In physics, tension is the force exerted by a rope, wire, cable, and the like on one or more objects. Anything that is pulled, hung, supported or swung is subject to the force of tension. Like any other force, tension can cause an object to accelerate or deform it. Being able to calculate tension is important not only for physics students but also for engineers and architects who, in order to build safe buildings, need to know whether the tension on a given rope or cable can withstand the strain caused by weight. of the object before it yields and breaks. Read on to learn how to calculate voltage in different physical systems.
Steps
Method 1 of 2: Determine the Tension on a Single Rope
Step 1. Define the forces of both ends of the rope
The tension in a given rope is the result of the forces pulling on the rope from both ends. A little reminder: force = mass × acceleration. Assuming the string is well pulled, any change in acceleration or mass in the objects supported by the string will cause a change in the string tension. Don't forget the gravitational acceleration constant - even if a system is isolated, its components are subject to this force. Take a given string, its tension will be T = (m × g) + (m × a), where "g" is the gravitational constant of each object supported by the string and "a" corresponds to any other acceleration on any other object supported by the rope.
- For most physical problems, we assume ideal threads - in other words, our string is thin, without mass, and cannot be stretched or broken.
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As an example, let's consider a system in which a weight is attached to a wooden beam by a single rope (see figure). The weight and the rope are immobile - the whole system does not move. With these prerogatives we know that, in order for the weight to be kept in balance, the force of tension must be equivalent to the force of gravity exerted on the weight. In other words, Voltage (Ft) = Force of gravity (Fg) = m × g.
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Suppose we have a weight of 10kg, the tension force will be 10kg × 9.8m / s2 = 98 Newton.
Calculate Tension in Physics Step 2 Step 2. Calculate the acceleration
Gravity is not the only force that affects tension in a rope, because any force relative to the acceleration of an object to which the rope is attached affects its tension. For example, if a suspended object is accelerated by a force on the rope or cable, the acceleration force (mass × acceleration) adds to the tension caused by the weight of the object.
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Let's take into account that, taking the previous example of the weight of 10 kg suspended with a rope, the rope, instead of being fixed to a wooden beam, is used to pull the weight upwards with an acceleration of 1 m / s2. In this case, we must also calculate the acceleration on weight, as well as the force of gravity, with the following formulas:
- F.t = Fg + m × a
- F.t = 98 + 10 kg × 1 m / s2
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F.t = 108 Newton.
Calculate Tension in Physics Step 3 Step 3. Calculate the rotational acceleration
An object rotated around a central point by the use of a rope (such as a pendulum) exerts tension on the rope due to the centripetal force. Centripetal force is the additional tension force that the rope exerts by "pulling" inward to keep an object moving within its arc and not in a straight line. The faster an object moves, the greater the centripetal force. The centripetal force (Fc) is equivalent to m × v2/ r where by "m" is meant the mass, by "v" the speed, while "r" is the radius of the circumference in which the arc of movement of the object is inscribed.
- As the direction and magnitude of the centripetal force changes as the object on the rope moves and changes speed, so does the total tension on the rope, which always pulls parallel to the rope towards the center. Also remember that the force of gravity constantly affects the object, "calling" it downward. Therefore, if an object is rotated or made to oscillate vertically, the total voltage is greater in the lower part of the arc (in the case of the pendulum, we speak of the point of balance) when the object moves at a greater speed and less in the upper bow when moving slower.
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Let's go back to our example and assume that the object is no longer accelerating upwards but that it is swinging like a pendulum. Let's say the rope is 1.5 meters long and our weight moves at 2 m / s as it passes the lowest point of the swing. If we want to calculate the point of maximum stress exerted on the lower part of the arc, we should first recognize that the stress due to gravity at this point is equal to when the weight was immobile - 98 Newton. To find the centripetal force to add, we need to use these formulas:
- F.c = m × v2/ r
- F.c = 10 × 22/1, 5
- F.c = 10 × 2, 67 = 26.7 Newtons.
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So our total tension will be 98 + 26, 7 = 124, 7 Newton.
Calculate Tension in Physics Step 4 Step 4. Know that the tension due to gravity changes as an object's arc oscillates
As we said before, both the direction and the magnitude of the centripetal force change when an object oscillates. However, although the force of gravity remains constant, the tension from gravity also changes. When a swinging object is not at the bottom of its arc (its point of balance), gravity pulls the object directly downward, but the tension pulls upward at a certain angle. Therefore, tension only has the function of partially neutralizing the force of gravity, but not completely.
- Dividing the force of gravity into two vectors can be useful to better visualize the concept. At any given point in the arc of a vertically oscillating object, the rope forms an angle "θ" with the line passing through the point of balance and the center point of rotation. When the pendulum swings, the force of gravity (m × g) can be divided into two vectors - mgsin (θ) which is the tangent of the arc in the direction of the equilibrium point and mgcos (θ) which is parallel to the tension force in the opposite direction. Tension responds only to mgcos (θ) - the force opposing it - not to the entire force of gravity (except at the point of equilibrium, where they are equivalent).
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Let's say that when our pendulum makes an angle of 15 degrees with the vertical, it moves at 1.5 m / s. We will find tension with these formulas:
- Tension generated by gravity (T.g) = 98cos (15) = 98 (0, 96) = 94, 08 Newtons
- Centripetal force (Fc) = 10 × 1, 52/ 1, 5 = 10 × 1, 5 = 15 Newtons
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Total voltage = T.g + Fc = 94, 08 + 15 = 109, 08 Newton.
Calculate Tension in Physics Step 5 Step 5. Calculate the friction
Any object attached to a rope that experiences a "drag" force due to friction against another object (or fluid) transfers this force to the tension in the rope. The force given by the friction between two objects is calculated as in any other condition - with the following equation: frictional force (generally denoted by Fr) = (mu) N, where mu is the coefficient of friction between two objects and N is the normal force between the two objects, or the force they exert on each other. Know that static friction - the friction generated by setting a static object in motion - is different from dynamic friction - the friction generated by wanting to keep an object in motion that is already in motion.
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Let's say our 10kg weight has stopped swinging and is now dragged horizontally across the floor by our rope. Let's say the floor has a dynamic friction coefficient of 0.5 and our weight is moving at a constant speed that we want to accelerate to 1 m / s2. This new problem presents two important changes - first, we no longer have to calculate the tension caused by gravity because the rope is not supporting the weight against its force. Second, we must calculate the tension caused by friction and that given by the acceleration of the mass of the weight. We use the following formulas:
- Normal force (N) = 10 kg × 9.8 (acceleration due to gravity) = 98 N.
- Force given by dynamic friction (Fr) = 0.5 × 98 N = 49 Newtons
- Force given by acceleration (Fto) = 10 kg × 1 m / s2 = 10 Newton
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Total voltage = Fr + Fto = 49 + 10 = 59 Newton.
Method 2 of 2: Calculate the Tension on Multiple Ropes
Calculate Tension in Physics Step 6 Step 1. Lift parallel and vertical loads using a pulley
Pulleys are simple machines consisting of a suspended disc that allows the tension force in a rope to change direction. In a simply prepared pulley, the rope or cable goes from one weight to the other passing through the suspended disc, thus creating two ropes with different lengths. In any case, the tension in both parts of the string is equivalent, although forces of different magnitudes are exerted on each end. In a system of two masses hanging from a vertical pulley, the tensions are equal to 2g (m1) (m2) / (m2+ m1), where "g" means gravitational acceleration, "m1"the mass of the object 1 and for" m2"the mass of the object 2.
- Know that physics problems usually involve ideal pulleys - pulleys without mass, without friction and that cannot be broken or deformed and are inseparable from the ceiling or the wire that supports them.
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Let's say we have two weights hanging vertically from a pulley, on two parallel ropes. Weight 1 has a mass of 10 kg, while weight 2 has a mass of 5 kg. In this case we will find the tension with these formulas:
- 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 Newton.
- Know that since one weight is heavier than the other, and it is the only condition that varies in the two parts of the pulley, this system will start to accelerate, the 10 kg will move downwards and the 5 kg upwards.
Step 2. Lift loads using a pulley with non-parallel ropes
Pulleys are often used to direct tension in a direction other than "up" and "down". If, for example, a weight is suspended vertically from the end of a rope while the other end of the rope is attached to a second weight with a diagonal inclination, the non-parallel pulley system will have the shape of a triangle whose vertices they are the first weight, the second weight and the pulley. In this case, the tension in the rope is affected both by the force of gravity on the weight and by the components of the return force parallel to the diagonal section of the rope.
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Let's take a system with 10 kg of weight (m1) which hangs vertically, connected via a pulley to a weight of 5kg (m2) on a 60 degree ramp (assume the ramp is frictionless). To find the tension in the rope, it is easier to first proceed with the calculation of the forces that accelerate the weights. Here's how to do it:
- The suspended weight is heavier and we don't have to deal with friction, so we know it accelerates downward. However, the tension in the rope pulls upwards, thereby accelerating according to the net force F = m1(g) - T, or 10 (9, 8) - T = 98 - T.
- We know that the weight on the ramp will accelerate as it travels upwards. Since the ramp is frictionless, we know that the tension pulls up the ramp and only your own weight pulls down. The component element of the force that pulls down on the ramp is given by mgsin (θ), so in our case we can say that it accelerates up the ramp due to the net force F = T - m2(g) sin (60) = T - 5 (9, 8) (, 87) = T - 42, 14.
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If we make these two equations equivalent, we have 98 - T = T - 42, 14. Isolating T we will have 2T = 140, 14, that is T = 70.07 Newtons.
Calculate Tension in Physics Step 8 Step 3. Use multiple ropes to hold a suspended object
To conclude, consider an object suspended in a system of "Y" ropes - two ropes are attached to the ceiling, and meet at a central point from which a third rope starts at the end of which a weight is attached. The tension in the third rope is obvious - it is simply the tension caused by the force of gravity, or m (g). The tensions in the other two ropes are different and must be added to the equivalent of the force of gravity for the vertical upward direction and to an equivalent zero for both horizontal directions, assuming we are in an isolated system. The tension in the ropes is affected by both the mass of the suspended weight and the angle that each rope forms when it meets the ceiling.
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Suppose our Y system weighs 10 kg lower and the top two strings meet the ceiling forming two angles of 30 and 60 degrees, respectively. If we want to find the tension in each of the two strings, we will have to consider for each the vertical and horizontal elements of tension. To solve the problem for T1 (the tension in the rope at 30 degrees) and T.2 (the tension in the rope at 60 degrees), proceed as follows:
- According to the laws of trigonometry, the relationship between T = m (g) and T1 or T2equals the cosine of the angle between each chord and the ceiling. To T1, cos (30) = 0, 87, while for T2, cos (60) = 0.5
- Multiply the voltage in the lower chord (T = mg) by the cosine of each angle to find T1 and T2.
- T.1 =.87 × m (g) =.87 × 10 (9, 8) = 85, 26 Newton.
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T.2 =.5 × m (g) =.5 × 10 (9, 8) = 49 Newton.
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