In physics, tension is the force exerted by a string, thread, cable, or other similar object on one or more objects. Any object that is pulled, hung, held, or swung by a rope, thread, etc. is subjected to a tension force. As with all forces, tension can accelerate an object or cause it to deform. The ability to calculate stresses is important not only for students studying physics, but also for engineers and architects. To build a safe building they must be able to determine whether the tension in a particular rope or cable can withstand the strain caused by the weight of an object before it stretches and breaks. See Step 1 to learn how to calculate stresses in some physical systems.
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Method 1 of 2: Determining the Tension at One End of the Rope
Step 1. Determine the tension at the end of the rope
The tension in a string is a reaction to the pulling force at each end of the string. As a reminder, force = mass × acceleration. Assuming the rope is pulled until it is tense, any change in the acceleration or mass of the object being held up by the rope will cause a change in the tension in the rope. Don't forget the constant acceleration due to gravityeven if a system is at rest; its components are subject to the force of gravity. The tension in the rope can be calculated by T = (m × g) + (m × a); "g" is the acceleration due to gravity on the object held by the rope and "a" is the other acceleration on the object held by the rope.
 In almost all problems of physics, we assume an ideal rope  in other words, a rope or cable, or something else, we think of as thin, massless, unstretched or damaged.
 For example, imagine a system; a weight is suspended from a wooden cross by a rope (see picture). Neither the object nor the string is movingthe whole system is at rest. Therefore, we can say that the load is in equilibrium so that the tension force must be equal to the gravitational force on the object. In other words, Voltage (F_{t}) = gravitational force (F_{g}) = m × g.

Assume a mass of 10 kg, then the tension in the string is 10 kg × 9.8 m/s^{2} = 98 Newtons.

Step 2. Calculate acceleration
Gravity isn't the only force that can affect the tension in a stringso any force that accelerates an object the string is holding on to can affect it. If, for example, an object hanging on a string is accelerated by a force on the rope or cable, the accelerating force (mass × acceleration) is added to the stress caused by the object's weight.
 For example, in our example an object with a mass of 10 kg is hanging by a rope instead of hanging from a wooden bar. The rope is pulled with an upward acceleration of 1 m/s.^{2}. In this case, we must take into account the acceleration experienced by the object other than the force of gravity with the following calculation:
 F_{t} = F_{g} + m × a
 F_{t} = 98 + 10 kg × 1 m/s^{2}

F_{t} = 108 Newtons.
Step 3. Calculate the angular acceleration
An object moving around a central point through a string (such as a pendulum) exerts tension on the string due to the centripetal force. The centripetal force is the additional tension in the string caused by the "pull" inward to keep the object moving in a circle instead of moving in a straight line. The faster the object is moving, the greater the centripetal force. Centripetal force (F_{c}) is equal to m × v^{2}/r; "m" is mass, "v" is velocity, and "r" is radius of circular motion of the object.
 Since the direction and magnitude of the centripetal force change as the suspended object moves and changes its speed, so does the total tension in the string, which is always parallel to the string pulling the object toward the center of rotation. Remember that the force of gravity always acts on objects downward. Thus, when the object rotates or swings vertically, the total stress is greatest at the lowest point of the arc (at the pendulum this point is called the equilibrium point) when the object is moving fastest and is lowest at the highest point of the arc when the object is moving the most. slow.
 In our example, the object does not continue to accelerate upward but swings like a pendulum. Suppose the length of the rope is 1.5 m in length and the object is moving with a speed of 2 m/s as it passes through the lowest point of the swing. If we want to calculate the stress at the lowest point of swing, i.e. the greatest stress, we must first know that the stress due to gravity at this point is the same as when the object is stationary98 Newtons. To find the additional centripetal force, we can calculate it as follows:
 F_{c} = m × v^{2}/r
 F_{c} = 10 × 2^{2}/1, 5
 F_{c} =10 × 2.67 = 26.7 Newtons.

So, the total stress is 98 + 26, 7 = 124, 7 Newtons.
Step 4. Understand that the stress due to gravity changes along the arc of the swing
As mentioned above, both the direction and magnitude of the centripetal force change as the object swings. However, although the gravitational force remains constant, the stress due to gravity also changes. When a swinging object is not at its lowest point of swing (its equilibrium point), gravity pulls it down, but tension pulls it up at an angle. Therefore, stress only reacts to a part of the force caused by gravity, not to all of it.
 Break the force of gravity into two vectors to help you visualize this concept. At any point in the motion of a vertically swinging object, the string makes an angle "θ" with the line passing through the equilibrium point and the center of the circular motion. As the pendulum swings, the gravitational force (m × g) can be split into two vectorsmgsin(θ) whose direction is tangent to the arc of the swinging motion and mgcos(θ) which is parallel and opposite to the tension force. The stress only needs to be against mgcos(θ)the force pulling itnot the entire gravitational force (except at the equilibrium point; they are the same value).
 For example, when a pendulum makes an angle of 15 degrees with the vertical axis, it moves with a speed of 1.5 m/s. The voltage can be calculated as follows:
 Stress due to gravity (T_{g}) = 98cos(15) = 98(0, 96) = 94, 08 Newton
 Centripetal force (F_{c}) = 10 × 1, 5^{2}/1, 5 = 10 × 1.5 = 15 Newtons

Total stress = T_{g} + F_{c} = 94, 08 + 15 = 109, 08 Newtons.
Step 5. Calculate friction
Each object is pulled by a rope which experiences a "resistance" force of friction against another object (or fluid) transferring this force to the tension in the string. The frictional force between two objects can be calculated as in any other casefollowing the following equation: The frictional force (usually written as F_{r}) = (mu)N; mu is the coefficient of friction between two objects and N is the normal force between the two objects, or the force that the two objects press against each other. Remember that static friction (that is, the friction that occurs when a stationary object moves) is different from kinetic friction (the friction that occurs when an object in motion keeps moving).
 For example, the original object with a mass of 10 kg is no longer hanging, but is pulled horizontally on the ground by a rope. For example, soil has a coefficient of kinetic friction of 0.5 and an object is moving at a constant speed, then accelerates by 1 m/s^{2}. This new problem presents two changesfirst, we don't need to calculate the stress due to gravity because the rope doesn't support the weight of the object. Second, we must take into account the stresses due to friction, in addition to those caused by the acceleration of a massed body. This problem can be solved as follows:
 Normal force (N) = 10 kg × 9.8 (acceleration of gravity) = 98 N
 The force from kinetic friction (F_{r}) = 0.5 × 98 N = 49 Newton
 Force from acceleration (F_{a}) = 10 kg × 1 m/s^{2} = 10 Newtons

Total stress = F_{r} + F_{a} = 49 + 10 = 59 Newtons.
Method 2 of 2: Calculating Tension in More Than One Rope
Step 1. Lift the vertical weight using a pulley
A pulley is a simple machine consisting of a suspended disk that allows a change in the direction of the tension force on a string. In a simple pulley configuration, a rope tied to an object is raised on a hanging pulley, then lowered back down so that it divides the rope into two hanging halves. However, the tension in the two ropes is the same, even when the two ends of the rope are pulled with different forces. For a system with two masses hanging on a vertical pulley, the stress is equal to 2g(m_{1})(m_{2})/(m_{2}+m_{1}); "g" is the acceleration due to gravity, "m_{1}" is the mass of object 1, and "m_{2}" is the mass of the object 2.
 Remember that physics problems assume an ideal pulley  a pulley that has no mass, has no friction, can't break, deform, or detach from hangers, ropes, or whatever holds it in place.
 Suppose we have two objects hanging vertically on a pulley with parallel strings. Object 1 has a mass of 10 kg, while object 2 has a mass of 5 kg. In this case, the voltage can be 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.
 Note that one object is heavier than the other, other things being equal, the system will accelerate, with a 10 kg object moving down and a 5 kg object moving up.
Step 2. Lift the weight using a pulley with the vertical ropes misaligned
Pulleys are often used to direct tension in a direction other than up or down. For example, a weight hangs vertically from one end of a rope while at the other end a second object hangs on an inclined slope; This nonparallel pulley system is in the form of a triangle whose points are the first object, the second object, and the pulley. In this case, the tension in the rope is affected by both the gravitational force on the object and the component of the pulling force on the rope parallel to the slope.
 For example, this system has a mass of 10 kg (m_{1}) hanging vertically is connected via a pulley to a second object of mass 5 kg (m_{2}) on an inclined slope of 60 degrees (assume the slope has no friction). To calculate the tension in a string, the easiest way is to find the equation for the object that causes the acceleration first. The process is as follows:
 The suspended object is heavier and has no friction, so we can calculate its acceleration downwards. The tension in the string pulls it upward so that it will have a resultant force F = m_{1}(g)  T, or 10(9, 8)  T = 98  T.
 We know that an object on a slope will accelerate up the slope. Since the slope has no friction, we know that the tension in the rope is pulling it up and only the weight itself is pulling it down. The component of the force pulling it down the slope is sin(θ); so in this case, the object will accelerate up the slope with the resultant force F = T  m_{2}(g)sin(60) = T  5(9, 8)(0, 87) = T  42, 63.
 The acceleration of these two objects is the same so that (98  T)/m_{1} = (T  42, 63) /m_{2}. By solving this equation, we will get T = 60, 96 Newtons.
Step 3. Use more than one string to hang objects
Finally, we'll look at an object hanging from the ceiling with a "Yshaped" rope system, at the knot point hanging a third rope holding the object. The tension in the third rope is quite obviousonly experiencing tension from the force of gravity, or m(g). The tensions in the other two ropes are different and when added together in the vertical direction must be equal to the gravitational force and equal to zero when added up in the horizontal direction, if the system is not moving. The tension in the rope is affected both by the weight of the hanging object and by the angle between the rope and the ceiling.
 For example, the Yshaped system is loaded with a mass of 10 kg on two ropes hanging from the ceiling at an angle of 30 degrees and 60 degrees. If we want to find the tension in the two upper ropes, we need to take into account the components of the tension in the vertical and horizontal directions, respectively. However, in this example, the two hanging strings form right angles, making it easier for us to calculate according to the definition of trigonometric functions as follows:
 Comparison between T_{1} or T_{2} and T = m(g) is equal to the sine of the angle between the two ropes holding the object and the ceiling. For T_{1}, sin(30) = 0, 5, while for T_{2}, sin(60) = 0.87
 Multiply the tension in the bottom string (T = mg) by the sine for each angle to calculate T_{1} and T_{2}.

T_{1} = 0.5 × m(g) = 0.5 × 10(9, 8) = 49 Newtons.

T_{2} = 0.87 × m(g) = 0.87 × 10(9, 8) = 85, 26 Newtons.