Equilibrium weights formula. Body balance

The section of mechanics in which the conditions for the equilibrium of bodies are studied is called statics. It follows from Newton's second law that if the vector sum of all forces applied to a body is zero, then the body keeps its speed unchanged. In particular, if the initial velocity is zero, the body remains at rest. The condition of the invariance of the speed of the body can be written as:

or in projections on the coordinate axes:

It is obvious that a body can be at rest only with respect to one particular coordinate system. In statics, the equilibrium conditions of bodies are studied precisely in such a system. The necessary equilibrium condition can also be obtained by considering the motion of the center of mass of a system of material points. Internal forces do not affect the movement of the center of mass. The acceleration of the center of mass is determined by the vector sum of the external forces. But if this sum is equal to zero, then the acceleration of the center of mass, and, consequently, the speed of the center of mass. If at the initial moment , then the center of mass of the body remains at rest.

Thus, the first condition for the equilibrium of bodies is formulated as follows: the speed of the body does not change if the sum of external forces applied at each point is zero. The resulting rest condition for the center of mass is a necessary (but not sufficient) condition for the equilibrium of a rigid body.

Example

It may be that all the forces acting on the body are balanced, however, the body will accelerate. For example, if you apply two equal and oppositely directed forces (they are called a pair of forces) to the center of mass of the wheel, then the wheel will be at rest if its initial speed was zero. If these forces are applied to different points, then the wheel will begin to rotate (Fig. 4.5). This is because the body is in equilibrium when the sum of all forces is zero at every point of the body. But if the sum of external forces is equal to zero, and the sum of all forces applied to each element of the body is not equal to zero, then the body will not be in equilibrium, possibly (as in the example considered) rotational motion. Thus, if a body can rotate about some axis, then for its equilibrium it is not enough that the resultant of all forces be equal to zero.

To obtain the second equilibrium condition, we use the equation of rotational motion , where is the sum of the moments of external forces about the axis of rotation. When , then b = 0, which means that the angular velocity of the body does not change . If at the initial moment w = 0, then the body will not rotate further. Consequently, the second condition for mechanical equilibrium is the requirement that the algebraic sum of the moments of all external forces about the axis of rotation be equal to zero:

In the general case of an arbitrary number of external forces, the equilibrium conditions can be represented as follows:

These conditions are necessary and sufficient.

Example

Equilibrium is stable, unstable and indifferent. The equilibrium is stable if, with small displacements of the body from the equilibrium position, the forces acting on it and the moments of forces tend to return the body to the equilibrium position (Fig. 4.6a). The equilibrium is unstable if the acting forces at the same time take the body even further from the equilibrium position (Fig. 4.6b). If, at small displacements of the body, the acting forces are still balanced, then the equilibrium is indifferent (Fig. 4.6c). 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.

An interesting example of the equilibrium of a body on a support is the leaning tower in the Italian city of Pisa, 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 radius of 7 m. The top of the tower is deviated from the vertical by 4.5 m.

The Leaning Tower of Pisa is famous for its steep slope. The tower is falling. The height of the tower is 55.86 meters from the ground on the lowest side and 56.70 meters on the highest side. Its weight is estimated at 14,700 tons. The current slope is about 5.5°. 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.

It was believed that the curvature of the tower was originally conceived by the architects - in order to demonstrate their outstanding skills. But something else is much more likely: the architects knew that they were building on an extremely unreliable foundation, and therefore laid in the design the possibility of slight deviation.

When there was a real threat of the collapse of the tower, modern engineers took it up. It was pulled into a steel corset of 18 cables, the foundation was weighted with lead blocks and at the same time the soil was strengthened by pumping concrete underground. With the help of all these measures, it was possible to reduce the angle of inclination of the falling tower by half a degree. Experts say that now it will be able to stand for at least another 300 years. From the point of view of physics, the measures taken mean that the equilibrium conditions of the tower have become more reliable.

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 (Fig. 4.7a). If the center of mass is located above the axis, the equilibrium state is unstable (Fig. 4.7b).

A special case of equilibrium 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. The body is in equilibrium if a vertical line drawn through the center of mass of the body passes through the support area, that is, inside the contour formed by lines connecting the support points. If this line does not cross the area of support, then the body overturns.

« Physics - Grade 10 "

Remember what a moment of force is.
Under what conditions is the body at rest?

If the body is at rest relative to the chosen frame of reference, then the body is said to be in equilibrium. Buildings, bridges, beams with supports, parts of machines, a book on a table and many other bodies are at rest, despite the fact that forces are applied to them from other bodies. The problem of studying the equilibrium conditions of bodies is of great practical importance for mechanical engineering, construction, instrument making, and other areas of technology. All real bodies under the influence of forces applied to them change their shape and size, or, as they say, deform.

In many cases that occur in practice, the deformations of bodies in their equilibrium are insignificant. In these cases, deformations can be neglected and the calculation can be carried out, considering the body absolutely solid.

For brevity, an absolutely rigid body will be called solid body or simply body. Having studied the equilibrium conditions of a rigid body, we will find the equilibrium conditions for real bodies in cases where their deformations can be ignored.

Remember the definition of a perfectly rigid body.

The branch of mechanics in which the conditions for the equilibrium of absolutely rigid bodies are studied is called static.

In statics, the dimensions and shape of bodies are taken into account; in this case, not only the value of the forces is significant, but also the position of the points of their application.

Let us first find out, using Newton's laws, under what condition any body will be in equilibrium. To this end, let us mentally divide the whole body into a large number of small elements, each of which can be considered as a material point. As usual, we call the forces acting on the body from other bodies, external, and the forces with which the elements of the body itself interact, internal (Fig. 7.1). So, force 1.2 is the force acting on element 1 from element 2. Force 2.1 acts on element 2 from element 1. These are internal forces; these also include forces 1.3 and 3.1, 2.3 and 3.2. It is obvious that the geometric sum of internal forces is equal to zero, since according to Newton's third law

12 = - 21 , 23 = - 32 , 31 = - 13 etc.

Statics is a special case of dynamics, since the rest of bodies, when forces act on them, is a special case of motion (= 0).

In general, each element can be acted upon by several external forces. Under 1 , 2 , 3 etc. we mean all external forces applied respectively to the elements 1, 2, 3, ... . In the same way, through " 1 , " 2 , " 3 etc. we denote the geometric sum of the internal forces applied to the elements 2, 2, 3, ... respectively (these forces are not shown in the figure), i.e.

" 1 = 12 + 13 + ... , " 2 = 21 + 22 + ... , " 3 = 31 + 32 + ... etc.

If the body is at rest, then the acceleration of each element is zero. Therefore, according to Newton's second law, the geometric sum of all forces acting on any element will also be equal to zero. Therefore, we can write:

1 + "1 = 0, 2 + "2 = 0, 3 + "3 = 0. (7.1)

Each of these three equations expresses the equilibrium condition for an element of a rigid body.

The first condition for the equilibrium of a rigid body.

Let us find out what conditions must be satisfied by external forces applied to a solid body in order for it to be in equilibrium. To do this, we add equations (7.1):

(1 + 2 + 3) + ("1 + "2 + "3) = 0.

In the first brackets of this equality, the vector sum of all external forces applied to the body is written, and in the second - the vector sum of all internal forces acting on the elements of this body. But, as you know, the vector sum of all internal forces of the system is equal to zero, since according to Newton's third law, any internal force corresponds to a force equal to it in absolute value and opposite in direction. Therefore, on the left side of the last equality, only the geometric sum of the external forces applied to the body will remain:

1 + 2 + 3 + ... = 0 . (7.2)

In the case of an absolutely rigid body, condition (7.2) is called the first condition for its equilibrium.

It is necessary, but not sufficient.

So, if a rigid body is in equilibrium, then the geometric sum of the external forces applied to it is equal to zero.

If the sum of external forces is equal to zero, then the sum of the projections of these forces on the coordinate axes is also equal to zero. In particular, for the projections of external forces on the OX axis, one can write:

F 1x + F 2x + F 3x + ... = 0. (7.3)

The same equations can be written for the projections of forces on the OY and OZ axes.

The second condition for the equilibrium of a rigid body.

Let us verify that condition (7.2) is necessary but not sufficient for the equilibrium of a rigid body. Let us apply to the board lying on the table, at different points, two equal in magnitude and oppositely directed forces, as shown in Figure 7.2. The sum of these forces is zero:

+ (-) = 0. But the board will still rotate. In the same way, two identical in magnitude and oppositely directed forces turn the steering wheel of a bicycle or car (Fig. 7.3).

What other condition for external forces, besides the equality of their sum to zero, must be satisfied in order for a solid body to be in equilibrium? We use the theorem on the change in kinetic energy.

Let us find, for example, the equilibrium condition for a rod hinged on a horizontal axis at point O (Fig. 7.4). This simple device, as you know from the elementary school physics course, is a lever of the first kind.

Let forces 1 and 2 be applied to the lever perpendicular to the rod.

In addition to forces 1 and 2 , the normal reaction force 3 directed vertically upwards acts on the lever from the side of the lever axis. When the lever is in equilibrium, the sum of all three forces is zero: 1 + 2 + 3 = 0.

Let us calculate the work done by external forces when the lever is rotated through a very small angle α. The points of application of forces 1 and 2 will go along the paths s 1 = BB 1 and s 2 = CC 1 (arcs BB 1 and CC 1 at small angles α can be considered straight segments). Work A 1 \u003d F 1 s 1 of force 1 is positive, because point B moves in the direction of the force, and work A 2 \u003d -F 2 s 2 of force 2 is negative, since point C moves in the direction opposite to the direction of force 2. Force 3 does no work, since the point of its application does not move.

The paths s 1 and s 2 traveled can be expressed in terms of the angle of rotation of the lever a, measured in radians: s 1 = α|BO| and s 2 = α|СО|. With this in mind, let's rewrite the expressions to work like this:

А 1 = F 1 α|BO|, (7.4)
A 2 \u003d -F 2 α | CO |.

The radii of BO and CO of arcs of circles described by the points of application of forces 1 and 2 are perpendiculars dropped from the axis of rotation on the line of action of these forces

As you already know, the arm of a force is the shortest distance from the axis of rotation to the line of action of the force. We will denote the arm of the force by the letter d. Then |BO| = d 1 - arm of force 1 , and |CO| \u003d d 2 - arm of force 2. In this case, expressions (7.4) take the form

A 1 \u003d F 1 αd 1, A 2 \u003d -F 2 αd 2. (7.5)

From formulas (7.5) it can be seen that the work of each of the forces is equal to the product of the moment of force and the angle of rotation of the lever. Consequently, expressions (7.5) for work can be rewritten in the form

A 1 = M 1 α, A 2 = M 2 α, (7.6)

and the total work of external forces can be expressed by the formula

A \u003d A 1 + A 2 \u003d (M 1 + M 2) α. α, (7.7)

Since the moment of force 1 is positive and equal to M 1 \u003d F 1 d 1 (see Fig. 7.4), and the moment of force 2 is negative and equal to M 2 \u003d -F 2 d 2, then for work A you can write the expression

A \u003d (M 1 - | M 2 |) α.

When a body is in motion, its kinetic energy increases. To increase the kinetic energy, external forces must do work, i.e. in this case A ≠ 0 and, accordingly, M 1 + M 2 ≠ 0.

If the work of external forces is equal to zero, then the kinetic energy of the body does not change (remains equal to zero) and the body remains motionless. Then

M 1 + M 2 = 0. (7.8)

Equation (7 8) is the second condition for the equilibrium of a rigid body.

When a rigid body is in equilibrium, the sum of the moments of all external forces acting on it about any axis is equal to zero.

So, in the case of an arbitrary number of external forces, the equilibrium conditions for an absolutely rigid body are as follows:

1 + 2 + 3 + ... = 0, (7.9)
M 1 + M 2 + M 3 + ... = 0.

The second equilibrium condition can be derived from the basic equation of the dynamics of the rotational motion of a rigid body. According to this equation where M is the total moment of forces acting on the body, M = M 1 + M 2 + M 3 + ..., ε is the angular acceleration. If the rigid body is motionless, then ε = 0, and, consequently, M = 0. Thus, the second equilibrium condition has the form M = M 1 + M 2 + M 3 + ... = 0.

If the body is not absolutely rigid, then under the action of external forces applied to it, it may not remain in equilibrium, although the sum of external forces and the sum of their moments about any axis are equal to zero.

Let us apply, for example, two forces equal in magnitude and directed along the cord in opposite directions to the ends of a rubber cord. Under the action of these forces, the cord will not be in equilibrium (the cord is stretched), although the sum of external forces is zero and zero is the sum of their moments about the axis passing through any point of the cord.

In order to judge the behavior of a body in real conditions, it is not enough to know that it is in equilibrium. We still need to evaluate this balance. There are stable, unstable and indifferent equilibrium.

The balance of the body is called sustainable if, when deviating from it, forces arise that return the body to the equilibrium position (Fig. 1, position 2). In stable equilibrium, the center of gravity of the body occupies the lowest of all close positions. The position of stable equilibrium is associated with a minimum of potential energy in relation to all close neighboring positions of the body.

The balance of the body is called unstable if, with the slightest deviation from it, the resultant of the forces acting on the body causes a further deviation of the body from the equilibrium position (Fig. 1, position 1). In the position of unstable equilibrium, the height of the center of gravity is maximum and the potential energy is maximum in relation to other close positions of the body.

The equilibrium at which the displacement of the body in any direction does not cause a change in the forces acting on it and the balance of the body is maintained is called indifferent(Fig. 1 position 3).

Indifferent equilibrium is associated with the constant potential energy of all close states, and the height of the center of gravity is the same in all sufficiently close positions.

A body that has an axis of rotation (for example, a homogeneous ruler that can rotate around an axis passing through point O, shown in Figure 2), is in equilibrium if a vertical line passing through the center of gravity of the body passes through the axis of rotation. Moreover, if the center of gravity C is above the axis of rotation (Fig. 2.1), then with any deviation from the equilibrium position, the potential energy decreases and the moment of gravity about the axis O deflects the body further from the equilibrium position. This is an unstable equilibrium. If the center of gravity is below the axis of rotation (Fig. 2.2), then the equilibrium is stable. If the center of gravity and the axis of rotation coincide (Fig. 2.3), then the equilibrium position is indifferent.

A body with a support area is in equilibrium if the vertical line passing through the center of gravity of the body does not go beyond the support area of this body, i.e. outside the contour formed by the points of contact of the body with the support. Equilibrium in this case depends not only on the distance between the center of gravity and the support (i.e., on its potential energy in the gravitational field of the Earth), but also on the location and size of the support area of this body.

Figure 2 shows a body shaped like a cylinder. If it is tilted at a small angle, then it will return to its original position 1 or 2. If it is deflected at an angle (position 3), then the body will tip over. For a given mass and area of support, the stability of the body is the higher, the lower its center of gravity is, i.e. the smaller the angle between the straight line connecting the center of gravity of the body and the extreme point of contact of the support area with the horizontal plane.

Class: 10

Presentation for the lesson

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Lesson Objectives: To study the state of equilibrium of bodies, to get acquainted with various types of equilibrium; find out the conditions under which the body is in equilibrium.

Lesson objectives:

Training: To study two conditions of equilibrium, types of equilibrium (stable, unstable, indifferent). Find out under what conditions bodies are more stable.
Developing: To promote the development of cognitive interest in physics. Development of skills to compare, generalize, highlight the main thing, draw conclusions.
Educational: To cultivate attention, the ability to express one's point of view and defend it, to develop the communication skills of students.

Lesson type: lesson learning new material with computer support.

Equipment:

Disk "Work and power" from "Electronic lessons and tests.
Table "Equilibrium conditions".
Prism inclined with a plumb line.
Geometric bodies: cylinder, cube, cone, etc.
Computer, multimedia projector, interactive whiteboard or screen.
Presentation.

During the classes

Today in the lesson we will learn why the crane does not fall, why the Roly-Vstanka toy always returns to its original state, why the Leaning Tower of Pisa does not fall?

I. Repetition and updating of knowledge.

Formulate Newton's first law. What is the status of the law?
What question does Newton's second law answer? Formula and wording.
What question does Newton's third law answer? Formula and wording.
What is the resultant force? How is she?
From the disk “Movement and interaction of bodies”, complete task No. 9 “The resultant of forces with different directions” (the rule of vector addition (2, 3 exercises)).

II. Learning new material.

1. What is called equilibrium?

Equilibrium is a state of rest.

2. Equilibrium conditions.(slide 2)

a) When is the body at rest? What law does this come from?

The first equilibrium condition: A body is in equilibrium if the geometric sum of the external forces applied to the body is zero. ∑ F = 0

b) Let two equal forces act on the board, as shown in the figure.

Will she be in balance? (No, she will turn)

Only the central point is at rest, while the others move. This means that for the body to be in equilibrium, it is necessary that the sum of all forces acting on each element be equal to 0.

The second equilibrium condition: The sum of the moments of forces acting clockwise must be equal to the sum of the moments of forces acting counterclockwise.

∑ M clockwise = ∑ M counterclockwise

Moment of force: M = F L

L - shoulder of force - the shortest distance from the fulcrum to the line of action of the force.

3. The center of gravity of the body and its location.(slide 4)

Center of gravity of the body- this is the point through which the resultant of all parallel gravity forces acting on individual elements of the body passes (at any position of the body in space).

Find the center of gravity of the following figures:

4. Types of balance.

but) (slides 5-8)

Output: Equilibrium is stable if, with a small deviation from the equilibrium position, there is a force tending to return it to this position.

The position in which its potential energy is minimal is stable. (slide 9)

b) The stability of bodies located on the fulcrum or on the fulcrum.(slides 10-17)

Output: For the stability of a body located on one point or line of support, it is necessary that the center of gravity be below the point (line) of support.

c) The stability of bodies on a flat surface.

(slide 18)

1) Support surface- this is not always a surface that is in contact with the body (but one that is limited by lines connecting the legs of the table, tripod)

2) Analysis of a slide from "Electronic lessons and tests", disk "Work and power", lesson "Types of balance".

Picture 1.

How are the stools different? (Square footing)
Which one is more stable? (with larger area)
How are the stools different? (Location of the center of gravity)
Which one is the most stable? (which center of gravity is lower)
Why? (Because it can be deflected to a larger angle without tipping over)

3) Experience with a deviating prism

We put a prism with a plumb line on the board and begin to gradually lift it over one edge. What do we see?
As long as the plumb line crosses the surface bounded by the support, the balance is maintained. But as soon as the vertical passing through the center of gravity begins to go beyond the boundaries of the support surface, the bookcase overturns.

Parsing slides 19–22.

Conclusions:

The body with the largest area of support is stable.
Of two bodies of the same area, the body whose center of gravity is lower is stable, because it can be deflected without overturning at a large angle.

Parsing slides 23–25.

Which ships are the most stable? Why? (For which the cargo is located in the holds, and not on the deck)

What cars are the most stable? Why? (To increase the stability of cars on turns, the roadbed is tilted in the direction of the turn.)

Conclusions: Equilibrium can be stable, unstable, indifferent. The stability of the bodies is greater, the larger the area of support and the lower the center of gravity.

III. Application of knowledge about the stability of bodies.

What specialties most need knowledge about the balance of bodies?
Designers and constructors of various structures (high-rise buildings, bridges, television towers, etc.)
Circus artists.
Drivers and other professionals.

(slides 28–30)

Why does Roly-Vstanka return to the equilibrium position at any inclination of the toy?
Why is the Leaning Tower of Pisa tilted and not falling?
How do cyclists and motorcyclists keep their balance?

Lesson takeaways:

There are three types of equilibrium: stable, unstable, indifferent.
The position of the body is stable, in which its potential energy is minimal.
The stability of bodies on a flat surface is the greater, the larger the area of support and the lower the center of gravity.

Homework: § 54 – 56 (G.Ya. Myakishev, B.B. Bukhovtsev, N.N. Sotsky)

Used sources and literature:

G.Ya. Myakishev, B.B. Bukhovtsev, N.N. Sotsky. Physics. Grade 10.
Filmstrip "Stability" 1976 (scanned by me on a film scanner).
Disk "Movement and interaction of bodies" from "Electronic lessons and tests".
Disk "Work and power" from "Electronic lessons and tests".

A body is at rest (or moves uniformly and in a straight line) if the vector sum of all forces acting on it is zero. The forces are said to balance each other. When we are dealing with a body of a certain geometric shape, when calculating the resultant force, all forces can be applied to the center of mass of the body.

The condition for the equilibrium of bodies

In order for a body that does not rotate to be in equilibrium, it is necessary that the resultant of all forces acting on it be equal to zero.

F → = F 1 → + F 2 → + . . + F n → = 0 .

The figure above shows the equilibrium of a rigid body. The block is in a state of equilibrium under the action of three forces acting on it. The lines of action of the forces F 1 → and F 2 → intersect at the point O. The point of application of gravity is the center of mass of the body C. These points lie on one straight line, and when calculating the resultant force F 1 → , F 2 → and m g → are reduced to point C .

The condition that the resultant of all forces be equal to zero is not enough if the body can rotate around some axis.

The shoulder of the force d is the length of the perpendicular drawn from the line of action of the force to the point of its application. The moment of force M is the product of the arm of the force and its modulus.

The moment of force tends to rotate the body around its axis. Those moments that rotate the body counterclockwise are considered positive. The unit of measurement of the moment of force in the international SI system is 1 Newton meter.

Definition. moment rule

If the algebraic sum of all the moments applied to the body relative to the fixed axis of rotation is equal to zero, then the body is in equilibrium.

M1 + M2 + . . + M n = 0

Important!

In the general case, for the equilibrium of bodies, two conditions must be met: the resultant force is equal to zero and the rule of moments is observed.

There are different types of equilibrium in mechanics. Thus, a distinction is made between stable and unstable, as well as indifferent equilibrium.

A typical example of an indifferent equilibrium is a rolling wheel (or ball), which, if stopped at any point, will be in a state of equilibrium.

Stable equilibrium is such an equilibrium of a body when, with its small deviations, forces or moments of forces arise that tend to return the body to an equilibrium state.

Unstable equilibrium - a state of equilibrium, with a small deviation from which the forces and moments of forces tend to bring the body out of balance even more.

In the figure above, the position of the ball is (1) - indifferent equilibrium, (2) - unstable equilibrium, (3) - stable equilibrium.

A body with a fixed axis of rotation can be in any of the described equilibrium positions. If the axis of rotation passes through the center of mass, there is an indifferent equilibrium. In stable and unstable equilibrium, the center of mass is located on a vertical line that passes through the axis of rotation. When the center of mass is below the axis of rotation, the equilibrium is stable. Otherwise, vice versa.

A special case of equilibrium is the equilibrium of a body on a support. In this case, the elastic force is distributed over the entire base of the body, and does not pass through one point. A body is at rest in equilibrium when a vertical line drawn through the center of mass intersects the area of support. Otherwise, if the line from the center of mass does not fall into the contour formed by the lines connecting the support points, the body overturns.

An example of the balance of a body on a support is the famous Leaning Tower of Pisa. According to legend, Galileo Galilei dropped balls from it when he conducted his experiments on the study of the free fall of bodies.

A line drawn from the center of mass of the tower intersects the base approximately 2.3 m from its center.

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