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In physics, a string tension is a force exerted by a string, cable or similar object on one or more other objects. Anything that is pulled, hung, powered, or swayed on a string generates tension. Like other forces, tension can change the speed of an object or deform it. String tension calculation is an important skill not only for students majoring in physics but also for engineers and architects who have to calculate to know if a string in use can withstand the tension of a string. impact object before letting go of support lever. Read step 1 to learn how to calculate tension in a multi-body system.

Steps

Method 1 of 2: Determine the tension force of a single wire

1. 

   
   
   Determine the tension at the ends of the string. The tension on a string is the result of being subjected to tension by both ends. Repeat the formula “force = mass × acceleration. Assuming the string is pulled very tight, any change in the object's weight or acceleration changes the tension. Don't forget the factor of acceleration caused by force - even if the system is at rest, everything in the system will still suffer from this force. We have the formula T = (m × g) + (m × a), where "g" is the acceleration due to the gravity of the objects in the system and "a" is the specific acceleration of the body.
   • In physics, to solve problems, we often hypothesize that the string is under "ideal conditions" - that is, the string in use is very strong, has no mass or negligible mass, and cannot elastic or break.
   • For example, consider a system of objects consisting of a weight hanging from a rope as shown in the picture. Both objects do not move because they are in a resting state. Position, we know that with the weight in equilibrium, the tension of the rope acting on it must be equal to the gravity. In other words, Force (F_t) = Gravity (F_g) = m × g.
     • Assuming a 10 k weight, the tension force is 10 kg × 9.8 m / s = 98 Newton.

2. 

   
   
   Now let's add the acceleration. While the force is not the only factor affecting the tension force, every other force related to the acceleration of the object that the string is holding has the same ability. For example, if we apply a force that changes the motion of a hanging object, the accelerating force of that object (mass × acceleration) will be added to the value of the tension force.
   • In our example: Let a 10 kg weight hang on the rope, but instead of previously fixed to the wooden beam we now pull the rope vertically with an acceleration of 1 m / s. In this case, the acceleration of the weight as well as the gravity must be included. The calculation is as follows:
     • F_t = F_g + m × a
     • F_t = 98 + 10 kg × 1 m / s
     • F_t = 108 Newtons.

3. 

   
   
   Calculate the acceleration of rotation. An object that is being rotated rotates at a fixed center through a string (such as a pendulum) produces tension based on the radial force. Radial force also plays an additional role in tension because it also "pulls" the object inward, but here instead of pulling in a straight direction, it pulls in an arc. The faster the object rotates, the greater the radial force. Radial force (F_c) is calculated using the formula m × v / r where "m" is the mass, "v" is the velocity, and "r" is the radius of the circle containing the arc of the object.
   • Since the direction and magnitude of the radial force change as the object moves, so does the total tension force, because this force pulls the object in a direction parallel to the string and toward the center. Also, remember that gravity always plays a role in the correct linear direction. In short, if an object is swinging in a straight direction, then the tension of the string will be maximizing at the lowest point of the arc (with the pendulum, we call it the equilibrium position), when we know that the object will move fastest there and brightest at the edges.
   • Still use the example of a weight and a rope, but instead of pulling, we swing the weight like a pendulum. Suppose the rope is 1.5 meters long and the weight moves at 2 m / s when it is in equilibrium. To calculate the tension in this case, we need to calculate the tension due to gravity as if it were not in motion as 98 Newtons, then calculate the additional radial force as follows:
     • F_c = m × v / r
     • F_c = 10 × 2/1.5
     • F_c = 10 × 2.67 = 26.7 Newtons.
     • So the total tension is 98 + 26.7 = 124.7 Newton.

4. 

   Understand that the tension in the string will be different at different positions of the object on the moving arc. As mentioned above, both the direction and magnitude of the radial force of an object change as the object moves. However, even though the gravity remains the same, the tension created by gravity will still change as usual! When the object is in equilibrium, the force of gravity will be vertical and so will the tension force, but when the object is in a different position, these two forces will create a certain angle together. Therefore, tension forces "neutralize" part of the gravity instead of fully fusing.
   • Dividing gravity into two vectors will help you see this definition better. At any point in the direction of an object's motion vertically, the string creates an angle "θ" with the path from the center to the equilibrium position of the object. When moving, gravity (m × g) will be divided into two vectors - mgsin (θ) asymptotic to the arc moving towards equilibrium position. And mgcos (θ) is parallel to the tension in the opposite direction. Thereby we see that tension must only be against mgcos (θ) - its reaction - and not all of gravity (Except when the object is in an equilibrium position, the forces are in the same direction and direction).
   • Now let through the shaker with the vertical angle of 15 degrees, moving at a speed of 1.5 m / s. So we calculate the tension as follows:
     • Tensile force created by gravity (T_g) = 98cos (15) = 98 (0.96) = 94.08 Newton
     • Radial force (F_c) = 10 × 1.5 / 1.5 = 10 × 1.5 = 15 Newtons
     • Total force = T_g + F_c = 94.08 + 15 = 109.08 Newton.

5. 

   Calculate the friction force. Any object being pulled creates a "drag" force by friction against the surface of another object (or liquid) and this force changes the tension force somewhat. The friction force of 2 objects in this case will also be calculated the way we usually do: Force that close (usually denoted as F_r) = (mu) N, where mu is the coefficient of friction where N is the force exerted by two objects, or the compressive force of one object on the other. Note that static friction is different from dynamic friction - static friction is the result of causing an object to move from resting to motion and that dynamic friction is produced while maintaining an object to continue its motion.
   • Suppose we have a 10 kg weight but now it is dragged across the floor horizontally. Let the coefficient of dynamic friction of the floor be 0.5 and the initial weight has a constant speed but now we are adding it with a 1 m / s acceleration. This new problem has two important changes - First, we no longer calculate the tension due to gravity, because now tension and gravity do not cancel each other. Second, we have to add friction and acceleration. The calculation looks like this:
     • Normal force (N) = 10 kg × 9.8 (acceleration of gravity) = 98 N
     • Dynamic friction force (F_r) = 0.5 × 98 N = 49 Newtons
     • Acceleration force (F_a) = 10 kg × 1 m / s = 10 Newtons
     • Total tension force = F_r + F_a = 49 + 10 = 59 Newton.
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Method 2 of 2: Determination of the tension force of a multi-string system

1. 

   Use pulleys to pull a package in a parallel direction. A pulley is a simple mechanical machine consisting of a circular disc that changes the direction of force. In a simple pulley system, the rope or cable runs up onto the pulley and then down again, forming a two-wire system. However, no matter how intense you are pulling a heavy object, the tension of the two "strings" is equal. In a system of 2 such weights and 2 such strings, the tension force is equal to 2g (m₁) (m₂) / (m₂+ m₁), where "g" is the acceleration of gravity, "m₁"is the mass of the object 1, and" m₂"is the mass of the object 2.
   • Note, normally in physics we will apply "ideal pulley" - no weight or insignificant mass, no friction, the pulley does not fail or fall off the machine. Such assumptions would be much easier to calculate.
   • For example we have 2 weights hanging vertically on 2 pulleys. Weight 1 weighs 10 kg, fruit 2 weighs 5 kg. The tension force is calculated as follows:
     • T = 2g (m₁) (m₂) / (m₂+ m₁)
     • T = 2 (9.8) (10) (5) / (5 + 10)
     • T = 19.6 (50) / (15)
     • T = 980/15
     • T = 65.33 Newtons.
   • Note, because there is one weight and one light, the system will move, the weight will move downwards and the light weight will be the opposite.
2. Use pulleys to pull a package in a non-parallel direction. Usually you use a pulley to adjust the direction of the object going up or down. But if, one weight is properly hanging at one end of the rope, the other is on an inclined plane, then we will have a non-parallel pulley system consisting of the pulley and two weights. Tensile force will now have an additional effect from gravity and drag on the inclined plane.
   • For a vertical weight of 10 kg (m₁) and a weight on an inclined plane weighing 5 kg (m₂), the inclined plane is created to the floor at an angle of 60 degrees (assuming the plane has negligible friction). To calculate the tension force, first find the calculation of the force of motion of the weights:
     • The straight-hanging weight is heavier, and since friction is not taken into account, the system will move downwards in the direction of the weight. The tension of the string now will pull it up, so the force of motion will have to subtract the tension: F = m₁(g) - T, or 10 (9.8) - T = 98 - T.
     • We know that the weight on the inclined plane will be pulled up. Since friction has been eliminated, tension pulls the weight up and only the weight of the weight pulls it down. The component pulling down the weight we set is sin (θ). So in this case, we calculate the force of pull of the weight as: F = T - m₂(g) sin (60) = T - 5 (9.8) (. 87) = T - 42.63.
     • The acceleration of two objects is equal, we have (98 - T) / m₁ = T - 42.63 / m₂. From there it is calculated T = 79.54 Newton.

3. 

   Where many wires hang the same object. Finally, consider a “Y”-shaped system of objects - two strings tied to the ceiling at the other end tied together and tied together with a third wire and one end of the third string hanging a weight. The tension of the third string is already right in front of us - it's simply gravity, T = mg. The tension force of the strings 1 and 2 is different and their total tension must be equal to the gravity in the vertical direction and zero if the horizontal, assuming the body is at rest. The tension of each wire is affected by the weight and the angle created by each rope to the ceiling.
   • Assume that our Y-shaped system is hanging through it weighs 10 kg, the angle made by 2 wires with the ceiling is 30 degrees and 60 degrees respectively. If we want to calculate the tension of each wire, we have to consider the horizontal and vertical tension of each component. Moreover, these two strings are perpendicular to each other, making it somewhat easier to calculate by applying the quantum system in the triangle:
     • Ratio T₁ or T₂ and T = m (g) is equal to the sine values ​​of the angles created by the wire corresponding to the ceiling. We get T₁, sin (30) = 0.5, and T₂, sin (60) = 0.87
     • Multiply the tension of the third wire (T = mg) by the sine value of each angle to find T₁ and T₂.
     • T₁ = .5 × m (g) = .5 × 10 (9.8) = 49 Newton.
     • T₂ = .87 × m (g) = .87 × 10 (9.8) = 85.26 Newton.
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