# Constraint Relations in Pulleys

> Constraint relations for pulleys and wedges in JEE Physics: constant string length, differentiation for velocity and acceleration, movable pulley 2:1 rule.

- Canonical URL: https://prepelephant.com/topics/jee/physics/constraint-relations-pulleys
- Exam / course: JEE · Subject: Physics
- Publisher: PrepElephant (https://prepelephant.com) — Prepared and reviewed by the PrepElephant Academic Review Team
- First published: 2026-10-02
- Last updated: 2026-10-02
- How to cite: "Constraint Relations in Pulleys", PrepElephant, https://prepelephant.com/topics/jee/physics/constraint-relations-pulleys

## Direct answer

Constraint relations are the kinematic equations that follow from geometry alone, chiefly the inextensibility of string: label each segment's length along the string, write total length = constant, and differentiate — once to relate velocities, twice to relate accelerations. For any string passing over a movable pulley, the acceleration of the pulley's axle equals the average of the accelerations of the two string ends: a_pulley = (a1 + a2)/2 with consistent sign convention, which is why one end pulled with 2 m/s² makes the other accelerate at 4 m/s² if the pulley is fixed. The same logic gives contact constraints (bodies touching share acceleration components perpendicular to the surface) and the rolling constraint v = ωR. Almost every multi-pulley JEE numerical is unsolvable without first writing these relations.

## What you must remember

- **Master method:** total string length is constant; differentiate L = Σ segments = constant to get Σ(velocity of each segment) = 0, and again for accelerations.
- **Movable pulley rule:** the pulley's acceleration is the mean of the two end accelerations: a_p = (a1 + a2)/2, the relation that powers every pulled-pulley numerical.
- **Fixed pulley rule:** both ends of a single string over a fixed, frictionless pulley have equal speeds — magnitude equal, directions along the string.
- **Sign convention:** define one direction along the string as positive and measure every segment change in it; reversing signs mid-solution is the single largest source of error.
- **The classic 2:1:** one string end tied to the ceiling, then passing under a movable pulley carrying a mass: if the free end moves x, the movable pulley rises x/2, so the free end always accelerates at twice the hanging mass's acceleration.
- **Contact (wedge) constraint:** a block pressed on a wedge cannot interpenetrate, so the components of their accelerations perpendicular to the contact surface are equal.
- **Rolling constraint:** for a wheel rolling without slipping, v_contact = 0 gives v = ωR and a = αR — the same geometry-first logic.

## Building the constraint step by step

Set up the archetype: a string is tied to the ceiling, drops under a movable pulley P (carrying mass m2 = 1 kg), rises over a fixed pulley, and hangs mass m1 = 2 kg at its free end. Call y1 the length from the fixed pulley down to m1 and yP the height of the movable pulley below the ceiling tie point. The string consists of two segments to the movable pulley (down and up, each yP) plus y1 plus constants: 2yP + y1 = constant. Differentiate twice: 2a_P + a1 = 0, so m1's downward acceleration is twice the pulley's upward acceleration — magnitudes a1 = 2a2.

Now finish with Newton's laws, taking g = 10 m/s². For m1: m1g − T = m1a1 = 20 − T = 2a1. For m2 with the same tension throughout the ideal string: T − m2g = m2a2, so T − 10 = a2. Substituting a1 = 2a2: 20 − T = 4a2 and T = 10 + a2, giving a2 = 2 m/s² upward, a1 = 4 m/s² downward, T = 12 N. The habit worth internalising: geometry first, then Newton per body, then simultaneous solution.

## Where marks are lost

The classic error is guessing the relation instead of differentiating it: students write a1 = a2 for a movable-pulley arrangement that actually demands 2a1 = a2 or worse. The second loss is signs: if both ends of a string over a fixed pulley are assigned downward-positive, the velocity relation reads v1 = −v2; assign directions along the string once, at the start, and never switch. Third, taut does not mean rigid: a string can only pull, so if your solved tension comes out negative, the string is slack and the constraint relation no longer holds — re-solve as free flight. JEE Main prefers the single movable pulley with a pulled end, where "pull the rope at 2 m/s and the load rises at 1 m/s" is the expected instant recognition.

## Frequently asked questions

### How do you derive the acceleration relation for a pulley system?

Write the total length of every string segment as a constant, differentiate twice with a fixed sign convention, and the acceleration relation falls out of pure geometry.

### What is the acceleration of a movable pulley in terms of its string ends?

a_pulley = (a1 + a2)/2 — the average of the two end accelerations measured along the string, so one fixed end forces the free end to move at twice the pulley's rate.

### Why do both ends of a string over a fixed pulley have equal acceleration?

The string is inextensible: whatever length one end surrenders, the other gains, so speeds and accelerations along the string match in magnitude.

### When does a constraint relation become invalid?

The moment a string goes slack (tension would need to be negative) or a contact is lost, the geometric link breaks and the body must be reanalysed as a projectile or free body.

### How is rolling without slipping also a constraint?

The contact point has zero relative velocity, giving v = ωR and a = αR, the rotational counterpart of the same geometry-first reasoning.