2.2.6.2. Pulley Systems

Table of Contents

Idealized pulley systems

A pulley system is a common application of Newton's laws in which ropes connect two or more objects so that the motion of one object affects the motion of the others. The main challenge is not learning a new force, but carefully applying familiar forces, especially tension and weight, together with the motion constraints created by the rope.

In introductory physics, pulley problems are usually treated with an ideal model. The rope is taken to be massless and inextensible, and the pulley is taken to be massless and frictionless. Under these assumptions, the tension has the same magnitude everywhere along a single continuous rope, and the rope length stays constant.

For an ideal pulley with a single massless rope, the tension is the same throughout that rope.

Why pulleys are useful

A pulley does not create energy or force from nothing. Its role is to change the direction of a force, or in more complicated arrangements, to allow a smaller force to act over a larger distance. In this chapter, the main focus is how pulleys connect the accelerations of different objects.

A simple fixed pulley changes only the direction of the tension force. If one mass goes down, the other goes up with the same speed and the same magnitude of acceleration.

A movable pulley can also change how the force is distributed. In that case, one object may be supported by two rope segments, so the upward pull can become $2T$ instead of $T$.

Two masses connected by a rope over a pulley

The most common pulley system has two masses, $m_1$ and $m_2$, connected by a rope passing over a pulley. This is often called an Atwood machine.

If $m_2 > m_1$, then $m_2$ accelerates downward and $m_1$ accelerates upward. Because the rope length is fixed, both masses have accelerations of equal magnitude.

Let the magnitude of the acceleration be $a$. Choose positive upward for $m_1$ and positive downward for $m_2$. Then Newton's second law gives

$$
T - m_1 g = m_1 a
$$

and

$$
m_2 g - T = m_2 a
$$

Adding the equations eliminates $T$:

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

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

Then the tension is found from either mass:

$$
T = m_1 g + m_1 a
$$

$$
T = m_2 g - m_2 a
$$

Combining these gives

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

For an ideal Atwood machine,
$$
a = \frac{(m_2 - m_1)g}{m_1 + m_2}
$$
with the heavier mass moving downward.

Visualizing the Atwood machine

Rendering scientific sketch…

Two masses over a fixed pulley

Motion constraint from the rope

The most important idea in pulley systems is often not the force equation, but the geometric constraint from the rope. If the rope length is constant, then the displacements of connected parts must be related.

For the simple two mass system, if one mass moves up by a distance $x$, the other moves down by the same distance $x$. Therefore,

v_1

v_2

, \qquad

a_1

a_2

The directions are opposite, but the magnitudes are equal.

In more complicated systems, this relation changes. For example, if one mass is attached to a movable pulley supported by two rope segments, the supported mass may move only half as far as the free end of the rope.

Always write the rope length constraint before solving a complex pulley problem. It determines how the accelerations are related.

A movable pulley

Suppose a load is attached to a pulley that is itself supported by two segments of the same rope. If the tension in each segment is $T$, then the total upward force on the load-pulley combination is $2T$.

If the load has mass $m$, and upward is positive, then

$$
2T - mg = ma
$$

This is one of the key differences between fixed and movable pulleys. A fixed pulley gives one upward rope force, while a movable pulley often gives two.

The displacement relation is also different. If the load rises by a distance $y$, each supporting rope segment shortens by $y$, so the free end of the rope must move by $2y$. Therefore,

$$
x_{\text{free end}} = 2y_{\text{load}}
$$

and similarly for speeds and accelerations,

$$
v_{\text{free end}} = 2v_{\text{load}}, \qquad a_{\text{free end}} = 2a_{\text{load}}
$$

in magnitude.

Visualizing a movable pulley

Rendering scientific sketch…

Movable pulley supporting a load

How to solve pulley problems

The method is systematic. First identify each object that may move. Then draw a separate free body diagram for each one. Next choose positive directions, usually in the expected direction of motion. Then write Newton's second law for each object. Finally add the rope constraint equations that relate the accelerations.

The unknowns are often tension and acceleration. If the pulley is ideal and only one rope is involved, the same $T$ appears in every segment of that rope. If there are multiple ropes, each rope may have its own tension.

A helpful summary is shown below.

Step	What to do
1	Identify all moving objects
2	Draw free body diagrams
3	Choose positive directions
4	Write $\sum F = ma$ for each object
5	Use rope length constraints
6	Solve the simultaneous equations

Common mistakes

A frequent mistake is assuming that the same tension means the same force on every object. The tension may be the same in each rope segment, but an object can be pulled by more than one segment. That is why a movable pulley can produce an upward force of $2T$.

Another common mistake is forgetting that connected objects may have different accelerations in compound pulley systems. Equal acceleration magnitude is true for the simple two mass pulley, but not for every pulley arrangement.

Sign errors are also common. The safest approach is to define the positive direction clearly for each object before writing the equations.

Same rope tension does not mean same net force. Count how many rope segments pull on each object.

A short worked example

Consider two masses, $m_1 = 2\,\text{kg}$ and $m_2 = 5\,\text{kg}$, connected by an ideal rope over an ideal pulley.

The acceleration is

$$
a = \frac{(5 - 2)g}{5 + 2} = \frac{3g}{7}
$$

Using $g = 9.8\,\text{m/s}^2$,

$$
a = \frac{3 \times 9.8}{7} = 4.2\,\text{m/s}^2
$$

approximately.

The tension is

$$
T = m_1(g+a) = 2(9.8 + 4.2) = 28\,\text{N}
$$

So the heavier mass moves downward at $4.2\,\text{m/s}^2$, the lighter mass moves upward at the same magnitude of acceleration, and the rope tension is $28\,\text{N}$.

Physical interpretation

Pulley systems are valuable because they turn a multi object motion problem into a set of linked force equations. Newton's laws still apply exactly as usual, but the rope creates a connection between the motions of different objects.

The key ideas are simple: use free body diagrams, remember that an ideal rope has uniform tension, and use the constant rope length to relate displacements, velocities, and accelerations.

In pulley systems, the full solution always combines two ingredients:
$$
\sum F = ma
$$
for each object, and the rope constraint from constant rope length.

Final remarks

Pulley systems are among the best examples of how Newton's laws handle interacting bodies. Once the forces are drawn correctly and the rope constraint is written carefully, even complicated systems can be solved in a clear and organized way.

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