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

Question

Given any pulley system (fixed, movable, or combination), how do we systematically set up the equations to find acceleration and tension?

Solution — Step by Step

First, classify each pulley:

Fixed pulley: attached to ceiling/wall, only changes direction of tension. Does NOT give mechanical advantage.
Movable pulley: attached to the mass or rope, moves with the system. Provides mechanical advantage — the tension supports twice.

Draw a free body diagram (FBD) for every mass and every movable pulley separately. Label tension in each rope segment.

The string is inextensible (length does not change). This gives us the constraint equation:

For a simple Atwood machine (two masses, one fixed pulley):

a_1 = -a_2

If mass $m_1$ goes down with acceleration $a$ , mass $m_2$ goes up with acceleration $a$ .

For a movable pulley system, if the pulley moves down by $x$ , each side of the string passing over it must shorten by $x$ . So:

a_{\text{pulley}} = \frac{a_1 + a_2}{2}

Atwood machine example ( $m_1 > m_2$ , connected over a fixed pulley):

For $m_1$ (moving down):

m_1 g - T = m_1 a

For $m_2$ (moving up):

T - m_2 g = m_2 a

Adding both equations:

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

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

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

For systems with multiple pulleys, follow this algorithm:

Assume a direction of motion for each mass
Write constraint equations using the string length method: sum of all string segments = constant, differentiate twice to get accelerations
Write $F = ma$ for each mass
Solve the simultaneous equations

flowchart TD
    A["Pulley Problem"] --> B["Step 1: Identify fixed vs movable pulleys"]
    B --> C["Step 2: Draw FBD for each mass and movable pulley"]
    C --> D["Step 3: Write string constraint equations"]
    D --> E["Step 4: Write F = ma for each mass"]
    E --> F["Step 5: Solve simultaneous equations"]
    F --> G["Get acceleration and tension"]

Why This Works

The string constraint is the key insight. An inextensible string means that the total length of string remains constant — so if one part lengthens, another must shorten by exactly the same amount. Differentiating this length equation once gives velocity relations, and differentiating again gives acceleration relations. Combined with Newton’s second law, we get a complete system of equations.

The beauty of this systematic approach is that it works for ANY pulley arrangement — no matter how complex.

Alternative Method

For quick solutions in MCQs, use the virtual work method: assume the system moves slightly, calculate the work done by all forces, and set the net virtual work equal to the total kinetic energy change. This avoids writing constraint equations entirely and is faster for competitive exams.

For the standard Atwood machine, memorise these results directly — they appear as sub-parts of larger problems in JEE Main. The acceleration formula $a = \frac{(m_1 - m_2)g}{m_1 + m_2}$ and tension formula $T = \frac{2m_1 m_2 g}{m_1 + m_2}$ save precious time.

Common Mistake

Students assume tension is the same throughout a string even when it passes over a pulley with mass. Tension is the same on both sides only for a massless, frictionless pulley. If the pulley has mass $M$ and radius $R$ , the net torque $(T_1 - T_2)R = I\alpha$ must be included. JEE Advanced frequently adds a “pulley of mass M” to catch students who skip this condition.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

Try These Next