Two blocks connected by string over pulley — Atwood machine problem

Question

Two blocks of masses $m_1 = 4\,\text{kg}$ and $m_2 = 6\,\text{kg}$ are connected by a light inextensible string passing over a smooth, massless pulley. Find the acceleration of the system and the tension in the string. (Take $g = 10\,\text{m/s}^2$ )

(JEE Main 2020, similar pattern)

Solution — Step by Step

For the heavier block ( $m_2 = 6\,\text{kg}$ , moves down):

Weight $m_2 g$ downward, tension $T$ upward
Net force: $m_2 g - T = m_2 a$ … (i)

For the lighter block ( $m_1 = 4\,\text{kg}$ , moves up):

Weight $m_1 g$ downward, tension $T$ upward
Net force: $T - m_1 g = m_1 a$ … (ii)

Both have the same acceleration $a$ (inextensible string) and the same tension $T$ (massless pulley).

Adding (i) and (ii):

$m_2 g - m_1 g = (m_1 + m_2)a$

a = \frac{(m_2 - m_1)g}{m_1 + m_2} = \frac{(6-4) \times 10}{4+6} = \frac{20}{10} = \mathbf{2\,\text{m/s}^2}

Substitute $a$ into equation (ii):

$T = m_1(g + a) = 4(10 + 2) = \mathbf{48\,\text{N}}$

Verify with equation (i): $T = m_2(g - a) = 6(10 - 2) = 48\,\text{N}$ ✓

\boxed{a = 2\,\text{m/s}^2, \quad T = 48\,\text{N}}

Why This Works

The Atwood machine is a classic constrained system. The string constraint forces both blocks to have the same magnitude of acceleration — one goes up, the other goes down. The heavier block “wins” and accelerates downward.

The acceleration $a = \frac{(m_2 - m_1)}{(m_1 + m_2)}g$ is always less than $g$ . When $m_1 = m_2$ , $a = 0$ (balanced). When one mass is much larger, $a \to g$ (free fall). The tension $T = \frac{2m_1 m_2}{m_1 + m_2}g$ is always between $m_1 g$ and $m_2 g$ .

Alternative Method — Using reduced mass

The system behaves as if a net force $(m_2 - m_1)g$ accelerates a total mass $(m_1 + m_2)$ :

a = \frac{F_{net}}{m_{total}} = \frac{(m_2 - m_1)g}{m_1 + m_2}

This is the “system approach” — treat both blocks as one system and find the net unbalanced force.

For the tension formula, memorise the symmetric form: $T = \frac{2m_1 m_2 g}{m_1 + m_2}$ . Notice this looks like the harmonic mean of $m_1 g$ and $m_2 g$ . This formula is directly applicable in MCQs without solving simultaneous equations.

Common Mistake

Students sometimes write both equations with gravity in the same direction: $m_1 g - T = m_1 a$ AND $m_2 g - T = m_2 a$ . This gives the wrong result because it assumes both blocks move downward. In an Atwood machine, one block goes up and the other goes down. The lighter block has $T - m_1 g = m_1 a$ (net force upward), not $m_1 g - T$ . Always be consistent with the direction of acceleration for each block.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using reduced mass

Common Mistake

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