Constraint Equations in Pulley Systems

What Constraint Equations Actually Mean

Pulley problems trip up almost every Class 11 student in the first month. The forces look obvious — gravity here, tension there, Newton’s second law for each block — and yet the answers come out wrong. The missing piece is the constraint equation: the geometric relationship between the accelerations of the bodies, set by the inextensible string.

Once you can write the constraint correctly, every pulley problem becomes a system of linear equations. We’ll walk through how to spot the constraint, why it works, and the patterns that show up in JEE Main, JEE Advanced and CBSE board exams.

The big idea: a string of fixed length can’t stretch. So if one end of it moves, the other end is forced to move in a coordinated way. That coordination is the constraint.

Key Terms & Definitions

Inextensible string — a string whose length stays constant no matter how much tension it carries. Real strings aren’t perfectly inextensible, but the model is good enough for nearly all introductory problems.

Light pulley — a pulley with negligible mass. Tension is the same on both sides of a light pulley.

Massless pulley + frictionless axle — together these guarantee equal tensions on both sides. If either fails, tensions differ.

Constraint equation — an algebraic relation between displacements (or velocities, or accelerations) of the bodies, derived from the geometric fact that string length is constant.

Degrees of freedom — number of independent coordinates needed to fully describe the system. Each constraint reduces the DOF by one.

Methods/Concepts

Method 1 — String length constancy

Pick a fixed reference (the ceiling, usually). Call $x_1$ the distance from the reference to block 1, $x_2$ the distance from the reference to block 2, and so on. Write the total string length in terms of these $x_i$ .

Then differentiate twice — once for velocity, twice for acceleration. Total string length is constant, so the time-derivative is zero. That gives the constraint.

If two blocks hang from a single fixed pulley with one string of length $L$ :

x_1 + x_2 = L

Differentiating: $v_1 + v_2 = 0$ , so $a_1 + a_2 = 0$ . The blocks have equal magnitude, opposite sign accelerations.

Method 2 — Velocity along the string

Sometimes the geometry is messy (blocks on inclined planes, strings at angles). Direct differentiation gets ugly. Instead, use the rule: the components of velocity along the string must be equal at the two ends (because the string is inextensible).

Decompose each block’s velocity vector into a component parallel to its string and a component perpendicular. The parallel components on the two ends must match.

For any segment of string with ends moving at velocities $\vec{v}_A$ and $\vec{v}_B$ , with $\hat{s}$ along the string from $A$ to $B$ :

\vec{v}_A \cdot \hat{s} = \vec{v}_B \cdot \hat{s}

Method 3 — Movable pulleys

A movable pulley adds an extra body to the system, plus a constraint linking its motion to the string. If a pulley is supported by a string with tension $T$ on both sides, and the pulley itself is massless, the net upward force is $2T$ , and the load attached to the pulley sees this $2T$ .

The constraint becomes: if the load attached to the movable pulley moves down by $x$ , each of the two string segments supporting it must lengthen by $x$ . So the free end of the string (wrapped over a fixed pulley) must move twice that distance.

Solved Example 1 — Atwood machine (Easy, CBSE)

Two blocks of mass $m_1 = 3$ kg and $m_2 = 5$ kg hang from a single light string over a frictionless fixed pulley. Find the acceleration and tension.

Both blocks share the same string over a single fixed pulley. Constraint: $a_1 = -a_2$ , magnitudes equal. Call $a$ the magnitude.

For $m_2$ (going down): $m_2 g - T = m_2 a$ .

For $m_1$ (going up): $T - m_1 g = m_1 a$ .

Add the equations: $(m_2 - m_1) g = (m_1 + m_2) a$ .

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

Tension: $T = m_1 (g + a) = 3 \times 12.5 = 37.5$ N.

Solved Example 2 — Movable pulley (Medium, JEE Main)

A block of mass $M$ hangs from a movable pulley. The string from the movable pulley passes over a fixed pulley and connects to a hand. The hand pulls down with constant velocity $v$ . Find the velocity of the block.

Let $x_M$ be the distance of $M$ below the fixed pulley, and $x_H$ be the length of string pulled by the hand (measured from fixed pulley downwards). Total string length: $2 x_M + x_H = L$ (the factor of 2 because the movable pulley has two string segments above it).

$2 \dot{x}_M + \dot{x}_H = 0 \Rightarrow \dot{x}_M = -\dot{x}_H / 2$ .

If hand moves down at $v$ , $\dot{x}_H = +v$ . So $\dot{x}_M = -v/2$ — block moves up at $v/2$ .

The block moves at half the speed of the hand. This is the classic “mechanical advantage” of a single movable pulley — half the speed but twice the force.

Solved Example 3 — Angled string on a wedge (Hard, JEE Advanced)

A block on a smooth incline of angle $\theta$ is connected by a string passing over a pulley at the top of the incline to a hanging block. Both blocks have mass $m$ . Find the acceleration.

The string passes over a single fixed pulley. The block on the incline moves along the incline; the hanging block moves vertically. Both speeds along the string are equal.

Block on incline: $T - mg \sin\theta = ma$ .

Hanging block: $mg - T = ma$ .

Adding: $mg(1 - \sin\theta) = 2ma$ , so $a = g(1 - \sin\theta)/2$ . Tension follows from either equation.

Exam-Specific Tips

JEE weightage: Constraint problems appear under “Laws of Motion” — typically 1-2 questions per JEE Main paper. Movable pulleys are a JEE Advanced favourite, often combined with friction or spring forces.

CBSE board: Atwood machine variants are standard. The board rarely tests movable pulleys, but Class 11 boards often include 3-mark problems on inclined-plane plus pulley setups.

NEET: Pulley questions are rare in NEET, but when they do appear, they are simple Atwood machines or single-incline + pulley combinations. Don’t over-prepare for NEET on this topic.

Common Mistakes to Avoid

Mistake 1: Assuming both blocks have the same magnitude of acceleration in every pulley problem. This is true only for a single-string single-fixed-pulley setup. Movable pulleys break this assumption.

Mistake 2: Using equal tension in every string. If two pulleys each carry their own string, the tensions in the two strings can differ — only tensions in the same string (across a frictionless light pulley) are equal.

Mistake 3: Forgetting the factor of 2 in movable-pulley problems. Two string segments support a movable pulley, so the pulley’s velocity is half the velocity of the free end of the string.

Mistake 4: Direction errors. Pick a sign convention (e.g., “down is positive”) and stick with it for every block. If you change conventions mid-problem, you’ll get sign errors that cascade.

Mistake 5: Confusing constraint on positions with constraint on accelerations. The constraint usually applies to accelerations directly only when the string is along the direction of motion. For inclined or angled strings, you may need the velocity-along-string rule.

Practice Questions

Q1. Two blocks of masses $4$ kg and $6$ kg are connected over a smooth fixed pulley. Find their common acceleration. (Take $g = 10$ m/s $^2$ .)

$a = \frac{(6-4) \times 10}{10} = 2$ m/s $^2$ . Heavier block accelerates downward, lighter upward.

Q2. A block of mass $M$ rests on a smooth horizontal table, connected by a string over a pulley at the edge to a hanging mass $m$ . Find acceleration.

Constraint gives equal acceleration magnitudes. For $m$ : $mg - T = ma$ . For $M$ : $T = Ma$ . Solve: $a = mg/(m+M)$ .

Q3. In a movable pulley system, a hand pulls one end of a string at $4$ m/s. The other end is attached to the ceiling, and the load hangs from the movable pulley. Find the load’s speed.

The load moves at $4/2 = 2$ m/s upward. Two segments support the pulley; pulling $4$ m of string lifts the load by $2$ m.

Q4. A wedge of angle $30°$ is fixed on the ground. A block on the wedge connects via a string over a pulley at the top to a hanging block. Both blocks have mass $5$ kg. Find acceleration.

$a = g(1 - \sin 30°)/2 = 10 \times 0.5 / 2 = 2.5$ m/s $^2$ .

Q5. Three blocks of mass $m$ each hang from two pulleys arranged so the middle block is supported by a movable pulley between two strings. Find the system’s acceleration.

This is a classic two-constraint problem. Tension in each outer string is $T$ , total upward force on middle pulley is $2T$ . Outer blocks: $mg - T = ma_1$ . Middle block: $2T - mg = ma_2$ . Constraint: $a_2 = (a_1 + a_3)/2$ with $a_1 = a_3$ by symmetry. Solve to get $a_1 = a_3 = g/5$ down, $a_2 = 3g/5$ up.

Q6. A string of length $\ell$ connects two blocks over a fixed pulley. Block 1 is at depth $x$ below the pulley. Find block 2’s depth.

By string length constancy: depth of block 2 $= \ell - x$ .

Q7. Two pulleys are stacked: a fixed pulley at the top and a movable pulley below it. A load of $12$ kg hangs from the movable pulley. Find the force needed to hold the load steady.

For a movable pulley, two string segments share the load: each carries $mg/2 = 60$ N. Force on the free end (over the fixed pulley) = $60$ N. Mechanical advantage of 2.

Q8. A car pulls a load up a smooth incline of angle $\theta$ via a rope passing over a pulley at the top. If the car moves horizontally at speed $v$ and the rope makes angle $\alpha$ with the horizontal at the car end, find the load’s speed up the incline.

Use the velocity-along-string rule. Component of car’s velocity along the rope $= v \cos\alpha$ . The load moves along the incline; rope at the load end is along the incline. So load’s speed along the incline $= v \cos\alpha$ .

FAQs

Q: Why is tension the same on both sides of a light pulley?

A: Apply Newton’s second law to the pulley itself. Net torque on a massless pulley with frictionless axle is zero, so tensions on the two sides must balance — i.e., be equal.

Q: When does tension differ across a pulley?

A: When the pulley has appreciable mass (and thus moment of inertia), or when there is friction at the axle. Then $T_1 - T_2 = I\alpha/r$ for the pulley.

Q: How do I decide the direction of acceleration before solving?

A: Make an educated guess. If your final answer comes out negative, the actual direction is opposite to your assumption. The magnitude is unaffected.

Q: What if the string is elastic?

A: Then the string can stretch, and the constraint $a_1 = -a_2$ no longer applies. You’d need to model the string as a spring with some stiffness $k$ . Beyond Class 11 / 12.

Q: Are constraint problems really common in JEE?

A: Very. Roughly $30\%$ of “Laws of Motion” JEE Main questions involve some constraint, usually a pulley or wedge. JEE Advanced sometimes combines pulleys with rotational motion or with springs.

Q: How is the velocity-along-string rule different from the length constraint?

A: They are the same physical idea expressed differently. Length constancy works when geometry is simple. Velocity-along-string is faster when strings are at angles. Master both — switch between them as the problem demands.

Q: Can two strings have different tensions in a single problem?

A: Yes, if they’re physically separate strings. Each string has its own tension, set by Newton’s laws on its endpoints. Only segments of the same string sharing a frictionless light pulley share tension.

Q: What does “two degrees of freedom” mean for a pulley system?

A: It means you need two independent coordinates (e.g., positions of two blocks) to fully describe the configuration. Each constraint cuts one DOF, so a system with $n$ blocks and $k$ independent strings has $n - k$ DOF.