The Tension Formula gives the pulling force transmitted through a string, rope, or cable when forces act on both ends. It is expressed as T = mg + ma for an accelerating system and T = mg for a body in equilibrium. This concept is covered in Class 11 Physics, Chapter 5 (Laws of Motion) under the NCERT curriculum. It is also a high-frequency topic in JEE Main, JEE Advanced, and NEET. This article covers the formula, derivation, a complete formula sheet, three solved examples, CBSE exam tips, common mistakes, and JEE/NEET applications.
Key Tension Formulas at a Glance
Quick reference for the most important tension formulas used in CBSE and competitive exams.
- Tension (equilibrium): \( T = mg \)
- Tension (upward acceleration): \( T = m(g + a) \)
- Tension (downward acceleration): \( T = m(g – a) \)
- Tension in Atwood machine: \( T = \frac{2m_1 m_2}{m_1 + m_2} g \)
- Tension on inclined plane: \( T = mg\sin\theta \)
- Tension with two masses (horizontal): \( T = \frac{m_2}{m_1 + m_2} F \)
- Tension in circular motion: \( T = \frac{mv^2}{r} + mg\cos\theta \)
What is the Tension Formula?
The Tension Formula describes the contact force that a string, rope, wire, or cable exerts on an object attached to it. Tension always acts along the length of the string and pulls the object toward the point of attachment. It is a pulling force, never a pushing force.
In NCERT Class 11 Physics, Chapter 5 (Laws of Motion), tension is introduced as an application of Newton’s Second Law. When a body of mass m hangs from a string, the string must support its weight. If the body accelerates upward, the string must exert a greater force. If the body accelerates downward, the string exerts a lesser force.
The general form of the Tension Formula is derived by applying Newton’s Second Law along the direction of the string. The formula changes depending on whether the system is in equilibrium, accelerating upward, accelerating downward, or moving on an inclined plane. Understanding each case is essential for scoring well in CBSE board exams and cracking JEE/NEET problems on connected bodies and pulleys.
Tension Formula — Expression and Variables
The most commonly used forms of the Tension Formula are listed below.
Case 1: Body in Equilibrium (no acceleration)
\[ T = mg \]
Case 2: Body accelerating upward
\[ T = m(g + a) \]
Case 3: Body accelerating downward
\[ T = m(g – a) \]
Case 4: Body in free fall (a = g)
\[ T = 0 \]
| Symbol | Quantity | SI Unit |
|---|---|---|
| T | Tension in the string | Newton (N) |
| m | Mass of the object | Kilogram (kg) |
| g | Acceleration due to gravity (9.8 m/s²) | m/s² |
| a | Acceleration of the system | m/s² |
| θ | Angle of incline with horizontal | Degrees or Radians |
| v | Speed of object in circular motion | m/s |
| r | Radius of circular path | Metre (m) |
Derivation of the Tension Formula
Consider a block of mass m hanging from a string. Two forces act on the block: weight mg downward and tension T upward.
Step 1: Apply Newton’s Second Law along the vertical direction.
\[ T – mg = ma \]
Step 2: Rearrange to find tension.
\[ T = m(g + a) \]
Step 3: For equilibrium, set acceleration a = 0.
\[ T = mg \]
This derivation directly follows from Newton’s Second Law as taught in NCERT Class 11, Chapter 5. The sign of a depends on the direction of acceleration relative to the weight.
Complete Physics Formula Sheet — Tension and Related Laws of Motion
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| Tension (equilibrium) | \( T = mg \) | m = mass, g = gravity | N | Class 11, Ch 5 |
| Tension (upward acceleration) | \( T = m(g + a) \) | a = acceleration upward | N | Class 11, Ch 5 |
| Tension (downward acceleration) | \( T = m(g – a) \) | a = acceleration downward | N | Class 11, Ch 5 |
| Tension (Atwood machine) | \( T = \frac{2m_1 m_2}{m_1 + m_2} g \) | m&sub1;, m&sub2; = masses on each side | N | Class 11, Ch 5 |
| Tension on inclined plane | \( T = mg\sin\theta \) | θ = angle of incline | N | Class 11, Ch 5 |
| Tension (two masses, horizontal) | \( T = \frac{m_2}{m_1 + m_2} F \) | F = applied force | N | Class 11, Ch 5 |
| Tension at bottom of circular path | \( T = mg + \frac{mv^2}{r} \) | v = speed, r = radius | N | Class 11, Ch 6 |
| Tension at top of circular path | \( T = \frac{mv^2}{r} – mg \) | v = speed, r = radius | N | Class 11, Ch 6 |
| Newton’s Second Law | \( F = ma \) | F = net force, m = mass, a = acceleration | N | Class 11, Ch 5 |
| Weight of a body | \( W = mg \) | W = weight, m = mass, g = gravity | N | Class 11, Ch 5 |
Tension Formula — Solved Examples
Example 1 (Class 9-10 Level) — Simple Hanging Mass
Problem: A block of mass 5 kg hangs from a string attached to the ceiling. Calculate the tension in the string. Take g = 10 m/s².
Given: m = 5 kg, g = 10 m/s², a = 0 (system in equilibrium)
Step 1: Since the block is stationary, use the equilibrium form of the Tension Formula:
\[ T = mg \]
Step 2: Substitute the values:
\[ T = 5 \times 10 = 50 \text{ N} \]
Answer
The tension in the string is 50 N. This equals the weight of the block because the system is in equilibrium.
Example 2 (Class 11-12 Level) — Accelerating Lift
Problem: A person of mass 60 kg stands in a lift. The lift accelerates upward at 3 m/s². Find the tension (normal reaction) in the floor of the lift acting on the person. Take g = 10 m/s².
Given: m = 60 kg, a = 3 m/s² (upward), g = 10 m/s²
Step 1: Identify the direction of acceleration. The lift moves upward, so use:
\[ T = m(g + a) \]
Step 2: Substitute the values:
\[ T = 60 \times (10 + 3) = 60 \times 13 = 780 \text{ N} \]
Step 3: Compare with the person’s actual weight:
\[ W = mg = 60 \times 10 = 600 \text{ N} \]
The apparent weight (780 N) is greater than the real weight (600 N). This is the “heavier feeling” in an accelerating lift.
Answer
The tension (normal force) acting on the person is 780 N.
Example 3 (JEE/NEET Level) — Atwood Machine
Problem: Two masses m&sub1; = 8 kg and m&sub2; = 4 kg are connected by a light, inextensible string over a frictionless pulley (Atwood machine). Find the tension in the string and the acceleration of the system. Take g = 10 m/s².
Given: m&sub1; = 8 kg, m&sub2; = 4 kg, g = 10 m/s²
Step 1: Find the acceleration of the Atwood machine system:
\[ a = \frac{(m_1 – m_2)}{(m_1 + m_2)} g = \frac{(8 – 4)}{(8 + 4)} \times 10 = \frac{4}{12} \times 10 = 3.33 \text{ m/s}^2 \]
Step 2: Use the Atwood machine Tension Formula:
\[ T = \frac{2m_1 m_2}{m_1 + m_2} g = \frac{2 \times 8 \times 4}{8 + 4} \times 10 \]
Step 3: Calculate the numerator and denominator:
\[ T = \frac{64}{12} \times 10 = 5.33 \times 10 = 53.3 \text{ N} \]
Step 4: Verify using Newton’s Second Law for m&sub2;:
\[ T – m_2 g = m_2 a \Rightarrow T = m_2(g + a) = 4 \times (10 + 3.33) = 4 \times 13.33 = 53.3 \text{ N} \checkmark \]
Answer
The acceleration of the system is 3.33 m/s² and the tension in the string is 53.3 N.
CBSE Exam Tips 2025-26
- Draw a Free Body Diagram (FBD) first. We recommend this as the single most important step. Mark all forces on each object separately before writing any equation.
- Identify the direction of acceleration. Always define a positive direction. Upward acceleration increases tension. Downward acceleration decreases tension.
- Use the correct formula for each case. The formula \( T = m(g + a) \) applies only when acceleration is upward. Do not use it blindly for all problems.
- Remember the weightlessness condition. When \( a = g \) (free fall), tension becomes zero. This is a favourite CBSE one-mark question in 2025-26 board exams.
- For connected bodies on a horizontal surface, find the common acceleration first using \( a = F / (m_1 + m_2) \), then find tension using \( T = m_2 \cdot a \).
- Check units at every step. Mass must be in kg, acceleration in m/s², and the answer in Newtons (N). Unit errors cost marks in CBSE 2025-26 board exams.
Common Mistakes to Avoid
- Mistake 1: Using T = mg for all cases. Many students use \( T = mg \) even when the object is accelerating. Always check whether the system is in equilibrium or accelerating before choosing the formula.
- Mistake 2: Wrong sign for acceleration. If the body moves downward with acceleration a, the correct formula is \( T = m(g – a) \), not \( T = m(g + a) \). Confusing the sign is the most common error in CBSE exams.
- Mistake 3: Forgetting that tension is the same throughout a light string. For a massless, inextensible string over a frictionless pulley, tension is identical on both sides. Students sometimes calculate two different tensions for the same string.
- Mistake 4: Not drawing a Free Body Diagram. Attempting tension problems without an FBD leads to missing forces. Always draw and label all forces before applying Newton’s Second Law.
- Mistake 5: Confusing weight and tension. Weight (W = mg) acts downward on the object. Tension acts upward along the string. They are equal only in equilibrium. They are different when the object accelerates.
JEE/NEET Application of the Tension Formula
In our experience, JEE aspirants encounter the Tension Formula in at least 2–3 questions per paper. NEET also tests tension in the context of connected bodies and circular motion. Here are the three most important application patterns.
Pattern 1: Atwood Machine Problems
The Atwood machine is a classic JEE topic. Two unequal masses hang over a pulley. You must find both the acceleration and the tension. Use:
\[ a = \frac{(m_1 – m_2)g}{m_1 + m_2}, \quad T = \frac{2m_1 m_2 g}{m_1 + m_2} \]
JEE Advanced sometimes adds friction at the pulley axle or a massive pulley. In those cases, you must apply torque equations alongside the Tension Formula.
Pattern 2: Circular Motion — Tension at Different Points
In JEE Main and NEET, a ball on a string moves in a vertical circle. The tension changes at every point. At the bottom of the circle:
\[ T_{\text{bottom}} = mg + \frac{mv^2}{r} \]
At the top of the circle:
\[ T_{\text{top}} = \frac{mv^2}{r} – mg \]
The minimum speed at the top for the string to remain taut is found by setting \( T_{\text{top}} = 0 \), which gives \( v_{\text{min}} = \sqrt{gr} \). This is a direct application of the Tension Formula tested in JEE Main almost every year.
Pattern 3: Connected Bodies on a Surface
Two or more blocks are connected by strings on a horizontal surface. An external force F is applied to one block. The system accelerates together. The Tension Formula for the connecting string is:
\[ T = \frac{m_2}{m_1 + m_2} \times F \]
In our experience, JEE aspirants who master all three patterns can solve tension-based problems in under 90 seconds. Practise each pattern with at least five problems before the exam.
For further reading on related force concepts, visit the official NCERT Class 11 Physics Chapter 5 on Laws of Motion on the official NCERT website.
FAQs on Tension Formula
Explore more related topics on ncertbooks.net. Read our detailed guide on the Buoyancy Formula to understand upward forces in fluids. Study the Bernoulli’s Equation Formula for fluid dynamics applications. You can also revisit our complete Physics Formulas hub for a full list of NCERT and competitive exam formulas.