Solving problems on the motion of a system of coupled bodies. Movement of a system of bodies Find the value of the tension force in the thread
The tension force is the one that acts on an object comparable to a wire, cord, cable, thread, and so on. It can be several objects at once, in which case the tension force will act on them and not necessarily evenly. An object of tension is any object suspended from all of the above. But who needs to know? Despite the specificity of the information, it can be useful even in everyday situations.
For example, when renovating a house or apartment. And, of course, to all people whose profession is related to calculations:
- engineers;
- architects;
- designers, etc.
Thread tension and similar objects
Why do they need to know this and what is the practical use of it? In the case of engineers and designers, knowledge of the power of tension will allow you to create sustainable structures. This means that structures, equipment and other structures will be able to maintain their integrity and strength longer. Conventionally, these calculations and knowledge can be divided into 5 main points in order to fully understand what is at stake.
Stage 1
Task: to determine the tension force at each end of the thread. This situation can be seen as the result of forces acting on each end of the thread. It is equal to the mass multiplied by the acceleration due to gravity. Let's assume that the thread is taut. Then any impact on the object will lead to a change in tension (in the thread itself). But even in the absence of active actions, the force of attraction will act by default. So, let's substitute the formula: T=m*g+m*a, where g is the acceleration of fall (in this case, a suspended object), and is any other acceleration acting from the outside.
There are many third-party factors that affect the calculations − the weight of the thread, its curvature, and so on. For simple calculations, we will not take this into account for the time being. In other words, let the thread be perfect from a mathematical point of view and “without flaws”.
Let's take a "live" example. A strong thread with a load of 2 kg is suspended on a beam. At the same time, there is no wind, swaying and other factors that somehow affect our calculations. Then the force of tension is equal to the force of gravity. In the formula, this can be expressed as follows: Fn \u003d Ft \u003d m * g, in our case it is 9.8 * 2 \u003d 19.6 newtons.
Stage 2
It concludes on the issue of acceleration. Let's add a condition to the existing situation. Its essence is that acceleration also acts on the thread. Let's take a simpler example. Imagine that our beam is now being lifted up at a speed of 3 m/s. Then, the acceleration of the load will be added to the tension and the formula will take the following form: Fn \u003d Ft + usk * m. Focusing on past calculations, we get: Fn \u003d 19.6 + 3 * 2 \u003d 25.6 newtons.
Stage 3
Here it is more difficult, since we are talking about about angular rotation. It should be understood that when the object is rotated vertically, the force acting on the thread will be much greater at the bottom point. But let's take an example with a slightly smaller swing amplitude (like a pendulum). In this case, the formula is needed for calculations: Fc \u003d m * v² / r. Here the desired value indicates the additional power of tension, v is the speed of rotation of the suspended load, and r is the radius of the circle along which the load rotates. The last value is actually equal to the length of the thread, even if it is 1.7 meters.
So, substituting the values, we find the centrifugal data: Fc=2*9/1.7=10.59 newtons. And now, in order to find out the total force of the thread tension, it is necessary to add the centrifugal force to the available data on the state of rest: 19.6 + 10.59 = 30.19 newtons.
Stage 4
Consideration should be given to the changing tension force as the load passes through the arc. In other words, regardless of the constant magnitude of attraction, the centrifugal (resultant) force changes as the suspended load swings.
To better understand this aspect, it is enough to imagine a weight tied to a rope that can be freely rotated around the beam to which it is attached (like a swing). If the rope is swung strong enough, then at the moment it is in the upper position, the force of attraction will act in the “reverse” direction relative to the tension in the rope. In other words, the load will become “lighter”, which will also weaken the tension on the rope.
Assume that the pendulum is deflected at an angle equal to twenty degrees from the vertical and moves at a speed of 1.7 m/s. The force of attraction (Fп) with these parameters will be equal to 19.6*cos(20)=19.6*0.94=18.424 N; centrifugal force (F c \u003d mv² / r) \u003d 2 * 1.7² / 1.7 \u003d 3.4 N; well, the total tension (Fpn) will be equal to Fp + Fc \u003d 3.4 + 18.424 \u003d 21.824 N.
Stage 5
Its essence lies in the force of friction between a load and another object, which together indirectly affects the tension of the rope. In other words, the friction force contributes to an increase in the tensile force. This is clearly seen in the example of moving objects on rough and smooth surfaces. In the first case, the friction will be large, and therefore it becomes harder to move the object.
The total tension in this case is calculated by the formula: Fn \u003d Ftr + Fy, where Ftr is friction, and Fu is acceleration. Ftr \u003d μR, where μ is the friction between objects, and P is the force of interaction between them.
To better understand this aspect, consider the problem. Let's say we have a load of 2 kg and the coefficient of friction is 0.7 with an acceleration of 4m/s of constant speed. Now we use all the formulas and get:
- The force of interaction is P=2*9.8=19.6 newtons.
- Friction - Ftr=0.7*19.6=13.72 N.
- Acceleration - Fu=2*4=8 N.
- The total tension force is Fn \u003d Ftr + Fy \u003d 13.72 + 8 \u003d 21.72 newtons.
Now you know more and can find and calculate the desired values yourself. Of course, for more accurate calculations, more factors need to be taken into account, but these data are quite enough to pass the coursework and abstract.
Video
This video will help you better understand this topic and remember it.
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 the body, subject to the action of opposing forces that attract it.
She is known as stretching force, which, when exposed to an elastic body, creates tension; This last concept has various definitions, which 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 talking about mechanical stress, to denote the force per unit area surrounded by 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 an 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. Therefore, the fluid resists by increasing its surface.
How to find the pull force
Knowing that force tension is force, with which a line or string is stretched, one can find the tension in a situation of a static type 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, tension is allowed knowing that the sum of the horizontal and vertical components of the forces involved must give zero.
The first step to do this calculation- draw a slope and place a block of mass M on it. To the right, the slope increases, and at one point it meets a wall, from which the line runs parallel to the first one. and tie the block, holding it in place and applying tension T. Next, you must identify the angle of inclination with the Greek letter, which can be "alpha", and the force it exerts on the block with the letter N, since we are talking about normal force .
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 the newton, which is denoted by the letter N. In the case of a normal force, it must be expanded in vertical and horizontal vectors using the angle it makes with the axis x: to compute the up vector g is equal to the cosine of the angle, and for the vector in the direction from the left, towards the bosom of this.
Finally, the left component of the normal force must be equated to the right side of the stress T, finally resolving the stress.
library science
To know well the term librarianship, which we are now occupied with, it is necessary to start 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 "tehe", which is synonymous with the word "box" or "the place where it is stored." -The suffix "-logía", which is used to denote "the science that studies". This is known as librarianship in 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 directional movement that a living being realizes in order to respond to a stimulus that it perceives. The taxi can be negative (when the living being moves away from the source of the stimulus) or positive (the living being approaches what the stimulus in question generates). To organize
definition
extension
Expansion, from the Latin expansĭo, is the action and effect of expanding or expanding (spreading, spreading, unfolding, unrolling, giving more 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 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 bar 2, causing it to move forward, but it also acts on bar 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 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 be moving 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 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.
Rice. 4. Illustration for task 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 forces bar 2 to 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 will get the answer - 1.5. This is the third answer.
On the table lies a bar of mass 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 friction coefficient of the bar on the table is 0.35.
Rice. 6. Illustration for task 3 ()
We write down a brief condition of the problem:
Rice. 7. Solution of 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. In the same way, 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's write down the equations of Newton's second law for each of these bodies, when added, the internal tension forces will decrease, and we get the usual equation, substituting the data into it, we get that the acceleration is .
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 of 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 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 the answer.At the top of two inclined planes making angles with the horizon and , a block is fixed. On the surface of the planes at a friction coefficient of 0.2, bars kg and move, connected by a thread thrown over the block (Fig. 8). Find the pressure force on the axis of the block.
Rice. 8. Illustration for task 4 ()
Let's make a brief note of the problem condition and an explanatory drawing (Fig. 9):
Rice. 9. Solution of problem 4 ()
We remember that if one plane makes an angle of 60 0 with the horizon, and the second plane makes an angle of 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 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 between themselves, these tension forces will be equal, they make up a right angle between themselves, therefore, when these forces are added, a square is obtained instead of an ordinary parallelogram. The desired force F d is the diagonal of the square. We see that for the result we need to find the tension in the thread. Let's analyze: in which direction does the system of two connected bars move? A more massive block, of course, will pull over a 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 the pressure force on the axis of the block should be found:
We found that the pressure force on the axis of the block 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 design and operation of those machines and mechanisms that you will have to deal with in production, in the army, and at home.
Bibliography
- Tikhomirova S.A., Yavorsky B.M. Physics (basic level) - M.: Mnemozina, 2012.
- Gendenstein L.E., Dick Yu.I. Physics grade 10. - M.: Mnemosyne, 2014.
- Kikoin I.K., Kikoin A.K. Physics-9. - M.: Enlightenment, 1990.
Homework
- What law do we use when writing equations?
- What quantities are the same for bodies connected by an inextensible thread?
- Internet portal Bambookes.ru ( ).
- Internet portal 10klass.ru ().
- Internet portal Festival.1september.ru ().
1. A 5 kg kettlebell is suspended from the ceiling on two identical ropes attached to the ceiling at two different points. The threads form an angle a = 60° with each other (see figure). Find the tension in each thread.
2. (e) A Christmas tree ball is suspended from a horizontal branch on two identical threads attached to the branch at two different points. The threads form an angle a = 90° with each other. Find the mass of the ball if the tension force of each thread is 0.1 N.
3. A large iron pipe is suspended by its ends from a crane hook on two identical cables, forming an angle of 120 ° with each other (see figure). The tension force of each cable is 800 N. Find the mass of the pipe.
4. (e) A concrete beam weighing 400 kg, suspended by the ends to a hook on two cables, is lifted by a tower crane with an upward acceleration of 3 m/s 2 . The angle between the cables is 120°. Find the tension in the ropes.
5. A weight of 2 kg is suspended from the ceiling on a thread, to which, on another thread, a weight of 1 kg is suspended (see Fig.). Find the tension in each thread.
6. (e) A weight of 500 g is suspended from the ceiling on a thread, to which, on another thread, another weight is suspended. The tension force of the lower thread is 3 N. Find the mass of the lower load and the tension force of the upper thread.
7. A load weighing 2.5 kg is lifted onto the threads with an acceleration of 1 m / s 2 directed upwards. To this load, on another thread, a second load is suspended. The tension force of the upper thread (i.e., which is pulled up) is 40 N. Find the mass of the second load and the tension force of the lower thread.
8. (e) A weight of 2.5 kg is lowered onto the strings with a downward acceleration of 3 m/s 2 . To this load, on another thread, a second load is suspended. The tension force of the lower thread is 1 N. Find the mass of the second load and the tension force of the upper thread.
9.
A weightless and inextensible thread is thrown through a fixed block attached to the ceiling. Weights with masses m 1 = 2 kg and m 2 = 1 kg are suspended from the ends of the thread (see Fig.). In which direction and with what acceleration does each of the loads move? What is the tension in the thread?10. (e) A weightless and inextensible thread is thrown over an immovable block attached to the ceiling. Weights are suspended from the ends of the thread. The mass of the first load m 1 \u003d 0.2 kg. It moves up with an acceleration of 3 m/s 2 . What is the mass of the second load? What is the tension in the thread?
11. A weightless and inextensible thread is thrown through a fixed block attached to the ceiling. Weights are suspended from the ends of the thread. The mass of the first load m 1 \u003d 0.2 kg. It moves upward, increasing its speed from 0.5 m/s to 4 m/s in 1 s. What is the mass of the second load? What is the tension in the thread?
12. (e) A weightless and inextensible thread is thrown over an immovable block attached to the ceiling. Weights with masses m 1 = 400 g and m 2 = 1 kg are suspended from the ends of the thread. They are held at rest and then released. With what acceleration does each of the loads move? How far will each of them travel in 1 second of movement?
13. A weightless and inextensible thread is thrown through a fixed block attached to the ceiling. Weights with masses m 1 = 400 g and m 2 = 0.8 kg are suspended from the ends of the thread. They are held at rest at the same level, and then released. What will be the distance between the loads (in height) after 1.5 s after the start of movement?
14. (e) A weightless and inextensible thread is thrown over an immovable block attached to the ceiling. Weights are suspended from the ends of the thread. The mass of the first load m 1 \u003d 300 g. The loads are kept at rest at the same level, and then released. After 2 s after the start of movement, the difference in heights at which the loads are located has reached 1 m. What is the mass m 2 of the second load and what is the acceleration of the loads?
Problems on a conical pendulum
15. A small ball weighing 50 g, suspended on a weightless inextensible thread 1 m long, moves in a circle in a horizontal plane. The thread makes an angle of 30° with the vertical. What is the tension in the thread? What is the speed of the ball?
16. (e) A small ball suspended on a weightless inextensible thread 1 m long moves in a circle in a horizontal plane. The thread makes an angle of 30° with the vertical. What is angular ball speed?
17. A ball of mass 100 g moves in a circle with a radius of 1 m, being suspended on a weightless and inextensible rope 2 m long. What is the tension in the rope? What angle does the rope make with the vertical? What is the speed of the ball?
18. (e) A ball of mass 85 g moves along a circle with a radius of 50 cm, being suspended on a weightless and inextensible rope 577 mm long. What is the tension in the rope? What angle does the rope make with the vertical? What is angular ball speed?
Section 17.
Body weight, support reaction force and weightlessness.
1. A person weighing 80 kg is in an elevator moving with an acceleration of 2.5 m / s 2 directed upwards. What is the weight of the person in the elevator?
2. (e) A person is in an elevator moving with an upward acceleration of 2 m/s 2 . What is the mass of a person if his weight is 1080 N?
3. A beam weighing 500 kg is lowered on a cable with an acceleration of 1 m/s 2 directed downwards. What is the weight of the beam? What is the tensile strength of the cable?
4. (e) A circus acrobat is lifted up on a rope with an acceleration of 1.2 m/s 2 , also directed upwards. What is the mass of the acrobat if the rope tension is 1050 N? What is the weight of the acrobat?
5. If the elevator moves with an acceleration equal to 1.5 m / s 2 directed upwards, then the weight of a person in the elevator is 1000 N. What will be the weight of a person if the elevator moves with the same acceleration, but directed downwards? What is the mass of a person? What is the weight of this person in a stationary elevator?
6. (e) If the elevator moves with acceleration directed upwards, then the weight of the person in the elevator is 1000 N. If the elevator moves with the same acceleration, but directed downwards, then the weight of the person is 600 N. What is the acceleration of the elevator and what is the mass of the person?
7. A person with a mass of 60 kg rises in an elevator moving uniformly upwards. The elevator at rest gained a speed of 2.5 m/s in 2 s. What is the person's weight?
8. (e) A person of mass 70 kg is going up in an elevator moving uniformly upwards. An elevator at rest traveled a distance of 4 m in 2 s. What is the weight of a person in this case?
9. The radius of curvature of a convex bridge is 200 m. A car with a mass of 1 ton moves along the bridge at a speed of 72 km/h. What is the weight of the car at the top of the bridge?
10. (e) The radius of curvature of a convex bridge is 150 m. A car with a mass of 1 ton is moving on the bridge. Its weight at the top of the bridge is 9500 N. What is the speed of the car?
11. The radius of curvature of a convex bridge is 250 m. A car is moving along the bridge at a speed of 63 km/h. Its weight at the top of the bridge is 20,000 N. What is the mass of the car?
12. (e) A car of mass 1 ton moves along a convex bridge at a speed of 90 km/h. The weight of the car at the top of the bridge is 9750 N. What is the radius of curvature of the convex surface of the bridge?
13. A tractor weighing 3 tons drives onto a horizontal wooden bridge, which sags under the weight of the tractor. The speed of the tractor is 36 km/h. The weight of the tractor at the lowest deflection point of the bridge is 30,500 N. What is the radius of the bridge surface?
14. (e) A 3 ton tractor drives onto a horizontal wooden bridge that sags under the weight of the tractor. The speed of the tractor is 54 km/h. The radius of curvature of the bridge surface is 120 m. What is the weight of the tractor?
15. A wooden horizontal bridge can withstand a load of 75,000 N. The mass of the tank that must pass over the bridge is 7200 kg. How fast can a tank move across the bridge if the bridge bends so that the radius of curvature of the bridge is 150 m?
16. (e) The length of a wooden bridge is 50 m. A truck moving at a constant modulo speed passes the bridge in 5 s. At the same time, the maximum deflection of the bridge is such that the radius of curvature of its surface is 220 m. The weight of the truck in the middle of the bridge is 50 kN. What is the weight of the truck?
17. A car moves along a convex bridge, the radius of curvature of which is 150 m. At what speed of the car will the driver feel weightlessness? What else will he feel (unless, of course, the driver is a normal person)?
18. (e) A car is moving on a convex bridge. Did the driver of the car feel that at the highest point of the bridge at a speed of 144 km / h the car loses control? Why is this happening? What is the radius of curvature of the bridge surface?
19. The spacecraft starts up with an acceleration of 50 m/s 2 . What kind of overload do the astronauts experience in the spacecraft?
20. (e) An astronaut can withstand a tenfold short-term overload. What should be the upward acceleration of the spacecraft at this time?
In technology, there is another type of stretched elements, in determining the strength of which the dead weight is important. These are the so-called flexible threads. This term refers to flexible elements in power lines, in cable cars, in suspension bridges and other structures.
Let (Fig.1) there is a flexible thread of constant section, loaded with its own weight and suspended at two points located at different levels. Under the influence of its own weight, the thread sags along a certain curve AOW.
The horizontal projection of the distance between the supports (the points of its attachment), denoted by , is called span.
The thread has a constant cross section, therefore, its weight is distributed evenly along its length. Typically, the sag of the thread is small compared to its span, and the length of the curve AOB differs little (no more than 10%) from the length of the chord AB. In this case, with a sufficient degree of accuracy, we can assume that the weight of the thread is uniformly distributed not along its length, but along the length of its projection onto the horizontal axis, i.e. along span l.
Fig.1. Calculation scheme of a flexible thread.
We will consider this category of flexible threads. Let us assume that the intensity of the load uniformly distributed over the span of the thread is equal to q. This load, which has the dimension force/length, can be not only the thread's own weight per unit span length, but also the weight of ice or any other load, also uniformly distributed. The assumption made about the law of load distribution greatly facilitates the calculation, but at the same time makes it approximate; if with an exact solution (the load is distributed along the curve) the sag curve will be a catenary, then in the approximate solution the sag curve turns out to be a square parabola.
We choose the origin of coordinates at the lowest point of the sagging thread ABOUT, whose position, unknown to us so far, obviously depends on the magnitude of the load q, on the relationship between the length of the thread along the curve and the length of the span, as well as on the relative position of the reference points. At the point ABOUT the tangent to the sag curve is obviously horizontal. Let's direct the axis to the right along this tangent.
We cut out two sections at the origin and at a distance from the origin (section m n) part of the thread length. Since the thread is assumed to be flexible, i.e., capable of resisting only stretching, the action of the discarded part on the remaining part is possible only in the form of a force directed tangentially to the sagging curve of the thread at the cut point; no other direction of this force is possible.
Figure 2 shows the cut out part of the thread with the forces acting on it. Uniformly distributed load intensity q directed vertically down. The impact of the left thrown part (horizontal force H) is directed, due to the fact that the thread is in tension, to the left. Action of the right thrown part, force T, is directed to the right tangent to the thread slack curve at that point.
Let us compose the equilibrium equation for the cut section of the thread. Take the sum of the moments of all forces relative to the point of application of force T and set it equal to zero. In this case, we will take into account, based on the assumption given at the beginning, that the resultant of the distributed load with intensity q will be , and that it is attached in the middle of the segment . Then
Fig.2. Fragment of the cut out part of the flexible thread
,It follows that the thread sagging curve is a parabola. When both suspension points of the thread are at the same level, then the Value in this case will be the so-called sag arrow. It's easy to define. Since in this case, due to symmetry, the lowest point of the thread is in the middle of the shed, then; substituting in equation (1) values and get:
Value H is called the horizontal tension of the thread.
and tension H, then by formula (2) we find the sagging arrow . At given and tension H is determined by formula (3). The connection of these quantities with the length of the thread along the sag curve is established using an approximate formula known from mathematics)
Let us compose one more equilibrium condition for the cut out part of the thread, namely, we equate to zero the sum of the projections of all forces on the axis :
From this equation we find the force T tension at an arbitrary point
Whence it follows that the force T increases from the lowest point of the thread to the supports and will be greatest at the suspension points where the tangent to the sagging curve of the thread makes the largest angle with the horizontal. With a small thread sag, this angle does not reach large values, therefore, with a degree of accuracy sufficient for practice, we can assume that the force in the thread is constant and equal to its tension. H. This value is usually used to calculate the strength of the thread. If, nevertheless, it is required to calculate the greatest force at the suspension points, then for a symmetrical thread, its value will be determined in the following way. The vertical components of the reactions of the supports are equal to each other and equal to half the total load on the thread, i.e. . The horizontal components are equal to the force H determined by formula (3). The total reactions of the supports will be obtained as the geometric sums of these components:
The strength condition for a flexible thread, if through F the cross-sectional area is indicated, has the form:
Replacing the tension H its value according to the formula (3), we get:
From this formula, given , , and you can determine the required sag . In this case, the solution will be simplified if only the own weight is included in; then , where is the weight of a unit volume of the thread material, and
i.e. value F will not be included in the calculation.
If the suspension points of the thread are at different levels, then, substituting the values and into equation (1), we find and :
From here, from the second expression, we determine the tension
and dividing the first by the second, we find:
Bearing in mind that , we get:
Substituting this value into the formula for a specific tension H, we finally determine:
The two digits in the denominator indicate that there can be two main forms of thread slack. First form at lower value H(plus sign in front of the second root) gives us the top of the parabola between the thread's supports. With more tension H(minus sign in front of the second root) the top of the parabola will be located to the left of the support A(Fig.1). We get the second shape of the curve. A third (intermediate between the two main) forms of sagging is also possible, corresponding to the condition ; then the origin is aligned with the point A. One form or another will be obtained depending on the relationship between the length of the thread along the sagging curve AOB(Fig.1) and chord length AB.
If, when the thread is suspended at different levels, the sagging arrows and are unknown, but the tension is known H, then it is easy to get the distances A And b and sag arrows, and . Difference h suspension levels is equal to:
Substitute in this expression the values and , and transform it, keeping in mind that:
and since then
It should be borne in mind that at , the first form of sagging of the thread will take place, at the second form of sagging and at the third form. Substituting the values and into the expressions for the sagging arrows and , we obtain the values and :
Now let's find out what will happen to a symmetrical thread spanning the span if, after hanging it at a temperature and load intensity, the temperature of the thread rise to and the load will increase to intensity (for example, due to its icing). In this case, suppose that in the first state either the tension or the sag is given (Knowing one of these two quantities, you can always determine the other.)
When counting deformations thread, which is a small value compared to the length of the thread, we make two assumptions: the length of the thread "is equal to its span, and the tension is constant and equal to H. With flat threads, these assumptions give a small error.
