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• Method 1 of 2: Determining the tension on a single cord
• Method 2 of 2: Calculating tension on multiple cords

In physics, tension is the force exerted by a rope, string, cable, or similar object on one or more other objects. Anything that is pulled, supported or swung on a rope or the like, is exposed to the force of tension. Like other forces, tension can accelerate objects or cause them to deform. Being able to calculate voltage is an important skill for physics students as well as engineers and architects. After all, in order to design safe buildings, they need to know exactly whether the voltage on a cable can withstand the load of an object. Read on at Step 1 to learn how to calculate stress in various physical systems.

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Method 1 of 2: Determining the tension on a single cord

1. Determine the forces on each side of the wire. The tension in a given thread of a rope is the sum of all the forces pulling the rope from both ends. Do not forget: force = mass × acceleration. Suppose the rope is stretched tight, any change in the acceleration or mass of the objects supporting the rope will cause a change in the tension in the rope. Don't forget the constant acceleration due to gravity - even when a system is at rest, every component is exposed to gravity. The tension in a particular rope can be expressed as T = (m × g) + (m × a), where "g" is the acceleration due to the force of gravity of any object supported by the rope, and "a" is any different gear on any object supported by the rope.
   • For the sake of simplicity, we can assume that we are dealing with a ideal thread In other words, that the rope, cable, etc. is thin and massless, and cannot stretch or break.
   • An example: suppose we have a system where a mass hangs from a wooden beam, attached with a single rope (see image). Still the masses still move the rope - the whole system is at rest. We now know that the mass is in equilibrium, where the tension force is equal to the force of gravity on the mass. In other words, Voltage (F._t) = Force or Gravity (F._g) = m × g.
     • Suppose we have a mass of 10 kg, then holds: tension = 10 kg × 9.8 m / s = 98 Newton.
2. Consider the acceleration. Gravity is not the only force that affects the tension in a rope - any force can be related to the acceleration of an object to which the rope is attached. If a hanging object is accelerated by a force on the rope or cable, then the force caused by the acceleration (mass × acceleration) is added to the tension caused by the mass of the object.
   • Suppose in our example, the mass of 10 kg hangs from a rope, which is not attached to a beam, but is used to lift the mass with an acceleration of 1 m / s. In these cases, we have to take into account not only the acceleration on the mass, but also the force of gravity, by solving this as follows:
     • F._t = F_g + m × a
     • F._t = 98 + 10 kg × 1 m / s
     • F._t = 108 Newton.
3. Also consider a circular gear. An object that is turned around a central point on a rope (such as a pendulum) exerts a tension on the rope caused by the centripetal force. Centripetal force is force that the rope exerts on an object by "pulling" it inward, so that the object continues to move in an arc, instead of going straight. The faster the object moves, the greater the centripetal force. Centripetal force (F._c) is equal to m × v / r where "m" is the mass, "v" is the speed, and "r" is the radius of the circle, which is the path in which the object is moving.
   • Since the direction and magnitude of the centripetal force changes as the object moves on the rope and the speed changes, so does the total tension in the rope, which always pulls parallel to the rope in the direction of the central point. Remember that gravity is constantly pulling on the object. So, if an object is thrown around in a vertical position, then the total tension is it greatest at the bottom of the object's trajectory (in the case of a pendulum this is also called the equilibrium), where the object moves fastest. The tension is least at the top of the circular motion, where the speed is lowest.
   • Suppose in the example that the object is swinging like a pendulum. The rope is 1.5 meters long and the mass moves at a speed of 2 m / s at the lowest point. If we want to calculate the stress at that point, the point at which the speed is highest, then we will first have to understand that the stress due to gravity at this point is the same as when the pendulum is at rest - 98 Newton. To find the centripetal force, we calculate as follows:
     • F._c = m × f / r
     • F._c = 10 × 2/1.5
     • F._c = 10 × 2.67 = 26.7 Newtons.
     • So, the total voltage is 98 + 26.7 = 124.7 Newtons.
4. Understand that gravity tension changes during the pendulum period. As mentioned earlier, both the direction and magnitude of the centripetal force change while an object is swinging. But while gravity remains constant, the tension due to gravity also change. Like a swinging object not at the bottom of the pendulum (the point of equilibrium), gravity pulls straight down, but the tension pulls the object at an angle. Because of this, the tension will cancel some of the gravity, but not completely.
   • By breaking gravity into two vectors, you may be able to better visualize this concept. At any point in the arc of the motion of a meandering object, the rope forms an angle of "θ" with the line through equilibrium and the central point of rotation. While the rope is swinging you can divide the gravity (m × g) into 2 vectors - mgsin (θ) is the tangent to the arc in the direction of equilibrium, and mgcos (θ), the parallel to the tension force in the opposite direction . The tension need only counteract mgcos (θ) - the force that opposes - not the full force of gravity (except at the point of equilibrium, when it is equal to the stress).
   • Suppose the pendulum forms an angle of 15 degrees with the vertical, and then has a speed of 1.5 m / s. We find the tension as follows:
     • Stress due to gravity (T._g) = 98cos (15) = 98 (0.96) = 94.08 Newtons
     • Centripetal force (F._c) = 10 × 1.5 / 1.5 = 10 × 1.5 = 15 Newtons
     • Total voltage = T_g + F_c = 94,08 + 15 = 109.08 Newtons.
5. Also think about the friction. Any object pulled by a rope and experiencing friction from another object (or liquid) transfers this frictional 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 - by the following equation: Force due to friction F_r = (mu) N, where mu is the coefficient of friction between the two objects and where N is the normal force between the two objects (the force with which they press against each other). Note that static friction - the friction that occurs when you want to move a stationary object - is different from kinetic friction - the friction that occurs when you try to keep a moving object moving.
   • Suppose the mass of 10 kg no longer swings, but is dragged along, horizontally over the ground and on a rope. Now we say that the ground has a kinetic coefficient of friction of 0.5 and that the mass moves at a constant speed, but we want to accelerate it at 1 m / s. This new exercise shows two important changes - first, we no longer need to calculate the tension due to gravity, because the rope no longer supports the mass and counteracts the force. We now have to take into account the frictional force and the resulting stress, as well as the stress caused by the acceleration of the object. We solve this as follows:
     • Normal force (N) = 10 kg × 9.8 (acceleration due to gravity) = 98 N
     • Force of kinetic friction (F._r) = 0.5 × 98 N = 49 Newtons
     • Force of the acceleration (F._a) = 10 kg × 1 m / s = 10 Newton
     • Total voltage = F._r + F_a = 49 + 10 = 59 Newton.

Method 2 of 2: Calculating tension on multiple cords

1. Lifting parallel vertical loads with a pulley. A pulley is a simple machine consisting of a suspended wheel that allows the direction of the force of tension in a rope to be changed. In a simple setup, the rope or cable runs from one hanging mass up through the pulley, then down to another mass, giving you two lengths of rope. But the tension in both parts of the rope is the same, even if there are masses of different sizes hanging from both ends of the rope. In a system of two masses suspended from a pulley, the tension is equal to 2g (m₁) (m₂) / (m₂+ m₁), where "g" is the acceleration of gravity, "m₁"the mass of object 1 and" m₂"the mass of object 2.
   • Note that we assume an "ideal pulley - no mass, no friction and pulleys that cannot break, deform or come off the ceiling.
   • Suppose we have two masses suspended from a pulley, on parallel ropes. Weight 1 has a mass of 10 kg and weight 2 has a mass of 5 kg. In this case we find the voltage 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 that because one mass is heavier than the other, the system will accelerate, moving 10 kg down and 5 kg up.
2. Lifting weights with a pulley on cords that are vertical but not parallel. Pulleys are often used to direct tension in a direction other than up or down. For example, if a mass hangs vertically from one end of the rope, while a second mass is attached to the other end on a slope, this non-parallel pulley system will take the form of a triangle with vertices being the first mass, the second mass and the pulley itself. In this case, the tension in the rope is determined by both the force of gravity on the mass and by the component of the tensile force acting parallel to the diagonal part of the rope.
   • Suppose we have a system with a mass of 10 kg (m₁), vertically connected, via a pulley, with a mass of 5 kg (m₂) on a slope of 60 degrees (we assume that the slope is frictionless). To find the tension in the rope, it is easier to be the first to establish equations for the forces that accelerate the masses. Please proceed as follows:
     • The hanging mass is heavier and we don't have to factor in friction, so we know there's a gear down. But the tension in the rope pulls the mass up, so we calculate the net force on the rope as follows: F = m₁(g) - T, or 10 (9.8) - T = 98 - T.
     • We know that the mass will accelerate up the slope. Because the slope is frictionless, we know that the stress pulls the mass up the slope, held back only by the weight's own mass. The force component that pulls the weight down is calculated by mgsin (θ), so in our case we can say that the weight accelerates up the slope by the net force F = T - m₂(g) sin (60) = T - 5 (9.8) (. 87) = T - 42.63.
     • The acceleration of the two masses is the same, so we have (98 - T) / m₁ = T - 42.63 / m₂. After some simple algebra we then get T = 61.09 Newtons.
3. Using multiple cords to hang an object. Finally, consider the case where an object hangs from a "Y-shaped" system of ropes - two ropes are attached to the ceiling, meeting at a central point, where a weight hangs from a third rope. The tension in the third rope is obvious - this is simply the resulting tension due to gravity. The tensions in the other two ropes are different and should add up to be equal to the force of gravity in an upward and vertical direction, and equal to zero in the horizontal direction (assume the system is at rest). The tension in the ropes is affected by both the mass of the hanging object and the angle of each rope with the ceiling.
   • Suppose that in this Y-shaped system, the object has a weight of 10 kg and the two top ropes are at an angle to the ceiling of 30 degrees and 60 degrees. If we want to find the tension in each of the top ropes, then we have to consider the vertical and horizontal components of the tension for each rope. The two ropes in this example are perpendicular to each other, making it easy to calculate these stresses, according to the definitions of the trigonometric functions. So as follows:
     • The ratio of T.₁ or T₂ and T = m (g) is equal to the sine of the angle between each supporting rope and the ceiling. For T₁ sin (30) = 0.5, while for T₂ it holds that sin (60) = 0.87.
     • Multiply the tension in the bottom rope (T = mg) by the sine of each corner to get T.₁ and T₂ to find.
     • T.₁ = 0.5 × m (g) = 0.5 × 10 (9.8) = 49 Newton.
     • T.₂ = 0.87 × m (g) = 0.87 × 10 (9.8) = 85.26 Newtons.