Constraint Relations

In short

A constraint relation is an equation that links the accelerations (or velocities) of connected bodies. For an inextensible string, the total string length L is constant — differentiate L = \text{const} once for velocity relations and twice for acceleration relations. In a simple Atwood machine, the constraint gives a_1 = -a_2. In a double Atwood machine with a movable pulley, the string geometry gives a_1 = 2a_2 — the block on one side accelerates twice as fast as the movable pulley.

A well in a village has a rope that runs over a pulley at the top. One end is tied to a bucket, the other end you hold in your hands. When you pull the rope down by one metre, the bucket rises by exactly one metre. Not approximately one metre — exactly one metre. The rope does not stretch, it does not compress, and it does not pass through the pulley like a ghost. These everyday facts — a rope keeps its length, a surface stays solid — are the physical basis of constraint relations, and they are the single most powerful tool for solving connected-body problems in mechanics.

If you have ever set up a free body diagram for two blocks connected by a string and wondered, "I have two unknowns for acceleration but only Newton's second law for each block — where is the missing equation?", the answer is the constraint relation. It is the equation that the string (or surface, or rod) gives you for free, because of the physical fact that these connectors do not change their length or shape.

What constraints are

A constraint is a physical restriction on how a body can move. The three most common constraints in mechanics are:

Inextensible string — the string does not stretch. Its total length stays constant no matter how the system moves.
Rigid surface — the block stays on the surface. It can slide along the surface, but it cannot sink through it or fly off it (as long as the normal force remains positive).
Rigid rod — the rod does not stretch or compress. The distance between its two ends is fixed.

Each constraint removes one degree of freedom from the system and, in return, gives you one equation relating the motions of the connected bodies.

The three common constraints in mechanics. Each one restricts motion and provides one equation linking the accelerations of the connected bodies.

Think of it this way: Newton's second law tells you what happens because of the forces. The constraint tells you what is physically impossible. Together, they give you enough equations to solve the problem.

The inextensible string constraint — the master technique

The most common constraint in JEE and board-exam problems involves a string that does not stretch. The technique for extracting the constraint equation is always the same, regardless of how complicated the pulley arrangement looks:

The method: Write down the total length of the string in terms of the position coordinates of the bodies it connects. Set L = \text{constant}. Differentiate once with respect to time to get a velocity relation. Differentiate again to get an acceleration relation.

This works because the string length is a geometric fact — it depends only on positions. Differentiating a geometric relationship with respect to time converts it into a kinematic relationship: positions become velocities, velocities become accelerations.

Why differentiating works

Suppose two blocks are connected by a string of fixed length L. You describe the position of block 1 by a coordinate x_1 (measured from some fixed reference point) and block 2 by x_2. The string length is some function of these positions:

L = f(x_1, x_2)

Since the string does not stretch, L does not change with time:

\frac{dL}{dt} = 0

By the chain rule:

\frac{\partial f}{\partial x_1} \cdot \dot{x}_1 + \frac{\partial f}{\partial x_2} \cdot \dot{x}_2 = 0

Why: the chain rule converts a position relationship into a velocity relationship. Each \dot{x} is a velocity.

Differentiate once more:

\frac{\partial f}{\partial x_1} \cdot \ddot{x}_1 + \frac{\partial f}{\partial x_2} \cdot \ddot{x}_2 + \text{(terms involving } \dot{x}^2 \text{)} = 0

Why: the second differentiation gives you accelerations (\ddot{x}). The extra terms involving \dot{x}^2 vanish for the simple cases (straight strings) and matter only when the geometry is curved. For pulleys with straight segments, you get a clean linear relation between accelerations.

This is the engine of constraint relations. Now let us see it in action.

The simple Atwood machine

The simplest connected-body problem: two blocks of masses m_1 and m_2 hang from the two ends of a single inextensible string that passes over a fixed, frictionless, massless pulley.

Assumptions: The string is massless and inextensible. The pulley is massless and frictionless (so the tension is the same throughout the string). Air resistance is negligible.

The simple Atwood machine. The coordinate $x_1$ measures how far $m_1$ is below the pulley centre (positive downward), and $x_2$ does the same for $m_2$. The string segment lengths are $\ell_1 = x_1$ and $\ell_2 = x_2$ (ignoring the arc over the pulley, which is constant).

Writing the string-length equation

Define x_1 as the distance of m_1 below the centre of the pulley (measured along the string), and x_2 as the distance of m_2 below the centre. Both are positive downward.

The total string length is:

L = x_1 + x_2 + \underbrace{C}_{\text{string wrapped over the pulley}}

Why: the string has three parts — a vertical segment of length x_1 on the left, a vertical segment of length x_2 on the right, and a curved portion that wraps around the pulley. The curved portion has a fixed length C = \pi R (half the pulley circumference) that does not change as the blocks move.

Differentiating for velocity

Since L and C are constants:

\frac{dL}{dt} = \frac{dx_1}{dt} + \frac{dx_2}{dt} + 0 = 0

v_1 + v_2 = 0

\boxed{v_1 = -v_2}

Why: if m_1 descends (positive v_1), m_2 must ascend at exactly the same speed (negative v_2, since upward is negative in our convention). The rope feeds from one side to the other — one block's loss of rope is the other block's gain.

Differentiating for acceleration

Differentiate the velocity relation:

\frac{dv_1}{dt} + \frac{dv_2}{dt} = 0

\boxed{a_1 = -a_2}

Why: the accelerations are equal in magnitude and opposite in sign. If you call the downward acceleration of m_1 as a, then m_2 accelerates upward at a. This is the constraint equation you need alongside Newton's second law to solve the Atwood machine.

This result feels obvious — "of course, if one block goes down, the other goes up at the same rate." But writing it formally as a differentiated string-length equation matters, because the same technique generalises to problems where the answer is not obvious at all. The double Atwood machine, which you will see shortly, is the proof.

Wedge constraints

Not every constraint involves a string. When a block sits on a wedge, the constraint is that the block must move along the wedge surface — it cannot pass through the incline or hover above it.

A block of mass $m$ on a wedge of mass $M$ and angle $\theta$. The wedge is free to slide on a frictionless floor. The block slides along the incline — it cannot lift off or dig into the surface. This geometric fact is the wedge constraint.

The constraint here is: the block must remain on the wedge surface. If the wedge moves horizontally by \Delta X to the right and the block slides a distance \Delta s down the incline (relative to the wedge), then the block's displacement in the lab frame has two components:

Horizontal: \Delta X + \Delta s \cos\theta (wedge carries the block, plus the block's own horizontal motion along the incline)
Vertical: \Delta s \sin\theta (the block descends along the incline)

The acceleration constraint follows by differentiating twice. If A is the wedge's horizontal acceleration and a is the block's acceleration along the incline relative to the wedge:

a_{x,\text{block}} = A + a\cos\theta

a_{y,\text{block}} = a\sin\theta

Why: these are just the components of the block's absolute acceleration. The A term comes from the block being carried along with the wedge. The a\cos\theta and a\sin\theta terms come from the block's motion along the incline. The constraint forces you to write the block's lab-frame acceleration in terms of only two unknowns (A and a) instead of three.

These two equations, combined with Newton's second law for the block (two component equations) and for the wedge (one horizontal equation), give you five equations for five unknowns (A, a, normal force N between block and wedge, and the block's two acceleration components). The system is solvable.

The key insight is that the wedge constraint reduces the number of independent degrees of freedom. Without the constraint, the block could move in any direction in the plane (2 degrees of freedom), and the wedge has 1 degree of freedom — a total of 3. The constraint says "the block stays on the surface," which eliminates 1 degree of freedom, leaving 2: the wedge position X and the block's position along the incline s.

Writing constraint equations — the systematic method

Here is the step-by-step procedure that works for every string-and-pulley problem:

Step 1. Choose a fixed reference point (usually the pulley axle or a fixed support).

Step 2. Define a position coordinate for each body, measured from the fixed reference. Choose the positive direction so that increasing the coordinate means the string on that side gets longer.

Step 3. Write the total string length in terms of these coordinates:

L = x_1 + x_2 + \text{constant segments}

Any part of the string whose length does not change (like the arc over a fixed pulley) is a constant — lump it into a single constant C.

Step 4. Differentiate once to get the velocity relation:

\dot{x}_1 + \dot{x}_2 = 0 \quad \text{(for a single string over one fixed pulley)}

Step 5. Differentiate again to get the acceleration relation:

\ddot{x}_1 + \ddot{x}_2 = 0

Step 6. Interpret physically: assign a_1 = \ddot{x}_1 and a_2 = \ddot{x}_2, and you have the constraint.

The power of this method is that it is mechanical — you do not need physical intuition to derive the constraint. You just need to carefully account for all the string segments. The geometry does the thinking for you.

Worked examples

Example 1: Simple Atwood machine — finding the acceleration

Two blocks of mass 3 kg and 5 kg hang from the ends of a light, inextensible string that passes over a fixed, frictionless, massless pulley. Find the acceleration of each block and the tension in the string.

The Atwood machine with $m_1 = 3$ kg and $m_2 = 5$ kg. Each block has two forces: tension $T$ upward and weight $mg$ downward. The coordinates $x_1$ and $x_2$ are measured downward from the pulley centre.

Step 1. Write the constraint equation.

The total string length is L = x_1 + x_2 + C, where C is the arc over the pulley.

Differentiate twice:

a_1 + a_2 = 0 \implies a_1 = -a_2

Why: the string is inextensible, so the total length cannot change. If the 5 kg block descends (positive a_2), the 3 kg block ascends (a_1 is negative, meaning upward). Call the magnitude a: then a_2 = a (5 kg goes down) and a_1 = -a (3 kg goes up).

Step 2. Apply Newton's second law to each block.

Since the heavier block (m_2 = 5 kg) descends, take its downward direction as positive acceleration a.

For m_2 (taking downward positive):

m_2 g - T = m_2 a \tag{1}

5(9.8) - T = 5a

For m_1 (taking upward positive, since it accelerates upward):

T - m_1 g = m_1 a \tag{2}

T - 3(9.8) = 3a

Why: for each block, the net force in the direction of acceleration equals ma. The tension is the same on both sides because the pulley is massless and frictionless.

Step 3. Solve the system.

Add equations (1) and (2):

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

a = \frac{(m_2 - m_1)}{(m_1 + m_2)} \cdot g = \frac{5 - 3}{5 + 3} \times 9.8 = \frac{2}{8} \times 9.8

\boxed{a = 2.45 \text{ m/s}^2}

Why: adding the equations eliminates T. The result shows that the acceleration depends only on the mass difference (m_2 - m_1) and the total mass (m_1 + m_2). If the masses were equal, a = 0 — the system would be in equilibrium.

Step 4. Find the tension.

Substitute back into equation (2):

T = m_1(g + a) = 3(9.8 + 2.45) = 3 \times 12.25

\boxed{T = 36.75 \text{ N}}

Why: the tension must be between m_1 g = 29.4 N and m_2 g = 49 N. It is closer to the lighter block's weight — this makes sense because the system accelerates at only 2.45 m/s^2, not 9.8 m/s^2. The string is "holding back" the heavy block while "pulling up" the light one.

What this shows: The constraint equation a_1 = -a_2 was the crucial link. Without it, you would have two equations (Newton's law for each block) but three unknowns (a_1, a_2, and T). The string constraint provides the third equation. This is a pattern you will see in every connected-body problem: Newton's laws provide as many equations as there are bodies, and constraints provide the remaining equations needed to close the system.

Example 2: Double Atwood machine (movable pulley) — the $a_1 = 2a_2$ constraint

A block of mass m_1 hangs from one end of a string. The string passes over a fixed pulley at the top, and its other end is attached to a small, massless, movable pulley. Two more blocks could hang from this movable pulley, but for now, focus on the geometry: the string connecting m_1 to the movable pulley. A second block of mass m_2 hangs from the axle of the movable pulley. Derive the constraint relating the accelerations of m_1 and the movable pulley.

The double Atwood machine. Block $m_1$ hangs on the left. The string passes over the fixed pulley and connects to a movable pulley, from whose axle $m_2$ hangs. The coordinate $x_1$ is the distance of $m_1$ below the fixed pulley; $x_2$ is the distance of the movable pulley below the fixed pulley. The string length is $L = x_1 + x_2 + C$.

Now here is where the double Atwood machine makes things interesting. The movable pulley can itself move up or down. When the movable pulley descends by a distance \Delta x_2, both segments of string on either side of the movable pulley must get longer to accommodate its new position. The string feeding from the fixed pulley down to the movable pulley gets longer by \Delta x_2. But the string that wraps around the movable pulley and goes back up also gets longer by \Delta x_2. So the total extra string consumed by the movable pulley's descent is 2\Delta x_2.

That extra string must come from somewhere, and the only source is the segment on m_1's side. So m_1 must rise by 2\Delta x_2.

Let us derive this rigorously.

Step 1. Identify all string segments.

Look at the single string that runs from m_1 to its attachment point. It has these parts:

Segment from m_1 up to the fixed pulley: length = x_1
Arc over the fixed pulley: constant = C_1
Segment from fixed pulley down to the top of the movable pulley: length = x_2

But wait — the string wraps around the movable pulley too. After going down the right side to the movable pulley, the string wraps under it and goes back up. The string on the far side of the movable pulley goes from the bottom of the movable pulley back up to a fixed attachment point. That far-side segment has length (h - x_2), where h is the height of the attachment point above the fixed pulley (a constant).

Actually, let us simplify. The key insight is that the total string passing over the movable pulley involves two segments that both change when the movable pulley moves. Let the position of the movable pulley below the fixed pulley be x_2. Then:

Left segment (from m_1 to fixed pulley): \ell_1 = x_1
Right segment (from fixed pulley to movable pulley): \ell_2 = x_2
Arc over fixed pulley: C_1 (constant)
Arc under movable pulley: C_2 (constant)
Far segment (from movable pulley back up to ceiling): \ell_3 = (H - x_2), where H is the ceiling height — but actually, this segment connects back to the ceiling, so its length also depends on x_2.

The clearest approach: the total string length is

L = x_1 + C_1 + 2x_2 + C_2

Wait — let me count more carefully. The string goes:

From m_1 up to the fixed pulley — length x_1
Over the fixed pulley — constant C_1
Down from the fixed pulley to the movable pulley — length x_2
Under the movable pulley — constant C_2
Back up from the movable pulley to a fixed point on the ceiling — length x_2 (because the fixed point is directly above the fixed pulley, and the movable pulley is at depth x_2... but this is getting complicated).

Here is the clean way to set it up. Define x_1 = distance of m_1 below the fixed pulley, x_2 = distance of the movable pulley below the fixed pulley. Assume the far end of the string is fixed to the ceiling directly above the fixed pulley.

The string has five parts:

m_1 to fixed pulley: x_1
Arc over fixed pulley: C_1
Fixed pulley to movable pulley: x_2
Arc under movable pulley: C_2
Movable pulley back up to ceiling attachment: x_2 (since the attachment is at the same height as the fixed pulley centre, this segment equals x_2)

L = x_1 + C_1 + x_2 + C_2 + x_2 = x_1 + 2x_2 + (C_1 + C_2)

Why: the movable pulley is "reached" by two segments of string — one coming down from the fixed pulley (x_2) and one going back up to the ceiling (x_2). Both change when the pulley moves. That factor of 2 is where the 2a_2 = a_1 relation comes from.

Step 2. Differentiate for velocity.

\frac{dL}{dt} = 0 \implies \dot{x}_1 + 2\dot{x}_2 = 0

v_1 + 2v_2 = 0

\boxed{v_1 = -2v_2}

Why: if the movable pulley descends at speed v_2, block m_1 must rise at speed 2v_2. The movable pulley consumes string from both sides, so m_1's side must feed string twice as fast.

Step 3. Differentiate for acceleration.

\ddot{x}_1 + 2\ddot{x}_2 = 0

\boxed{a_1 = -2a_2}

Why: the magnitude of m_1's acceleration is twice the magnitude of the movable pulley's acceleration. This is the famous a_1 = 2a_2 constraint of the double Atwood machine (with the sign indicating opposite directions).

Physical check: Imagine pulling the free end of the string by 10 cm. The movable pulley rises by only 5 cm — you can verify this with a real rope and a real pulley. Each side of the movable pulley gains 5 cm, consuming the full 10 cm of slack. This 2:1 ratio is exactly what our constraint equation predicts.

What this shows: The "differentiate the string length" method works even when the answer is not obvious. In the simple Atwood machine, you might guess a_1 = -a_2 from intuition. In the double Atwood machine, the factor of 2 is not intuitive — it emerges from carefully counting how many string segments depend on each coordinate. The method is systematic, reliable, and generalises to any number of pulleys.

Common confusions

"The accelerations of two blocks connected by a string must be equal." Not always. This is true only when the string runs directly between the two blocks (or over fixed pulleys). When a movable pulley is involved, the constraint is a_1 = na_2 where n depends on how many string segments are attached to the movable pulley. Always count the segments — do not assume n = 1.
"The constraint relates the accelerations of the two blocks." Be careful about which blocks. In a movable-pulley problem, the constraint relates the acceleration of m_1 to the acceleration of the pulley, not directly to the acceleration of the block hanging from the pulley (unless the pulley and block move together).
"I can use the constraint equation from Problem A in Problem B." No. Every constraint equation is specific to its string geometry. A new pulley arrangement means a new string-length equation and a new constraint. Always derive it from scratch by counting string segments.
"Differentiating once gives me the acceleration." No — differentiating once gives velocity, differentiating twice gives acceleration. The chain is: L = \text{const} \to v_1 + v_2 = 0 \to a_1 + a_2 = 0.
"The string wraps around the pulley, so I need to account for the pulley's rotation." For a massless pulley, you do not. The arc length over the pulley is constant (it equals \pi R) and drops out when you differentiate. Pulley rotation matters only when the pulley has mass (then you need \tau = I\alpha as an additional equation).

If you are comfortable with single-string and double-string constraints and can set them up reliably, you have what you need for most board-exam and JEE Mains problems. The material below covers more complex systems — multi-string configurations and the general virtual-work method — for JEE Advanced preparation.

Multi-string systems

Some problems use more than one string. Each string gives one independent constraint equation. If you have two strings, you get two constraint equations. The rule is: one string, one constraint.

Consider a system where block A is connected by string 1 to a movable pulley P, and block B is connected by string 2 to the same movable pulley P, with block C hanging from P. Here you have two strings and three blocks. String 1 gives one equation relating a_A and the pulley acceleration a_P. String 2 gives another equation relating a_B and a_P. Together with Newton's second law for each of the three blocks (and the massless pulley condition \Sigma F = 0 on the pulley), you have enough equations.

The procedure is the same: for each string, write its total length, set the derivative to zero, differentiate twice. Then solve the combined system.

The virtual displacement method

There is an alternative to differentiating string lengths: the method of virtual displacements. You imagine giving one body a small displacement \delta x_1 and ask: what displacement must every other connected body undergo, consistent with the constraints?

For the simple Atwood machine: if m_1 moves down by \delta x_1, the string on m_1's side gets longer by \delta x_1. Since the total string length is fixed, the string on m_2's side must get shorter by \delta x_1, so m_2 moves up by \delta x_1. Thus \delta x_2 = -\delta x_1.

For the movable pulley: if the movable pulley moves down by \delta x_2, both segments attached to it get longer by \delta x_2 each, for a total of 2\delta x_2. This string must come from m_1's side, so \delta x_1 = -2\delta x_2.

The virtual displacement method is equivalent to differentiating the string length — it is the same physics, just expressed as "what-if" reasoning instead of calculus. Some students find one approach more intuitive than the other; use whichever comes naturally.

When the constraint is nonlinear

All the examples so far used strings running over pulleys, giving linear constraint equations. But when the geometry is curved — for instance, a bead sliding on a circular wire while attached by a string to a hanging weight — the string length is a nonlinear function of the coordinates. Differentiating still works, but the second derivative produces extra terms involving \dot{x}^2 (the squared velocities). These terms represent the centripetal effects of the curved geometry.

For example, if a rod of length L has one end sliding along the floor (position x) and the other end sliding down a wall, the constraint is:

x^2 + y^2 = L^2

Differentiate: 2x\dot{x} + 2y\dot{y} = 0, so x v_x + y v_y = 0.

Differentiate again: x a_x + y a_y + v_x^2 + v_y^2 = 0.

The v_x^2 + v_y^2 term is new — it did not appear in the straight-string problems. It arises because the geometry is curved, and it means the acceleration constraint depends on the current velocities, not just the positions. This is a genuinely harder class of problems, and it shows why the systematic differentiation method is essential: you cannot guess these extra terms by intuition.

Where this leads next

Pulleys and String Constraints — more complex pulley arrangements with multiple movable pulleys.
Wedge and Block Systems — detailed treatment of blocks sliding on accelerating wedges.
Free Body Diagrams — the companion skill: draw the forces before you write the constraints.
Applications: Connected Bodies and Systems — full worked problems combining FBDs, constraints, and Newton's laws.
Newton's Laws of Motion — the foundation: the laws that the constraint equations supplement.