In short
An ideal pulley is massless and frictionless — it changes the direction of tension without changing its magnitude. A fixed pulley gives no mechanical advantage but lets you pull downward to lift something upward. A movable pulley gives a mechanical advantage of 2: the load is supported by two segments of the same rope, so each carries half the weight. In systems with multiple pulleys and masses, the inextensibility of the string imposes constraint relations that link the accelerations of all connected bodies.
A woman at a village well wraps a rope around a small wheel at the top of the frame and pulls down to lift a bucket of water. She is pulling down — using her own weight to help — yet the bucket moves up. The wheel does not add any force. It does not store energy. It does not push the bucket. All it does is change the direction of the rope's tension. And that single trick — redirecting force — is the entire idea behind a pulley.
But pulleys can do more than redirect. At a construction site, a worker attaches a pulley to the load itself, threads the rope through it and back up to a fixed beam, and lifts a 100 kg cement bag by pulling with only 50 kg of force. The load has not gotten lighter. Physics has not been violated. The rope is simply arranged so that two segments share the weight, and each segment carries half. That is a movable pulley, and it gives you a mechanical advantage of 2.
This article builds everything from these two physical pictures — the village well and the construction site — and then shows you how to handle any system of pulleys, ropes, and masses by writing constraint equations that come from one simple fact: the rope does not stretch.
The ideal pulley — what you assume and why
Before diving into problems, name the assumptions explicitly.
Assumptions for an ideal pulley: The pulley is massless (it has no rotational inertia to overcome). The axle is frictionless (no torque is needed to turn it). The rope is massless and inextensible (it transmits force instantaneously without stretching or sagging).
These assumptions have a powerful consequence: the tension is the same on both sides of the pulley. Here is why. If the pulley has no mass, it has no moment of inertia (I = 0). By Newton's second law for rotation, \tau = I\alpha, any net torque — no matter how small — would cause infinite angular acceleration. Since the pulley does not spin up to infinite speed, the net torque must be zero. The only torques come from the tension on each side of the rope, acting at the same radius R. For zero net torque:
Why: the massless, frictionless assumption forces the tension to be uniform throughout the rope. This is not an approximation you can take lightly — it is the foundation of every calculation in this article.
Fixed vs movable pulleys
The distinction is simple but important.
Fixed pulley
A fixed pulley is attached to a rigid support — a ceiling, a beam, the frame of a well. It does not move. Its only job is to change the direction of the applied force. You pull down, the load goes up. The mechanical advantage is 1: you must pull with a force equal to the load's weight.
Why bother, if there is no force advantage? Because pulling downward is easier than lifting upward. You can use your body weight to help. The woman at the well does not need to reach below and push the bucket up — she stands at ground level and pulls the rope down. That directional advantage is worth the pulley even without any force multiplication.
Movable pulley
A movable pulley is attached to the load itself and moves with it. The rope passes under it and both ends go upward — one end is fixed to the support, the other is the free end you pull. Now look at the forces on the pulley: two segments of rope pull upward, each with tension T, while the load's weight Mg pulls downward. For equilibrium (or uniform acceleration):
Why: the load hangs from the pulley, and the pulley is supported by two segments of the same rope. Each segment carries T, and together they support the full weight. So you pull with force T = Mg/2 — half the weight. The mechanical advantage is 2.
But there is a trade-off. When you pull the free end of the rope by a length d, the load rises by only d/2. The rope has to shorten on both sides of the movable pulley, so the total rope you pull is twice the distance the load travels. You gain in force, you pay in distance — and the work done (F \times d) is the same either way. Energy is conserved; the pulley is not a magic machine.
The string constraint — the key to pulley problems
Every rope in an ideal pulley system is inextensible: its total length does not change. This single fact — the string length is constant — gives you a constraint equation that relates the positions, velocities, and accelerations of every object in the system.
Here is the method. It works for any configuration of pulleys and strings, no matter how complicated.
Step 1. Choose a reference line (usually the centre of the fixed pulley at the top).
Step 2. Define position variables for each mass and each movable pulley, measured downward from the reference. Call them x_1, x_2, \ldots
Step 3. Write the total length of each string in terms of these position variables. Since each string passes over pulleys, its length is the sum of the segments between pulleys and masses.
Step 4. Set the time derivative of the string length to zero (because the length is constant). This gives you a velocity constraint. Differentiate once more to get the acceleration constraint.
The key insight: differentiating the string-length equation twice gives a linear relationship between the accelerations of all connected bodies. This relationship is the constraint equation you need to solve the system.
A single movable pulley — the constraint derived
Take the simplest movable-pulley system: a fixed pulley at the top, a movable pulley carrying a load of mass M, and a person pulling the free end of the rope. Let x_p be the distance of the movable pulley below the fixed pulley, and let x_f be the length of rope the person has pulled.
The total rope length is:
Why: the rope goes from the fixed end down to the movable pulley (x_p), under the movable pulley, and back up to the fixed pulley (another x_p), then from the fixed pulley down to the person's hand (x_f). The wrap around the pulleys adds a constant (half-circumferences), which drops out on differentiation.
Since L is constant, differentiate with respect to time:
Why: if the movable pulley descends at speed v, the person must pull the rope at speed 2v to take up the slack on both sides. The negative sign means that when the pulley moves down (positive \dot{x}_p), the free end moves up (the person pulls rope in).
Differentiate again:
If the load accelerates downward at a, the free end of the rope accelerates upward at 2a. This is the constraint: the acceleration of the free end is twice the acceleration of the movable pulley.
Finding acceleration in a two-mass system
Now connect real masses and solve. This is the classic Atwood-machine variant with a movable pulley: one mass hangs directly from the rope, and the other hangs from a movable pulley.
Consider a fixed pulley at the top. A rope passes over it. On one side, a mass m_1 hangs directly. On the other side, the rope goes down to a movable pulley, under it, and back up to the ceiling. A mass m_2 hangs from the movable pulley.
The constraint equation
The rope goes from m_1 up to the fixed pulley (length x_1), across the pulley, down to the movable pulley (length x_2), under the movable pulley, and back up to the ceiling (also x_2, since the fixed hook is at the same height as the fixed pulley). So:
The constant accounts for the rope wrapped around the pulleys. Differentiate twice:
Why: if mass m_1 accelerates downward (positive a_1), the movable pulley must accelerate upward at half that rate (a_2 = -a_1/2). The factor of 2 comes from the two rope segments supporting the movable pulley.
The force equations
Draw free body diagrams for each mass.
For m_1: Weight m_1 g downward, tension T upward.
For m_2 (and the movable pulley): Weight m_2 g downward, two rope segments pull upward with tension T each.
Why: the movable pulley is massless, so its weight is zero. The net upward force on the pulley-plus-m_2 system is 2T - m_2 g. If m_2 accelerates upward, a_2 is negative (since we defined downward as positive).
Solving the system
From the constraint (1): a_1 = -2a_2, so a_2 = -a_1/2.
Substitute into equation (3):
From equation (2): T = m_1 g - m_1 a_1. Substitute into (3'):
Multiply numerator and denominator by 2:
Why: if 2m_1 > m_2, mass m_1 accelerates downward and m_2 accelerates upward. If 2m_1 < m_2, the signs reverse. The factor of 2m_1 (not m_1) appears because the movable pulley gives m_2 a mechanical advantage — it effectively acts as if it were twice as heavy in balancing m_1.
And the tension:
Worked examples
Example 1: Fixed pulley + movable pulley system
A 6 kg mass m_1 hangs from one side of a fixed pulley. The other side of the rope goes down to a movable pulley carrying a 8 kg mass m_2, then back up to the ceiling (as in the diagram above). Find the acceleration of each mass and the tension in the rope. Take g = 10 m/s^2.
Step 1. Write the constraint equation.
Why: this is equation (1) from the derivation above. If m_1 accelerates downward, the movable pulley (and m_2) accelerates upward at half the rate.
Step 2. Write Newton's second law for m_1 (taking downward as positive).
Step 3. Write Newton's second law for m_2 (taking downward as positive for consistency; m_2 accelerates upward, so a_2 is negative).
Why: two rope segments pull m_2 upward, so the upward tension is 2T. Using the constraint, a_2 = -a_1/2, so m_2 a_2 = 8 \times (-a_1/2) = -4a_1.
Step 4. Solve the system. From (i): T = 60 - 6a_1. Substitute into (ii):
Why: the positive value confirms m_1 accelerates downward, which makes sense because 2m_1 = 12 > 8 = m_2. The movable pulley effectively doubles m_1's pull on the system.
Step 5. Find the remaining quantities.
Step 6. Verify. Check Newton's second law for m_2:
Result: m_1 accelerates downward at 2.5 m/s^2. m_2 accelerates upward at 1.25 m/s^2 (half the rate, as the constraint requires). The tension in the rope is 45 N.
What this shows: The movable pulley halves the acceleration relationship: m_2 moves at half the rate of m_1 but is supported by twice the tension. Even though m_2 = 8 kg is heavier than m_1 = 6 kg, mass m_1 still goes down because the movable pulley gives it an effective mechanical advantage — the system behaves as if m_1 were competing against m_2/2 = 4 kg.
Example 2: Compound Atwood machine — three masses, two pulleys
A fixed pulley P_F is attached to the ceiling. String 2 passes over it: on one end hangs m_3 = 6 kg, and the other end is tied to the axle of a movable pulley P_M. Over this movable pulley, string 1 supports m_1 = 4 kg on one side and m_2 = 2 kg on the other. Find the acceleration of each mass and the tensions. Take g = 10 m/s^2.
Step 1. Set up coordinates and constraints. Let all positions be measured downward from the fixed pulley (positive downward). Let y be the position of the movable pulley P_M, and x_3 the position of m_3.
String 2 connects m_3 and the movable pulley over P_F:
Why: if the movable pulley goes down, m_3 goes up by the same amount, since the string connecting them is inextensible.
String 1 passes over the movable pulley. The positions of m_1 and m_2 relative to P_M sum to a constant. But in absolute terms, the acceleration of each mass is the pulley's acceleration plus its acceleration relative to the pulley. This gives:
Why: the average of a_1 and a_2 equals the acceleration of the pulley they hang from. If m_1 goes down faster than the pulley descends, m_2 must go up relative to the pulley by the same amount.
Combining: a_y = -a_3, so a_1 + a_2 = -2a_3.
Massless pulley condition: The net force on P_M is zero (since it has no mass). String 1 pulls downward on both sides with tension T_1 each. String 2 pulls upward with tension T. So:
Step 2. Write Newton's second law for each mass (downward positive).
Why: m_1 and m_2 each have weight pulling down and one strand of string 1 pulling up. Mass m_3 has weight pulling down and string 2 (tension T = 2T_1) pulling up.
Step 3. Solve. From (i): T_1 = 40 - 4a_1. From (ii): T_1 = 20 - 2a_2. Set equal:
From (iii) with T_1 = 40 - 4a_1:
Using the constraint a_3 = -(a_1 + a_2)/2:
From (A): a_2 = 2a_1 - 10. Substitute into (B):
Why: m_1 is the heaviest mass on the inner Atwood machine, so it accelerates downward.
Step 4. Find the remaining accelerations and tension.
Step 5. Verify for m_3: m_3 g - T = 60 - 56.5 = 3.5 N. And m_3 a_3 = 6 \times 10/17 = 60/17 \approx 3.5 N. \checkmark
Result:
- m_1 (4 kg): 50/17 \approx 2.94 m/s^2 downward
- m_2 (2 kg): 70/17 \approx 4.12 m/s^2 upward
- m_3 (6 kg): 10/17 \approx 0.59 m/s^2 downward
- T_1 \approx 28.2 N (string 1), T \approx 56.5 N (string 2)
What this shows: The constraint a_1 + a_2 = -2a_3 links all three accelerations through the movable pulley. The massless-pulley condition T = 2T_1 doubles the tension — so the 6 kg mass is effectively fighting the combined inertia of the 4 kg and 2 kg masses. The method is the same as Example 1: write constraint from string lengths, write Newton's law for each mass, and solve.
Common confusions
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"A pulley multiplies force." Only a movable pulley does, and even then, it is not creating force from nothing. You pull with less force but over a longer distance. The work done is the same. A fixed pulley gives no force advantage at all — mechanical advantage of 1.
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"The tension is different on the two sides of a pulley." Not for an ideal (massless, frictionless) pulley. The tension is the same throughout the rope. If the pulley has mass or friction, then yes, the tensions differ — but that is a more advanced problem (see the going-deeper section).
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"Constraint equations are just for pulleys." Not at all. Constraint equations come from any geometric restriction: a block sliding on a wedge, a rod pivoting at a hinge, a wheel rolling without slipping. The pulley constraint is just the simplest and most common example. The underlying idea — differentiate a geometric length equation — works everywhere.
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"The mechanical advantage of a system with n rope segments supporting the load is always n." This is true for an ideal system in static equilibrium. In dynamic problems (where masses accelerate), the effective "advantage" is more nuanced — the constraint relates accelerations, and the tension depends on the masses and their motion, not just the number of rope segments.
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"If the masses are equal in an Atwood machine, nothing happens." Correct for a standard Atwood machine — but in a compound system with movable pulleys, equal masses can still produce net acceleration because the constraints introduce asymmetry. The movable pulley effectively changes the mass ratio.
If you came here to solve pulley problems — write constraints, draw FBDs, solve — you have the full method. What follows is for readers who want the virtual work approach, the effect of pulley mass, and a general strategy for complex systems.
Virtual work and the pulley
The principle of virtual work offers an elegant alternative to free body diagrams for finding equilibrium conditions (and, with d'Alembert's principle, for dynamics too).
The idea: imagine giving the system a small virtual displacement consistent with the constraints. If the system is in equilibrium, the total virtual work done by all forces is zero.
For the movable pulley: if the load M moves down by a small amount \delta y, the free end of the rope must be pulled by 2\delta y (from the constraint). The work done by gravity on the load is Mg \cdot \delta y (positive, since force and displacement are in the same direction). The work done by the applied force F is -F \cdot 2\delta y (negative, since the force opposes the displacement of the rope end). For equilibrium:
Why: virtual work gives the same result as Newton's laws but without drawing a free body diagram. The constraint is built into the virtual displacement itself. For equilibrium problems, this is often faster than resolving forces.
For dynamic problems, d'Alembert's principle extends this: replace the real forces with the forces minus the "inertial forces" (-ma for each mass) and set the virtual work to zero. This is equivalent to Newton's second law but in a form that automatically handles the constraints.
What changes if the pulley has mass?
If the pulley has mass M_p and radius R, it has a moment of inertia I = \frac{1}{2}M_p R^2 (for a uniform disc). Now the tensions on the two sides are not equal. The difference in tension provides the torque that angularly accelerates the pulley:
This adds one more equation and one more unknown (T_1 \neq T_2) to the system. The method is the same — write constraints, write force equations, write the torque equation — but the algebra gets heavier.
General strategy for complex pulley systems
For a system with n masses and k strings:
- Identify all strings. Each inextensible string gives one constraint equation.
- Define position coordinates for every mass and every movable pulley, measured from fixed reference points (usually the fixed pulleys).
- Write the length of each string in terms of the position coordinates. The length is constant, so its second derivative is zero. This gives k constraint equations.
- Draw a free body diagram for each mass and each movable pulley. A massless pulley gives an algebraic relation between tensions (like T = 2T_1). A massive pulley gives a torque equation.
- Count equations and unknowns. You need n force equations (one per mass) plus k constraint equations. The unknowns are n accelerations and the tensions. If the system is properly constrained, the number of equations equals the number of unknowns.
- Solve the linear system. Substitution works for 2-3 masses. For larger systems, matrix methods are cleaner.
The key insight that makes this tractable: every pulley problem, no matter how complicated, reduces to a system of linear equations. The constraints are linear (they come from differentiating a sum of position coordinates). Newton's second law is linear in acceleration and tension. So the entire problem is a set of simultaneous linear equations, and you can always solve it.
Where this leads next
- Constraint Relations — the general framework for deriving acceleration relationships from geometric restrictions, including wedges, inclines, and rolling.
- Applications: Connected Bodies and Systems — more complex problems involving multiple bodies connected by strings, springs, and contact forces.
- Wedge and Block Systems — constraint equations for a block sliding on a wedge that is itself free to move.
- Tension in Strings and Ropes — what happens when the string has mass, and how tension varies along a heavy rope.
- Work and Energy — the energy perspective on pulley systems, including the virtual work principle in full generality.