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Tension, commonly referred to as a force in physics, is a fundamental concept used to analyze and understand the behavior of objects under various circumstances. It is a force that acts to stretch or pull objects, and plays a crucial role in determining the equilibrium and stability of structures. In the field of physics, tension is a concept encountered in numerous scenarios, including mechanics, engineering, and even everyday situations. By understanding the principles behind tension and learning how to calculate it accurately, we can gain a deeper appreciation for the forces at play in our physical world. This article will provide a comprehensive guide on how to calculate tension in various contexts, offering step-by-step explanations and real-world examples to enhance our clarity and practical understanding. Whether you are a student seeking to grasp the fundamentals of tension or simply curious about unraveling the mysteries of physics, this guide will equip you with the necessary knowledge to confidently tackle even the most complex tension calculations.

This article was co-written by Bess Ruff, MA. Bess Ruff is a graduate student in geography at Florida. She received her Master’s degree in Environmental Science and Management from the Bren School of Environmental Science & Management, UC Santa Barbara in 2016. She has conducted survey work for marine spatial planning projects in the coastal area. Caribbean and support research as a contributor to the Sustainable Fisheries Group.

This article has been viewed 299,310 times.

In physics, string tension is a force exerted by a rope, cable, or similar object on one or more other objects. Anything that is pulled, suspended, powered, or swayed on a rope produces tension. Like other forces, string tension can change the speed of an object or deform it. Calculating rope tension is an important skill not only for physics students but also for engineers and architects, who must calculate whether the rope in use can withstand the tension of the rope. the object does not act before letting go of the support rod. Read step 1 to learn how to calculate string tension in a multibody system.

## Steps

### Determine the string tension of a single string

**Determine the tension at the ends of the rope.**The tension in the rope is the result of its two ends being pulled. Repeat the formula “force = mass × acceleration

**. Assuming the rope is stretched very taut, any change in the weight or acceleration of the object causes the tension in the string to change. Don’t forget about the acceleration caused by force – even if the system is at rest, everything in the system will still be subject to this force. We have the string tension formula T = (m × g) + (m × a), where “g” is the acceleration due to 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 in “ideal conditions” – that is, the rope in use is very strong, has no mass or negligible mass, and cannot flex or break.
- As an example, consider a system consisting of a weight hanging on a string as shown in the figure. Both objects do not move because they are at rest. Position, we know that with the weight in equilibrium, the tension on the string must be equal to gravity. In other words, Force (F
_{t}) = Gravity (F_{g}) = m × g.- Assuming a mass of 10 k, the string tension is 10 kg × 9.8 m/s
^{2}=**98 Newtons.**

- Assuming a mass of 10 k, the string tension is 10 kg × 9.8 m/s

**Now we add the acceleration.**While force is not the only factor affecting the tension in the string, any other force related to the acceleration of the object the rope is holding is equally likely. For example, if we apply a force that changes the motion of a suspended object, the force on the object’s acceleration (mass × acceleration) will be added to the value of the tension in the string.

- In our example: Let a 10 kg weight hang on a rope, but instead of before the rope was fixed to the wooden beam, now we pull the rope vertically with an acceleration of 1 m/s
^{2}. In this case, we have to include the acceleration of the mass as well as gravity. The calculation is as follows:- F
_{t}= F_{g}+ m × a - F
_{t}= 98 + 10 kg × 1 m/s^{2} -
_{Ft}=**108 Newtons.**

- F

**Add rotational acceleration.**A spinning object rotating a fixed center through a string (e.g. a pendulum) will produce tension based on the centripetal force. The centripetal force also acts in addition to the tension in the string 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 centripetal force. The centripetal force (F

_{c}) is calculated using the formula m × v

^{2}/r where “m” is the mass, “v” is the velocity and “r” is the radius of the circle containing the arc of motion of the body.

- Since the direction and magnitude of the centripetal force changes as the object moves, so does the total tension in the string, because this force pulls the object in a direction parallel to the rope and toward the center. Also, remember that gravity is always acting in the right direction. In short, if an object is swinging in a straight line, the tension in the string will be maximum at the lowest point of the arc of motion (with a pendulum, we call it the equilibrium position), when we know that The object will move the fastest there and the fastest at the two edges.
- Still taking the example of a weight and a string, but instead of pulling, we let the weight swing like a pendulum. Assume the rope is 1.5 meters long and the weight is moving with a velocity of 2m/s when it is in equilibrium. To calculate the tension in the string in this case, we need to calculate the tension in the string due to gravity as if it were not moving at 98 Newtons, then calculate the additional centripetal force as follows:
- F
_{c}= m × v^{2}/r - F
_{c}= 10 × 2^{2}/1.5 -
_{Fc}= 10 × 2.67 = 26.7 Newtons. - So the total tension in the string is 98 + 26.7 =
**124.7 Newtons.**

- F

**You should understand that the string tension will be different at different positions of the object on the arc of motion.**As noted above, both the direction and magnitude of the centripetal force of an object will change as the object moves. However, even though gravity remains constant, the string tension created by gravity will still change as usual! When the object is in an equilibrium position, gravity will act vertically and so will the tension in the string, but when the object is in a different position, these two forces will create a certain angle with each other. So the tension in the string “neutralizes” part of the gravity instead of integrating it all.

- Dividing gravity into two vectors will give you a better idea of this definition. At any point on the vertical arc of motion, the string makes an angle “θ” with the path from the center to the equilibrium position of the body. When moving, the gravitational force (m × g) will be divided into 2 vectors – mgsin(θ) asymptotically with the motion arc towards the equilibrium position. And mgcos(θ) is parallel to the tension in the string in the opposite direction. Thereby we see that the tension in the string only has to oppose mgcos(θ) – its reaction – and not the whole force of gravity (Except when the object is in equilibrium, those forces are in the same direction and direction).
- Now let the swing make an angle of 15 degrees with the vertical, moving with a speed of 1.5m/s. Then we calculate the tension as follows:
- The tension in the string due to gravity (T
_{g}) = 98cos(15) = 98(0.96) = 94.08 Newton - Radial force (F
_{c}) = 10 × 1.5^{2}/1.5 = 10 × 1.5 = 15 Newton - Total force = T
_{g}+ F_{c}= 94.08 + 15 =**109.08 Newtons.**

- The tension in the string due to gravity (T

**Calculate additional friction force.**Any object that is pulled produces a “drag” force by friction against the surface of another object (or liquid) and this force changes the tension in the string somewhat. The frictional force of the two objects in this case will also be calculated the way we usually do: The force that friction (usually denoted as F

_{r}) = (mu)N, where mu is the coefficient of friction where N is 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 bringing an object from rest to motion while dynamic friction is produced by keeping an object in its motion.

- Suppose we have a 10 kg ball but now it is dragged across the floor horizontally. Let the coefficient of dynamic friction of the floor be 0.5 and the weight initially has a constant velocity but now we add an acceleration of 1 m/s
^{2}. This new problem has two important changes – First, we no longer calculate the tension due to gravity, because now the string tension and gravity do not cancel each other. Second, we have to add friction and acceleration. The calculation will be as follows:- Normal force (N) = 10 kg × 9.8 (acceleration of gravity) = 98 N
- Dynamic Friction Force (F
_{r}) = 0.5 × 98 N = 49 Newton - Acceleration force (F
_{a}) = 10 kg × 1 m/s^{2}= 10 Newton - Total tension in the string = F
_{r}+ F_{a}= 49 + 10 =**59 Newtons.**

### Determine the string tension of a multi-string system

**Use a pulley 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 the force. In a simple pulley system, the rope or cable runs up the pulley and then down again, forming a two-string system. However, no matter how hard you are pulling the heavy object, the tension between the two “strings” is equal. In such a system of 2 weights and 2 strings, the tension in the string is 2g(m

_{1})(m

_{2})/(m

_{2}+ m

_{1}), where “g” is the acceleration due to gravity, “m

_{1 )}” is the mass of object 1, and “m

_{2}” is the mass of object 2.

- Note, usually in physics we will apply the “ideal pulley” – massless or negligible mass, no friction, the pulley does not fail or fall off the machine. Such an assumption would be much easier to calculate.
- For example, we have 2 weights hanging vertically on 2 pulleys. Fruit 1 weighs 10 kg, fruit 2 weighs 5 kg. The tension in the string is calculated as follows:
- 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.**

- T = 2g(m
- Note, because there is a heavy and a light body, the system will move, the heavy body will move downwards and the light body will do the opposite.

**Use a pulley to pull a package in a non-parallel direction. Usually we use a pulley to adjust the direction of the object going up or down.**But if, one weight is hanging straight at one end of the rope, the other is on an inclined plane, then we will now have a non-parallel pulley system consisting of a pulley and two weights. The tension in the rope will now have an additional effect from gravity and the pulling force on the inclined plane.

- Given a vertically hanging weight 10 kg (m
_{1}) and a weight on an inclined plane weighing 5 kg ( m_{2}), the inclined plane makes an angle of 60 degrees with the floor (assuming the plane has negligible friction) . To calculate the tension in the string, first find the calculation of the moving forces of the weights:- The weight hanging straight up is heavier, and since we don’t take into account friction, the system will move downwards in the direction of the weight. The tension in the string will now act to pull it up, so the moving force will have to subtract the tension in the string: F = m
_{1}(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, the tension in the string pulls the weight up and only the weight of the weight pulls it back down. The component that pulls the weight down is set as sin(θ). So in this case, we can calculate the force pulling the weight as: F = T – m
_{2}(g)sin(60) = T – 5(9.8)(.87) = T – 42.63. - The accelerations of the two objects are equal, we have (98 – T)/m
_{1}= T – 42.63 /m_{2}. From there, calculate**T = 79.54 Newton**.

- The weight hanging straight up is heavier, and since we don’t take into account friction, the system will move downwards in the direction of the weight. The tension in the string will now act to pull it up, so the moving force will have to subtract the tension in the string: F = m

**In the case of many ropes hanging the same object.**Finally, consider a “Y”-shaped system – 2 ropes tied to the ceiling at the other end and tied together with a 3rd rope and a 3rd string hanging from the other end. The tension on the 3rd rope is already right in front of us – Simply gravity, T = mg. The tension in ropes 1 and 2 is different and their total tension must be equal to gravity in the vertical direction and zero in the horizontal direction, assuming the system is at rest. The tension in each string is exerted by the mass of the weight and the angle made by each string to the ceiling.

- Assuming that our Y-system is hanging over a weight of 10 kg, the angle made by the 2 ropes to the ceiling is 30 degrees and 60 degrees respectively. If we want to calculate the tension in each string, we must consider what is 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 trigonometric equation:
- The ratio T
_{1}or T_{2}and T = m(g) is equal to the sine value of the angles formed by the rope corresponding to the ceiling. We get T_{1}, sin(30) = 0.5, and T_{2}, sin(60) = 0.87 - Multiply the tension on the 3rd string (T = mg) by the sine of each angle to find T
_{1}and T_{2}. - T
_{1}= .5 × m(g) = .5 × 10(9.8) =**49 Newtons.** - T
_{2}= .87 × m(g) = .87 × 10(9.8) =**85.26 Newtons.**

- The ratio T

This article was co-written by Bess Ruff, MA. Bess Ruff is a graduate student in geography at Florida. She received her Master’s degree in Environmental Science and Management from the Bren School of Environmental Science & Management, UC Santa Barbara in 2016. She has conducted survey work for marine spatial planning projects in the coastal area. Caribbean and support research as a contributor to the Sustainable Fisheries Group.

This article has been viewed 299,310 times.

In physics, string tension is a force exerted by a rope, cable, or similar object on one or more other objects. Anything that is pulled, suspended, powered, or swayed on a rope produces tension. Like other forces, string tension can change the speed of an object or deform it. Calculating rope tension is an important skill not only for physics students but also for engineers and architects, who must calculate whether the rope in use can withstand the tension of the rope. the object does not act before letting go of the support rod. Read step 1 to learn how to calculate string tension in a multibody system.

In conclusion, calculating tension in physics is an essential skill that allows us to better understand and analyze various systems and objects in motion. By understanding the concept of tension, we can solve a wide range of problems and predict the behavior of objects in different scenarios. The tension is influenced by factors such as mass, gravity, and acceleration, which should be taken into account when applying the relevant equations and formulas. Through practice and familiarity with the fundamental principles of tension, we can enhance our problem-solving abilities and gain a deeper insight into the dynamics of the physical world. Overall, mastering the calculation of tension empowers us to unravel the complexities of physics and expand our knowledge of how objects interact with their surroundings.

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