How to find the tension force knowing the acceleration of the mass. weightless block thread tension friction acceleration load mass find force

A thread is thrown through a weightless block, connecting body 3 with body 2, to which body 1 is suspended. The mass of each body is 2 kg. Find the acceleration of body 1 and the tension in the thread connecting it with body 2.

task 12431

In the installation (Fig. 3) the angle α = 50° inclined plane with the body mass horizon m 1 = 0.15 kg and m 2 = 0.5 kg. Considering the thread and the block to be weightless, and neglecting the forces of friction, determine the acceleration with which the bodies will move if the body of mass m 2 is lowered.

task 13039

two loads ( m 1 = 500 g and m 2 = 700 g) are tied with a weightless thread and lie on a smooth horizontal surface. To cargo m 1 applied horizontally directed force F\u003d 6 N. Neglecting friction, determine 1) acceleration of loads; 2) the tension force of the thread.

task 13040

The simplest Atwood machine used to study laws uniformly accelerated motion, represents two loads with unequal masses m 1 and m 2 (for example m 1 > m 2), which are suspended on a light thread thrown over a fixed block. Considering the thread and the block to be weightless and neglecting the friction in the axis of the block, determine 1) the acceleration of the loads; 2) thread tension force T; 3) strength F acting on the axis of the block.

task 13041

Loads with masses m 1 = 200 g and m 2 = 500 g are suspended from the system of blocks (see Fig.). The load m 1 rises, the movable block with m 2 falls, the blocks and the thread are weightless, there are no friction forces. Determine: 1) the tension force of the thread T; 2) cargo acceleration.

task 13042

Bodies with masses m 1 = 200 g and m 2 = 150 g are connected by a weightless thread. The angle α between the inclined plane and the horizon is 20°. Neglecting the forces of friction and assuming the block is weightless, determine the acceleration with which the bodies move, assuming that the body m 2 moves down.

task 13043

On a horizontal table there is a body A with a mass M \u003d 2 kg, connected by threads using blocks with bodies B (m 1 \u003d 0.5 kg) and C (m 2 \u003d 0.3 kg). Assuming blocks and threads to be weightless and neglecting friction forces, find: 1) the acceleration with which these bodies move; 2) the difference in the tension forces of the threads.

task 13044

The angles between the inclined planes and the horizon are given: α=30° and β=45°. A weightless thread connecting bodies with masses m 1 = 0.45 kg and m 2 = 0.5 kg is thrown over a weightless block. Find: 1) acceleration of the motion of the body; 2) the force of the thread tension. Ignore the forces of friction.

task 13052

The load lying on the table is connected by a thread thrown over a weightless block on the edge of the table with a hanging load of the same mass (m 1 \u003d m 2 \u003d 0.5 kg). The coefficient of friction of the load m 2 on the table f = 0.15. Find: 1) cargo acceleration; 2) the tension force of the thread. Ignore block friction.

task 13055

The angle α between the plane and the horizon is 30°, the masses of the bodies are the same in m = 1 kg. A body lies on the plane, the coefficient of friction between which and the plane is f = 0.1. Neglecting friction in the axis of the block and considering the block and the thread to be weightless, determine the force of pressure on the axis.

task 13146

A weightless thread, at the ends of which bodies with masses m 1 = 0.35 kg and m 2 = 0.55 kg are tied, is thrown over a fixed block in the form of a continuous homogeneous cylinder with mass m = 0.2 kg. Find: 1) acceleration of loads; 2) the ratio of T 2 /T 1 tension forces of the threads. Neglect the friction in the axis of the block.

task 13147

Using a block in the form of a thin-walled hollow cylinder, a body with a mass of m 1 = 0.25 kg is connected by a weightless thread to a body with a mass of m 2 = 0.2 kg. The first body slides on the surface of a horizontal table with a friction coefficient f equal to 0.2. Block mass m = 0.15 kg. Neglecting friction in bearings, determine: 1) acceleration a of bodies; 2) tension forces T 1 and T 2 of the thread on both sides of the block.

task 14495

Two weights with masses m 1 = 2 kg and m 2 = 1 kg are connected by a thread and thrown over a weightless block. Find the acceleration a with which the weights move and the tension force T in the thread. Ignore the friction in the block.

task 14497

The weightless block is fixed at the top of the inclined plane making an angle α = 30° with the horizon. Weights 1 and 2 of the same mass m 1 = m 2 = 1 kg are connected by a thread and thrown over the block. Find the acceleration a with which the weights move and the tension force T in the thread. Neglect the friction of the weight on the inclined plane and the friction in the block.

task 14499

The weightless block was fixed on the top of two inclined planes, which made angles α = 30° and β = 45° with the horizon, respectively. Weights 1 and 2 of the same mass m 1 = m 2 = 1 kg were connected by a thread thrown over the block. Find the acceleration a with which the weights move and the force of tension in the thread T. The friction of the weights on inclined planes and the friction in the block can be neglected.

task 15783

The simplest Atwood machine (Fig. 1), used to study uniformly accelerated motion, consists of two loads with masses m 1 \u003d 0.5 kg and m 2 \u003d 0.2 kg, which are suspended on a light thread thrown over a fixed block. Considering the thread and the block to be weightless and neglecting the friction in the axis of the block, determine: 1) the acceleration of the loads; 2) the tension force of the thread.

task 15785

The simplest Atwood machine (Fig. 1), used to study uniformly accelerated motion, consists of two loads with masses m 1 \u003d 0.6 kg and m 2 \u003d 0.2 kg, which are suspended on a light thread thrown over a fixed block. Considering the thread and the block to be weightless and neglecting the friction in the axis of the block, determine: 1) the acceleration of the loads; 2) the tension force of the thread.

task 15787

The simplest Atwood machine (Fig. 1), used to study uniformly accelerated motion, consists of two loads with masses m 1 \u003d 0.8 kg and m 2 \u003d 0.15 kg, which are suspended on a light thread thrown over a fixed block. Considering the thread and the block to be weightless and neglecting the friction in the axis of the block, determine: 1) the acceleration of the loads; 2) the tension force of the thread.

task 15789

The simplest Atwood machine (Fig. 1), used to study uniformly accelerated motion, consists of two loads with masses m 1 \u003d 0.35 kg and m 2 \u003d 0.55 kg, which are suspended on a light thread thrown over a fixed block. Considering the thread and the block to be weightless and neglecting the friction in the axis of the block, determine: 1) the acceleration of the loads; 2) the tension force of the thread.

task 15791

The simplest Atwood machine (Fig. 1), used to study uniformly accelerated motion, consists of two loads with masses m 1 \u003d 0.8 kg and m 2 \u003d 0.2 kg, which are suspended on a light thread thrown over a fixed block. Considering the thread and the block to be weightless and neglecting the friction in the axis of the block, determine: 1) the acceleration of the loads; 2) the tension force of the thread.

task 15796

In the installation (Fig. 3), the angle α = 30° of the inclined plane with the horizon of the body mass m 1 = 300 g and m 2 = 0.8 kg. Considering the thread and the block to be weightless, and neglecting the forces of friction, determine the acceleration with which the bodies will move if the body of mass m 2 is lowered.

task 15798

In the installation (Fig. 3), the angle α = 60° of the inclined plane with the horizon of the body mass m 1 = 500 g and m 2 = 0.6 kg. Considering the thread and the block to be weightless, and neglecting the forces of friction, determine the acceleration with which the bodies will move if the body of mass m 2 is lowered.

task 15800

In the installation (Fig. 3), the angle α = 20° of the inclined plane with the horizon of the body mass m 1 = 350 g and m 2 = 0.2 kg. Considering the thread and the block to be weightless, and neglecting the forces of friction, determine the acceleration with which the bodies will move if the body of mass m 2 is lowered.

task 15802

In the installation (Fig. 3), the angle α = 60° of the inclined plane with the horizon of the body mass m 1 = 100 g and m 2 = 0.2 kg. Considering the thread and the block to be weightless, and neglecting the forces of friction, determine the acceleration with which the bodies will move if the body of mass m 2 is lowered.

task 17126

In the installation (Fig. 2.13), the angles α and β with the horizon are respectively equal to 45 ° and 30 ° of the mass of bodies m 1 = 0.5 kg and m 2 = 0.45 kg. Considering the thread and the block to be weightless and neglecting the forces of friction, determine: 1) the acceleration with which the bodies move; 2) the tension force of the thread.

task 17211

Bodies with masses m 1 = 5 kg and m 2 = 3 kg are connected by a weightless thread thrown over a block with mass m = 2 kg and radius r = 10 cm, lie on conjugate inclined planes with inclination angles β = 30°. A vertical force F equal to 15 acts on the body m 2

task 40125

Loads of the same mass (m 1 \u003d m 2 \u003d 0.5 kg) are connected by a thread and thrown over a weightless block mounted at the end of the table. The coefficient of friction of the load m 2 on the table µ = 0.15. Neglecting friction in the block, determine: a) the acceleration with which the loads move; b) the force of the thread tension.

task 40126

A weightless thread is thrown through a block in the form of a homogeneous disk with a mass of 80 g, to the ends of which weights with masses m 1 \u003d 100 g and m 2 \u003d 200 g are attached. Find the acceleration with which the weights will move? Ignore friction.

task 40482

Two different weights are attached to the ends of a weightless thread thrown over a block with a radius of 0.4 m with a moment of inertia of 0.2 kg·m 2 . The moment of friction forces during the rotation of the block is 4 Nm. Find the difference in the tension in the thread on both sides of the block rotating at a constant angular acceleration 2.5 rad/s 2 .

task 40499

At the top of two inclined planes forming angles α = 28° and β = 40° with the horizon, a block is fortified. Loads with the same masses are attached to the thread thrown over the block. Assuming that the thread and block are weightless and neglecting friction, determine the acceleration a of the weights.

task 40602

The free end of a thin and weightless thread is attached to the ceiling of an elevator going down with an acceleration a l, which is wound on a thin-walled hollow cylinder of mass m. Find the acceleration of the cylinder relative to the elevator and the tension in the thread. The thread is considered vertical.

task 40620

Loads of masses 19 kg and 10 kg were connected by a thread thrown over a weightless block fixed to the ceiling. Neglecting friction in the block, determine the tension of the thread.

task 40623

An inclined plane, on top of which a weightless block is fixed, forms an angle of 19 degrees with the horizon. Two weights equal mass 5 kg are attached to the ends of the thread thrown over the block. In this case, one of the weights moves along an inclined plane, while the other hangs vertically on a thread without touching the plane. Find the thread tension. Neglect friction in the block and friction on the plane.

Movement of a system of bodies

Dynamics: motions of a system of connected bodies.

Projection of the forces of several objects.

The action of Newton's second law on bodies that are fastened with a thread

If you, my friend, have forgotten how to project a powerhouse, I advise you to refresh it in your little head.

And for those who remember everything, let's go!

Problem 1. On a smooth table there are two bars connected by a weightless and inextensible thread with a mass of 200 g of the left and a mass of the right of 300 g. A force of 0.1 N is applied to the first, a force of 0.6 N is applied to the left in the opposite direction. cargo?

Movement occurs only on the X axis.

Because 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 tasks, if there are no conditions, the tension force at different bodies the same T₁ and T₂.

Let's express the acceleration:

Task 2. Two bars connected by an inextensible thread are on 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 of the bars on the plane are the same and equal to μ. The forces F₁ and F₂ are less than the gravity of the bars. The system moves to the left.

The system moves to the left, but the axis can be directed in any direction (it's only 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, but we love cons! Let's project the forces onto Ox (if this is difficult).

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

Let's add the equations and express the acceleration:

Let's 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 fixed block, to which three identical weights are suspended (two on one side and one on the other) with a mass of 5kg each. Find the acceleration of the system. What distance will the goods travel in the first 4 seconds of movement?

In this problem, we can imagine that 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 that starting speed is zero, we use the path formula for uniformly accelerated motion:

Problem 4. Two weights of masses 4 kg and 6 kg are connected by a light inextensible thread. Coefficients of friction between load and tableμ = 0.2. Determine the acceleration with which the loads will move.

Let's write down the movement of bodies on the axis, from Oy we will find N for the friction force (Ftr = μN):

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

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

Let's express the acceleration:

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

We direct the axis arbitrarily and assume that the right weight outweighs the left one and lifts it up the inclined plane.

From the equation for the Y axis, we express N for the friction force on the X axis (Ftr = μN):

We solve the system by taking the equation for the left body along the X axis and for the right body along the Y axis:

We express the acceleration so that only one unknown T remains, and find it:

The system will be in balance. This means that the sum of all forces acting on each of the bodies will be equal to zero:

We got a negative coefficient of friction, which means that you chose the movement of the system incorrectly (acceleration, friction force). You can check this by substituting the thread tension T in any equation and finding the acceleration. But it's okay, the values ​​remain the same in modulus, but opposite in direction.

Means, right direction forces should look like this, and the friction coefficient at which the system will be in equilibrium is 0.06.

Task 6. On two inclined planes, there is a load of mass 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.

In this problem, all equations on both axes for each body will already be required:

Find N in both cases, substitute them for friction and write together the equations for the X axis of both bodies:

Add up the equations, reduce by mass:

Let's express the acceleration:

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

And now we will overcome the last point and deal with the mass ratio. The sum of all forces acting on any of the bodies is equal to zero in order for the system to be in equilibrium:

Let's add the equations

Everything with the same mass will be transferred to one part, everything else to the other part of the equation:

We got that the ratio of the masses should be as follows:

However, if we assume that the system can move in a different direction, that is, the right weight will outweigh the left one, the direction of acceleration and friction forces will change. The equations will remain the same, but the signs will be different, and then the mass ratio will turn out like this:

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

Many may be under the impression that the mass ratio should be some specific value, and not an interval. This is true if there is no friction 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, no movement will begin.

In physics, tensile force is the force acting on a rope, cord, cable, or similar object or group of objects. Anything stretched, suspended, supported, or swayed by a rope, cord, cable, and so on, is subject to tension. Like all forces, tension can accelerate objects or cause them to deform. The ability to calculate the tension force is an important skill not only for students of the Faculty of Physics, but also for engineers and architects; those who build stable houses need to know if a certain rope or cable will withstand the pulling force from the object's weight so that it does not sag or collapse. Start reading the article to learn how to calculate the tension force in some physical systems.

Steps

Determination of the tension force on one thread

1. Determine the forces at each end of the string. The tensile force of a given thread, a rope, is the result of the forces pulling on the rope at each end. We remind you 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 in the rope itself. Don't forget about constant acceleration gravity - even if the system is at rest, its components are objects of gravity. We can assume that the tensile 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.
   • For example, let's consider a system in which a load is suspended from a wooden beam with a single rope (see image). Neither the load nor the rope is moving - the system is at rest. As a result, we know that for a load to be in equilibrium, the tension force must be equal to the force of gravity. In other words, Tension force (F t) = Gravity force (F g) = m × g.
     • 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 on a rope - any force applied to an object on a rope with acceleration does the same. 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 tensile force generated by the object's weight.

   • Let's assume that in our example a load of 10 kg is suspended from a rope, and instead of being attached to a wooden beam, it is pulled upwards 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 \u003d 98 + 10 kg × 1 m / s 2
     • F t = 108 newtons.

3. Consider angular acceleration. An object on a rope that revolves around a point considered to be the center (like a pendulum) exerts tension on the rope through centrifugal force. Centrifugal force is the extra tension that the rope causes by "pushing" it inward so that the load continues to move in an arc instead of a straight line. The faster an object moves, the greater the centrifugal force. The 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 moves.

   • Since the direction and magnitude of the centrifugal force change as the object moves and changes its speed, the full tension of the rope is always parallel to the rope at the center point. Remember that gravity is constantly acting on an object and pulling it down. So if the object is swinging vertically, the total tension the strongest at the lowest point of the arc (for a pendulum this is called the equilibrium point) when the object reaches top speed, and weakest at the top of the arc when 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 moving at a speed of 2 m/s as it passes through the bottom of the swing. If we need to calculate the tension force at the bottom point of the arc, when it is greatest, then first we need to find out whether the load is experiencing an equal pressure of gravity at this point, as well as at rest - 98 Newtons. To find the additional centrifugal force, we need to solve the following:
     • F c \u003d m × v 2 / r
     • F c = 10 × 2 2 /1.5
     • F c \u003d 10 × 2.67 \u003d 26.7 Newtons.
     • Thus, the total tension will be 98 + 26.7 = 124.7 newtons.

4. Note that the pulling force due to gravity changes as the load passes through the arc. As noted above, the direction and magnitude of the centrifugal force change as the object wobbles. In any case, although the force of gravity remains constant, net tensile force due to gravity also changes. When the swinging object is not at the bottom point of the arc (balance 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, and 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 point of balance and the center of rotation. As soon as the pendulum begins to swing, the force of gravity (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 (excluding the equilibrium point where all forces are the same).
   • Let's assume that when the pendulum is deflected 15 degrees from the vertical, it is moving at a speed of 1.5 m/s. We will find the tension force by the following steps:
     • The ratio of tension to gravity (T g) = 98cos(15) = 98(0.96) = 94.08 Newtons
     • Centrifugal force (F c) = 10 × 1.5 2 / 1.5 = 10 × 1.5 = 15 Newtons
     • Full tension = T g + F c = 94.08 + 15 = 109.08 Newtons.

5. Calculate friction. Any object that is pulled by the rope and experiences a "drag" force from the friction of another object (or fluid) imparts that force to the tension in the rope. The force of friction between two objects is calculated in the same way as in any other situation - according to the following equation: Force of friction (usually written as F r) = (mu)N, where mu is the coefficient of the 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, the friction that results from trying to set an object at rest in motion, is different from motion friction, the friction that results from trying to keep a moving object moving.

   • Let's assume that our 10 kg load is no longer swinging, it is now being towed on a horizontal plane with a rope. Let's assume that the coefficient of friction of the earth's movement is 0.5 and our load is moving with constant speed, but we need to give it an acceleration of 1m / s 2. This problem introduces two important changes - first, we no longer need to calculate tension in relation to gravity, since our rope is not holding the weight. Second, we will have to calculate the tension due to friction as well as due to the acceleration of the load mass. We need to decide the following:
     • Normal force (N) = 10 kg & × 9.8 (acceleration due to gravity) = 98 N
     • Motion friction force (F r) = 0.5 × 98 N = 49 Newtons
     • Acceleration force (F a) = 10 kg × 1 m/s 2 = 10 Newtons
     • Total tension = F r + F a = 49 + 10 = 59 newtons.

   Calculation of tension force on several threads

   1. Lift vertical parallel weights with a pulley. Blocks are simple mechanisms, consisting of a suspension disk, which allows you to change the direction of the rope tension force. In a simple pulley configuration, a rope or cable runs from a suspended weight up to the pulley, then down to another weight, thus 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 pulled by forces of different magnitudes. For a system of two masses suspended vertically in a block, the tension force is 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 "- the mass of the second object.

      • We note the following, physical tasks suggest that blocks are perfect- have no mass, no friction, they do not break, deform or separate from the rope that supports them.
      • Let's assume that we have two weights suspended vertically at parallel ends of a rope. One load 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 \u003d 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 start to accelerate, hence the weight of 10 kg will move down, causing the second weight to go up.

   2. Hang the weights using blocks with non-parallel vertical threads. Pulleys are often used to direct tension in a direction other than up or down. If, for example, the load is suspended vertically from one end of the rope, and the other end holds the load in a diagonal plane, then the non-parallel system of blocks takes the form of a triangle with corners at points with the first load, the second, and the block itself. In this case, the tension in the rope depends on both the force of gravity and 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) weight suspended vertically, connected to a 5 kg (m 2) weight placed on a 60 degree inclined plane (this slope is considered frictionless). To find tension in a rope, the easy way will first make equations for the forces accelerating the loads. Next, we act like this:
        • The suspended load is heavier, there is no friction, so we know that it is accelerating downward. The tension in the rope pulls upward so that it accelerates with respect to the net force F = m 1 (g) - T, or 10(9.8) - T = 98 - T.
        • We know that a load on an inclined plane accelerates upward. Since it has no friction, we know that tension pulls the load up on 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 he is accelerating with respect to the net 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 several threads 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 comes a third rope with a load. The pull on the third rope is obvious - a simple pull due to gravity or m(g). The tensions on the other two ropes are different and should add up to the force equal to strength gravity upwards in the vertical position and are zero in both horizontal directions, assuming that the system is at rest. The tension in the rope depends on the mass of the suspended loads and on the angle at which each of the ropes deviates from the ceiling.

      • Let's assume that in our Y-system the bottom weight has a mass of 10 kg and is suspended from two ropes, one of which is at a 30-degree angle with the ceiling, and the other is at a 60-degree angle. If we need to find the tension in each of the ropes, we need to calculate the horizontal and vertical components of the tension. To find T 1 (the tension in the rope with a 30 degree slope) and T 2 (the tension in the rope with a 60 degree slope), 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 \u003d 0.87 × m (g) \u003d 0.87 × 10 (9.8) \u003d 85.26 Newtons.
        • T 2 \u003d 0.5 × m (g) \u003d 0.5 × 10 (9.8) \u003d 49 newtons.