How to calculate tensile force in physics

Content

Steps
Method 1 of 2: Determining the tensile force on a single strand
Method 2 of 2: Calculating the tensile force on multiple strands

In physics, a pulling force is a force acting on a rope, cord, cable, or a similar object or group of objects. Anything that is pulled, suspended, supported, or swayed by a rope, cord, cable, and so on, is subject to a pulling force. Like all forces, tension can accelerate objects or cause them to deform.The ability to calculate the tensile force is an important skill not only for physics students, but also for engineers, architects; Those who build stable houses need to know if a particular rope or cable will withstand the pulling force of the object's weight so that it does not sag or collapse. Start reading this article to learn how to calculate the tensile force in some physical systems.

Steps

Method 1 of 2: Determining the tensile force on a single strand

1 Determine the forces at each end of the thread. The pulling force of a given thread, rope, is the result of the forces pulling the rope at each end. We remind you force = mass × acceleration... Assuming the rope is taut, any change in the acceleration or mass of an object suspended from the rope will change the tension in the rope itself. Do not forget about the constant acceleration of gravity - even if the system is at rest, its components are objects of the action of gravity. We can assume that the pulling force of a given rope is T = (m × g) + (m × a), where “g” is the acceleration of gravity of any of the objects supported by the rope, and “a” is any other acceleration, acting on objects.
- To solve many physical problems, we assume perfect rope - in other words, our rope is thin, has no mass and cannot stretch or break.
- As an example, let's consider a system in which a load is suspended from a wooden beam using a single rope (see image). Neither the load itself nor the rope moves - the system is at rest. As a consequence, we know that the load is in balance, the tensile force must be equal to the force of gravity. In other words, Pulling force (F_t) = Gravity (F_g) = m × g.
  - Suppose the load has a mass of 10 kg, therefore the tensile force is 10 kg × 9.8 m / s = 98 Newtons.
2 Consider acceleration. Gravity is not the only force that can affect the pulling force of a rope - any force applied to an object on the rope with acceleration produces the same effect. If, for example, an object suspended on a rope or cable is accelerated by a force, then the acceleration force (mass × acceleration) is added to the tensile force generated by the weight of that object.
- Suppose in our example a 10 kg load is suspended on a rope, and instead of being attached to a wooden beam, it is pulled upward with an acceleration of 1 m / s. In this case, we need to account for the acceleration of the load, as well as the acceleration of gravity, as follows:
  - F_t = F_g + m × a
  - F_t = 98 + 10 kg × 1 m / s
  - F_t = 108 Newtons.
3 Consider angular acceleration. An object on a rope revolving around a point considered to be the center (like a pendulum) exerts tension on the rope through centrifugal force. Centrifugal force is the additional pulling force that the rope creates by “pushing” it inward so that the load continues to move in an arc rather than in a straight line. The faster the object moves, the greater the centrifugal force. Centrifugal 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 along which the load moves.
- Since the direction and value of the centrifugal force changes depending on how the object moves and changes its speed, the total tension on the rope is always parallel to the rope at the center point. Remember that gravity constantly acts on the object and pulls it down. So if the object is swinging vertically, full tension the strongest at the lowest point of the arc (for a pendulum this is called the equilibrium point), when the object reaches its maximum speed, and the weakest at the top of the arc as the object slows down.
- Let's assume that in our example, the object is no longer accelerating upward, but swinging like a pendulum. Let our rope be 1.5 m long, and our load moves at a speed of 2 m / s, when passing through the lowest point of swing.If we need to calculate the tension force at the lowest point of the arc, when it is greatest, then first we need to find out whether the load is experiencing equal gravity pressure at this point, as in the state of rest - 98 Newtons. To find additional centrifugal force, we need to solve the following:
  - F_c = m × v / r
  - F_c = 10 × 2/1.5
  - F_c = 10 × 2.67 = 26.7 Newtons.
  - Thus, the total tension will be 98 + 26.7 = 124.7 Newtons.
4 Note that the pulling force due to gravity changes as the load travels through the arc. As noted above, the direction and magnitude of the centrifugal force changes as the object sways. In any case, although the force of gravity remains constant, net tensile force due to gravity changes too. When the swinging object is not at the lowest point of the arc (equilibrium point), gravity pulls it down, but the pulling force pulls it up at an angle. For this reason, the pulling force must resist part of the force of gravity, and not its entirety.
- Dividing the force of gravity into two vectors can help you visualize this state. At any point in the arc of a vertically swinging object, the rope makes an angle "θ" with a line through the equilibrium point and the center of rotation. As soon as the pendulum begins to swing, the gravitational force (m × g) is divided into 2 vectors - mgsin (θ), acting tangentially to the arc in the direction of the equilibrium point and mgcos (θ), acting parallel to the tension force, but in the opposite direction. The tension can only resist mgcos (θ) - the force directed against it - not all the gravitational force (except for the equilibrium point, where all the forces are the same).
- Let's assume that when the pendulum is tilted 15 degrees from the vertical, it moves at a speed of 1.5 m / s. We will find the tensile force by the following actions:
  - The ratio of the pulling force to the gravitational force (T_g) = 98cos (15) = 98 (0.96) = 94.08 Newtons
  - Centrifugal force (F_c) = 10 × 1.5 / 1.5 = 10 × 1.5 = 15 Newtons
  - Full tension = T_g + F_c = 94,08 + 15 = 109.08 Newtons.
5 Calculate the friction. Any object that is pulled by the rope and experiences a "braking" force from the friction of another object (or fluid) transfers this effect to the tension in the rope. The friction force between two objects is calculated in the same way as in any other situation - using the following equation: Friction force (usually written as F_r) = (mu) N, where mu is the coefficient of the friction force between objects and N is the usual force of interaction between objects, or the force with which they press against each other. Note that friction at rest - friction that occurs as a result of trying to bring an object at rest into motion - is different from friction of motion - friction that results from trying to force a moving object to keep moving.
- Let's assume that our 10kg load no longer sways, now it is being towed horizontally with a rope. Suppose that the coefficient of friction of the movement of the earth is 0.5 and our load is moving at a constant speed, but we need to give it an acceleration of 1m / s. This problem introduces two important changes - first, we no longer need to calculate the pulling force in relation to gravity, since our rope does not support the weight. Second, we will have to calculate the tension due to friction as well as due to the acceleration of the mass of the load. We need to decide the following:
  - Ordinary Force (N) = 10kg & × 9.8 (Acceleration by Gravity) = 98 N
  - Frictional force of movement (F_r) = 0.5 × 98 N = 49 Newtons
  - Acceleration force (F_a) = 10 kg × 1 m / s = 10 Newtons
  - Total tension = F_r + F_a = 49 + 10 = 59 Newtons.

Method 2 of 2: Calculating the tensile force on multiple strands

1 Lift vertical parallel weights with a pulley. Blocks are simple mechanisms consisting of a suspended disc that allows the direction of the rope's pulling force to be reversed. In a simple block configuration, the rope or cable runs from the suspended load up to the block, then down to another load, thus creating two sections of rope or cable. In any case, the tension in each of the sections will be the same, even if both ends are pulled by forces of different magnitudes. For a system of two masses suspended vertically in a block, the tensile force is 2g (m₁) (m₂) / (m₂+ m₁), where "g" is the acceleration of gravity, "m₁"Is the mass of the first object," m₂»Is the mass of the second object.
- Note the following, physical problems assume that blocks are perfect - do not have mass, friction, they do not break, do not deform and do not separate from the rope that supports them.
- Let's suppose that we have two weights suspended vertically at the parallel ends of the rope. One load has a mass of 10 kg, and the other has a mass of 5 kg. In this case, we need to calculate the following:
  - 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, since one weight is heavier, all other elements are equal, this system will begin to accelerate, therefore, a 10 kg weight will move downward, forcing the second weight to go up.
2 Suspend weights using blocks with non-parallel vertical strands. Blocks are often used to direct pulling force in a direction other than up or down. If, for example, a load is suspended vertically from one end of the rope, and the other end holds the load in a diagonal plane, then the non-parallel system of blocks takes the form of a triangle with angles at points with the first load, the second and the block itself. In this case, the tension in the rope depends both on the force of gravity and on the component of the pulling force, which is parallel to the diagonal part of the rope.
- Let's suppose that we have a system with a load of 10 kg (m₁), suspended vertically, connected to a load of 5 kg (m₂) located on an inclined plane of 60 degrees (it is believed that this slope does not give friction). To find the tension in the rope, the easiest way is to first write equations for the forces that accelerate the weights. Next, we act like this:
  - The suspended load is heavier, there is no friction, so we know that it is accelerating downward. The tension in the rope pulls upward so that it accelerates with respect to the resultant force F = m₁(g) - T, or 10 (9.8) - T = 98 - T.
  - We know that the load on an inclined plane accelerates upward. Since it has no friction, we know that tension pulls the load up the plane, and pulls it down only your own weight. The component of the force pulling down the inclined one is calculated as mgsin (θ), so in our case we can conclude that it is accelerating with respect to the resultant force F = T - m₂(g) sin (60) = T - 5 (9.8) (0.87) = T - 42.14.
  - If we equate these two equations, we get 98 - T = T - 42.14. Find T and get 2T = 140.14, or T = 70.07 Newtons.
3 Use multiple strands to hang the object. To conclude, let's imagine the object is suspended from a "Y-shaped" rope system - two ropes are fixed to the ceiling and meet at the center point from which the third rope with a load comes. The pulling force of the third rope is obvious - a simple pull due to gravity or m (g). The tensions on the other two ropes are different and should add up to a force equal to the upward gravity in the vertical position and zero in both horizontal directions, assuming the system is at rest. The tension in the rope depends on the weight of the suspended loads and on the angle by which each rope is deflected from the ceiling.
- Let's assume that in our Y-system, the bottom weight has a mass of 10 kg and is suspended by two ropes, one of which is 30 degrees from the ceiling and the other 60 degrees. If we need to find the tension in each of the ropes, we need to calculate the horizontal and vertical components of the tension. To find T₁ (tension in the rope, the slope of which is 30 degrees) and T₂ (tension in that rope, the slope of which is 60 degrees), you need to decide:
  - According to the laws of trigonometry, the relationship between T = m (g) and T₁ and T₂ equal to the cosine of the angle between each of the ropes and the ceiling. For T₁, cos (30) = 0.87, as for T₂, cos (60) = 0.5
  - Multiply the tension in the bottom rope (T = mg) by the cosine of each angle to find T₁ and T₂.
  - T₁ = 0.87 × m (g) = 0.87 × 10 (9.8) = 85.26 Newtons.
  - T₂ = 0.5 × m (g) = 0.5 × 10 (9.8) = 49 Newtons.