The most important application of Newton's motion Laws is the explanation of tension. Ropes and strings are very useful in machines and mechanical systems. These are used to push or pull heavy loads. We can see the tension in using of ropes or cables. If we consider a load which is pulled by a rope, a person is exerting a force at one end of the rope who is not directly in the contact of the block. Thus, the force which is felt by block through the use of rope is called tension force.

In case of mass-less rope in classical mechanics, the force is transmitted from one side to the other side. These mass-less ropes feel two equal tension forces but in the opposite direction. We can take an example of string, force is applied in string musical instruments. The string tension is static force which is acted on these instruments. Here we discuss more about this tension force and its mathematical formulation in detail.

Just the force that exist either within or applied by a string or wire. Every piece of the rope feels a Tension force in both directions, except the end points, which feel a tension on one side and a force on the other side from whatever object is attached to the end. In some cases the tension may vary along the rope. Tension in Physics is the intermolecular force that exist in inside a rope, string or any kind of elastic material. The Purpose is to maintain integrity of the system

By definition, Tension is a pulling force exerted by a chain or string on another object.For simplicity we will assume the following.

By definition, Tension is a pulling force exerted by a chain or string on another object.For simplicity we will assume the following.

- The strings, cable and rope are considered as mass-less.
- For the entire rope the tension is the same.

$T$ = $(\frac{2m_{1}m_{2}}{m_{1}+m_{2}})$g

The frequency of transverse vibration of a stretched string is directly proportional to the square root of the tension (T) applied to the string provided the vibrating length $l$ and mass per unit length $m$ are kept constant. That is

$\nu \propto \sqrt{T}$, if $l$ and $m$ are constant.

Units: The unit of tension is Newton (N).

Two objects of mass $m_{1}$ and $m_{2}$, with $m_{2} > m_{1}$ are connected by a cord and hung over a pulley with less friction shown in figure. Both cord and pulley have negligible mass. We can find the magnitude of the acceleration of the system and the tension in the cord.

The acceleration of the masses $m_{1}$ and $m_{2}$ are given as $a_{1}$ and $a_{2}$. They are equal in magnitude, but opposite in direction,

Each mass is acted on by a Tension Force $\vec{T}$ in the upwards direction and a force of gravity in the downwards direction.

$a_{1}$ = $-a_{2}$

Each mass is acted on by a Tension Force $\vec{T}$ in the upwards direction and a force of gravity in the downwards direction.

By applying Newton's second law to each of the masses individually,

$m_{1}a_{1} = T – m_{1}g$ ...(1)

$m_{2}a_{2} = T – m_{2}g$ ...(2)

Substituting $a_{2}= -a_{1}$ in the second equation, and multiply both the sides by $-1$,

$m_{2}a_{1} = -T +m_{2}g$ ...(3)

Add equation (1) and (3), and hence solve for $a_{1}$

$(m_{1}+m_{2})a_{1}$ = $m_{2}g – m_{1}g$

$a_{1}$ = $(\frac{m_{2}-m_{1}}{m_{1}+m_{2}})$g

Substitute this result into equation (1) to find $T$:

$T$ = $(\frac{2m_{1}m_{2}}{m_{1}+m_{2}})$g

- The acceleration both the blocks are equal but opposite in direction.
- When $m_{2}$ gets very large compared with $m_{1}$, the acceleration of the system approaches $g$.

To find the torque on the pulley, we first note that the tension in the rope will make the pulley rotate clockwise. According to our convention, the torque is negative. The rotation axis is the axle of the pulley, so the torque equals the tension multiplied by the radius of the pulley.

That is,

$\tau$ = $TR_{pulley}$

Torque is the amount of force it takes to install and Tension is the stretching of a bolt that provides the clamping force (clamp load). For an object on a string at the top of a circle, there are two forces that can point towards the center of the circle and, therefore, contribute towards a net centripetal force in the direction. One is the tensional force from the string, and other is the gravitational force.

The tensional force pluse the component of the gravitational force that point towards the center of the circle will always add up the centripetal force required to move the object at that speed around a circle with that radius.

When a cord or rope pulls on an object, it is said to be under tension and the force it exerts is called a tension force FT.

When a cord or rope pulls on an object, it is said to be under tension and the force it exerts is called a tension force FT.

The following are the problems:

Given $m_{1}$ = 5 kg

$m_{2}$ = 2 kg

$g = 9.8 m/s$

We have

$T$ = $(\frac{2m_{1}m_{2}}{m_{1}+m_{2}})$g

$T$ = $(\frac{2m_{1}m_{2}}{m_{1}+m_{2}})$g

= $(\frac{2\times 5 \times 2}{5 +2})$9.8

$T$ = $28$ N

Given $m_{1}$ = 10 kg

$m_{2}$ = 15 kg

$g = 9.8 m/s$

We have,

$T$ = $(\frac{2m_{1}m_{2}}{m_{1}+m_{2}})$g

$T$ = $(\frac{2m_{1}m_{2}}{m_{1}+m_{2}})$g

= $(\frac{2\times 10 \times 15}{10 +15})$9.8

$T$ = $117.6$ N

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