The main task of dynamics when considering the motion of connected bodies in different frames of reference is to explain the reasons that determine the nature of the motion. In this case, it becomes necessary to understand under what conditions the systems of bodies move in a straight line, in which case their trajectory is a curve, as a result of what causes the bodies move uniformly, accelerated or slowed down.
When studying the behavior of systems of connected bodies with speeds much less than the speed of light, the classical laws of Newton are used:
If the bodies do not interact with other bodies or the action of other bodies is compensated, then the speed of the system does not change either in absolute value or in direction. The system moves evenly and in a straight line.
The force ($ \\ overline (F) $) causing the acceleration of the system of bodies ($ \\ overline (a) $), in the inertial reference frame, is proportional to the mass ($ m $) of the bodies multiplied by its acceleration:
\\ [\\ overline (F) \u003d m \\ overline (a) \\ left (1 \\ right). \\]
The forces of interaction of bodies are equal in magnitude, directed along one straight line and have opposite directions.
Unless otherwise indicated, bonds, usually strands, are considered inextensible and weightless. In this case, when considering the motion of connected bodies, it is necessary to remember that the acceleration of motion of bodies in the system is the same (the result of the action of links). The tension force of the threads is considered equal along the entire length of the thread.
In some cases, you can choose different coordinate systems when considering the motion of different bodies in the system.
Scheme for solving a typical problem when moving connected bodies
- We graphically depict the situation described in the task. We draw a drawing, depict forces, speeds of movement of bodies, accelerations. Choosing and depicting frames of reference.
- We write down the basic law of the dynamics of translational motion (Newton's second law) in vector form, the necessary kinematic equations of motion, other necessary laws and formulas, for example, the basic law of the dynamics of rotational motion, the formula for the friction force, etc.
- We design vector equations on the axes of coordinate systems.
- We solve equations.
- We carry out the necessary calculations, after making sure that all quantities are written in a single system of units.
Examples of tasks with a solution
Example 1
The task. On the horizontal surface there is a trolley with mass $ M $. A weightless, inextensible thread is tied to it. The thread is thrown over a weightless block. A weight of mass $ m $ is attached to the second end of the thread (Fig. 1). With what acceleration the cart will move. Do not take friction forces into account.
Decision. Let's depict the forces that act on the trolley and the load in Fig. 1. and acceleration of the motion of the bodies of the system. Remember that there are no frictional forces. Note that the accelerations of the connected bodies (trolley and load) will be the same, besides this, the thread tension forces ($ \\ overline (N) $) acting on the trolley and on the load are equal in magnitude (weightless block), but have different directions (Fig. 1). Let's write Newton's second law for the cart:
\\ [(\\ overline (F)) _ R + M \\ overline (g) + \\ overline (N) \u003d M \\ overline (a) \\ left (1.1 \\ right). \\]
The basic law of dynamics for cargo is:
Let us connect the reference system to the Earth, write down the projections of equation (1.1) on the coordinate axis:
\\ [\\ left \\ (\\ begin (array) (c) X: N \u003d Ma \\\\ Y: Mg \u003d F_R \\ end (array) \\ right. \\ left (1.3 \\ right). \\]
In projections onto the same coordinate axes, equation (1.2) will give one scalar equation:
\\ [\\ left \\ (\\ begin (array) (c) X: mg-N \u003d ma \\\\ Y: 0 \\ end (array) \\ right. \\ left (1.4 \\ right). \\]
From the first equation of system (1.3) and equation (1.4) we have:
Let us express the required acceleration from (1.5):
Answer. $ a \u003d \\ frac (m) (M + m) $
Example 2
The task. Two weights of masses $ m_1 \\ and \\ m_2, $ connected by a weightless thread, slide on a smooth surface. A load of mass $ m_1 \\ $ acts with a force F directed horizontally (Fig. 2). What is cargo acceleration? What will be the tensile force of the thread connecting these weights? Do not take into account the force of friction of loads on the surface. \\ textit ()
Decision. In fig. 2 we will depict the forces acting on a load of mass $ m_1 $ (Fig. 2).
In accordance with Newton's second law, we write:
Let's connect the frame of reference with the Earth, choose the directions of the coordinate axes (Fig. 2).
To solve the problem, we need the projection of equation (2.1) only on the Y axis:
In equation (2.2), we have two unknown quantities: the thread tension force ($ T $) and the body acceleration ($ a $). To find the acceleration with which the first body and the entire system of connected bodies move, let us find out what forces act on the system, if both bodies are considered one whole. Then one force $ \\ overline (F) $ acts on the system with mass $ m_1 + m_2 $. In this case, Newton's second law takes the form:
\\ [\\ left (m_1 + m_2 \\ right) \\ overline (a) \u003d \\ overline (F) + \\ left (m_1 + m_2 \\ right) \\ overline (g) + \\ overline (N) \\ left (2.3 \\ right) . \\]
In the projection onto the Y-axis of formula (2.3), we obtain:
\\ [\\ left (m_1 + m_2 \\ right) a \u003d F \\ \\ left (2.4 \\ right). \\]
From (2.4) the acceleration of bodies is equal to:
From (2.2) and (2.5) we obtain the thread tension equal to:
Answer.$ \\ 1) a \u003d \\ frac (F) (m_1 + m_2). $ 2) $ T \u003d F \\ left (1- \\ frac (m_1) (m_1 + m_2) \\ right) $
In this problem, it is necessary to find the ratio of the tensile force to
Figure: 3. Solution to problem 1 ()
The stretched thread in this system acts on bar 2, forcing it to move forward, but it also acts on bar 1, trying to impede its movement. These two pulling forces are equal in magnitude, and we just need to find this pulling force. In such problems, it is necessary to simplify the solution as follows: we consider that the force is the only external force that makes the system of three identical bars move, and the acceleration remains unchanged, that is, the force makes all three bars move with the same acceleration. Then the tension always moves only one bar and will be equal to ma according to Newton's second law. will be equal to twice the product of mass and acceleration, since the third bar is on the second and the tension thread should already move two bars. In this case, the ratio to will be 2. The correct answer is the first.
Two bodies with mass and, connected by a weightless inextensible thread, can slide without friction on a smooth horizontal surface under the action of a constant force (Fig. 4). What is the ratio of the thread tension forces in cases a and b?
Choice of answer: 1. 2/3; 2. 1; 3. 3/2; 4. 9/4.
Figure: 4. Illustration for problem 2 ()
Figure: 5. Solution to problem 2 ()
The same force acts on the bars, only in different directions, therefore the acceleration in case "a" and case "b" will be the same, since the same force causes the acceleration of two masses. But in case "a" this tension force also makes bar 2 move, in case "b" it is bar 1. Then the ratio of these forces will be equal to the ratio of their masses and we get the answer - 1.5. This is the third answer.
On the table lies a block weighing 1 kg, to which a thread is tied, thrown over a fixed block. A weight of 0.5 kg is suspended from the second end of the thread (Fig. 6). Determine the acceleration with which the bar moves if the coefficient of friction of the bar on the table is 0.35.
Figure: 6. Illustration for problem 3 ()
We write down a short statement of the problem:
Figure: 7. Solution to problem 3 ()
It must be remembered that the forces of tension and as vectors are different, but the magnitudes of these forces are the same and equal. Similarly, we will have the same accelerations of these bodies, since they are connected by an inextensible thread, although they are directed in different directions: - horizontally, - vertically. Accordingly, we choose our own axes for each of the bodies. Let us write down the equations of Newton's second law for each of these bodies, when adding up, the internal tension forces will decrease, and we get the usual equation, substituting the data into it, we get that the acceleration is equal to.
To solve such problems, you can use the method that was used in the last century: the driving force in this case is the resultant external forces applied to the body. The gravity force of the second body makes this system move, but the friction force of the bar on the table interferes with the movement, in this case:
Since both bodies are moving, the driving mass will be equal to the sum of the masses, then the acceleration will be equal to the ratio of the driving force to the driving mass 
So you can immediately come to an answer.
A block is fixed at the apex of two inclined planes that make angles with the horizon. On the surface of the planes with a friction coefficient of 0.2 kg and bars move, connected by a thread thrown over the block (Fig. 8). Find the force of pressure on the block axis.
Figure: 8. Illustration for problem 4 ()
Let's make a short record of the problem statement and an explanatory drawing (Fig. 9):
Figure: 9. Solution to problem 4 ()
We remember that if one plane makes an angle of 60 0 with the horizon, and the second plane makes 30 0 with the horizon, then the angle at the vertex will be 90 0, this is an ordinary right-angled triangle. A thread is thrown through the block, to which the bars are suspended, they pull down with the same force, and the action of the tension forces F n1 and F n2 leads to the fact that their resulting force acts on the block. But these tensile forces will be equal to each other, they make up a right angle to each other, therefore, when these forces are added, a square is obtained instead of the usual parallelogram. The sought force F d is the diagonal of the square. We see that for the result we need to find the thread tension. Let's analyze: in which direction is the system of two connected bars moving? A more massive bar, naturally, will pull a lighter one, bar 1 will slide down, and bar 2 will move up the slope, then the equation of Newton's second law for each of the bars will look like:
The solution of the system of equations for coupled bodies is performed by the addition method, then we transform and find the acceleration:
This acceleration value must be substituted into the formula for the tension force and find the pressure force on the block axis:
We found out that the force of pressure on the block axis is approximately 16 N.
We have considered various ways of solving problems that will be useful to many of you in the future in order to understand the principles of the structure and operation of those machines and mechanisms with which you will have to deal in production, in the army, in everyday life.
Bibliography
- Tikhomirova S.A., Yavorsky B.M. Physics (basic level) - M .: Mnemosina, 2012.
- Gendenshtein L.E., Dick Yu.I. Physics grade 10. - M .: Mnemosina, 2014.
- Kikoin I.K., Kikoin A.K. Physics-9. - M .: Education, 1990.
Homework
- What law do we use when writing equations?
- What quantities are the same for bodies connected by an inextensible thread?
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The movement of a system of bodies
Dynamics: motion of a system of connected bodies.
Projecting the forces of multiple objects.
Action of Newton's second law on bodies that are held together by a thread
If you, my friend, have forgotten how to project a silushka, I advise you to refresh it in your head.
And for those who remember everything, let's go!
Problem 1. On a smooth table lie two bars connected by a weightless and inextensible thread with a mass of 200 g of the left and a mass of 300 g of the right. A force of 0.1 N is applied to the first, a force of 0.6 N is applied to the left - in the opposite direction. With what acceleration do they move cargo?
Movement occurs only on the X axis.
Because a large force is applied to the right weight, the movement of this system will be directed to the right, so we will direct the axis in the same way. Acceleration from both bars will be directed in the same direction - the direction of greater force.
Let's add the upper and lower equations. In all problems, if there are no conditions, the tensile force for different bodies is the same T₁ and T.
Let's express the acceleration:
Answer: 1 m / s²
Objective 2. Two bars, connected by an inextensible thread, are on a horizontal plane. The forces F₁ and F₂ are applied to them, making angles α and β with the horizon. Find the system acceleration and thread tension. The coefficients of friction of the bars on the plane are the same and equal to μ. The forces F₁ and F₂ are less than the gravity of the bars. The system moves to the left.
The system moves to the left, but the axis can be directed in any direction (it's just the signs, you can experiment at your leisure). For a change, let's direct it to the right, against the movement of the entire system, we love the cons! Let's project the forces onto Oh (if this is difficult -).
According to II z. Newton, we will project the forces of both bodies on Ox:
Let's add the equations and express the acceleration:
Let us express the thread tension. To do this, we equate the acceleration from both equations of the system:
Objective 3. A thread is thrown through the non-suspended block, to which three identical weights (two on one side and one on the other) weighing 5kg each. Find system acceleration. What way will the goods go in the first 4 s of movement?
In this problem, you can imagine that the two left weights are fastened together without a thread, this will save us from projecting mutually equal forces.
Subtract the second from the first equation:
Knowing the acceleration and the fact that the initial speed is zero, we use the path formula for uniformly accelerated motion:
Answer: 26.64 m
Problem 4. Two loads weighing 4 kg and 6 kg are connected with a light inextensible thread. Coefficients of friction between load and tableμ = 0.2. Determine the acceleration with which the loads will move.
We write down the motion of bodies on the axis, from Oy we find N for the friction force (Ffr \u003d μN):
(If it is difficult to understand what equations are needed to solve the problem, it is better to write down everything)
We add the two lower equations so that T can be canceled:
Let's express the acceleration:
Answer: 2.8 m / s²
Objective 5. On an inclined plane with an angle of inclination of 45 ° lies a block weighing 6 kg. A weight weighing 4 kg is attached to the bar with a thread and thrown over the block. Determine the thread tension if the coefficient of friction of the bar against the plane μ \u003d 0.02. At what values \u200b\u200bof μ will the system be in equilibrium?
We direct the axis arbitrarily and assume that the right weight outweighs the left one and lifts it up the inclined plane.
From the equation for the Y axis, we express N for the friction force on the X axis (Ffr \u003d μN):
Let's solve the system by taking the equation for the left body along the X axis and for the right body along the Y axis:
Let us express the acceleration so that one unknown T remains, and find it:
The system will be in balance. This means that the sum of all the forces acting on each of the bodies will be equal to zero:
We received a negative coefficient of friction, which means that we chose the movement of the system incorrectly (acceleration, friction force). You can check this by substituting the thread tension T into any equation and finding the acceleration. But it's okay, the values \u200b\u200bremain the same in magnitude, but opposite in direction.
This means that the correct direction of the forces should look like this, and the coefficient of friction at which the system will be in equilibrium is 0.06.
Answer: 0.06
Problem 6. On two inclined planes there is a load of 1 kg. The angle between the horizontal and the planes is α\u003d 45 ° and β \u003d 30 °. The coefficient of friction for both planes μ= 0.1. Find the acceleration with which the weights are moving and the tension in the thread. What should be the ratio of the masses of the goods so that they are in equilibrium.
In this problem, all equations on both axes for each body are already required:
Let us find N in both cases, substitute them in the friction force, and write together the equations for the X axis of both bodies:
Let us add the equations and reduce by mass:
Let's express the acceleration:
Substituting the found acceleration into any equation, we find T:
And now we will overcome the last point and deal with the mass ratio. The sum of all the forces acting on any of the bodies is zero in order for the system to be in equilibrium:
Add the equations
We will transfer everything with the same mass to one part, everything else to the other part of the equation:
We got that the mass ratio should be as follows:
However, if we assume that the system can move in a different direction, that is, the right weight will outweigh the left one, the direction of acceleration and friction force will change. The equations will remain the same, but the signs will be different, and then the mass ratio will be as follows:
Then, with a mass ratio of 1.08 to 1.88, the system will be at rest.
Many people may get the impression that the mass ratio should be some specific value, and not a gap. This is true if frictional force is absent. To balance the forces of gravity at different angles, there is only one option when the system is at rest.
In this case, the friction force gives a range in which, until the friction force is overcome, the movement will not begin.
Answer: 1.08 to 1.88
Find the acceleration of a weight $ 3m $ in a system consisting of a stationary block and a moving block. Neglect the masses of the blocks and the friction in their axes. The thread thrown over the blocks is weightless and inextensible. Free fall acceleration $ g $.
Decision
Let us choose a frame of reference associated with a fixed block. We select the coordinate system as shown in the figure (the $ Oy $ coordinate axis is highlighted in red). It is an inertial frame of reference, since it is stationary in relation to the Earth. It obeys Newton's laws.
1) Let's draw the forces acting on a body with mass $ m $ (in the figure they are marked in blue): $ m \\ vec (g _ ()) $ - gravity, always directed vertically downward; $ \\ vec (T _ ()) $ - thread tension force directed along the thread from the body.
2) Let's draw the forces acting in the system, consisting of a body with a mass of $ 3m $ and a movable block (marked in green in the figure): $ 3m \\ vec (g _ ()) $ - gravity; $ \\ vec (T _ ()) $ - thread tension force directed along the thread from the body.
Suppose the loads are moving as shown in the figure. We denote the acceleration of a load with a mass of $ 3m $ by $ \\ vec (a_1) $, and a load with a mass of $ m $ - $ \\ vec (a_2) $.
3) According to Newton's second law for a body of mass $ m $: $ \\, \\ vec (R_2) \u003d m \\ vec (a_2) $, that is, $ m \\ vec (g _ ()) + \\ vec (T_ ()) \u003d m \\ vec (a_2) $.
Projected onto the $ Oy $ axis:
$ T-mg \u003d ma_2 \\, \\, (1). $
4) According to Newton's second law for a body of mass $ 3m $: $ \\, \\ vec (R_1) \u003d 3m \\ vec (a_1) $, that is, $ 3m \\ vec (g _ ()) + 2 \\ vec (T_ ()) \u003d 3m \\ vec (a_1) $.
Projected onto the $ Oy $ axis:
$ 2T-3mg \u003d -3ma_1 \\, \\, (2). $
5) To find the relationship between the accelerations $ a_1 $ and $ a_2 $, you need to deal with the kinematic relationship of bodies. The movable block gives a twofold gain in strength. According to the golden rule of mechanics, how many times we win in effort, how many times we lose in distance. This means that if a body with mass $ 3m $ descends to a distance $ x $, then a body with mass $ m $ will rise $ 2x $, hence.
Objectives (for students):
1. Systematization of knowledge about the resultant of all forces applied to the body.
2. Interpretation of Newton's laws regarding the concept of resultant forces.
3. Perception of these laws.
4. Application of the knowledge gained to a familiar and new situation in solving physical problems.
Lesson Objectives (for the teacher):
Educational:
1. To clarify and expand the knowledge about the resultant force and the ways of finding it when a system of bodies moves.
2. To form the ability to apply the concept of resultant force to the substantiation of the laws of motion (Newton's laws)
3. Reveal the level of mastering the topic.
4. Continue the formation of skills of introspection of the situation and self-control.
Educational:
1. To contribute to the formation of the ideology of the cognizability of the phenomena and properties of the surrounding world;
2. Emphasize the importance of modulation in the cognizability of matter;
a) efficiency;
b) independence;
c) accuracy;
d) discipline;
e) a responsible attitude towards learning.
Developing:
1. To carry out the mental development of children;
2. Work on the formation of skills to compare phenomena, draw conclusions, generalizations;
3.To teach:
b) analyze the situation,
c) make logical inferences based on this analysis and available knowledge;
4. Check the level of independent thinking of the student on the application of existing knowledge in various situations.
Equipment: board, chalk, handouts.
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Preview:
Lesson topic: "The motion of a system of connected bodies."
Objectives (for students):
1. Systematization of knowledge about the resultant of all forces applied to the body.
2. Interpretation of Newton's laws regarding the concept of resultant forces.
3. Perception of these laws.
4. Application of the knowledge gained to a familiar and new situation in solving physical problems.
Lesson objectives (for the teacher):
Educational:
1. To clarify and expand knowledge about the resultant force and the ways of finding it when the system of bodies moves.
2. To form the ability to apply the concept of resultant force to the substantiation of the laws of motion (Newton's laws)
3. Reveal the level of mastering the topic.
4. Continue the formation of skills of introspection of the situation and self-control.
Educational:
1. To contribute to the formation of the worldview idea of \u200b\u200bthe cognizability of the phenomena and properties of the surrounding world;
2. Emphasize the importance of modulation in the cognizability of matter;
3. Pay attention to the formation of universal human qualities:
a) efficiency;
b) independence;
c) accuracy;
d) discipline;
e) a responsible attitude towards learning.
Developing:
1. To carry out the mental development of children;
2. Work on the formation of skills to compare phenomena, draw conclusions, generalizations;
3.To teach:
a) highlight signs of similarity in the description of phenomena,
b) analyze the situation,
c) make logical inferences based on this analysis and available knowledge;
4. Check the level of independent thinking of the student on the application of existing knowledge in various situations.
Equipment: board, chalk, handouts.
During the classes
Teacher: let us recall the words of R. Feynman: "A physicist is one who sees the solution of a problem without solving it." This can be achieved by solving several thousand problems. This is a little, 3 - 4 problem books.
In this lesson, we have to learn how to solve physical problems based on Newton's laws. These tasks are also referred to as dynamic tasks.
On the first sheet in front of you, you can consider such tasks.
All tasks requiring the application of Newton's laws are solved according to one algorithm. Let's remember this algorithm.
Students try to remember the problem solving algorithm.
Teacher: let's read and analyze this algorithm on sheet 2.
1. Having carefully read the condition of the problem (if you knew how many errors occur from inattentive reading of the condition of the problem!), Find out the physical content of the problem, what processes and phenomena are included in its condition. Having familiarized yourself with the condition of the problem, one should not try to immediately find the desired value. Remember, the goal of the solution is to reduce the problem from physical to mathematical by writing down its condition using formulas.
2. Find out what forces act on the bodies, the movement of which we are interested in. All known forces must be depicted in the figure. In this case, one must clearly imagine from which bodies the forces under consideration act. Indicate all quantities that characterize this phenomenon. It should not be forgotten that the action of one body on another is reciprocal. We should not talk about the action of bodies, but about their interaction, which obeys Newton's third law.
3. Choose a frame of reference with respect to which you will consider the movement of bodies. The choice of the frame of reference does not affect the answer to the problem, but a well-chosen direction of the axes can make it easier to solve the problem. In the case of rectilinear motion, it is convenient to direct one of the axes along the direction of acceleration, and the other perpendicular to it.
4. Using physical laws and formulas, establish a mathematical relationship between all quantities. As a result, one or more equations will be obtained - a physical problem is reduced to a mathematical one.
5. Solve the compiled system of equations, making sure that the number of equations is equal to the number of unknowns.
6. Analyze the obtained result and numerical calculation.
Teacher: now fill in the table on sheet number 3.
Pupils independently fill out the table with an explanation.
Teacher : make the drawing arbitrary (either on a horizontal surface or vertically).
Teacher : To complete the third column, first answer the questions below the table.
Review task:
- The basic Law - newton's second law.
- Forces acting on bodies -friction, elasticity, gravity, support reaction, thread tension, traction, Archimedes.
- How to guide the coordinate axis? -in the direction of acceleration.
- What are the external forces -friction, elasticity, gravity, support reaction, traction, Archimedes.
- What are the internal forces -thread tension.
- What is the weight of the bodies? -thread tension.
Teacher : now we will solve the first problem, when analyzing the problem, use the memo on sheet No. 4:
Problem number 1. On a smooth horizontal surface there are two bodies connected by a weightless, inextensible thread. Left body mass m2 \u003d 1 kg, right - m 1 \u003d 2kg. A force F \u003d 3N is applied to the right weight, directed along the thread. Determine the thread tension.
The first stage: analysis of the problem (analysis of a physical phenomenon).
Teacher: what kind of motion does a system of connected bodies perform under the action of an external force F? Do friction forces act in this case?
Pupil: the motion will be rectilinear uniformly accelerated. By condition, the surface is smooth, which means that the friction force can be neglected.
Teacher: what forces arise between bodies connected by an inextensible thread? Do bodies receive the same accelerations?
Pupil: interaction forces arise between bodies, which will be equal according to Newton's III law. Since the thread is not stretching, the acceleration of both bodies is the same.
Teacher: what do you think. Will the tensile forces be the same in the first and second cases? In which case will these forces be greater?
Pupil: tensile forces will be greater when an external force is applied to a body with a smaller mass. (If the students find it difficult, we proceed to the solution)
Step two: plan for a solution One student solves a problem on the board.
Teacher: let's complete the drawing. Let's select ISO (one axis associated with the support). Let us introduce the appropriate designations. Let us write down brief conditions of the problem.
The general idea of \u200b\u200bthe solution is to describe the motion of two material points using Newton's laws. Taking into account the third law, this description can be divided into two: a description of the movement of one material point and a description of the movement of another. Thus, we get a system of two equations.This concludes the stage of setting the problem. We have done the work of a physicist. Now you need to go to the state of a mathematician and solve the system of equations we have obtained. This is the second stage, mathematical.
Stage three: implementation of the plan, or decision
Fourth stage: discussion of the solution (analysis, reflection)
Having received the solution to the problem in general, it is necessary to go to the state of physicists and start analyzing the solution. First of all, you need to check the dimension.
Conclusions. At the center of the solution to any problem is the mathematical description (modeling) of physical phenomena. That is why, firstly, it is necessary to select the necessary physical phenomena, and, secondly, to describe them by physical laws. At the first and second stages of solving the problem, there is a preparation for mathematical modeling of a physical phenomenon. At the third stage, work with a mathematical model. It is important here to correctly and skillfully perform all the necessary mathematical operations: to compose systems of equations, project them on the axis of the reference system, perform algebraic transformations, express the required physical quantity and find its numerical value. It is clear that when performing all actions, you need to be careful - a mistake in any action makes all the rest of the work in vain. That is why it is necessary to gradually, accurately carry out a drawing, mathematical operations, etc. Successful solution of any problems requires these qualities. If, for example, some force is not indicated on the drawing, then the equation will be incorrectly drawn up, the work to solve it will be in vain.
Problem number 2. Two weights weighing 90 g and 110 g are suspended at the ends of the thread thrown over a fixed block. Initially, they are at the same level. With what acceleration do bodies move? How much will the larger load go down in 2 s?
Task analysis:
- If the masses of the loads are the same, then what will the acceleration of the loads be equal to? (zero).
- How will the goods move? (uniformly or at rest).
- What will be the tension force of the thread and the weight of each of the weights? (mg).
- If the mass of the second load is much greater than the mass of the first load, then what is the acceleration of the loads in this case? (a \u003d g modulo).
- Where will the acceleration of a more massive load be directed? (down).
Problem number 3. Three bodies with masses m1 \u003d m2 \u003d m and M \u003d 2m are fixed at the ends and in the middle of a long spring of rigidity k (see Fig.). All bodies are placed on a smooth horizontal table. A horizontal external force is applied to a body of mass M, the modulus of which is F. Find the extension of the entire spring.
The weights in the horizontal direction are subject to tension forces from the spring. Considering the spring to be weightless, we write the equation of Newton's second law for each body:
Hence, after substitutions, we obtain
The same forces stretch the spring halves:
- stiffness of the spring halves.
The total elongation of the spring is:
Summarizing
Teacher: Let's sum up the lesson. We repeated Newton's laws, solved qualitative and quantitative problems on the application of laws.
Conclusion: Newtonian mechanics was the first complete theory in the history of physics that correctly describes a vast class of phenomena - the motion of bodies. One of Newton's contemporaries expressed his admiration for this theory in verses, which, translated by S. Ya. Marshak, sound like this (epigraph on the blackboard).
"This world was shrouded in deep darkness.
Let there be light. And then Newton appeared. "
The laws of physics allow, in principle, to solve any problem in mechanics.