of problem No. 108 by Grigory Oster from the problem book “Physics”).

The grandfather, holding the turnip, develops a traction force up to 600 N, the grandmother - up to 100 N, the granddaughter - up to 50 N, the Bug - up to 30 N, the cat - up to 10 N and the mouse - up to 2 N. What is the resultant of all these forces, pointing in the same straight line in the same direction? Would this company handle the turnip without a mouse if the forces holding the turnip in the ground are 791 N?

(Manipulation on a board with polyethylene-based arrows).

Answer. The module of the resultant force, equal to the sum of the modules of forces with which the grandfather pulls the turnip, the grandmother pulls the grandfather, the granddaughter pulls the grandmother, the Bug pulls the granddaughter, the cat pulls the Bug, and the mouse pulls the cat, will be equal to 792 N. The contribution of the muscle force of the mouse to this mighty impulse is 2 N. Without Myshkin's Newtons, things will not work.

Task number 2.

And if the forces acting on the body are directed at right angles to each other? (Manipulation on a board with polyethylene-based arrows).

(We write down the rules p. 104 Shatalov “Support notes”).

Task number 3.

Let's try to find out if I.A. is right in the fable. Krylov.

If we assume that the traction force of the three animals described in the fable is the same and comparable (or more) with the weight of the cart, and also exceeds the static friction force, then, using Figure 2 (1) for Problem 3, after constructing the resultant, we obtain that And .BUT. Krylov, of course, is right.

If we use the data below, prepared by the students in advance, then we get a slightly different result (see Figure 2 (1) for task 3).

The power developed by bodies during uniform rectilinear motion, which is possible when the traction force and the resistance force are equal, can be calculated using the following formula.