Thread tension moment. Thread tension

Problem 10048

A disk-shaped block with a mass of m = 0.4 kg rotates under the action of the tension force of a thread, to the ends of which weights of masses m 1 = 0.3 kg and m 2 = 0.7 kg are suspended. Determine the tension forces T 1 and T 2 of the thread on both sides of the block.

Problem 13144

A light thread is wound on a homogeneous solid cylindrical shaft of radius R = 5 cm and mass M = 10 kg, to the end of which a load of mass m = 1 kg is attached. Determine: 1) the dependence s(t), according to which the load moves; 2) the tension force of the thread T; 3) dependence φ(t), according to which the shaft rotates; 4) angular velocity ω of the shaft t = 1 s after the start of movement; 5) tangential (a τ) and normal (a n) accelerations of points located on the surface of the shaft.

Problem 13146

A weightless thread is thrown through a stationary block in the form of a homogeneous solid cylinder with a mass m = 0.2 kg, to the ends of which bodies with masses m 1 = 0.35 kg and m 2 = 0.55 kg are attached. Neglecting friction in the axis of the block, determine: 1) acceleration of the load; 2) the ratio T 2 /T 1 of the thread tension forces.

Problem 40602

A thread (thin and weightless) is wound around a hollow thin-walled cylinder of mass m. Its free end is attached to the ceiling of an elevator moving downward with acceleration a l. The cylinder is left to its own devices. Find the acceleration of the cylinder relative to the elevator and the tension force of the thread. During movement, consider the thread vertical.

Problem 40850

A mass weighing 200 g is rotated on a thread 40 cm long in a horizontal plane. What is the tension force of the thread if the load makes 36 revolutions in one minute?

Problem 13122

A charged ball of mass m = 0.4 g is suspended in the air on a silk thread. A charge q of different and equal magnitude is brought from below to it at a distance of r = 2 cm. As a result, the tension force of the thread T increases by n = 2.0 times. Find the amount of charge q.

Problem 15612

Find the ratio of the modulus of the tension force of the thread of the mathematical pendulum in the extreme position with the modulus of the tension force of the thread of the conical pendulum; the lengths of the threads, the masses of the weights and the angles of deflection of the pendulums are the same.

Problem 16577

Two small identical balls, each weighing 1 μg, are suspended on threads of equal length and touching. When the balls were charged, they separated by a distance of 1 cm, and the tension force on the thread became equal to 20 nN. Find the charges of the balls.

Problem 19285

Establish a law according to which the tension force F of the thread of a mathematical pendulum changes over time. The pendulum oscillates according to the law α = α max cosωt, its mass m, length l.

Problem 19885

The figure shows a charged infinite plane with a surface plane of charge σ = 40 μC/m 2 and a similarly charged ball with mass m = l g and charge q = 2.56 nC. The tension force of the thread on which the ball hangs is...

Movement of a system of bodies

Dynamics: motion of a system of connected bodies.

Projection of forces of several objects.

The action of Newton's second law on bodies that are held together by a thread

If you, my friend, have forgotten how to project, I advise you to refresh your little 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 on the left and a mass on the right 300 g. A force of 0.1 N is applied to the first, and a force of 0.6 N is applied to the left in the opposite direction. With what acceleration are they moving? cargo?

Movement occurs only on the X axis.

Because If a large force is applied to the right load, the movement of this system will be directed to the right, so we will direct the axis in the same way. The acceleration of both bars will be directed in one direction - the side of greater force.

Let's add the upper and lower equations. In all problems, unless there are some conditions, the tension force of different bodies is the same T₁ and T₂.

Let's express the acceleration:

Answer: 1 m/s²

Task 2. Two bars connected by an inextensible thread are located on a horizontal plane. Forces F₁ and F₂ are applied to them, making angles α and β with the horizon. Find the acceleration of the system and the tension in the thread. The coefficients of friction between the bars and the plane are the same and equal to μ. The forces F₁ and F₂ are less than the gravity force 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 a matter of signs, you can experiment at your leisure). For a change, let's point to the right, against the movement of the entire system, we love minuses! Let's project forces onto Oh (if there are difficulties with this -).

According to II. Newton, we project the forces of both bodies onto Ox:

Let's add up the equations and express the acceleration:

Let us express the tension of the thread. To do this, we equate the acceleration from both equations of the system:

Task 3. A thread is thrown through a stationary block, from which three identical weights (two on one side and one on the other) with a mass of 5 are suspended.kg each. Find the acceleration of the system. How far will the loads travel in the first 4 seconds of movement?

In this problem, we 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 masses of masses 4 kg and 6 kg are connected by a light inextensible thread. Friction coefficients between load and tableμ = 0.2. Determine the acceleration with which the loads will move.

Let us write down the motion of the bodies on the axis, and from Oy we find N for the friction force (Ftr = μN):

(If it is difficult to understand what equations will be needed to solve the problem, it is better to write everything down)

Let's add the two lower equations so that T is reduced:

Let's express the acceleration:

Answer: 2.8 m/s²

Task 5. A block of mass 6 kg lies on an inclined surface with an angle of inclination of 45°. A mass of 4 kg is attached to a block using a thread and thrown over the block. Determine the tension of the thread if the coefficient of friction of the bar on the plane is μ = 0.02. At what values of μ will the system be in equilibrium?

Let's direct the axis arbitrarily and assume that the right load 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 (Ftr = 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's express the acceleration so that there is only one unknown T left, and find it:

The system will be in equilibrium. This means that the sum of all 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 tension force of the thread T into any equation and finding the acceleration. But it’s okay, the values remain 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 equal to 0.06.

Answer: 0.06

Problem 6. On two inclined planes there is a load with masses of 1 kg. The angle between the horizontal and the planes is α= 45° and β = 30°. Friction coefficient for both planes μ= 0.1. Find the acceleration with which the weights move and the tension in the string. What should be the ratio of the masses of the loads so that they are in equilibrium.

This problem will require all the equations on both axes for each body:

Let's find N in both cases, substitute them for the force of friction and write together the equations for the X axis of both bodies:

Let's add up the equations and reduce by mass:

Let's express the acceleration:

Substituting the found acceleration into any equation, we find T:

Now let’s overcome the last point and figure out the mass ratio. The sum of all forces acting on any of the bodies is equal to zero in order for the system to be in equilibrium:

Let's add up the equations

Let’s move everything that has the same mass into one part, and everything else into the other part of the equation:

We found 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 load 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 like this:

Then, with a mass ratio from 1.08 to 1.88, the system will be at rest.

Many may have the impression that the mass ratio should be some specific value, and not a gap. This is true if there is no frictional force. 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, movement will not begin.

Answer: from 1.08 to 1.88

popular definition

Strength is action, which can change the state of rest or movement body; therefore, it can accelerate or change the speed, direction, or direction of motion of a given body. Against, tension- this is the state of a body subject to the action of opposing forces that attract it.

She is known as tensile force, which, when exposed to an elastic body, creates tension; This last concept has various definitions that depend on the branch of knowledge from which it is analyzed.

Ropes, for example, allow forces to be transferred from one body to another. When two equal and opposite forces are applied at the ends of a rope, the rope becomes taut. In short, tensile forces are each of these forces that supports the rope without breaking .

Physics And engineering talk about mechanical stress, to denote the force per unit area surrounding a material point on the surface of a body. Mechanical stress can be expressed in units of force divided by units of area.

Voltage is also a physical quantity that drives electrons through a conductor into a closed electrical circuit that causes electric current to flow. In this case, the voltage can be called voltage or potential difference .

On the other side, surface tension of a liquid is the amount of energy required to reduce its surface area per unit area. Consequently, the liquid exerts resistance, increasing its surface area.

How to find the tension force

Knowing that force tension is force, with which a line or string is tensioned, the tension can be found in a static type situation if the angles of the lines are known. For example, if the load is on a slope and a line parallel to the slope prevents the load from moving downward, the tension is resolved, knowing that the sum of the horizontal and vertical components of the forces involved must add up to zero.

First step to do this calculation- draw a slope and place a block of mass M on it. The slope increases on the right, and at one point it meets a wall, from which a line runs parallel to the first. and tie the block, holding it in place and creating a tension T. Next you should identify the angle of inclination with the Greek letter, which may be "alpha", and the force it exerts on the block with the letter N, since we are talking about normal strength .

From the block vector should be drawn perpendicular to the slope and up to represent the normal force, and one down (parallel to the axis y) to display gravity. Then you start with formulas.

To find strength F = M is used. g , Where g is his constant acceleration(in the case of gravity this value is 9.8 m/s^2). The unit used for the result is newton, which is denoted by N. In the case of a normal force, it must be expanded into vertical and horizontal vectors using the angle it makes with the axis x: to calculate the up vector g is equal to the cosine of the angle, and for the vector in the direction to the left, towards the bosom of this.

Finally, the left-hand side of the normal force must be equal to the right-hand side of the stress T, finally resolving the stress.

library science

In order to know well the term librarianship, which now occupies us, it is necessary to begin by clarifying its etymological origin. In this case, we can say that this word comes from Greek, since it is formed by the sum of several elements of this language: - The noun “biblion”, which can be translated as “book”. - The word "teche", which is synonymous with the word "box" or "place where it is stored." -The suffix "-logía", which is used to denote "the science that studies". This is known as librarianship, a discipline focused on

definition
taxismo

Taxiism is not a term accepted by the Royal Spanish Academy (RAE) in its dictionary. The concept is used with reference to the directed movement that a living thing implements to respond to a stimulus that it perceives. Taxi can be negative (when the living thing moves away from the source of the stimulus) or positive (the living thing moves closer to what generates the stimulus in question). To organi

definition
extension

Expansion, from the Latin expansĭo, is the action and effect of expanding or extending (spreading, spreading, spreading out, spreading out, giving greater amplitude, or making something take up more space). Expansion can be the territorial growth of a nation or empire from the conquest and annexation of new lands. For example: “The American expansion of the nineteenth century was very important and affected Mexi

definition
In physics, tension is the force acting on a rope, cord, cable or similar object or group of objects. Anything that is pulled, suspended, supported, or swings by a rope, cord, cable, etc., is the object of a tension force. Like all forces, tension can accelerate objects or cause them to deform. The ability to calculate tensile force is an important skill not only for students of the Faculty of Physics, but also for engineers and architects; those who build stable homes need to know whether a particular rope or cable will withstand the tension force of the object's weight without sagging or collapsing. Start reading this article to learn how to calculate the tension force in some physical systems.

Steps

Determination of tension on one thread
1. Determine the forces at each end of the thread. The tension in a given thread or rope is the result of forces pulling on the rope at each end. We remind you that force = mass × acceleration. Assuming the rope is taut, any change in the acceleration or mass of an object suspended from the rope will result in a change in the tension force in the rope itself. Don't forget about the constant acceleration of gravity - even if the system is at rest, its components are subject to gravity. We can assume that the tension force of a given rope is T = (m × g) + (m × a), where "g" is the acceleration due to gravity of any of the objects supported by the rope, and "a" is any other acceleration, acting on objects.
  - To solve many physical problems, we assume perfect rope- in other words, our rope is thin, has no mass and cannot stretch or break.
  - As an example, let's consider a system in which a load is suspended from a wooden beam using a single rope (see image). Neither the load itself nor the rope moves - the system is at rest. As a result, we know that in order for the load to be in equilibrium, the tension force must be equal to the force of gravity. In other words, Tension (F t) = Gravity (F g) = m × g.
    - Let's assume that the load has a mass of 10 kg, therefore the tension force is 10 kg × 9.8 m/s 2 = 98 Newtons.
2. Consider acceleration. Gravity is not the only force that can affect the tension of a rope - the same effect is produced by any force applied to an object on a rope with acceleration. If, for example, an object suspended from a rope or cable is accelerated by a force, then the acceleration force (mass × acceleration) is added to the tension force generated by the weight of the object.
  - In our example, suppose that a 10 kg load is suspended from a rope and, instead of being attached to a wooden beam, it is pulled upward with an acceleration of 1 m/s 2 . In this case, we need to take into account the acceleration of the load as well as the acceleration of gravity, as follows:
    - F t = F g + m × a
    - F t = 98 + 10 kg × 1 m/s 2
    - F t = 108 Newtons.
3. Consider angular acceleration. An object on a rope rotating about a point considered the center (like a pendulum) exerts tension on the rope through centrifugal force. Centrifugal force is the additional tension force caused by the rope, "pushing" it inward so that the load continues to move in an arc rather than in a straight line. The faster an object moves, the greater the centrifugal force. Centrifugal force (F c) is equal to m × v 2 /r where “m” is the mass, “v” is the speed, and “r” is the radius of the circle along which the load is moving.
  - Since the direction and magnitude of centrifugal force changes depending on how the object moves and changes its speed, the total tension in the rope is always parallel to the rope at the center point. Remember that the force of gravity is constantly acting on an object and pulling it down. So if the object is swinging vertically, the full tension strongest at the bottom of the arc (for a pendulum this is called the equilibrium point) when the object reaches its maximum speed, and weakest at the top of the arc as the object slows down.
  - Let's assume that in our example the object is no longer accelerating upward, but is swinging like a pendulum. Let our rope be 1.5 m long, and our load move at a speed of 2 m/s when passing through the lower point of the swing. If we need to calculate the tension force at the bottom point of the arc, when it is greatest, then we first need to find out whether the pressure of gravity is experienced by the load at this point, as at rest - 98 Newtons. To find the additional centrifugal force, we need to solve the following:
    - F c = m × v 2 /r
    - F c = 10 × 2 2 /1.5
    - F c =10 × 2.67 = 26.7 Newtons.
    - So the total tension will be 98 + 26.7 = 124.7 Newton.
4. Please note that the tension force due to gravity changes as the load passes through the arc. As noted above, the direction and magnitude of centrifugal force changes as the object swings. In any case, although gravity remains constant, net tension force due to gravity is also changing. When the swinging object is Not at the bottom of the arc (equilibrium point), gravity pulls it down, but tension pulls it up at an angle. For this reason, the tension force must counteract part of the force of gravity, not all of it.
  - Dividing the force of gravity into two vectors can help you visualize this state. At any point in the arc of a vertically swinging object, the rope makes an angle "θ" with a line passing through the equilibrium point and the center of rotation. As soon as the pendulum begins to swing, the gravitational force (m × g) is divided into 2 vectors - mgsin(θ), acting tangentially to the arc in the direction of the equilibrium point and mgcos(θ), acting parallel to the tension force, but in the opposite direction. Tension can only resist mgcos(θ) - the force directed against it - not the entire force of gravity (except at the equilibrium point, where all forces are equal).
  - Let's assume that when the pendulum is tilted at an angle of 15 degrees from the vertical, it moves at a speed of 1.5 m/s. We will find the tension force by the following steps:
    - Ratio of tension force to gravitational force (T g) = 98cos(15) = 98(0.96) = 94.08 Newton
    - Centrifugal force (F c) = 10 × 1.5 2 /1.5 = 10 × 1.5 = 15 Newtons
    - Total tension = T g + F c = 94.08 + 15 = 109.08 Newtons.
5. Calculate the friction. Any object that is pulled by a rope and experiences a "braking" force from the friction of another object (or fluid) transfers this force to the tension in the rope. The friction force between two objects is calculated in the same way as in any other situation - using the following equation: Friction force (usually written as F r) = (mu)N, where mu is the coefficient of friction force between objects and N is the usual force of interaction between objects, or the force with which they press on each other. Note that static friction, which is the friction that results from trying to force an object at rest into motion, is different from motion friction, which is the friction that results from trying to force a moving object to continue moving.
  - Let's assume that our 10 kg load is no longer swinging, but is now being towed along a horizontal plane using a rope. Let's assume that the coefficient of friction of the earth's motion is 0.5 and our load is moving at a constant speed, but we need to give it an acceleration of 1 m/s 2 . This problem introduces two important changes - first, we no longer need to calculate the tension force in relation to gravity, since our rope is not holding a weight suspended. Second, we will have to calculate the tension due to friction as well as that due to the acceleration of the mass of the load. We need to decide the following:
    - Normal force (N) = 10 kg & × 9.8 (gravity acceleration) = 98 N
    - Motion friction force (F r) = 0.5 × 98 N = 49 Newtons
    - Acceleration force (F a) = 10 kg × 1 m/s 2 = 10 Newton
    - Total tension = F r + F a = 49 + 10 = 59 Newtons.
  Calculation of tension force on several threads
  1. Lift vertical parallel weights using a block. Pulleys are simple mechanisms consisting of a suspended disk that allows you to change the direction of the tension force on the rope. In a simple pulley configuration, a rope or cable runs from a suspended weight up to a pulley, then down to another weight, thereby creating two sections of rope or cable. In any case, the tension in each of the sections will be the same, even if both ends are tensioned by forces of different magnitudes. For a system of two masses suspended vertically in a block, the tension force is equal to 2g(m 1)(m 2)/(m 2 +m 1), where “g” is the acceleration of gravity, “m 1” is the mass of the first object, “ m 2 ” – mass of the second object.
    - Note the following: physical problems assume that the blocks are perfect- have no mass, no friction, they do not break, are not deformed and do not separate from the rope that supports them.
    - Let's assume that we have two weights suspended vertically at parallel ends of a rope. One weight has a mass of 10 kg, and the second has a mass of 5 kg. In this case, we need to calculate the following:
      - T = 2g(m 1)(m 2)/(m 2 +m 1)
      - T = 2(9.8)(10)(5)/(5 + 10)
      - T = 19.6(50)/(15)
      - T = 980/15
      - T= 65.33 Newtons.
    - Note that since one weight is heavier, all other elements are equal, this system will begin to accelerate, hence the 10 kg weight will move down, causing the second weight to go up.
  2. Hang weights using pulleys with non-parallel vertical strings. Blocks are often used to direct the tension force in a direction other than down or up. If, for example, a load is suspended vertically from one end of a rope, and the other end holds the load in a diagonal plane, then the non-parallel system of pulleys takes the shape of a triangle with corners at the points of the first load, the second and the pulley itself. In this case, the tension in the rope depends both on gravity and on the component of the tension force that is parallel to the diagonal part of the rope.
    - Let's assume that we have a system with a 10 kg (m 1) load suspended vertically, connected to a 5 kg (m 2) load placed on a 60 degree inclined plane (this inclination is assumed to be frictionless). To find the tension in a rope, the easiest way is to first set up equations for the forces accelerating the loads. Next we proceed like this:
      - The suspended weight is heavier, there is no friction, so we know it is accelerating downward. The tension in the rope pulls upward, so that it accelerates with respect to the resultant force F = m 1 (g) - T, or 10(9.8) - T = 98 - T.
      - We know that a mass on an inclined plane accelerates upward. Since it has no friction, we know that tension pulls the load up along the plane, and pulls it down only your own weight. The component of the force pulling down the slope is calculated as mgsin(θ), so in our case we can conclude that it is accelerating with respect to the resultant force F = T - m 2 (g)sin(60) = T - 5( 9.8)(0.87) = T - 42.14.
      - If we equate these two equations, we get 98 - T = T - 42.14. We find T and get 2T = 140.14, or T = 70.07 Newtons.
  3. Use multiple strings to hang the object. Finally, let's imagine that the object is suspended from a "Y-shaped" system of ropes - two ropes are fixed to the ceiling and meet at a central point from which a third rope with a weight extends. The tension on the third rope is obvious - simple tension due to gravity or m(g). The tensions on the other two ropes are different and must add up to a force equal to the force of gravity upward in the vertical position and zero in both horizontal directions, assuming the system is at rest. The tension in a rope depends on the mass of the suspended loads and on the angle at which each rope is tilted from the ceiling.
    - Let's assume that in our Y-shaped system the bottom weight has a mass of 10 kg and is suspended on two ropes, one of which makes an angle of 30 degrees with the ceiling, and the second of which makes an angle of 60 degrees. If we need to find the tension in each of the ropes, we will need to calculate the horizontal and vertical components of the tension. To find T 1 (tension in the rope whose inclination is 30 degrees) and T 2 (tension in that rope whose inclination is 60 degrees), you need to solve:
      - According to the laws of trigonometry, the ratio between T = m(g) and T 1 and T 2 is equal to the cosine of the angle between each of the ropes and the ceiling. For T 1, cos(30) = 0.87, as for T 2, cos(60) = 0.5
      - Multiply the tension in the bottom rope (T=mg) by the cosine of each angle to find T 1 and T 2 .
      - T 1 = 0.87 × m(g) = 0.87 × 10(9.8) = 85.26 Newtons.
      - T 2 =0.5 × m(g) = 0.5 × 10(9.8) = 49 Newtons.
In this problem it is necessary to find the ratio of the tension force to

Rice. 3. Solution of problem 1 ()

The stretched thread in this system acts on block 2, causing it to move forward, but it also acts on block 1, trying to impede its movement. These two tension forces are equal in magnitude, and we just need to find this tension force. In such problems, it is necessary to simplify the solution as follows: we assume 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 block 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 located on the second and the tension thread should already move two bars. In this case, the ratio to will be equal to 2. The correct answer is the first one.

Two bodies of mass and , connected by a weightless inextensible thread, can slide without friction along 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?

Selected answer: 1. 2/3; 2. 1; 3. 3/2; 4. 9/4.

Rice. 4. Illustration for problem 2 ()

Rice. 5. Solution of problem 2 ()

The same force acts on the bars, only in different directions, so 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 block 2 move, in case “b” it is block 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.

A block weighing 1 kg lies on the table, to which a thread is tied, thrown over a stationary block. A load weighing 0.5 kg is suspended from the second end of the thread (Fig. 6). Determine the acceleration with which the block moves if the coefficient of friction of the block on the table is 0.35.

Rice. 6. Illustration for problem 3 ()

Let's write down a brief statement of the problem:

Rice. 7. Solution to problem 3 ()

It must be remembered that the tension forces and as vectors are different, but the magnitudes of these forces are the same and equal. Likewise, 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 select our own axes for each body. Let's write down the equations of Newton's second law for each of these bodies; when added, the internal tension forces are reduced, and we get the usual equation, substituting the data into it, we find 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 force of gravity of the second body forces this system to move, but the force of friction of the block on the table prevents 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 This way you can immediately come to the answer.

A block is fixed at the top of two inclined planes making angles and with the horizon. On the surface of the planes with a friction coefficient of 0.2, bars kg and , connected by a thread thrown over a block, move (Fig. 8). Find the pressure force on the block axis.

Rice. 8. Illustration for problem 4 ()

Let's make a brief statement of the problem conditions and an explanatory drawing (Fig. 9):

Rice. 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 triangle. A thread is thrown across the block, from which the bars are suspended; they pull down with the same force, and the action of the tension forces F H1 and F H2 leads to the fact that their resultant force acts on the block. But these tension forces will be equal to each other, they form a right angle with each other, so when adding these forces, you get a square instead of a regular parallelogram. The required force F d is the diagonal of the square. We see that for the result we need to find the tension force of the thread. Let's analyze: in which direction does the system of two connected bars move? The more massive block will naturally pull the lighter one, block 1 will slide down, and block 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 that the pressure force on the block axis is approximately 16 N.

We looked at various ways to solve problems that many of you will find useful in the future in order to understand the principles of the design and operation of those machines and mechanisms that you will have to deal with in production, in the army, and in everyday life.

Bibliography
1. Tikhomirova S.A., Yavorsky B.M. Physics (basic level) - M.: Mnemosyne, 2012.
2. Gendenshtein L.E., Dick Yu.I. Physics 10th grade. - M.: Mnemosyne, 2014.
3. Kikoin I.K., Kikoin A.K. Physics-9. - M.: Education, 1990.
Homework
1. What law do we use when composing equations?
2. What quantities are the same for bodies connected by an inextensible thread?
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