You are familiar with the term tension force. One of the major example that explains tension between two object is Tug of war. Have you ever played the game tug of war?

**It is a game in which the people of two groups pulls the rope to their side to win. The force exerted by the people is equal and opposite that transmitted along the rope which is in contact between the two groups, this force is nothing but the **__tension__**. We can calculate the tension between these two groups, that we are going to discuss in this article.**

**How to calculate tension between two objects: **

**T****ension between two objects under horizontal pull:**

Assume that two objects of mass, m_{1} and m_{2}, which are held by the force of tension through a rope. The force applied by m_{1} is equal to m_{2} but is in the opposite direction. But there is a force called friction which opposes the objects to pull it acts on the body with a friction coefficient µ. Then the tension between the two objects can be calculated as follows.

In this case, the net force involves in the pulling force, tension force along with friction force. The formula for friction force is given by;

F_{f} = µ(mg)

µ is the coefficient of friction force, and mg is the net force acting on the system.

Friction for the whole system is given by

F_{f }= µ (m_{1} + m_{2}) g

In block 1 the tension force and the friction force are acting in the opposite direction, so the net force is

F_{net} = T – F_{f} = T -µm_{1}g

T= µm_{1}g + m_{1}a

But in block 2 the applied force and the friction force is acting in the same direction so that the net force is given as;

F_{net}= F_{a} – T – F_{f}_{}

Where, F_{a} is the applied force

F_{net} = m_{2}a

T = F_{a} – m_{2}a -µm_{2}g

On subtracting the equations of motion,we get the acceleration as;

µm_{1}g + m_{1}a – (F_{a} – m_{2}a -µm_{2}g) =0

µm_{1}g + m_{1}a – F_{a}+m_{2}a+µm_{2}g = 0

µ(m_{1}+m_{2})g – F_{a}+a(m_{1}+m_{2}) = 0

a(m_{1}+m_{2}) = F_{a}– µ(m_{1}+m_{2})g

To find out tension, substitute the value of ‘a’ in the equations of motion;

and

Since both the equations are equals to the acceleration,

we can equate the above equations.

Tm_{2} -µm_{1}m_{2}g = F_{a}m_{1} -µm_{1}m_{2}g – Tm_{1}

Tm_{1} +Tm_{2} = F_{a}m_{1} – µm_{1}m_{2}g+µm_{1}m_{2}g

T(m_{1}+m_{2}) = F_{a}m_{1}

This is the equation for tension between the two objects under friction force.

**Tension between two objects in the circular path:**

Assume that you are swung an object in a circular path, which is tied with the help of a rope. The tension between the object and you along the circular path can be calculated by resolving all the forces that involves in the rotation of the object.

In this case there are three forces acting. One is the** tension force** on the rope, and the **gravity**, which tries to pull the object downwards, and another one is the **centripetal force** which pulls the objects towards the center.

The object is rotating in a circular orbit of radius ‘r’ and moving with velocity ‘v’.

The centripetal force is given as

The net force acting is

F_{net} = F_{c} +T

Since the object is rotating in the circular orbit, the tension is acting along T_{x }and T_{y}.

The total tension is given by,

But T_{x} is the centripetal force acting towards the center, given as;

And Ty = mg; it is nothing but net force acting on the system.

And Ty = mg; it is nothing but net force acting on the system.

This gives the equation for tension between two objects in circular a motion.

**Tension in an elevator:**

Have you ever been in an elevator? The elevator is also affected by the tension force as the cable of the elevator is moving downwards as the elevator moves upwards. The tension between the elevator can be calculated as below.

When the elevator is at rest, the tension and the net force are equal.

T = mg

When the elevator is accelerating in the upward direction, the tension is;

T – mg = ma

T = mg + ma

T = m(g+a)

When the elevator is accelerating in the downward direction, the tension is;

mg – T = ma

T = mg – ma

T = m(g-a)

**Solved problems on tension between two objects:**

**T****wo teams A and B, are playing tug of war. ****E****ach team has two members. ****B****oth exert a force of 830 and 850 N, respectively, to win the game. The mass of team A is 83 kg and B is 79 kg. ****C****alculate the tension acting on the rope involved in the tug of war.**

**Solution:**

Given: F_{1} = 830 N

F_{2} = 850 N

We know that T – F_{1} = m_{1}a

And F_{2} – T = m_{2}a

The acceleration is given by

But we can not consider the mass because if it has mass, then the tension would get two different results which is contradicting to the force of tension, which should have only one value.

So m_{1 }= m_{2} = 0 , it gives the infinite acceleration.

However, there is a force acting between the two teams, which is equal and opposite, exerted by both teams F_{1} = F_{2}._{ }_{}

the tension is given as;

T = -T.

**A**** ball of mass 3kg is attached to a cable of length 54m is rotating in a circular orbit. ****I****t is rotating at a speed of 14m/s. Calculate the tension acting on the cable.**

**Solution:**

Given: Mass of the ball (m) = 3kg

Speed of the ball (v) = 14 cm/s

The length of the cable is nothing but the radius r = 54m.

The formula to calculate the tension is;

T = 10.88 + 29.4

T = 40.28 N

**C****alculate the tension between the two blocks of mass 4kg and 6kg. ****T****he applied force on the first block is 25N, and the applied force on the second block is 30N.**

**Solution:**

Given: The mass of the blocks= m_{1} = 4kg

m_{2} = 6kg

Force acting on the blocks = F_{1} = 25N

F_{2} = 30N

T – F_{1} = m_{1}a

F_{2} – T = m_{2}a

a = 0.5 m/s^{2}

T = F_{1} + m_{1}a

T= F_{2 }-m_{2}a

2T = F_{1} + m_{1}a +F_{2} -m_{2}a

T = 27N.

**T****wo blocks of mass, 16kg and 19kg, are pulling each other in the opposite direction. ****T****here is a friction force with the coefficient µ= 0.32. ****Suppose**** the applied force is 125N. Calculate the acceleration of the two-body system.**

**Solution:**

Mass of the blocks m_{1} and m_{2} are 16kg and 24kg, respectively.

Acceleration due to gravity (g) is 10 m/s^{2} Applied force by both the side is 125N.

The coefficient of friction (µ) is 0.32

The acceleration is given by the formula,

a = 0.37 m/s^{2}.

**A**** shotput is tied to a rope and is rotating in a circular motion with a speed of 2.5m/s and mass of the object is 12kg. ****T****he centripetal force along the circular orbit is 63N. ****C****alculate the tension and also find the length of the rope. ( Take acceleration due to gravity as 10m/s**^{2}**) **

^{2}

**Solution:**

Given that mass is 12kg

The velocity is 2.5m/s

Centripetal force (F_{c}) along the circular path = 63N

To calculate the tension, we need to find out the radius of the circle.

The tension acting along the circular path is given by

Substituting the values in the above equation,

T = 63 + 12(10)

T = 63 + 120

T = 183N

The centripetal force equation is

On rearranging, we can find the radius, i.e., the length of the rope.

r = 1.19m.

**Frequently asked questions.**

**H****ow does friction affect the tension?**

Tension may change regularly if the friction involves in the system.

**When the friction force is involved between the body and cable, tension does not remain the same. If there is maximum friction and the net force, which is acting in the downward direction then the tension will be zero.**

**W****hat is meant by the coefficient of friction?**

It is the measure of interaction between the two surfaces under a contact force.

**The coefficient of friction can be defined as the ratio of force that is needed to maintain the relative motion between two surfaces to the force which is holding the surface to be in contact.**

**H****ow do the tension and the acceleration dependent on each other?**

Tension and acceleration are inversely depends on each other.

**Tension depends on the acceleration; as acceleration increases, it contributes the net force to increase, which proportionally increases the tension.**

**D****oes the elasticity increases when the tension increases? **

No, elasticity reduces as the tension increases.

**As the tension on the string increases, the wave which is propagating on the system makes the particles to bound more rigidly on the string, which reduces the elasticity.**

**H****ow does the tension and gravity are related?**

Tension balances the object to hang on the string.

**When an object is suspended by a string or cable, tension acts in the upward direction; meanwhile, gravity acts in the downward direction. If the tension could not act opposite to the gravity, then the object would certainly accelerate downwards due to gravity**.

## **D****oes the tension changes while rotating in the circular path?**

The tension may change while moving in the circular path.

**The tension will be greater than gravity at the bottom of the trajectory of the circular path so that the net force will be in upward at the bottom of trajectory and vice versa.**