Pulley system problems — how to set up equations for any pulley combination

Question

Two blocks of masses $m_1 = 5$ kg and $m_2 = 3$ kg are connected by a light inextensible string passing over a frictionless pulley (Atwood’s machine). Find the acceleration and the tension in the string. Take $g = 10$ m/s².

(CBSE Class 11 / JEE Main / NEET pattern)

Solution — Step by Step

For $m_1$ (heavier, moves down): Weight $m_1 g$ downward, tension $T$ upward. For $m_2$ (lighter, moves up): Weight $m_2 g$ downward, tension $T$ upward.

Since the string is inextensible and the pulley is frictionless, both blocks have the same acceleration $a$ and the tension is the same throughout the string.

flowchart TD
    A["Draw FBD for\neach block"] --> B["Assign acceleration\ndirections using\nstring constraint"]
    B --> C["Write Newton's 2nd law\nfor each block\nalong motion direction"]
    C --> D["Apply string constraint:\nif one goes up by x,\nother goes down by x"]
    D --> E["Solve simultaneous\nequations for a and T"]
    E --> F["Check: a should be\npositive, T should be\nbetween m₁g and m₂g"]

For $m_1$ (net force downward): $m_1 g - T = m_1 a$ … (1)

For $m_2$ (net force upward): $T - m_2 g = m_2 a$ … (2)

Adding equations (1) and (2) to eliminate $T$ :

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

a = \frac{(m_1 - m_2)g}{m_1 + m_2} = \frac{(5-3) \times 10}{5+3} = \frac{20}{8} = \mathbf{2.5 \text{ m/s}^2}

Substituting $a$ back into equation (2):

T = m_2(g + a) = 3(10 + 2.5) = 3 \times 12.5 = \mathbf{37.5 \text{ N}}

Sanity check: $T$ should lie between $m_2 g = 30$ N and $m_1 g = 50$ N. Our answer of 37.5 N passes this check.

Why This Works

The key insight in any pulley problem is the constraint: an inextensible string means whatever length one side gains, the other side loses. For a simple Atwood machine, this means both blocks have equal and opposite accelerations. The string transmits force (tension) but the tension is the same on both sides only when the pulley is massless and frictionless.

The formula $a = \frac{(m_1 - m_2)g}{m_1 + m_2}$ is worth memorising for speed — it appears directly in many MCQs. Notice that if $m_1 = m_2$ , acceleration is zero (equilibrium).

Alternative Method — Using Energy Conservation

For finding speed after block $m_1$ falls by height $h$ :

\frac{1}{2}(m_1 + m_2)v^2 = (m_1 - m_2)gh

v = \sqrt{\frac{2(m_1 - m_2)gh}{m_1 + m_2}}

This avoids finding acceleration and tension entirely — useful when the question asks only for velocity.

For complex pulleys (multiple pulleys, movable pulleys), the constraint changes. If a movable pulley supports a block, the block moves at half the acceleration of the free end of the string. Write the string length as a constant, differentiate twice, and you get the constraint equation. This technique handles any pulley system.

Common Mistake

Students often write the equation for $m_1$ as $T - m_1 g = m_1 a$ (putting tension first). This is wrong if $m_1$ is the heavier block moving downward — the net force is $m_1 g - T$ , not $T - m_1 g$ . Always take the positive direction as the direction of motion for each block. If $m_1$ moves down, “down is positive” for $m_1$ , so the equation is $m_1 g - T = m_1 a$ .

Question

Solution — Step by Step

Why This Works

Alternative Method — Using Energy Conservation

Common Mistake

Try These Next