Tension Formula: Explanation and Solved Examples

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The Tension Formula is commonly used to calculate the tension force exerted on any object.

Push, pull, thrust, lift, weight, friction, and tension are all names for forces.
Tension is the pulling force that is transmitted along the length of a flexible object, such as a string, cable, or chain.
It can also be described as the force that is exerted on each end of a rigid object, such as a rod or beam.
In both cases, tension is an action-reaction pair of forces, meaning that the force exerted on one end of the object is equal and opposite to the force exerted on the other end.

Table of Content

Tension
Tension Formula
Solved Examples
Things to Remember
Sample Questions

Key Terms: Tension, Force, Gravitational force, Action-reaction force, Tension force formula, Weight, Push or pull

Tension

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The pulling force transmitted axially by a string, rope, chain, or similar objects, or by each end of a rod, truss member, or similar three-dimensional object, is referred to as tension.

In Latin, the word "tension" means "stretching".
All physical things that come into contact with one another can exert forces on one another.
Depending on the types of objects in touch, we call these contact forces by different names.
The force tension is what we call it when one of the things applying the force is a rope, string, chain, or cable.
It is essential to know that tension is a pulling force because ropes cannot effectively push.
When you try to pull anything with a rope, it becomes slack and loses the tension that allowed it to pull in the first place.
This may seem self-evident, but when drawing the forces acting on an object, students frequently draw the force of tension in the wrong direction.
Remember that tension can only pull on an object.

Tension Formula

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Tension is a force acting along the length of a medium, particularly one that is flexible, such as a rope or cable.

The formula of tension is given by

\(T=mg\pm ma\)

Where

T is the tension force
m is the mass of the body
g is the acceleration due to gravity
a is the acceleration of the moving body
mg = W is the weight of the body

Cases:

When the body is moving upward, then the tension will be T = mg + ma
When the body is moving downward, then the tension will be T = mg – ma
When the tension is equivalent to the weight of the body T = mg.

Tension Formula

Also Read:

Surface Tension	Pressure of Ideal Gas

Solved Examples

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Ques. A light and inextensible string supports a body of mass of 25 kg hanging from its lower end. If the upper end of the string is firmly attached to a hook on the roof, then what is the tension in the string?

Ans. Since the mass attached to the string is not moving, therefore the tension on the string is given by

T = mg

Given the mass of the body, m = 25 kg

⇒ T = 25 x 9.8 = 245 N

Ques. A body of mass 10 kg is dangling at the end of a string. If the acceleration of the body is

5 m/s² in the upward direction
5 m/s² in a downward direction

Find the tension in the string.

Ans. Given, the mass of the body, m = 10 kg

If the body moves in the upward direction, then tension in the string is given by

T = mg + ma

⇒ T = (10 x 9.8) + (10 x 5) = 148 N

If the body moves in the downward direction, then tension in the string is given by

T = mg - ma

⇒ T = (10 x 9.8) - (10 x 5) = 48 N

Also Read:

Surface Tension	Buoyant Force	Effects of Forces: Types (Contact& Non Contact), Effects
Non Contact Force	Pseudo Force	Central Force

Things to Remember

Tension is a force acting along the length of a medium, particularly one that is flexible, such as a rope or cable.
Tension force is always a pulling force.
The net force acting on an object can be calculated as F_net = T − W = 0
The formula of tension force is given by T = W ± ma = mg ± ma
If the body is moving upward, the tension will be T = W + ma.
When the body is moving downward, the tension equals T = W – ma.
T = W if the tension is equal to the weight of the body.

Sample Questions

Ques. What is the force of tension? (1 Mark)

Ans. Tension force is the pulling force exerted axially along the length of a rope or cable.

Ques. A 15-kilogram body is suspended from the bottom end of the light and an inextensible cord. What is the tension in the string if the top end of the rope is firmly secured to a hook on the roof? (2 Marks)

Ans. Because the body is hung and not moving, the tension in the string will be equal to the body's weight. 15 kg = m

T = W = mg = 15 × 9.8 = 147 N

Ques. Is tension a non-contact or contact force? (1 Mark)

Ans. When a surface comes into direct contact with a rope, tension is created. As a result, it's a contact force.

Ques. A mass of 8 kilograms is dangling from the end of a thread. If the mass accelerates in the
(a) upper direction at a rate of 3 m/s².
(b) 3 m/s² in the direction of descent.
Then determine the tension of the thread. (3 Marks)

Ans. The following parameters are provided:

m = 8 kg, g = 9.8 m/s.

(a) Assuming a = 3m/s².

The tension force is T = mg + ma = 8×9.8+8×1.5 = 90.4 N if the body is moving upward.

(b) Assume a = 3m/s².

The tension force is T = mg – ma = 8×9.8–8×1.5 = 66.4 N, if the body is moving downward.

Ques. A monkey weighing ten kilograms climbs at a speed of two metres per second up a light vertical string strung from a hook. Determine the string's tension (take g = 10 m/s²). (3 Marks)

Ans. Given, m =10 kg, g = 10 m/s², and a = 2 m/s²

The tension in the rope will be equal to the apparent weight of the monkey as the monkey accelerates up.

i.e., T = m (g + a) = 10 (10 + 2) = 120 N

Ques. A ten-kilogram mass is dangling on a thread. If the mass's acceleration is as follows:
(a) A speed of 5 metres per second in the uphill direction.
(b) A downward velocity of 5 metres per second.
Calculate the thread's tension. (3 Marks)

Ans. We've been given the following information:

m = 10 kg, g = 9.8 m/s², a = 5 m/s²

(a) The tension force is T=mg+ma when the body accelerates upwards.

T =10(9.8+5)

T =148N

(b) The tension force is T = mg – ma while the body is moving downwards.

T=10(9.8–5)

T = 48 N

Ques. What are the applications of tension force? (3 Marks)

Ans. The following are the applications of tension force

It is used to pick up lightweight things in a crane machine.
It is used to pull automobiles or cars on a busy highway or road.
Weighing balances employ tension forces.
They are also utilized for a variety of gym equipment.

Ques. What are the different fundamental forces? (3 Marks)

Ans. The following are the fundamental forces

Gravitational force
Electromagnetic force
Strong nuclear force
Weak nuclear force

Ques. What is the unit of tension? (1 Mark)

Ans. The unit of tension is Newton (N).

Ques. List five non-fundamental forces. (2 Marks)

Ans. The following are the non-fundamental forces

Friction
Tension
Normal reaction
Centrifugal force
Centripetal force

Also Read:

Class 11 & 12 PCMB Study Guides
NCERT Solutions for Class 11 Chemistry	NCERT Solutions for Class 12 Chemistry	Important Chemistry Formulas
Comparison topics in Chemistry	Chemistry Study Notes	Class 12 Chemistry Practicals
Physics Study Notes	NCERT Solutions Class 6 to 12	Class 12 Physics Practicals
Topics with relation in physics	NCERT Solutions for Class 12 Maths	Topics for Comparison in Physics
Physics MCQs	SI units in Physics	Important Physics Constants and Units
Important Derivations in Physics	Formulas in Physics	NCERT Solutions for Class 12 Biology
NCERT Solutions for Class 12 English	Class 12 Physics Book PDF	Class 12 Physics Notes
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Class 12 PCMB Notes	NCERT Class 12 Biology Book	NCERT Class 12 Maths Book
Chemistry MCQs	NCERT Solutions for Class 11 Maths	Maths MCQs
Class 12 Chemistry Notes	NCERT Solutions for Class 11 Biology	Biology MCQs
NCERT Solutions for Class 11 Physics	NCERT Class 11 Physics Book	NCERT Class 11 Biology Book
Class 11 PCMB Syllabus	Comparison Topics in Maths	NCERT Solutions for Class 12 Physics
Important Maths Formulas	Comparison Topics in Biology	Class 12 Biology Notes

Tension Formula: Explanation and Solved Examples

Tension

Tension Formula

Solved Examples

Things to Remember

Sample Questions

CBSE CLASS XII Related Questions

1.
Two point charges \( q_1 = 16 \, \mu C \) and \( q_2 = 1 \, \mu C \) are placed at points \( \vec{r}_1 = (3 \, \text{m}) \hat{i}\) and \( \vec{r}_2 = (4 \, \text{m}) \hat{j} \). Find the net electric field \( \vec{E} \) at point \( \vec{r} = (3 \, \text{m}) \hat{i} + (4 \, \text{m}) \hat{j} \).

3.
Three batteries E1, E2, and E3 of emfs and internal resistances (4 V, 2 \(\Omega\)), (2 V, 4 \(\Omega\)) and (6 V, 2 \(\Omega\)) respectively are connected as shown in the figure. Find the values of the currents passing through batteries E1, E2, and E3.

4.
A current carrying circular loop of area A produces a magnetic field \( B \) at its centre. Show that the magnetic moment of the loop is \( \frac{2BA}{\mu_0} \sqrt{\frac{A}{\pi}} \).

6.
The ends of six wires, each of resistance R (= 10 \(\Omega\)) are joined as shown in the figure. The points A and B of the arrangement are connected in a circuit. Find the value of the effective resistance offered by it to the circuit.

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Tension Formula: Explanation and Solved Examples

Tension

Tension Formula

Solved Examples

Things to Remember

Sample Questions

CBSE CLASS XII Related Questions

1.Two point charges \( q_1 = 16 \, \mu C \) and \( q_2 = 1 \, \mu C \) are placed at points \( \vec{r}_1 = (3 \, \text{m}) \hat{i}\) and \( \vec{r}_2 = (4 \, \text{m}) \hat{j} \). Find the net electric field \( \vec{E} \) at point \( \vec{r} = (3 \, \text{m}) \hat{i} + (4 \, \text{m}) \hat{j} \).

3.Three batteries E1, E2, and E3 of emfs and internal resistances (4 V, 2 \(\Omega\)), (2 V, 4 \(\Omega\)) and (6 V, 2 \(\Omega\)) respectively are connected as shown in the figure. Find the values of the currents passing through batteries E1, E2, and E3.

4.A current carrying circular loop of area A produces a magnetic field \( B \) at its centre. Show that the magnetic moment of the loop is \( \frac{2BA}{\mu_0} \sqrt{\frac{A}{\pi}} \).

6.The ends of six wires, each of resistance R (= 10 \(\Omega\)) are joined as shown in the figure. The points A and B of the arrangement are connected in a circuit. Find the value of the effective resistance offered by it to the circuit.

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
Two point charges \( q_1 = 16 \, \mu C \) and \( q_2 = 1 \, \mu C \) are placed at points \( \vec{r}_1 = (3 \, \text{m}) \hat{i}\) and \( \vec{r}_2 = (4 \, \text{m}) \hat{j} \). Find the net electric field \( \vec{E} \) at point \( \vec{r} = (3 \, \text{m}) \hat{i} + (4 \, \text{m}) \hat{j} \).

3.
Three batteries E1, E2, and E3 of emfs and internal resistances (4 V, 2 \(\Omega\)), (2 V, 4 \(\Omega\)) and (6 V, 2 \(\Omega\)) respectively are connected as shown in the figure. Find the values of the currents passing through batteries E1, E2, and E3.

4.
A current carrying circular loop of area A produces a magnetic field \( B \) at its centre. Show that the magnetic moment of the loop is \( \frac{2BA}{\mu_0} \sqrt{\frac{A}{\pi}} \).

6.
The ends of six wires, each of resistance R (= 10 \(\Omega\)) are joined as shown in the figure. The points A and B of the arrangement are connected in a circuit. Find the value of the effective resistance offered by it to the circuit.