In the previous paragraph, the conditions for equilibrium of a body in the absence of rotation were clarified. But how is the absence of rotation of a body ensured, i.e. its balance, when forces act on it?
To answer this question, consider a body that cannot perform translational motion, but can turn or rotate. To make it impossible forward movement body, it is enough to secure it at one point in the same way as you can, for example, secure a board on a wall by nailing it with one nail; the forward movement of such a “nailed” board becomes impossible, but the board can rotate around the nail, which serves as its axis of rotation.
Let us find out under what conditions a body at rest with a fixed axis will not rotate under the influence of forces applied to it. Let us imagine some body to which different points two forces are applied: (Fig. 163, a). To find the resultant of these forces, we move the points of their application to point A (Fig. 163, b), at which the lines of action of both forces intersect. By constructing a parallelogram on the forces, we obtain their resultant
Now suppose that at some point O on the line along which the resultant is directed there passes a fixed axis perpendicular to the plane of the drawing. We can imagine, for example, that at point O a nail driven into a stationary wall passes through the body. The body in this case will be at rest, because the resultant is balanced by the reaction force (elasticity) from the side of the fixed axis (nail): both of them are directed along the same straight line, equal in absolute value and opposite in direction.
Let us now assume that one of the forces, for example, has ceased to act, so that the body is subject to the action of only one force (Fig. 163, c). From the figure it is clear that this force will cause the body to rotate around the O axis clockwise. If, on the contrary, we eliminate
force, then the remaining force will cause counterclockwise rotation (Fig. 163, d). This means that each of the forces has a rotating effect, and these actions are characterized by in opposite directions. But when both forces act together, their rotating actions cancel each other out: together they do not cause rotation. It must therefore be considered that, although the forces themselves are different both in magnitude and direction, their rotating effects are the same, but opposite in direction.
Let's try to find a value that characterizes the rotating effect of the force. We only know so far that it should have the same numerical values for both forces:
Let us turn to the figure The forces are not the same in absolute values: more But the distance from point O (axis) to the line of action of the force is less than the distance from the axis to the line of action of the force Thus, but
Perhaps the products are equal to each other
If this is so, then we can say that the quantity equal to the product force on the length of the perpendicular lowered from the fixed axis to the line of action of the force, precisely characterizes the rotating action of the force.
It is not difficult to prove that the equality
actually fulfilled. To do this, let us draw in Figure 163, d auxiliary straight lines OS and OB, parallel to the forces of similarity of triangles ABO and it follows that
From here, taking into account that AB = OS, we get:
Let us now consider the triangles OVK and These triangles are similar, like rectangular ones with equal angles at vertices C and B (they complement equal angles ASO and ABO up to 180°). From their similarity it follows that
Comparing proportions (1) and (2), we get:
The above assumption was justified.
The above rather long geometric reasoning allowed us to find a value that is the same for both forces and characterizes the rotating effect of the force. This quantity is the product of the force and the distance from its line. actions up to the axis of rotation. This quantity has a somewhat strange name - moment of force or torque about the axis passing through point O.
Lesson objectives:
Educational. Study two conditions for the equilibrium of bodies, types of equilibrium (stable, unstable, indifferent). Find out under what conditions bodies are more stable.
Educational: Promote development cognitive interest to physics, develop the ability to make comparisons, generalize, highlight the main thing, and draw conclusions.
Educational: cultivate discipline, attention, and the ability to express one’s point of view and defend it.
Lesson plan:
1. Updating knowledge
2. What is statics
3. What is balance. Types of balance
4. Center of mass
5. Problem solving
Progress of the lesson:
1. Updating knowledge.
Teacher: Hello!
Students: Hello!
Teacher: We continue to talk to you about forces. There's a body in front of you irregular shape(stone) suspended by a thread and attached to inclined plane. What forces are acting on this body?
Students: The body is acted upon by: the tension force of the thread, the force of gravity, the force tending to tear off the stone, which is opposite to the tension force of the thread, and the support reaction force.
Teacher: We have found the strength, what do we do next?
Students: We write Newton's second law.
There is no acceleration, so the sum of all forces is zero.
Teacher: What does this mean?
Students: This indicates that the body is at rest.
Teacher: Or we can say that the body is in a state of equilibrium. The balance 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: "Conditions for the equilibrium of bodies. Types of equilibrium."
2. Formation of new knowledge and methods of action.
Teacher: The branch of mechanics in which the equilibrium of absolutely rigid bodies is studied is called statics. There is not a single body around us that is not affected by forces. Under the influence of these forces, bodies are deformed.
When determining the equilibrium conditions of deformed bodies, it is necessary to take into account the magnitude and nature of the deformation, which complicates the problem put forward. Therefore, to clarify the basic laws of equilibrium, for convenience, the concept of absolutely solid.
An absolutely rigid body is a body in which the deformations arising under the influence of forces applied to it are negligible. Write down the definitions of statics, equilibrium of bodies and an absolutely rigid body from the screen (slide 2).
And what we have found out is that the body is in equilibrium if geometric sum of all forces applied to it is equal to zero is the first condition for equilibrium. Write down 1 equilibrium condition:
If the sum of forces is zero, then the sum of the projections of these forces on the coordinate axes is also zero. In particular, for projections of external forces onto the X axis, we can write .
The equality to zero of the sum of external forces acting on a solid body is necessary for its equilibrium, but not sufficient. For example, two forces of equal magnitude and opposite directions were applied to the board at different points. The sum of these forces is zero. Will the board be in equilibrium?
Students: The board will turn, for example, like the steering wheel of a bicycle or car.
Teacher: Right. In the same way, two forces of equal magnitude and opposite directions 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 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 balance. Therefore, we need to find out one more condition for the equilibrium of bodies. To do this, let's conduct an experiment. (Two students are called). One of the students applies the force closer to the axis of rotation of the door, the other student applies the force closer to the handle. They put effort into different sides. What happened?
Students: The one who exerted the force closest to the handle won.
Teacher: Where is the line of action of the force applied by the first student?
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 door handle.
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 does the result of the force depend on?
Students: The result 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 in this case?
Students: According to the rule of equilibrium of a lever, the forces acting on it are inversely proportional to the arms of these forces. .
Teacher: What is the product of the modulus of the force rotating the body and its shoulder?
Students: Moment of power.
Teacher: This means that the moment of force applied to the first students is equal to , and the moment of force applied to the second students is equal to
Now we can formulate the second equilibrium condition: A rigid body is in equilibrium if algebraic sum moments of external forces acting on it relative to any axis is equal to zero. (slide 3)
Let us introduce the concept of 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 forces gravity acting on individual elements body). There is also the concept of center of mass.
The center of mass of a system of material points is called 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 the types of balance using a block with depressions and slides and a ball.)
On slide 5 you see the same thing you saw in experience. Write down the conditions for equilibrium stability 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.The bodies are in a state 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 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 conditions of sustainability in all three cases.
Students: In the first case, if the support point 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 look at bodies that have a support area. The support area is the area of contact between the body and the support. (slide 10).
Let's consider how the position of the line of action of gravity changes relative to the axis of rotation of the body when the body having a support area is tilted. (slide 11)
Please 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. Thus, the inclined bodies will be in a state of stable equilibrium as long as the line of action of gravity passes through the area of support. Look at slide 12.
If, when a body that has a support area deviates, the center of gravity increases, then the equilibrium will be stable. At 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 support area, but different heights, have different limit angle tilt If this angle is exceeded, the bodies tip over. (slide 13)
At a lower center of gravity, it is necessary to spend great job to tip over the body. Therefore, the work of overturning can serve as a measure of its stability. (Slide 14)
Thus, tilted structures are in a position of stable equilibrium, because the line of action of gravity passes through the area of their support. For example, the Leaning Tower of Pisa.
The swaying or tilting of a person's body when walking is also explained by the desire to maintain a stable position. The area of support is determined by the area inside the line drawn around extreme points body touching the support. when a person is standing. The line of gravity passes through the support. When a person raises his leg, in order to maintain balance, he bends over, transferring the line 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 position of the structure’s center of gravity is lowered, making a powerful support, or the support area is increased and, at the same time, the structure’s center of gravity is lowered.
The sustainability of transport is determined by the same conditions. Thus, of the two types of transport, a car and a bus, a car is more stable on an inclined road.
With the same inclination of these types of transport, the line of gravity of the bus passes closer to the edge of the support area.
Problem solving
Problem: Material points with masses m, 2m, 3m and 4m are located at the vertices of a rectangle with sides 0.4 m and 0.8 m. Find the center of gravity of the system of these material points.
x s -? u s -?
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 in which the material point with mass is located 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, this point is located at the origin of coordinates. The coordinate is also zero, the coordinates of the points with masses on the OX axis are the same and equal to the length of the side of the rectangle. Substituting the coordinate values we get
The coordinate on the OY axis of a point with mass is zero, =0. The coordinates of the 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
1. Conditions for body equilibrium?
1 equilibrium condition:
A rigid body is in equilibrium if the geometric sum of external forces applied to it is equal to zero.
2 Equilibrium condition: A rigid body is in equilibrium if the algebraic sum of the moments of external forces acting on it relative to any axis is equal to zero.
2. Name the types of equilibrium.
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.
List of used literature:
1.
Physics. 10th grade: textbook. for general education institutions: basic and profile. levels / G. Ya. Myakishev, B. B. Bukhovtsev, N. N. Sotsky; edited by V. I. Nikolaeva, N. A. Parfentieva. - 19th ed. - M.: Education, 2010. - 366 p.: ill.
2.
Maron A.E., Maron E.A. "Collection quality tasks in physics 10th grade, M.: Prosveshchenie, 2006
3.
L.A. Kirik, L.E.Gendenshtein, Yu.I.Dik. Methodological materials for teacher 10th grade, M.: Ilexa, 2005.-304с:, 2005
4.
L.E.Gendenshtein, Yu.I.Dik. Physics 10th grade.-M.: Mnemosyne, 2010
Moment of power. Equilibrium condition for a body having an axis of rotation
A moment of power is a quantity that can cause and change the rotation of a body. In this case, the moment of force is isolated relative to the point (center) and relative to the axis.
Rice. 4.2
Moment of force relative fixed point ABOUT represents the vector defined vector product radius vector drawn from the point ABOUT to the point N application of force on force rice. 4.2:
where is the modulus of the moment of force M =Fr sina= F× l(l¾shoulder of force, that is, shortest distance between the line of action of the force and the point ABOUT). The vector is directed perpendicular to the plane passing through the center ABOUT and the force to the side where the rotation caused by the force is visible counterclockwise.
Example. Let a point load of mass m suspended on an inextensible and weightless thread of length R to a nail driven into the ceiling, commits fluctuations near the equilibrium position, Fig. 4.3.
Rice. 4.3
For the moment in time under consideration, when the load returns to the equilibrium position, the vector of the moment of force coincides in direction with the vector angular velocity its modulus is M 0 =mgl=mgR sina; moment of thread tension T Always equal to zero, because the shoulder of this force equals zero.
Moment of force relative fixed axis z is an algebraic quantity, equal projection to this axis of the vector of the moment of force, defined relative to an arbitrary point ABOUT on the axis z, rice. 4.4.
Rice. 4.4
To solve the usual school tasks it is sufficient to consider the moment of force relative to the axis z, perpendicular to the plane, which contains the vectors and Fig. 4.5.
The direction of the axis is chosen in such a way that the moment is positive if it causes clockwise rotation.
Rice. 4.5
Any body can be affected by moments various forces, however, for its equilibrium, in the presence of a fixed axis of rotation z, it is necessary that the algebraic sum of the moments of all forces acting on the body relative to this axis was equal to zero
or, to put it more in simple language, moments of all strength Mz rotating the body clockwise must be equal to the moments of all forces rotating it counterclockwise. In this case, the body will either be at rest or rotate uniformly around its axis.
If a body does not have a fixed axis of rotation, for its equilibrium it is necessary and sufficient to satisfy conditions (4.1) and (4.6) with respect to any possible axis.
Equilibrium conditions are often used to measure unknown forces by comparing them with by known forces. For example, the magnitude of various forces (gravitational, electrostatic, magnetic) is measured by comparing them with the elastic force. In particular, the force of gravity acting on the body can be determined from the readings of a spring dynamometer.
An important task statics is the determination of the center of gravity of a body or system of bodies. Center of gravity is point of application of the resultant of all gravity forces acting on a body at any position in space(usually found by crossing the body's suspension lines). The sum of the moments of all elementary forces of gravity about any axis that passes through the center of gravity is equal to zero.
For a homogeneous body, the center of gravity is located on the axis of symmetry and the intersection of the axes of symmetry, and it may be outside the body itself (for example, near a ring).
Example. Two people, weighing m 1 = 60 kg and m 2 = 100 kg are in equilibrium at different ends horizontally located homogeneous rectangular board, length l= 3 m and mass m 3 = 30 kg, having the same thickness and located on a fallen tree, Fig. 4.6. At what distance X from the right edge of the board is the center of gravity of the system consisting of a board and two people or, in other words, the point of contact of the board with the tree?
Rice. 4.6
Solution. According to condition (4.2), the resultant of gravity 
modulo equal to the modulus of the vector i.e. m 1 g+m 2 g+m 3 g=N. This expression useful for general reasoning and correct construction of the figure, but to solve the problem it is quite enough to use condition (4.6).
Let us find out under what conditions a body at rest relative to some inertial system 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 must 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 be at rest. For example, if the body has 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 begin to rotate.
In cases where something at rest in starting moment the body can be considered as a material point; the first condition of equilibrium is sufficient for the body to remain at rest. Let's look at examples.
Let a load of mass m be suspended on three cables and at rest (Fig. 35.1). Node A, connecting the cables, can be considered material point, which is in equilibrium. 
Consequently, the vector sum of the thread tension forces applied to node A is equal to zero (Fig. 35.2): 
We will show two ways to use this equation to solve problems.
We use vector projections. Let us select the coordinate axes and designate 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 = 0,
Oy: T 1 cos α 1 + T 2 cos α 2 – T 3 = 0,
T 3 = mg.
Use this system of equations to complete 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 is 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 equal to if the tension force of each cable: a) is equal to the weight of the load? b) 10 times more weight cargo?
So, the forces acting on the suspensions can be many times greater than the weight of the load!
Let's take advantage of the fact that three vectors whose sum is zero “close” into a triangle (Fig. 35.3). Let's look at 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 body equilibrium (the rule of moments)
Let us verify from experience that the first condition of equilibrium alone is not enough for the body to remain at rest.
Let's put experience
Attach two threads to a piece of cardboard and pull them opposite sides with forces equal in magnitude (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 rotate. 
Condition of equilibrium of a body fixed on an axis
The second condition for the equilibrium of a body is a generalization of the condition for the equilibrium of 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.) Let us recall it.
Let forces 1 and 2 act on a body fixed on the O axis (Fig. 35.6). A body can be in equilibrium only if
F 1 l 1 = F 2 l 2 (2)
Here l 1 and l 2 are the arms of the forces, then the distances from the axis of rotation O to the line of action of forces 1 and 2.
To find the arm of the force, you need to take 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 power.
6. Transfer Figure 35.7 to your notebook. One cell corresponds to 1 m. What are the arms of forces 1, 2, 3, 4 equal to?
The rotating effect 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 the force and its arm. The moment of force is considered positive if the force tends to rotate the body counterclockwise, and negative if it rotates clockwise. (Thus, the sign of the moment of force rotating the body in some direction coincides with the one familiar to you from school course mathematics sign of the angle of rotation in the same direction on the unit circle.)
For example, the moments of the forces shown in Figure 35.8 relative to point O are as follows:
M 1 = F 1 l 1 ; M 2 = –F 2 l 2 .
Moment of force is measured in newton * meters (N * m).
7. What are the moments of the forces shown in Figure 35.7 relative to point O? One cell corresponds to a distance of 1 m and a force of 1 N.
Let's rewrite relation (2) using moments of forces:
M 1 + M 2 = 0. (3)
This relationship 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 if 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 be at rest. If the algebraic sum of the moments of the forces applied to the body is 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 a raised bicycle or spinning top. After this, they will rotate for quite a long time: only a small friction force will slow them down. And our Earth has been rotating around its axis for billions of years, although no forces rotate the Earth around its axis!
Equilibrium condition for a body not fixed to an axis
Let us now take into account the force acting on a body fixed to an axis from the side of the axis. Thus, the body discussed above (Fig. 35.6) is actually in equilibrium under the influence of three forces: 1, 2 and 3 (Fig. 35.9, a). 
Now 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 relative to 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, at rest under the influence of forces 1, 2 and 3 (Fig. 35.9, b), can be fixed with a needle in arbitrary point O 1. The body “will not notice” the new axis of rotation O 1: it will remain at rest as it was.
When solving problems, the axis about which the moments of forces are found is often drawn through the point of application of a force or forces that are not specified in the condition: then their moments about 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 the second equilibrium condition alone 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 relative to any axis are equal to zero. (Strictly speaking, for this it is also necessary that the equilibrium be stable (see § 36).)
8. The upper end of a stationary light rod of length L is held by a horizontal cable (Fig. 35.10). The lower end of the rod is fixed in a hinge (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. The friction at the hinge can be neglected. Draw on the drawing the weight of the load m and the tension force of the cable that act on the rod. What are they equal to:
a) shoulder and moment of gravity relative to point O?
b) arm and moment of force about point O?
c) force modulus?
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). 
In this case:
- the vector sum of forces acting on the body will not change;
- the moment of this force relative to any axis will not change, because the arm 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 force of the cable.
b) Find graphically the direction of the force acting on the rod from 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 is directed along the rod?
3. Center of gravity
The center of gravity is the point of application of gravity. We will denote the center of gravity by the letter C. The center of gravity of a homogeneous body of a 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 point of intersection 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 free form The position of the center of gravity is found experimentally:
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 suspension point are not on the same vertical, then the algebraic sum of the moments of gravity and the force acting from the suspension will not be equal to zero (for example, relative to the center of gravity).
The algebraic sum of the moments of gravity acting on all parts of the body relative to the center of gravity of the body is zero. (Otherwise it would be impossible 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 attached. At what distance from the first ball is the center of gravity of this system?
12. A horizontally located uniform beam 1 m long and weighing 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 cable is fixed at a distance of 30 cm from its right end. Draw on the drawing the forces acting on the beam and their shoulders relative to the center of gravity of the beam. What are they equal to:
a) shoulders of strength? b) the tension force of the cables?
Additional questions and tasks
13. The ends of an inextensible cable 2 m long are fixed at the same height at a distance of 1 m from each other. Which maximum weight can the load be suspended from the middle of the cable so that the tension force of the cable 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 equal to?
b) What are the moments of these forces (taking into account the sign)?
c) Can cardboard be left alone? And if not, then 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 hold the pipe if the load on his shoulder is 100 N?
17. A light rod 1 m long is fixed to horizontal axis. If a certain load is suspended from the left end of the rod, and a weight weighing 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 16 kg is suspended from its left end.
a) What is the mass of the load?
b) At what distance from the center of the rod is the axis?
