Find tension in string connecting two blocks over frictionless pulley.

Tension in String Between Two Blocks — Pulley Problem

The Atwood machine is one of the most frequently tested problems in JEE Main and NEET. The setup looks simple — two blocks, one string, one pulley — but students routinely get the tension wrong because they don't isolate the bodies correctly. We'll do this properly with full FBDs.

Question

Two blocks of mass 3 kg and 5 kg are connected by a light inextensible string over a frictionless, massless pulley. The system is released from rest. Find:

The acceleration of the system
The tension in the string

(Take $g = 10$ m/s²)

Solution — Step by Step

Let $m_1 = 3$ kg (lighter block, goes up) and $m_2 = 5$ kg (heavier block, goes down). Let $T$ be the tension and $a$ be the acceleration.

Since the string is inextensible, both blocks have the same magnitude of acceleration $a$ .

Step 1: Apply Newton's Second Law to each block separately.

For the 5 kg block (moving down, take down as positive):

$m_2 g - T = m_2 a$

$50 - T = 5a \quad \cdots (i)$

For the 3 kg block (moving up, take up as positive):

$T - m_1 g = m_1 a$

$T - 30 = 3a \quad \cdots (ii)$

Step 2: Add equations (i) and (ii) to eliminate $T$ .

$50 - 30 = 5a + 3a$

$20 = 8a$

$a = 2.5 \text{ m/s}^2$

Step 3: Substitute back to find $T$ .

From equation (ii): $T = 30 + 3a = 30 + 3 \times 2.5 = 30 + 7.5 = 37.5$ N

Answers:

Acceleration $=$ 2.5 m/s²
Tension $=$ 37.5 N

Why This Works

The key insight is that tension is an internal force for the two-block system combined, but it becomes visible when we isolate each block. If we treat both blocks as one system, tension cancels out and we get $a$ directly. Then we go back to one block to find $T$ .

Notice that tension (37.5 N) is between the two weights: between 30 N ( $m_1 g$ ) and 50 N ( $m_2 g$ ). This makes physical sense — tension is never less than the lighter weight or more than the heavier weight in an Atwood machine.

📌 Note

The general formula for Atwood machine acceleration is:

$a = \frac{(m_2 - m_1)g}{m_1 + m_2}$

And tension:

$T = \frac{2m_1 m_2 g}{m_1 + m_2}$

These are worth memorizing for quick MCQ solving, but you should derive them at least once so you understand where they come from.

Alternative Method — System Approach

Treat both blocks as one system. Net external force on the system is the unbalanced gravitational force:

$F_{net} = m_2 g - m_1 g = (5 - 3) \times 10 = 20 \text{ N}$

Total mass of system $= m_1 + m_2 = 8$ kg

$a = \frac{F_{net}}{M_{total}} = \frac{20}{8} = 2.5 \text{ m/s}^2$

Now isolate $m_1$ to find tension:

$T - m_1 g = m_1 a$

$T = m_1(g + a) = 3 \times (10 + 2.5) = 37.5 \text{ N}$

This approach is faster for MCQs — find $a$ using the system, then use one block for $T$ .

Common Mistake

⚠️ Common Mistake

The single most common error: writing the same equation for both blocks without flipping the sign, i.e., writing $m_2 g - T = m_2 a$ AND $m_1 g - T = m_1 a$ for both. The second equation is wrong. For the lighter block going up, weight acts downward while tension acts upward, so the net upward force is $T - m_1 g = m_1 a$ .

Always define a consistent positive direction for each block based on how it actually moves. The heavier block moves down — take down as positive for that block. The lighter block moves up — take up as positive for that block.

🎯 Exam Insider

JEE Main variants of this problem include: one block on a horizontal surface (with friction), one block on an incline, or three blocks connected in series. The method is the same — FBD each body, write Newton's second law, solve the simultaneous equations.

Question

Solution — Step by Step

Why This Works

Alternative Method — System Approach

Common Mistake

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