With the simultaneous action of several forces on one body, the body moves with an acceleration, which is the vector sum of the accelerations that would arise under the action of each force separately. The forces acting on the body, applied to one point, are added according to the rule of addition of vectors.

The vector sum of all forces acting simultaneously on a body is called resultant force.

The straight line passing through the force vector is called the line of action of the force. If the forces are applied to different points of the body and act not parallel to each other, then the resultant is applied to the point of intersection of the lines of action of the forces. If the forces act parallel to each other, then there is no point of application of the resulting force, and the line of its action is determined by the formula: (see figure).

Moment of power. Lever balance condition

The main sign of the interaction of bodies in dynamics is the occurrence of accelerations. However, it is often necessary to know under what conditions a body, which is acted upon by several different forces, is in a state of equilibrium.

There are two types of mechanical movement - translation and rotation.

If the trajectories of movement of all points of the body are the same, then the movement progressive. If the trajectories of all points of the body are arcs of concentric circles (circles with one center - the point of rotation), then the movement is rotational.

Equilibrium of non-rotating bodies: a non-rotating body is in equilibrium if the geometric sum of the forces applied to the body is zero.

Equilibrium of a body with a fixed axis of rotation

If the line of action of the force applied to the body passes through the axis of rotation of the body, then this force is balanced by the elastic force from the side of the axis of rotation.

If the line of action of the force does not cross the axis of rotation, then this force cannot be balanced by the elastic force from the side of the axis of rotation, and the body rotates around the axis.

The rotation of a body around an axis under the action of one force can be stopped by the action of a second force. Experience shows that if two forces separately cause the rotation of the body in opposite directions, then with their simultaneous action the body is in equilibrium if the condition is met:

, where d 1 and d 2 are the shortest distances from the lines of action of the forces F 1 and F 2. The distance d is called shoulder of strength, and the product of the modulus of force by the arm is moment of force:

.

If a positive sign is assigned to the moments of forces that cause the body to rotate around an axis clockwise, and a negative sign to the moments of forces that cause counterclockwise rotation, then the equilibrium condition for a body with an axis of rotation can be formulated as moment rules: 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:

The SI unit of torque is a moment of force of 1 N, the line of action of which is at a distance of 1 m from the axis of rotation. This unit is called newton meter.

The general condition for the equilibrium of a body:a body is in equilibrium if the geometric sum of all forces applied to it and the algebraic sum of the moments of these forces about the axis of rotation are equal to zero.

Under this condition, the body is not necessarily at rest. It can move uniformly and rectilinearly or rotate.

Lesson objectives:

Educational. To study two conditions for the equilibrium of bodies, 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, to develop the ability to make comparisons, generalize, highlight the main thing, draw conclusions.

Educational: to cultivate discipline, attention, the ability to express their point of view and defend it.

Lesson plan:

1. Knowledge update

2. What is static

3. What is balance. Types of balance

4. Center of gravity

5. Problem solving

Lesson progress:

1.Updating knowledge.

Teacher: Hello!

Students: Hello!

Teacher: We continue to talk about forces. In front of you is an irregularly shaped body (stone), suspended on a thread and attached to an inclined plane. What forces are acting on this body?

Students: The body is affected by: the tension force of the thread, the force of gravity, the force tending to tear off the stone, opposite to the tension force of the thread, the reaction force of the support.

Teacher: Forces found, what do we do next?

Students: Write down Newton's second law.

There is no acceleration, so the sum of all forces is zero.

Teacher: What does it say?

Students: This indicates that the body is at rest.

Teacher: Or you can say that the body is in a state of equilibrium. The equilibrium of a body is the state of rest of that body. Today we will talk about the balance of bodies. Write down the topic of the lesson: "Equilibrium conditions for bodies. Types of equilibrium."

2. Formation of new knowledge and methods of action.

Teacher: The section of mechanics that studies the equilibrium of absolutely rigid bodies is called statics. There is not a single body around us that would not be affected by forces. Under the influence of these forces, the bodies are deformed.

When elucidating the equilibrium conditions for deformed bodies, it is necessary to take into account the magnitude and nature of the deformation, which complicates the task put forward. Therefore, in order to clarify the basic laws of equilibrium, for convenience, the concept of an absolutely rigid body was introduced.

An absolutely rigid body is a body in which the deformations that occur under the action of forces applied to it are negligible. Write down the definitions of statics, balance of bodies and absolutely rigid body from the screen (slide 2).

And the fact that we have found out that the body is in equilibrium, if the geometric sum of all the forces applied to it is equal to zero, is the first condition for equilibrium. Write down 1 equilibrium condition:

If the sum of the 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 X axis, we can write .

Equality to zero of the sum of external forces acting on a rigid body is necessary for its equilibrium, but not enough. For example, two equal and oppositely directed forces were applied to the board at different points. The sum of these forces is zero. Will the board be in balance?

Students: The board will turn, for example, like the steering wheel of a bicycle or car.

Teacher: Right. In the same way, two identical in magnitude and oppositely directed forces turn the steering wheel of a bicycle or car. Why is this happening?

Students: ???

Teacher: Any body is in equilibrium when the sum of all forces acting on each of its elements is equal to zero. But if the sum of external forces is equal to zero, then the sum of all forces applied to each element of the body may not be equal to zero. In this case, the body will not be in equilibrium. Therefore, we need to find out one more condition for the equilibrium of bodies. To do this, we will conduct an experiment. (Two students are called.) One of the students applies force closer to the axis of rotation of the door, the other student - closer to the handle. They apply forces in different directions. What happened?

Students: The one that applied the force closer to the handle won.

Teacher: Where is the line of action of the force applied by the first disciple?

Students: Closer to the axis of rotation of the door.

Teacher: Where is the line of action of the force applied by the second student?

Students: Closer to the doorknob.

Teacher: What else can we notice?

Students: That the distances from the axis of rotation to the lines of application of forces are different.

Teacher: So what else determines the result of the action of force?

Students: The result of the action of the force depends on the distance from the axis of rotation to the line of action of the force.

Teacher: What is the distance from the axis of rotation to the line of action of the force?

Students: Shoulder. The shoulder is a perpendicular drawn from the axis of rotation to the line of action of this force.

Teacher: How do forces and shoulders relate to each other in this case?

Students: According to the rule of equilibrium of a lever, the forces acting on it are inversely proportional to the shoulders of these forces. .

Teacher: What is the product of the modulus of the force that rotates the body and its arm?

Students: Moment of power.

Teacher: So the moment of force applied to the first students is , and the moment of force applied to the second students is

Now we can formulate the second equilibrium condition: A solid body is in equilibrium if the algebraic sum of the moments of external forces acting on it about any axis is zero. (Slide 3)

Let us introduce the concept of the center of gravity. The center of gravity is the point of application of the resultant force of gravity (the point through which the resultant of all parallel gravity forces acting on individual elements of the body passes). There is also the concept of the center of mass.

The center of mass of a system of material points is called a geometric point, the coordinates of which are determined by the formula:

; same for .

The center of gravity coincides with the center of mass of the system if this system is in a uniform gravitational field.

Look at the screen. Try to find the center of gravity of these figures. (slide 4)

(Demonstrate with the help of a bar with recesses and slides and a ball types of balance.)

On slide 5 you see what you saw in experience. Write down the equilibrium stability conditions from slides 6,7,8:

1. Bodies are in a state of stable equilibrium if, at the slightest deviation from the equilibrium position, a force or moment of force arises that returns the body to the equilibrium position.

2. Bodies are in a state of unstable equilibrium if, at the slightest deviation from the equilibrium position, a force or moment of force arises that removes the body from the equilibrium position.

3. Bodies are in a state of indifferent equilibrium if, at the slightest deviation from the equilibrium position, neither a force nor a moment of force arises that changes the position of the body.

Now look at slide 9. What can you say about the stability conditions in all three cases.

Students: In the first case, if the fulcrum is higher than the center of gravity, then the equilibrium is stable.

In the second case, if the fulcrum coincides with the center of gravity, then the equilibrium is indifferent.

In the third case, if the center of gravity is higher than the fulcrum, the balance is unstable.

Teacher: Now let's consider bodies that have a support area. The area of ​​support is understood as the area of ​​contact of the body with the support. (slide 10).

Let us consider how the position of the line of action of the force of gravity changes with respect to the axis of rotation of the body when the body with the area of ​​support is tilted. (slide 11)

Note that as the body rotates, the position of the center of gravity changes. And any system always tends to lower the position of the center of gravity. So inclined bodies will be in a state of stable equilibrium, while the line of action of gravity will pass through the area of ​​support. Look at slide 12.

If the deflection of a body having an area of ​​support increases the center of gravity, then the balance will be stable. In stable equilibrium, a vertical line passing through the center of gravity will always pass through the area of ​​support.

Two bodies that have the same weight and area of ​​support, but different heights, have different limiting angles of inclination. If this angle is exceeded, then the bodies overturn. (slide 13)

With a lower center of gravity, more work must be expended to tip the body. Therefore, the work of overturning can serve as a measure of its stability. (Slide 14)

So tilted structures are in a position of stable equilibrium, because the line of action of gravity passes through the area of ​​\u200b\u200bits support. For example, the Leaning Tower of Pisa.

The swaying or tilting of the human body when walking is also explained by the desire to maintain a stable position. The support area is determined by the area inside the line drawn around the extreme points of contact with the support body. when the person is standing. The line of action of gravity passes through the support. When a person lifts his leg, in order to maintain balance, he bends over, transferring the line of action of gravity to a new position so that it again passes through the area of ​​​​support. (slide 15)

For the stability of various structures, the support area is increased or the center of gravity of the structure is lowered, making a powerful support, or the support area is increased and, at the same time, the center of gravity of the structure is lowered.

The stability of transport is determined by the same conditions. So, of the two modes of transport, a car and a bus, a car is more stable on an inclined road.

With the same inclination of these modes of transport near the bus, the line of gravity runs closer to the edge of the support area.

Problem solving

Task: Material points with masses m, 2m, 3m and 4m are located at the vertices of a rectangle with sides 0.4m and 0.8m. Find the center of gravity of the system of these material points.

x s -? at with -?

Finding the center of gravity of a system of material points means finding its coordinates in the XOY coordinate system. Let us align the origin of coordinates XOY with the vertex of the rectangle containing the material point of mass m, and direct the coordinate axes along the sides of the rectangle. The coordinates of the center of gravity of the system of material points are equal to:

Here, is the coordinate on the OX axis of a point with mass . As follows from the drawing, because this point is located at the origin. The coordinate is also equal to zero, the coordinates of points with masses on the OX axis are the same and equal to the length of the side of the rectangle. Substituting the values ​​of the coordinates, we get

The coordinate on the OY axis of a point with mass is zero, =0. The coordinates of points with masses on this axis are the same and equal to the length of the side of the rectangle. Substituting these values, we get

Test questions:

1. Conditions for the equilibrium of the body?

1 equilibrium condition:

A rigid body is in equilibrium if the geometric sum of the external forces applied to it is zero.

2 Equilibrium condition: A solid body is in equilibrium if the algebraic sum of the moments of external forces acting on it about any axis is equal to zero.

2. Name the types of balance.

Bodies are in a state of stable equilibrium if, at the slightest deviation from the equilibrium position, a force or moment of force arises that returns the body to the equilibrium position.

Bodies are in a state of unstable equilibrium if, at the slightest deviation from the equilibrium position, a force or moment of force arises that removes the body from the equilibrium position.

Bodies are in a state of indifferent equilibrium if, at the slightest deviation from the equilibrium position, neither a force nor a moment of force arises that changes the position of the body.

Homework:

List of used literature:

1. Physics. Grade 10: textbook. for general education institutions: basic and profile. levels / G. Ya. Myakishev, B. B. Bukhovtsev, N. N. Sotsky; ed. V. I. Nikolaev, N. A. Parfenteva. - 19th ed. - M.: Enlightenment, 2010. - 366 p.: ill.
2. Maron A.E., Maron E.A. "Collection of qualitative problems in physics 10 cells, M.: Enlightenment, 2006
3. L.A. Kirik, L.E.Gendenshtein, Yu.I.Dik. Methodical materials for the teacher grade 10, M.: Ileksa, 2005.-304s:, 2005
4. L.E.Gendenshtein, Yu.I.Dik. Physics grade 10.-M.: Mnemosyne, 2010

In physics for grade 9 (I.K. Kikoin, A.K. Kikoin, 1999),
task №6
to chapter " LABORATORY WORKS».

The purpose of the work: to establish the ratio between the moments of forces applied to the arms of the lever when it is in equilibrium. To do this, one or more weights are suspended from one of the arms of the lever, and a dynamometer is attached to the other (Fig. 179).

This dynamometer measures the modulus of force F, which must be applied in order for the lever to be in balance. Then, with the help of the same dynamometer, the modulus of the weight of the goods P is measured. The lever arm lengths are measured with a ruler. After that, the absolute values ​​of the moments M 1 and M 2 of the forces P and F are determined:

The conclusion about the error of the experimental verification of the moment rule can be made by comparing with unity

relation:

Measuring:

1) ruler; 2) dynamometer.

Materials: 1) tripod with clutch; 2) lever; 3) a set of goods.

Work order

1. Mount the arm on a tripod and balance it in a horizontal position using the sliding nuts located at its ends.

2. Hang a load at some point on one of the arms of the lever.

3. Attach a dynamometer to the other arm of the lever and determine the force to be applied.

live towards the lever so that it is in balance.

4. Use a ruler to measure the length of the lever arms.

5. Using a dynamometer, determine the weight of the load R.

6. Find the absolute values ​​of the moments of forces P and F

7. Enter the found values ​​in the table:

				M 1 \u003d Pl 1, N⋅m

8. Compare the ratio

with unity and draw a conclusion about the error of the experimental verification of the moment rule.

The main purpose of the work is to establish the relationship between the moments of forces applied to a body with a fixed axis of rotation at its equilibrium. In our case, we use a lever as such a body. According to the rule of moments, for such a body to be in equilibrium, it is necessary that the algebraic sum of the moments of forces about the axis of rotation be equal to zero.

Consider such a body (in our case, a lever). Two forces act on it: the weight of the loads P and the force F (the elasticity of the spring of the dynamometer), so that the lever is in balance and the moments of these forces must be equal in absolute value to each other. The absolute values ​​of the moments of forces F and P will be determined respectively:

Conclusions about the error of the experimental verification of the moment rule can be made by comparing the ratio with unity:

Measuring instruments: ruler (Δl = ±0.0005 m), dynamometer (ΔF = ±0.05 H). The mass of weights from the set in mechanics is assumed to be (0.1 ± 0.002) kg.

Completing of the work

Definition

The equilibrium of the body is called such a state when any acceleration of the body is equal to zero, that is, all actions on the body of forces and moments of forces are balanced. In this case, the body can:

• be in a state of calm;
• move evenly and in a straight line;
• rotate uniformly around an axis that passes through its center of gravity.

Body equilibrium conditions

If the body is in equilibrium, then two conditions are simultaneously satisfied.

1. The vector sum of all forces acting on the body is equal to the zero vector : $\sum_n((\overrightarrow(F))_n)=\overrightarrow(0)$
2. The algebraic sum of all moments of forces acting on the body is equal to zero: $\sum_n(M_n)=0$

The two equilibrium conditions are necessary but not sufficient. Let's take an example. Consider a wheel rolling uniformly without slipping on a horizontal surface. Both equilibrium conditions are met, but the body is moving.

Consider the case when the body does not rotate. In order for the body not to rotate and be in balance, it is necessary that the sum of the projections of all forces on an arbitrary axis be equal to zero, that is, the resultant of the forces. Then the body is either at rest, or moves uniformly and rectilinearly.

A body that has an axis of rotation will be in equilibrium if the rule of moments of forces is followed: the sum of the moments of forces that rotate the body clockwise must be equal to the sum of the moments of forces that rotate it counterclockwise.

To get the right moment with the least effort, you need to apply force as far as possible from the axis of rotation, increasing the same arm of the force and, accordingly, reducing the value of the force. Examples of bodies that have an axis of rotation are: a lever, doors, blocks, a brace, and the like.

Three types of balance of bodies that have a fulcrum

1. stable equilibrium, if the body, being removed from the equilibrium position to the neighboring nearest position and left in peace, returns to this position;
2. unstable equilibrium, if the body, being removed from the equilibrium position to a neighboring position and left at rest, will deviate even more from this position;
3. indifferent equilibrium - if the body, being brought to a neighboring position and left in peace, remains in its new position.

Balance of a body with a fixed axis of rotation

1. stable, if in the equilibrium position the center of gravity C occupies the lowest position of all possible near positions, and its potential energy will have the smallest value of all possible values ​​in neighboring positions;
2. unstable if the center of gravity C occupies the highest of all nearby positions, and the potential energy has the greatest value;
3. indifferent if the center of gravity of the body C in all nearby possible positions is at the same level, and the potential energy does not change during the transition of the body.

Task 1

A body A with mass m = 8 kg is placed on a rough horizontal table surface. A thread is tied to the body, thrown over block B (Figure 1, a). What weight F can be tied to the end of the thread hanging from the block so as not to disturb the balance of the body A? Friction coefficient f = 0.4; ignore the friction on the block.

Let's define body weight ~A: ~G = mg = 8$\cdot $9.81 = 78.5 N.

We assume that all forces are applied to body A. When the body is placed on a horizontal surface, only two forces act on it: the weight G and the oppositely directed reaction of the support RA (Fig. 1, b).

If we apply some force F acting along a horizontal surface, then the reaction RA, which balances the forces G and F, will begin to deviate from the vertical, but the body A will be in equilibrium until the modulus of the force F exceeds the maximum value of the friction force Rf max , corresponding to the limit value of the angle $(\mathbf \varphi )$o (Fig. 1, c).

Having decomposed the reaction RA into two components Rf max and Rn, we obtain a system of four forces applied to one point (Fig. 1, d). Projecting this system of forces onto the x and y axes, we obtain two equilibrium equations:

$(\mathbf \Sigma )Fkx = 0, F - Rf max = 0$;

$(\mathbf \Sigma )Fky = 0, Rn - G = 0$.

We solve the resulting system of equations: F = Rf max, but Rf max = f$\cdot $ Rn, and Rn = G, so F = f$\cdot $ G = 0.4$\cdot $ 78.5 = 31.4 H; m \u003d F / g \u003d 31.4 / 9.81 \u003d 3.2 kg.

Answer: Mass of cargo m = 3.2 kg

Task 2

The system of bodies shown in Fig. 2 is in a state of equilibrium. Weight of cargo tg=6 kg. Angle between vectors $\widehat((\overrightarrow(F))_1(\overrightarrow(F))_2)=60()^\circ $. $\left|(\overrightarrow(F))_1\right|=\left|(\overrightarrow(F))_2\right|=F$. Find the mass of weights.

The resultant force $(\overrightarrow(F))_1and\ (\overrightarrow(F))_2$ is equal in absolute value to the weight of the load and opposite to it in direction: $\overrightarrow(R)=(\overrightarrow(F))_1+(\overrightarrow (F))_2=\ -m\overrightarrow(g)$. By the law of cosines, $(\left|\overrightarrow(R)\right|)^2=(\left|(\overrightarrow(F))_1\right|)^2+(\left|(\overrightarrow(F) )_2\right|)^2+2\left|(\overrightarrow(F))_1\right|\left|(\overrightarrow(F))_2\right|(cos \widehat((\overrightarrow(F)) _1(\overrightarrow(F))_2)\ )$.

Hence $(\left(mg\right))^2=$; $F=\frac(mg)(\sqrt(2\left(1+(cos 60()^\circ \ )\right)))$;

Since the blocks are movable, $m_g=\frac(2F)(g)=\frac(2m)(\sqrt(2\left(1+\frac(1)(2)\right)))=\frac(2 \cdot 6)(\sqrt(3))=6.93\ kg\ $

Answer: The mass of each weight is 6.93 kg.

Let us find out under what conditions a body at rest with respect to some inertial frame of reference will remain at rest.

If the body is at rest, then its acceleration is zero. Then, according to Newton's second law, the resultant of the forces applied to the body should also be equal to zero. Therefore, the first equilibrium condition can be formulated as follows:

If the body is at rest, then the vector sum (resultant) of the forces applied to it is equal to zero:

Note that condition (1) alone is not enough for the body to rest. For example, if the body had an initial speed, then it will continue to move at the same speed. In addition, as we will see later, even if the vector sum of the forces applied to a body at rest is zero, it can start to rotate.

In cases where the body at rest at the initial moment can be considered as a material point, the first equilibrium condition is sufficient for the body to remain at rest. Consider examples.

Let a load of mass m be suspended on three cables and rest (Fig. 35.1). Node A, connecting the cables, can be considered a material point that is in equilibrium.

Therefore, the vector sum of the thread tension forces applied to node A is zero (Fig. 35.2):

Let us show two ways of applying this equation in solving problems.

We use vector projections. We choose the coordinate axes and denote the angles between cables 1, 2 and the vertical, as shown in Figure 35.2.

1. Explain why the following equations are valid in this case:

Ox: -T 1 sin α 1 + T 2 sin α 2 \u003d 0,
Oy: T 1 cos α 1 + T 2 cos α 2 - T 3 = 0,
T3 = mg.

Use this system of equations for the following tasks.

2. What is the tension force of each cable, if m = 10 kg, α 1 = α 2 = 30º?

3. It is known that T 1 = 15 N, α 1 = 30º, α 2 = 45º. What are equal to: a) the tension force of the second cable T 2 ? 5) mass of cargo m?

4. Let α 1 = α 2 . What are these angles if the tension force of each cable: a) is equal to the weight of the load? b) 10 times the weight of the load?

So, the forces acting on the suspensions can many times exceed the weight of the load!

Let's take advantage of the fact that three vectors, the sum of which is equal to zero, "close" into a triangle (Fig. 35.3). Consider an example.

5. A lantern of mass m is suspended on three cables (Fig. 35.4). Let us denote the modules of the tension forces of the cables T 1 , T 2 , T 3 . Angle α ≠ 0.
a) Draw the forces acting on node A and explain why T 3 > mg and T 3 > T 2 .
b) Express T 3 in terms of m, g and T 2 .
Clue. Force vectors 1 , 2 and 3 form a right triangle.

2. The second condition for the equilibrium of the body (the rule of moments)

Let us be convinced by experience that the first equilibrium condition alone is not enough for the body to remain at rest.

Let's put experience
We attach two threads to a piece of cardboard and pull them in opposite directions with equal forces (Fig. 35.5). The vector sum of the forces applied to the cardboard is zero, but it will not remain at rest, but will begin to turn.

The condition for the balance of a body fixed on an axis

The second equilibrium condition for a body is a generalization of the equilibrium condition for a body fixed on an axis. It is familiar to you from the basic school physics course. (This condition is a consequence of the law of conservation of energy in mechanics.) Recall it.

Let forces 1 and 2 act on a body fixed on the O axis (Fig. 35.6). A body can only be in equilibrium if

F 1 l 1 \u003d F 2 l 2 (2)

Here l 1 and l 2 are the shoulders of forces, then the distances from the axis of rotation O to the line of action of forces 1 and 2.

To find the shoulder of the force, you need the line of action of the force and lower the perpendicular from the axis of rotation to this line. Its length is the shoulder of strength.

6. Transfer figure 35.7 to your notebook. One cell corresponds to 1 m. What are the arms of forces 1 , 2 , 3 , 4 ?

The rotating action of a force is characterized by a moment of force. The modulus of the moment of force is equal to the product of the modulus of force and its arm. The moment of force is considered positive if the force tends to rotate the body counterclockwise, and negative if it is clockwise. (Thus, the sign of the moment of force rotating the body in one direction coincides with the sign of the angle of rotation in the same direction on the unit circle familiar to you from the school mathematics course.)

For example, the moments of the forces shown in Figure 35.8 relative to the point O are as follows:

M 1 \u003d F 1 l 1; M 2 \u003d -F 2 l 2.

The moment of force is measured in newtons * meters (N * m).

7. What are the moments of the forces shown in Figure 35.7 about the point O? One cell corresponds to a distance of 1 m, as well as a force of 1 N.

Let us rewrite relation (2) using the moments of forces:
M1 + M2 = 0. (3)
This relation is called the rule of moments.

If several forces act on a body at rest, fixed on an axis, then it will remain at rest only under the condition that the algebraic sum of the moments of all these forces is equal to zero:

M 1 + M 2 + ... + M n = 0.

Note that this condition alone is not enough for the body to rest. If the algebraic sum of the moments of the forces applied to the body is equal to zero, but at the initial moment the body is rotating, then it will continue to rotate with the same angular velocity.

To verify this, spin the bicycle wheel of an elevated bike or top. After that, they will rotate for quite a long time: only a small friction force will slow them down. Yes, and our Earth for billions of years rotates around its axis, although no forces rotate the Earth around the axis!

The equilibrium condition for a body not fixed on an axis

Let us now take into account the force acting on the body fixed on the axis from the side of the axis. So, the body considered above (Fig. 35.6) is actually in equilibrium under the action of three forces: 1, 2 and 3 (Fig. 35.9, a).

And now we note that a body at rest does not rotate around any axis.

Therefore, the second equilibrium condition for a body not fixed on an axis can be formulated as follows:

in order for the body to remain at rest, it is necessary that the algebraic sum of the moments of all forces applied to the body about any axis be equal to zero:

M 1 + M 2 + … + M n = 0. (4)

(We assume that all forces applied to the body lie in the same plane.)

For example, a piece of cardboard resting under the action of forces 1, 2 and 3 (Fig. 35.9, b) can be fixed with a needle at an arbitrary point O 1. The body "does not notice" the new axis of rotation O 1: it will remain at rest as it was.

When solving problems, the axis relative to which the moments of forces are found is often drawn through the point of application of the force or forces that are not specified in the condition: then their moments relative to this axis are equal to zero. For example, in the following task, it is convenient to take the lower end of the rod as such an axis.

Note that one second equilibrium condition is also not enough for the body to remain at rest.

A body at rest at the initial moment will remain at rest only if both the resultant of the forces applied to the body and the algebraic sum of the moments of these forces about any axis are equal to zero. (Strictly speaking, this also requires that the equilibrium be stable (see § 36).)

8. The upper end of a light rod at rest with a length L is held by a horizontal cable (Fig. 35.10). The lower end of the rod is hinged (the rod can rotate around the lower end). The angle between the rod and the vertical is α. A load of mass m is suspended from the middle of the rod. Friction in the hinge can be neglected. Draw in the drawing the weight of the load m and the tension force of the cable, which act on the rod. What are equal to:
a) the shoulder and the moment of gravity relative to the point O?
b) arm and moment of force relative to point O?
c) modulus of force?

How can you move the point of application of force?

Let's move the point of application of forces from A to B along the line of action of the force (Fig. 35.11).

Wherein:
- the vector sum of the forces acting on the body will not change;
- the moment of this force relative to any axis will not change, because the shoulder l of this force has not changed.

So, the point of application of force can be transferred along the line of its action without disturbing the balance of the body.

9. Explain why a body can be at rest under the action of three non-parallel forces only if their lines of action intersect at one point (Fig. 35.12).

Please note: the point of intersection of the lines of action of these forces can be (and often is!) Outside the body.

10. Let's return to task 8 (Fig. 35.10).
a) Find the point of intersection of the lines of action of the weight of the load and the tension of the cable.
b) Find graphically the direction of the force acting on the rod from the side of the hinge.
c) Where should the attachment point of the horizontally directed cable be moved so that the force acting on the rod from the hinge side is directed along the rod?

3. Center of gravity

The center of gravity is the point at which gravity is applied. We will denote the center of gravity by the letter C. The center of gravity of a homogeneous body of regular geometric shape coincides with its geometric center.

For example, the center of gravity of a homogeneous:

• the disk coincides with the center of the disk (Fig. 35.13, a);
• a rectangle (in particular, a square) coincides with the intersection point of the diagonals (Fig. 35.13, b);
• a rectangular parallelepiped (in particular, a cube) coincides with the intersection point of the diagonals connecting opposite vertices;
• thin rod coincides with its middle (Fig. 35.13, c).

For bodies of arbitrary shape, the position of the center of gravity is found empirically:

if a body suspended at one point is in equilibrium, then its center of gravity lies on the same vertical with the point of suspension(Fig. 35.13, d).

Indeed, if the center of gravity and the point of suspension are not on the same vertical, then the algebraic sum of the moments of gravity and the force acting from the side of the suspension will not be equal to zero (for example, relative to the center of gravity).

The algebraic sum of the moments of the forces of gravity acting on all parts of the body, relative to the center of gravity of the body, is equal to zero. (Otherwise it would not be possible to hang it at one point.)

This is used when calculating the position of the center of gravity.

11. At the ends of a light rod of length l, balls of mass m1 and m2 are fixed. At what distance from the first ball is the center of gravity of this system?

12. A horizontally located homogeneous beam with a length of 1 m and a mass of 100 kg hangs on two vertical cables. The blue cable is fixed at a distance of 20 cm from the left end of the beam, and the green one at a distance of 30 cm from its right end. Draw in the drawing the forces acting on the beam and their shoulders relative to the center of gravity of the beam. What are equal to:
a) shoulders of forces? b) tension forces of cables?

Additional questions and tasks

13. At the same height at a distance of 1 m from each other, the ends of an inextensible cable 2 m long are fixed. What is the maximum mass of the load that can be suspended from the middle of the cable so that the cable tension does not exceed 100 N?

14. The lantern is suspended on two cables. The tension forces of the cables are 10 N and 20 N, and the angle between the cables is 120º. What is the mass m of the lantern?
Clue. If the sum of three vectors is zero, then they form a triangle.

15. Forces 1 and 2 are applied to a piece of cardboard fixed on the O axis at points A 1 and A 2 (Fig. 35.14). It is known that OA 1 = 15 cm, OA 2 = 20 cm, F 1 = 20 N, F 2 = 30 N, α = 60º, β = 30º.

a) What are the arms of forces 1 and 2?
b) What are the moments of these forces (taking into account the sign)?
c) Can the cardboard stay still? And if not, in which direction will it start to rotate?

16. Two people carry a cylindrical pipe with a mass of 30 kg and a length of 4 m. The first person holds the pipe at a distance of 1.2 m from the end. At what distance from the other end does the second person, eyelid, hold the pipe if the load on his shoulder is 100 N?

17. A light rod 1 m long is fixed on a horizontal axis. If a weight is suspended from the left end of the rod, and a weight of mass 1 kg is suspended from the right end, then the rod will be in equilibrium. And if the same load is suspended from the right end of the rod, then the rod will be in equilibrium if a weight of mass 16 kg is suspended from its left end.
a) What is the weight of the load?
b) How far from the center of the rod is the axis?