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It is the force generated when a load is applied at the ends of an object, normally to the cross-section of it. It can also be called the pulling force, stress, or tension.
This type of force is only exerted when there is contact between a cable and an object. Tension also allows force to be transferred across relatively large distances.
Tension when there is no acceleration
Let's assume we have a body of mass (m) on a piece of string, as shown below. Gravity is pulling it down, which makes its weight:
Tension in string
For the string not to accelerate downwards because of its mass, it must be pulled back upwards with an equal force. This is what we call tension. If it is not accelerating, we can say that T = mg.
Tension when there is acceleration
When we have tension in an object that is accelerating upwards, e.g. an elevator taking people to the top floors of a building, tension cannot be the same as the weight of the load – it will definitely be more. So, where does the addition come from? Tension = force to balance + extra force to accelerate. That is modelled mathematically as:
\[T = mg + ma\]
\[T = m (g + a)\]
It is a different scenario when the elevator is descending downwards. The tension won't be equal to 0, which would make it in free fall. It will be slightly less than the weight of the object. So to put that equation into words, Tension = force needed to balance - force let off. Mathematically that will be \(T = mg - ma\), \(T = m (g - a)\).
Worked examples
Let's look at a couple of worked examples.
When particles are released from rest in the diagram below, what is the tension in the string that holds them?
Tension in string example
Answer:
In a situation like this, the particle with the highest mass will be the one to drop, and the particle with the lowest mass will rise. Let's take the particle with 2kg mass as particle a and the one with 5kg mass as particle b.
To clarify the weight of each particle, we have to multiply its mass with gravity.
Weight of a = 2g
Weight of b = 5g
Now you can model an equation for each particle's acceleration and tension.
T -2g = 2a [Particle a] [Equation 1]
5g -T = 5a [Particle b] [Equation 2]
You now solve this simultaneously. Add both equations to eliminate the T variable.
3g = 7a
If you take 9.8 ms-2 gas
\(a = 4.2 ms^{-2}\)
You can substitute acceleration into any of the equations to give you tension.
Substitute acceleration into equation 1.
\(T = -2g = 2 \cdot 4.2 \rightarrow T -19.6 = 8.4 \rightarrow T = 28 N\)
There are two particles, one with a 2kg mass sitting on a smooth table and the other with a 20kg mass hanging on the side of the table over a pulley connecting both particles – demonstrated below. These particles have been held in place all this time, and they are now released. What will happen next? What is the acceleration and tension in the string?
Tension in a string with one particle on a smooth table
Answer: Let us add to the diagram to see what we are working with.
Tension in a string with one particle on a smooth table
Take particle with 2kg mass to be particle A.
And particle with 20kg mass to be particle B.
Now let's resolve particle A horizontally.
T = ma [equation 1]
Resolving particle B vertically
mg -T = ma [Equation 2]
We substitute the figures in them:
T = 2a [Equation 1]
20g - T = 20a [Equation 2]
We can now add both equations to cancel tensions.
20g = 22a
\(a = \frac{98}{11} = 8.9 ms^{-2}\)
Now factorise acceleration into either of the equations. We would do the first.
\(T = 2 \cdot \frac{98}{11} = 17.8 N\)
Tension at an angle
We can calculate for tension in a rope attached to a weight at an angle. Let's take an example to see how this is done.
Find the tension in each part of the string in the diagram below.
Tension at an angle
Answer: what we will need to do is to make two equations out of the entire diagram – one for the vertical forces and another for the horizontal. So what we are going to do is resolve tension for both strings into their respective vertical and horizontal components.
Tension at an angle
\(T_1 \sin 20 = T_2 \sin 30 \space [Equation \space 2] [Horizontal]\)
Since we have two equations and two unknowns here, we are going to use the simultaneous equation procedure to do this by substitution.
Now we will rearrange the second equation and substitute it into the first equation.
\(T_1 = \frac{T_2 \sin 30}{\sin 20}\)
\((\frac{0.5T_2}{0.342}) = \cos 20 + T_2 \cos 30 = 50\)
\((\frac{0.5T_2}{0.342})0.94 + 0.866 \space T_2 = 50\)
\(1.374 \space T_2 + 0.866 \space T_2 = 50\)
\(2.24 T_2 = 50\)
\(T_2 = 22.32 N\)
Now that we have a value for T2, we can go ahead to substitute that into any of the equations. Let's use the second.
\(T_1 \sin 20 = 22.32 \space \sin 30\)
\(T_1 = \frac{11.16}{0.342} = 32.63\)
Tension in strings - Key takeaways
- A tension force is a force developed in a rope, string, or cable when stretched under an applied force.
- When there is no acceleration, tension is the same as the weight of a particle.
- Tension can also be called the pulling force, stress, or tension.
- This type of force is only exerted when there is contact between a cable and an object.
- When there is acceleration present, tension is equal to the force required to balance plus the extra force needed to accelerate.
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Frequently Asked Questions about Tension in Strings
How do you find tension in a string?
The equation for tension is:
T = mg + ma
What is tension in a string?
A tension force is a force developed in a rope, string, or cable when stretched under an applied force.
How do you find tension in a string between two blocks?
Explore and resolve all forces acting on each block. Write equations for each block and substitute known figures into them. Find the unknowns.
How do you find tension in a pendulum string?
When tension is in instantaneous equilibrium position, it can be certain tension is constant. The degree of the angle the string is displaced is primary to finding your solution. Resolve the force using trigonometry, and substitute the known values into the equation to find tension.
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