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:
Tension between two objects under horizontal pull:
Assume that two objects of mass, m1 and m2, which are held by the force of tension through a rope. The force applied by m1 is equal to m2 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.
Ff = 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
Ff = µ (m1 + m2) g
In block 1 the tension force and the friction force are acting in the opposite direction, so the net force is
Fnet = T – Ff = T -µm1g
T= µm1g + m1a
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;
Fnet= Fa – T – Ff
Where, Fa is the applied force
Fnet = m2a
T = Fa – m2a -µm2g
On subtracting the equations of motion,we get the acceleration as;
µm1g + m1a – (Fa – m2a -µm2g) =0
µm1g + m1a – Fa+m2a+µm2g = 0
µ(m1+m2)g – Fa+a(m1+m2) = 0
a(m1+m2) = Fa– µ(m1+m2)g
To find out tension, substitute the value of ‘a’ in the equations of motion;
Since both the equations are equals to the acceleration,
we can equate the above equations.
Tm2 -µm1m2g = Fam1 -µm1m2g – Tm1
Tm1 +Tm2 = Fam1 – µm1m2g+µm1m2g
T(m1+m2) = Fam1
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
Fnet = Fc +T
Since the object is rotating in the circular orbit, the tension is acting along Tx and Ty.
The total tension is given by,
But Tx 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:
Two teams A and B, are playing tug of war. Each team has two members. Both 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. Calculate the tension acting on the rope involved in the tug of war.
Given: F1 = 830 N
F2 = 850 N
We know that T – F1 = m1a
And F2 – T = m2a
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 m1 = m2 = 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 F1 = F2.
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. It is rotating at a speed of 14m/s. Calculate the tension acting on the cable.
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
Calculate the tension between the two blocks of mass 4kg and 6kg. The applied force on the first block is 25N, and the applied force on the second block is 30N.
Given: The mass of the blocks= m1 = 4kg
m2 = 6kg
Force acting on the blocks = F1 = 25N
F2 = 30N
T – F1 = m1a
F2 – T = m2a
a = 0.5 m/s2
T = F1 + m1a
T= F2 -m2a
2T = F1 + m1a +F2 -m2a
T = 27N.
Two blocks of mass, 16kg and 19kg, are pulling each other in the opposite direction. There is a friction force with the coefficient µ= 0.32. Suppose the applied force is 125N. Calculate the acceleration of the two-body system.
Mass of the blocks m1 and m2 are 16kg and 24kg, respectively.
Acceleration due to gravity (g) is 10 m/s2 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/s2.
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. The centripetal force along the circular orbit is 63N. Calculate the tension and also find the length of the rope. ( Take acceleration due to gravity as 10m/s2)
Given that mass is 12kg
The velocity is 2.5m/s
Centripetal force (Fc) 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.
How 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.
What 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.
How 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.
Does 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.
How 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.
Does 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.