Lesson 4 · 40 min

Connected Particles and Pulleys

Hoists, elevators and conveyors join bodies with cords that run over pulleys. Each body obeys \(\sum\Fvec = m\avec\) on its own, but they are not independent: the cord's tension acts on both, and the cord's length ties their accelerations together. Write one free-body diagram per body, one equation per direction, and one equation for the cord.

Learning objectives

Draw a separate free-body diagram for each body in a connected system, with the cord's tension acting on each.
Use the ideal cord and pulley: the same tension on both sides of a light, smooth pulley.
Solve for the common acceleration and the tension of two bodies joined by a cord.
Relate the accelerations of bodies joined through a movable pulley by writing the cord's length in position coordinates.

Ideal cords and pulleys

Unless a problem says otherwise, a cord is light (its mass is negligible) and inextensible (it does not stretch), and a pulley is light and turns on a smooth axle. Two results follow.

One tension. Apply \(\sum F = ma\) to a piece of a light cord: with \(m = 0\), the forces on its ends must balance. So the tension is the same all along the cord, and the same on both sides of a light, smooth pulley.
One length. The cord's total length does not change. Write it in terms of the bodies' positions, differentiate twice, and you have an equation between their accelerations.

A cord can only pull. Its tension acts on each body it is tied to, along the cord and away from the body, so on a free-body diagram it starts where the cord is tied. If the analysis gives a negative tension, the cord has gone slack, and the assumption that it stays taut has failed.

A block on a table, pulled by a hanging block

Block \(A\) rests on a rough table. A cord tied to it runs over a pulley at the edge to block \(B\), which hangs. Before you open Figure 4.1, make a prediction.

Figure 4.1 Block \(A\) on a rough table, joined by a cord over a light, smooth pulley to block \(B\), which hangs. The forces on each block are drawn on it, to scale: the weights \(\colW{W}\) end at the centers of gravity, the table's push \(N_A\) ends on the bottom of \(A\), and friction \(\colF{F_f}\) and the cord's tension \(T\) start where they act. (Friction is never drawn shorter than \(0.25\ \text{m}\), so you can always see it.) The bars treat the two blocks as one system along the cord: \(\colW{W_B}\) drives it, friction resists, and the tension, internal to the system, does not appear.

Take each block's positive direction along its motion: to the right for \(A\), down for \(B\). The cord is inextensible, so both have the same acceleration \(a\).

\[ A{:}\quad \textstyle\sum F_y = 0{:}\ \ N_A = m_A g \qquad \sum F_x = m_A a{:}\ \ T - \mu_k m_A g = m_A a \] \[ B{:}\quad \textstyle\sum F_y = m_B a{:}\ \ m_B g - T = m_B a \]

Adding the two equations removes the tension

\[ \colMA{a} = \frac{\colW{m_B g} - \colF{\mu_k m_A g}}{m_A + m_B} \qquad\qquad T = m_B(g - a) \]

While \(B\) accelerates downward, \(T < m_B g\): the cord holds \(B\) back with less than its weight. The blocks start to move only if \(m_B g > \mu_s m_A g\); otherwise friction holds them, and \(T = m_B g\).

Adding the equations is the same as treating \(A\), \(B\) and the cord as one system, whose mass \(m_A + m_B\) is driven along the cord by \(W_B\) and held back by friction. That shortcut gives \(a\) quickly, but only a free-body diagram of one block on its own shows the tension.

Example 4.1 — The table and the hanging block

In Figure 4.1, \(m_A = 8\ \text{kg}\), \(m_B = 4\ \text{kg}\), \(\mu_s = 0.25\) and \(\mu_k = 0.2\). The blocks are released from rest. Find their acceleration, the tension in the cord, and their speed after they have moved \(1.5\ \text{m}\).

Show solution

Do they move? \(W_B = 39.24\ \text{N}\) is more than the most static friction can hold, \(\mu_s m_A g = 19.62\ \text{N}\), so yes.

\[ a = \frac{4(9.81) - 0.2(8)(9.81)}{8 + 4} = \frac{39.24 - 15.70}{12} = 1.962\ \text{m/s}^2 \] \[ T = m_B(g - a) = 4(9.81 - 1.962) = 31.39\ \text{N} \]

Check with block \(A\): \(T - 15.70 = 8(1.962)\) gives \(T = 31.39\ \text{N}\) too. The acceleration is constant, so \(v = \sqrt{2as} = \sqrt{2(1.962)(1.5)} = 2.43\ \text{m/s}\).

The Atwood machine

Hang two blocks from the ends of one cord over a fixed pulley, and the heavier one descends while the lighter one rises. With \(m_B > m_A\), up positive for \(A\) and down positive for \(B\):

\[ A{:}\quad T - m_A g = m_A a \qquad\qquad B{:}\quad m_B g - T = m_B a \]

Atwood machine

\[ \colMA{a} = \frac{m_B - m_A}{m_A + m_B}\,g \qquad\qquad T = \frac{2m_A m_B}{m_A + m_B}\,g \]

Two nearly equal masses give a small acceleration that is easy to time, which is how George Atwood measured \(g\) in 1784.

Example 4.2 — Two blocks over a pulley

Blocks of \(3\ \text{kg}\) and \(5\ \text{kg}\) hang from the ends of a cord over a light, smooth pulley. Find their acceleration, the tension, and the force on the pulley's support.

Show solution

\[ a = \frac{5 - 3}{3 + 5}(9.81) = 2.45\ \text{m/s}^2 \qquad T = \frac{2(3)(5)}{8}(9.81) = 36.8\ \text{N} \]

The cord pulls down on the pulley on both sides, so the support carries \(2T = 73.6\ \text{N}\): less than the blocks' total weight, \(78.5\ \text{N}\). The difference is the net force that accelerates the system, \((5 - 3)(2.45) = 4.9\ \text{N}\), since one block speeds up downward and the other upward.

Dependent motion: a movable pulley

When a cord wraps around a pulley that moves, the bodies no longer share one acceleration. Measure each body's position from a fixed datum, along its own line of motion, and write the cord's length in those position coordinates. Parts of the cord whose length never changes, such as the parts wrapped around pulleys, add only a constant.

Figure 4.2 Block \(A\), on a smooth table, is tied to a cord that runs over a fixed pulley, down and around a movable pulley that carries block \(B\), and up to a fixed support. \(s_A\) is measured from the fixed pulley's axle to \(A\), and \(s_B\) from the same axle down to the movable pulley's axle. The dashed outlines show where the blocks started: \(A\) always moves twice as far as \(B\).

Two lengths of cord, each \(s_B\) long, hold the movable pulley, so the cord's length is

The cord's length, in position coordinates

\[ s_A + 2s_B = \text{constant} \quad\Rightarrow\quad v_A = -2v_B \quad\Rightarrow\quad a_A = -2a_B \]

As \(B\) drops (\(s_B\) grows), \(A\) moves toward the pulley (\(s_A\) shrinks), twice as fast.

The forces change too. Two lengths of cord pull up on the movable pulley, so the free-body diagram of the pulley and \(B\) has \(2T\) up and \(m_B g\) down, while \(A\) feels only \(T\).

Example 4.3 — Twice as far, twice as fast

In Figure 4.2, \(m_A = 10\ \text{kg}\) and \(m_B = 20\ \text{kg}\); the pulleys are light and the table is smooth. Find the acceleration of each block and the tension in the cord.

Show solution

Take \(A\)'s positive direction toward the pulley and \(B\)'s downward; then the constraint reads \(a_A = 2a_B\).

\[ A{:}\quad T = m_A a_A = 10(2a_B) \qquad\qquad B{:}\quad m_B g - 2T = m_B a_B \;\Rightarrow\; 196.2 - 2(20\,a_B) = 20\,a_B \] \[ a_B = \frac{196.2}{60} = 3.27\ \text{m/s}^2 \qquad a_A = 6.54\ \text{m/s}^2 \qquad T = 65.4\ \text{N} \]

The general result is \(a_B = m_B g/(m_B + 4m_A)\). The movable pulley doubles the force on \(A\) for a given load, at the cost of making \(A\) travel twice as far.

Check your understanding

Key takeaways

Draw one free-body diagram per body. The cord's tension acts on each body it is tied to, starting where it is tied.
A light cord over a light, smooth pulley has the same tension on both sides.
A cord of fixed length ties the accelerations together: the same \(a\) for a cord over fixed pulleys, \(a_A = -2a_B\) through one movable pulley.
While a hanging block accelerates downward, the tension is less than its weight.
Treating the bodies as one system gives the acceleration quickly; the tension needs one body on its own.
Next, Lesson 5 handles forces that change with time, speed or position.