Home Nutrition Find the magnitude of the thread tension. The tension of the thread and the application of the formula in everyday situations. Sample stiffness. Young's modulus

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Find the magnitude of the thread tension. The tension of the thread and the application of the formula in everyday situations. Sample stiffness. Young's modulus

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...

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

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.
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.
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.
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.
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.

It is necessary to know the point of application and direction of each force. It is important to be able to determine which forces act on the body and in what direction. Force is denoted as , measured in Newtons. In order to distinguish between forces, they are designated as follows

Below are the main forces operating in nature. It is impossible to invent forces that do not exist when solving problems!

There are many forces in nature. Here we consider the forces that are considered in the school physics course when studying dynamics. Other forces are also mentioned, which will be discussed in other sections.

Gravity

Every body on the planet is affected by Earth's gravity. The force with which the Earth attracts each body is determined by the formula

The point of application is at the center of gravity of the body. Gravity always directed vertically downwards.

Friction force

Let's get acquainted with the force of friction. This force occurs when bodies move and two surfaces come into contact. The force occurs because surfaces, when viewed under a microscope, are not as smooth as they appear. The friction force is determined by the formula:

The force is applied at the point of contact of two surfaces. Directed in the direction opposite to movement.

Ground reaction force

Let's imagine a very heavy object lying on a table. The table bends under the weight of the object. But according to Newton's third law, the table acts on the object with exactly the same force as the object on the table. The force is directed opposite to the force with which the object presses on the table. That is, up. This force is called the ground reaction. The name of the force "speaks" support reacts. This force occurs whenever there is an impact on the support. The nature of its occurrence at the molecular level. The object seemed to deform the usual position and connections of the molecules (inside the table), they, in turn, strive to return to their original state, “resist.”

Absolutely any body, even a very light one (for example, a pencil lying on a table), deforms the support at the micro level. Therefore, a ground reaction occurs.

There is no special formula for finding this force. It is denoted by the letter , but this force is simply a separate type of elasticity force, so it can also be denoted as

The force is applied at the point of contact of the object with the support. Directed perpendicular to the support.

Since the body is represented as a material point, force can be represented from the center

Elastic force

This force arises as a result of deformation (change in the initial state of the substance). For example, when we stretch a spring, we increase the distance between the molecules of the spring material. When we compress a spring, we decrease it. When we twist or shift. In all these examples, a force arises that prevents deformation - the elastic force.

Hooke's law

The elastic force is directed opposite to the deformation.

Since the body is represented as a material point, force can be represented from the center

When connecting springs in series, for example, the stiffness is calculated using the formula

When connected in parallel, the stiffness

Sample stiffness. Young's modulus.

Young's modulus characterizes the elastic properties of a substance. This is a constant value that depends only on the material and its physical state. Characterizes the ability of a material to resist tensile or compressive deformation. The value of Young's modulus is tabular.

Body weight

Body weight is the force with which an object acts on a support. You say, this is the force of gravity! The confusion arises in the following: indeed, often the weight of a body is equal to the force of gravity, but these forces are completely different. Gravity is a force that arises as a result of interaction with the Earth. Weight is the result of interaction with support. The force of gravity is applied at the center of gravity of the object, while weight is the force that is applied to the support (not to the object)!

There is no formula for determining weight. This force is designated by the letter.

The support reaction force or elastic force arises in response to the impact of an object on the suspension or support, therefore the weight of the body is always numerically the same as the elastic force, but has the opposite direction.

The support reaction force and weight are forces of the same nature; according to Newton’s 3rd law, they are equal and oppositely directed. Weight is a force that acts on the support, not on the body. The force of gravity acts on the body.

Body weight may not be equal to gravity. It may be more or less, or it may be that the weight is zero. This condition is called weightlessness. Weightlessness is a state when an object does not interact with a support, for example, the state of flight: there is gravity, but the weight is zero!

It is possible to determine the direction of acceleration if you determine where the resultant force is directed

Please note that weight is force, measured in Newtons. How to correctly answer the question: “How much do you weigh”? We answer 50 kg, not naming our weight, but our mass! In this example, our weight is equal to gravity, that is, approximately 500N!

Overload- ratio of weight to gravity

Archimedes' force

Force arises as a result of the interaction of a body with a liquid (gas), when it is immersed in a liquid (or gas). This force pushes the body out of the water (gas). Therefore, it is directed vertically upward (pushes). Determined by the formula:

In the air we neglect the power of Archimedes.

If the Archimedes force is equal to the force of gravity, the body floats. If the Archimedes force is greater, then it rises to the surface of the liquid, if less, it sinks.

Electrical forces

There are forces of electrical origin. Occurs in the presence of an electrical charge. These forces, such as the Coulomb force, Ampere force, Lorentz force, are discussed in detail in the section Electricity.

Schematic designation of forces acting on a body

Often a body is modeled as a material point. Therefore, in diagrams, various points of application are transferred to one point - to the center, and the body is depicted schematically as a circle or rectangle.

In order to correctly designate forces, it is necessary to list all the bodies with which the body under study interacts. Determine what happens as a result of interaction with each: friction, deformation, attraction, or maybe repulsion. Determine the type of force and correctly indicate the direction. Attention! The amount of forces will coincide with the number of bodies with which the interaction occurs.

The main thing to remember

1) Forces and their nature;
2) Direction of forces;
3) Be able to identify the acting forces

There are external (dry) and internal (viscous) friction. External friction occurs between contacting solid surfaces, internal friction occurs between layers of liquid or gas during their relative motion. There are three types of external friction: static friction, sliding friction and rolling friction.

Rolling friction is determined by the formula

The resistance force occurs when a body moves in a liquid or gas. The magnitude of the resistance force depends on the size and shape of the body, the speed of its movement and the properties of the liquid or gas. At low speeds of movement, the drag force is proportional to the speed of the body

At high speeds it is proportional to the square of the speed

Let's consider the mutual attraction of an object and the Earth. Between them, according to the law of gravity, a force arises

Now let's compare the law of gravity and the force of gravity

The magnitude of the acceleration due to gravity depends on the mass of the Earth and its radius! Thus, it is possible to calculate with what acceleration objects on the Moon or on any other planet will fall, using the mass and radius of that planet.

The distance from the center of the Earth to the poles is less than to the equator. Therefore, the acceleration of gravity at the equator is slightly less than at the poles. At the same time, it should be noted that the main reason for the dependence of the acceleration of gravity on the latitude of the area is the fact of the Earth’s rotation around its axis.

As we move away from the Earth's surface, the force of gravity and the acceleration of gravity change in inverse proportion to the square of the distance to the center of the Earth.

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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 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 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.

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