a) Yes, you can.
b) No, you can’t.
IN WHICH OF THE CASES INDICATED IN FIGURE 1, THE TRANSFER OF FORCE FROM POINT A TO POINTS B, C OR D WILL NOT CHANGE THE MECHANICAL STATE OF THE SOLID BODY?
IN FIG. 1, b SHOW TWO FORCES, THE LINES OF ACTION OF WHICH LIE IN THE SAME PLANE. IS IT POSSIBLE TO FIND THEIR EQUAL ACTION BY THE PARALLELOGRAM RULE?
b) It is impossible.
5. Find a correspondence between the formula for determining the resultant of two forces F 1 and F 2 and the value of the angle between the lines of action of these forces
CONNECTIONS AND THEIR REACTIONS
IN WHICH RELATIONS LISTED BELOW ARE THE REACTIONS ALWAYS DIRECTED NORMAL (PERPENDICULAR) TO THE SURFACE?
a) Smooth plane.
b) Flexible connection.
c) Rigid rod.
d) Rough surface.
WHAT IS THE SUPPORT REACTION APPLIED TO?
a) To the support itself.
b) To the supporting body.
STANDARD ANSWERS
| Issue No. | |||||||
| No. |
FLAT SYSTEM OF CONVERGING FORCES
Choose the correct answer
8. AT WHAT VALUE OF THE ANGLE BETWEEN THE FORCE AND THE AXIS IS THE FORCE PROJECTION EQUAL TO ZERO?
IN WHICH OF THE CASES IS THE FLAT SYSTEM OF CONVERGING FORCES BALANCED?
A) å Fix = 40 H; å F iy = 40 H.
b) å Fix = 30 H; å F iy = 0 .
V) å Fix = 0 ; å F iy = 100 H.
G) å Fix = 0; å F iy = 0 .
10. WHICH OF THE SYSTEMS OF EQUILIBRIUM EQUATIONS LISTED BELOW IS FAIR FOR THE SYSTEM SHOWN IN FIGURE. 2 SYSTEMS OF CONVERGING FORCES?
A) å Fix = 0; F 3 cos 60° + F 4 cos 30° + F 2 = 0;
å F iy = 0; F 3 cos 30° - F 4 cos 60° + F 1 = 0.
b) å Fix = 0; - F 3 cos 60° - F 4 cos 30° + F 2 = 0;
å F iy = 0; F 3 cos 30° - F 4 cos 60° - F 1 = 0.
INDICATE WHICH VECTOR OF THE FORCE POLYGON IN FIG. 3, and IS AN EQUAL FORCE.
WHICH OF THE POLYGONS PRESENTED IN FIG. 3, CORRESPONDING TO A BALANCED SYSTEM OF CONVERGING FORCES?
c) none of them correspond.
STANDARD ANSWERS
| Issue No. | |||||
| No. |
PAIR OF FORCES AND MOMENTS OF FORCES
Choose the correct answer
DETERMINE WHICH FIGURE SHOWS A PAIR OF FORCES
THE EFFECT OF A PAIR OF FORCES DETERMINES
a) Product of force on the shoulder.
b) The moment of the couple and the direction of rotation.
A PAIR OF FORCES CAN BE BALANCED
a) By force alone.
b) A couple of forces.
THE EFFECT OF A PAIR OF FORCES ON A BODY FROM ITS POSITION IN THE PLANE
a) depends.
b) does not depend.
17. The body is affected by three pairs of forces applied in one plane: M 1 = - 600 Nm; M 2 = 320 Nm; M 3 = 280 Nm. UNDER THE INFLUENCE OF THESE THREE PAIRS OF FORCES
a) the body will be in equilibrium.
b) the body will not be in equilibrium.
IN FIG. 4 THE LEVER OF FORCE F RELATIVE TO POINT O IS A SEGMENT
MOMENT OF FORCE F RELATIVE TO POINT K IN FIG. 4 DETERMINED FROM THE EXPRESSION
a) Mk = F∙AK.
b) Mk = F∙ВK.
VALUE AND DIRECTION OF THE MOMENT OF FORCE RELATIVE TO A POINT FROM THE RELATIVE POSITION OF THIS POINT AND THE LINE OF ACTION OF THE FORCE
a) do not depend.
b) depend.
Choose all correct answers
2.1.6 Axiom 6, solidification axiom
If a deformable (not absolutely solid) body is in equilibrium under the influence of some system of forces, then its equilibrium is not disturbed even after it hardens (becomes absolutely solid).
The principle of solidification leads to the conclusion that the imposition of additional connections does not change the equilibrium of the body and makes it possible to consider deformable bodies (cables, chains, etc.) that are in equilibrium as absolutely rigid bodies and to apply static methods to them.
Exercises Consultations
| 6. The figure shows five equivalent systems of forces. On the basis of what axioms or properties of forces proven on their basis, the transformations of the initial (first) system of forces into each of the subsequent ones (first into second, first into third, etc.) were carried out? | 6.1The system of forces (1.) is transformed into a system of forces (2.) based on the axiom of joining or discarding systems of mutually balanced forces and . When such systems of forces are added or rejected, the resulting system of forces remains equivalent to the original system of forces and the kinematic state of the body does not change. 6.2 The system of forces (1.) is transformed into a system of forces (3.) based on the property of force: force can be transferred along its line of action within a given body to any point, while the kinematic state of the body or the equivalence of the force system does not change. 6.3 The system of forces (1.) is transformed into a system of forces (4.) by transferring forces along their line of action to a point WITH, and therefore the systems of forces (1.) and (4.) are equivalent. 6.4The system of forces (1.) is transformed into a system of forces (5.) by moving from the system of forces (1.) to the system of forces (4.) and adding forces at the point WITH based on the axiom about the resultant of two forces applied at one point. |
| 7. Calculate the resultant of two forces R 1 and R 2 if: 7 A) R 1 = P 2 = 2 N, φ = 30º; 7 b) R 1 = P 2 = 2 N, φ = 90º. | 7. Modulus of the resultant forces R 1 and R 2 is determined by the formula: 7, A) ;
 R = 3,86 N.
7,b) cos 90º = 0;   |
| 8. Make a drawing and find the resultant for cases: 8 A) R 1 = P 2 = 2 N, φ = 120º; 8 b) R 1 = P 2 = 2 N, φ = 0º; 8 V) R 1 = P 2 = 2 N, φ = 180º. | 8 A)  ;R= 2H. 
8 b) cos 0º = 1; R = P 1 +R 2 = 4 N. 
8V) cos 180º = –1; R = P 2 –R 1 = 2 – 2 = 0.
 Note: If R 1 ≠Р 2 and R 1 > R 2, then R directed in the same direction as the force R 1 .
|
Main:
1). Yablonsky A.A., Nikiforova V.L. Course of theoretical mechanics. M., 2002. p. 8 – 10.
2). Targ S.M. Short course in theoretical mechanics. M., 2002. p. 11 – 15.
3). Tsyvilsky V.L. Theoretical mechanics. M., 2001. p. 16 – 19.
4) Arkusha A.I. Guide to solving problems in theoretical mechanics. M., 2000. p. 4 – 20.
Additional:
5). Arkusha A.I. Technical mechanics. M., 2002. p. 10 – 15.
6). Chernyshov A.D. Statics of a rigid body. Krasn-k., 1989. p. 13 – 20.
7). Erdedi A.A. Theoretical mechanics. Strength of materials. M., 2001. p. 8 – 12.
8) Olofinskaya V.P. Technical mechanics. M., 2003. p. 5 – 7.
Questions for self-control
1. Give examples illustrating the axioms of statics .
2. Explain the situation: the axioms of statics are established experimentally.
3. Give examples of the application of the axioms of statics in technology.
4. Formulate an axiom about the balance of two forces.
5. Name the simplest system of forces equivalent to zero.
6. What is the essence of the axiom of inclusion and exclusion of a balanced system of forces?
7. What is the physical meaning of the axiom of solidification?
8. Formulate the rule of parallelogram of forces.
9. What does the axiom of inertia express?
10. Are the equilibrium conditions of an absolutely rigid body necessary and sufficient for the equilibrium of deformable bodies?
11. Give the formulation of the axiom of equality of action and reaction.
12. What is the fundamental error in the expression “action and reaction are balanced”?
13. How is the resultant R of the system of forces directed if the sum of the projections of these forces onto the axis OY equal to zero?
14. How is the projection of force on the axis determined?
15. State the algorithm (order) for determining the module of the resultant Fz, if given:
a) module and direction of one component F, as well as the direction of the other component F 2 and resultant;
b) the modules of both components and the direction of the resultant;
c) the directions of both components and the resultant.
Tests on the topic
| 1. | The figure shows two forces whose lines of action lie in the same plane. Is it possible to find their resultant using the parallelogram rule?  Can i. b) It is impossible. |
| 2. | Fill in the missing word. The projection of a vector onto an axis is... a quantity. a) vector; b) scalar. |
| 3. | In which of the cases indicated in figures a), b) and c), the transfer of force from the point A to points IN, WITH or D will not change the mechanical state of the solid?  a B C) |
| 4. | In Fig. b) (see point 3) two forces are depicted, the lines of action of which lie in the same plane. Is it possible to find their resultant using the parallelogram rule? Can i; b) It is impossible. |
| 5. | At what value of the angle between two forces F 1 and F 2 is their resultant determined by the formula F S = F 1 + F 2? a) 0°; b) 90°; c) 180°. |
| 6. | What is the projection of force on the y-axis?  a) F×sina; b) -F×sina; c) F×cosa; d) – F×cosa. |
| 7. | If two forces are applied to an absolutely rigid body, equal in magnitude and directed along one straight line in opposite directions, then the equilibrium of the body: a) will be disrupted; b) Will not be violated. |
| 8. | At what value of the angle between two forces F 1 and F 2 is their resultant determined by the formula F S = F 1 - F 2? a) 0°; b) 90°; c) 180°. |
| 9. | Determine the direction of the force vector if it is known: P x = 30N, P y = 40N. a) cos = 3/4; cos = 0. b) cos = 0; cos = 3/4. c) cos = 3/5; cos = 4/5. d) cos = 3/4; cos = 1/2. |
| 10. | What is the modulus of the resultant of the two forces?  A) ; b) ; V) ; G) . |
| 11. | Specify the correct expression for calculating the projection of force on the x-axis if the force modulus P = 100 N,  ;  . A)  N. b)  N.c) N.d)  N. e) There is no correct solution. |
| 12. | Can a force applied to a rigid body be transferred along the line of action without changing the effect of the force on the body? a) You can always. b) It is impossible under any circumstances. c) It is possible if no other forces act on the body. |
| 13. | The result of adding vectors is called... a) geometric sum. b) an algebraic sum. |
| 14. | Can a force of 50 N be divided into two forces, for example, 200 N each? Can i. b) It is impossible. |
| 15. | The result of subtracting vectors is called... a) geometric difference. b) algebraic difference. |
| 16. |  a) F x = F×sina. b) F x = -F×sina. c) F x = -F×cosa. d) F x = F×cosa. |
| 17. | Is force a sliding vector? a) Is. b) It is not. |
| 18. | The two systems of forces balance each other. Is it possible to say that their resultants are equal in magnitude and directed along the same straight line? a) Yes. b) No. |
| 19. | Determine the force modulus P if the following are known: P x = 30 N, P y = 40 N. a) 70 N; b) 50 N; c) 80 N; d) 10 N; e) There is no correct answer. |
| 20. | What is the projection of force on the y axis?  a) Р y = P×sin60°; b) Р y = P×sin30°; c) Р y = - P×cos30°; d) P y = -P×sin30°; e) There is no correct answer. |
| 21. | Do the modulus and direction of the resultant depend on the order in which the added forces are deposited? a) Depends; b) Do not depend. |
| 22. | At what value of the angle a between the force vector and the axis is the projection of the force onto this axis equal to 0? a) a = ; b) a = 9°; c) a = 180°; d) a = 6°; e) There is no correct answer. |
| 23. | What is the projection of force on the x axis?  a) -F×sina; b) F×sina; c) -F×cosa; d) F×cosa. |
| 24. | Determine the magnitude of the force if its projections on the x and y axes are known. A) ; b) ; V)  ; G)  .
|
| 25. | Can action and reaction forces cancel each other out? a) They can’t; b) They can. |
| 26. | An absolutely rigid body is in equilibrium under the action of two equal forces F 1 and F 2. Will the balance of the body be disrupted if these forces are transferred as shown in the figure? a) Will be violated; b) Will not be violated. |
| 27. | The projection of the vector onto the axis is equal to: a) the product of the modulus of the vector and the cosine of the angle between the vector and the positive direction of the coordinate axis; b) the product of the modulus of the vector and the sine of the angle between the vector and the positive direction of the coordinate axis. |
| 28. | Why can't action and reaction forces balance each other? a) These forces are not equal in magnitude; b) They are not directed in one straight line; c) They are not directed in opposite directions; d) They are applied to different bodies. |
| 29. | In what case can two forces acting on a rigid body be replaced by their geometric sum? a) At rest; b) In any case; c) When moving; d) Depending on additional conditions. |
2.5 Tasks for independent work of students
1). Explore subsection 2.1 this methodological instruction, having worked through the proposed exercises.
2) Answer self-control questions and tests for this section.
3). Make additions to your lecture notes, also referring to the recommended literature.
4). Study and make a brief summary of the next section “D” action on vectors"(4, pp. 4-20), (7, pp. 13,14):
1. Addition of vectors. Rules for parallelogram, triangle and polygon. Decomposition of a vector into two components. Vector difference.
3. Addition and decomposition of vectors using a graphic-analytical method.
4. Solve the following problem numbers yourself (4, pp. 14-16, 19): 6-2 ,8-2 ,9-2 ,10-2 ,13-3 ,14-3 .
Connections and their reactions
Relationship Concepts
As already noted, in mechanics bodies can be free and unfree. Systems of material bodies (points), positions and movements, which are subject to some geometric or kinematic restrictions, given in advance and independent of the initial conditions and given forces, are called not free. These restrictions imposed on the system and making it non-free are called connections. Communications can be carried out using various physical means: mechanical connections, liquids, electromagnetic or other fields, elastic elements.
Examples of non-free bodies are a load lying on a table, a door hanging on hinges, etc. The connections in these cases will be: for the load – the plane of the table, which prevents the load from moving vertically downwards; for the door - hinges that prevent the door from moving away from the jamb. Connections also include cables for loads, bearings for shafts, guides for sliders, etc.
Movably connected machine parts can come into contact along a flat or cylindrical surface, along a line or at a point. The most common contact between moving parts of machines is along a plane. This is how, for example, the slider and guide grooves of the crank mechanism, the tailstock of a lathe and the guide frames come into contact. Along the line, rollers come into contact with bearing rings, support rollers with the cylindrical frame of the trolley tipper, etc. Point contact occurs in ball bearings between balls and rings, between sharp bearings and flat parts.
- The elastic force arises due to deformation of the body, that is, a change in its shape. The elastic force is due to the interaction of the particles that make up the body.
- The force acting on the body from the support is called the normal reaction force.
- Two forces balance each other if these forces are equal in magnitude and directed in opposite directions. For example, the force of gravity and the force of normal reaction acting on a book lying on the table balance each other.
- The force with which a body presses on a support or stretches a suspension due to the attraction of the body by the Earth is called the weight of the body.
- The weight of a body at rest is equal to the force of gravity acting on this body: for a body at rest with mass m, the weight module P = mg.
- The weight of the body is applied to the support or suspension, and the force of gravity is applied to the body itself.
- The state in which the body weight is zero is called the state of weightlessness. In a state of weightlessness there are bodies on which only the force of gravity acts.
Questions and tasks
- What is elastic force? Give some examples of such power. What causes this force to arise?
- What is normal reaction force? Give an example of such power.
- When do two forces balance each other?
- What is body weight? What is the weight of a body at rest?
- What is your approximate weight?
- What common mistake does a person make when they say they weigh 60 kilograms? How to fix this error?
- Andrey's mass is 50 kg, and Boris weighs 550 N. Which of them has the greater mass?
Second level
- Give your own examples of cases when the deformation of the body, causing the appearance of elastic force, is noticeable to the eye and when it is invisible.
- What is the difference between weight and gravity and what do they have in common?
- Draw the forces acting on the block lying on the table. Do these forces balance each other?
- Draw the forces with which a block lying on a table acts on the table, and the table acts on the block. Why can’t we assume that these forces balance each other?
- Is the weight of a body always equal to the force of gravity acting on this body? Justify your answer with an example.
- What mass could you lift on the Moon?
- What is the state of weightlessness? Under what condition is a body in a state of weightlessness?
- Is it possible to be in a state of weightlessness near the surface of the Moon?
- Compose a problem on the topic “Weight” so that the answer to the problem is: “On the Moon I could, but on Earth I couldn’t.”
First level
Home laboratory
- What forces and from what bodies act on you when you stand? Do you feel these forces at work?
- Try being in a state of weightlessness.
There are a lot of differences between opposition and symbiosis. Opposition suggests that two forces or two sides neutralize or balance each other, while symbiosis describes a situation in which both organisms live together in harmony.
This reminded me of a theme running through Hayao Miyazaki's Kaze no Tani no Nausicaa (Warriors of the Wind), a fantasy film set in the distant future. In the film, humans coexist with the Omu, a species similar to a giant woodlice. Contrary to most people, the heroine, Nausicaa, believes that humanity should strive for balance with nature, including the Omu, rather than trying to destroy the "enemy".
Can Go, a game with more than 3,000 years of history, reflect such values? Certainly! In Go there is exactly this - a situation called seki.
Seki
One type of seki is shown in Diagram 1, where neither white nor black can play "A" or "B" to resolve a position involving marked stones.
D.2 presents another type of seki, in which each marked group has an eye, but neither side can capture the other with the move "A".
In D.3, the marked black stones do not have eyes, but the two groups of marked white stones do. However, White cannot capture Black's stones because both move “A” and move “B” will be suicidal.
D.4. Neither blacks nor whites can capture each other. What happens if White first covers all the outer queens marked with crosses, and then plays “A” or “B”? D.5 shows this situation.
Result on D.6. If White plays 3, then Black plays 4, and vice versa. This means that Black survived, and White's stones in the corner on D.5 were captured.
D 7. Black can capture the three marked stones starting with move 1, White plays tenuki (somewhere else on the board), and Black captures 3. But then White immediately moves inside Black's territory (D.8) and captures the entire black group. Consequently, if Black begins to capture the three stones marked in Diagram 5, he will die.
Diagrams 5-8 explain why D.4 is actually a seki situation, in which whoever plays first loses.
Solving the problems of the previous article
S.1A. After move b.1, it becomes urgent to prevent White “A” from sliding. Move part 2 does the job. Before move 10, Black defends his territory on the left with 2 and 8 and builds new territory on the right with 4, 6 and 10. Even after move 9, the White group has not yet completely freed itself from oppression.
S.1.B. Game 1-3 is more aggressive. Until move 14, White had more or less stabilized, while Black again gained territory on both sides.
S.2.A. From a local point of view, Black 1's invasion is done correctly. To prevent Black from sliding into A and preventing him from building a base, White plays 2 and 4 - good moves. But Black improves his position by extending 5.
S.2.B. The above result is too good for Black. Consequently, White will try to approach from the other side and pincer 2 first. After Black enters the center, defense 6 becomes paramount to maintain the base and prevent Black from building eyes on the bottom side. With moves 7 and 9, Black goes outside, leaving for the future the threat of cutting part “A”, b. “B”, part “C”. After strengthening his position, Black may have a move in the "D" area in mind.
R.2. Simply chasing Black with moves 2 and 4 leaves a weakness in White's position, which Black quickly emphasizes with moves 5 and 7. After Black moves to the center 9, White is left without sufficient guaranteed space to build eyes, and Black is eyeing move "A" which will create a miai cuts "B" and "C". Not a good position for White.
Tasks
Problem 1. I gave this task two weeks ago. Now that you have read the last two articles, you will be able to solve it. Black just played 1. How can White guarantee his life?
Problem 2. Black cannot capture white stones, but how can they build sacks?
Simple and complex in Go
In Go, it's better to give your opponent more choices to give him more ways to make a mistake. In other words, there is no need to make moves that allow you to make the obvious, correct answer.
D.1. Black's marked rim-shaped stones are cut in the most brutal way, while White's stones are optimally positioned.
D 2. This position is better for Black. At least they have the ability to fight and connect all their stones.
D.3. Before Black played tsuke (sticking) 1, White's lone stone had four dame. Until move 6, Black had only succeeded in increasing White's number of queens to 7. With moves 7-15, Black kept White's number of outside queens to no more than seven, but White maneuvered 8-16 to escape. At the end of the diagram, Black was left with four cutting points “A”-“D”, which they created for themselves. What was done wrong?
D.4. After Black saw that White's number of queens was increasing step by step, he tried playing 1 (7 on D.3). As a result, part 1's stone and Black's marked stone formed a border, and when White played 2, their stone, together with the white marked stone, was positioned optimally to cut Black's border. Compare this position with D.1.
D.5. Black then played 3, once again forming a keim with the black stone marked. But when White went 4, his stone teamed up with the marked white stone to cut the black border in the most efficient way possible. Then Black repeated this process several times, and got a catastrophic result for himself.
In other words, Black forced White to make good moves. Worse, White had no choice but to respond in the best possible way.
On D.6. one of the joseki is shown. Moves up to 7 are common. Now White can play tenuki (move elsewhere on the board), but if there is a Black stone at the top left, then 9 will be a strong move. B.10 – standard answer. 14-18 guarantees White an exit to the center with a sequence up to 22.
D 7. For a strong player, the sequence shown in the previous diagram looks natural, but I want to draw your attention to move part 11. Black could also play "A". White would answer 12, after which Black's A and 1 would form a border divided by White's 10 and 12. This is the reason why Black retreated 11. Through experience and diligent study, strong players know that move b. 12 is the best in this situation, which is not obvious to beginners. A less experienced player can play "A", which is not too bad. But move “B” is bad.
Solving last week's problems
S.1A. For move part 1, the best answer would be 2. Now Black can build sacks in sequence 3-7. Look at D.4-D.8 to understand why this position is a sack.
S.1.B. White's answer 2 is worse because White ends in gote, i.e. lose initiative. Move 9 becomes sente, forcing White to build seki 10.
R.1A. White cannot play 2 (or 4) because the combination of Black 3 and 5 takes over the group (if White starts with 4, then Black reverses the sequence of moves 3 and 5).
R.1B. To understand why White perishes in the previous diagram, let’s imagine that Black closed all the outside queens. With 8, White captures five stones. The result is shown in Problem 1 below.
Problem 1. Black moves and captures White.
S.2. Move part 1 is correct. After move 5 – sack.
R.2A. Answer b.2 seems more aggressive, but after part 5 White has nowhere to go, and Black can start with A at any moment convenient for him, pushing White into big trouble.
R.2B. It is wrong to start with part 1, because 2-6 will give White an eye, and Black will not be able to play “A”. This means that White can capture the impostor at any moment convenient for him, starting the fight 2. Black cannot win this ko. Therefore, White does not need to start it. The black stones died.
1. FA = ft. If FA = Ft, the forces balance each other, the body floats inside the liquid at any depth. In this case: FA= ?zhVg; Ft = ?tVg. Then from the equality of forces it follows: ?l = ?m, i.e., the average density of the body is equal to the density of the liquid. Fa. Ft.
Slide 5 from the presentation "Swimming conditions for bodies". The size of the archive with the presentation is 795 KB.Physics 7th grade
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“Speed of uniform rectilinear motion” - Rectilinear uniform motion. Equation for uniform motion. Types of trajectories for rectilinear motion. Speed graph. What is a trajectory? Types of trajectories. Requirements for knowledge and skills. Repetition. Develop an interest in physics. Moving. A visual experiment. Quantities. Rectilinear movement. Trajectory. Speed of uniform linear motion. Moving with uniform linear motion.
“Physics 7th grade “Atmospheric pressure”” - Temperature. Let us verify the existence of atmospheric pressure. We lower a cylinder with a piston into a vessel with water and lift the piston. Atmospheric pressure is the pressure of atmospheric air. Atmosphere pressure. Causes that create atmospheric pressure. The random movement of molecules and the effect of gravity on them. "Magdeburg hemispheres" in the human body. Glass of water. Atmospheric pressure exists. Lower layers of the atmosphere.
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