Connected bodies and pulleys

Two bodies, one string

When a light, taut string joins two bodies, they move together: the same speed and the same size of acceleration. The string pulls each of them with the same tension $T$.

A method that always works

1. The whole system. Only the forces from outside the system drive it; the tension is inside and cancels. So $a = \dfrac{\text{net outside force}}{\text{total mass}}$.
2. One body alone. Draw its free-body diagram and apply $F = ma$ to find $T$.
3. Check with the other body.

The Atwood machine

Two masses $m_1 > m_2$ hang over a smooth pulley. The net pull is $(m_1 - m_2)g$, and it must speed up both masses:

For the lighter mass, rising: $T - m_2g = m_2a$, which gives $T = \dfrac{2m_1m_2g}{m_1 + m_2}$.

You should find that nearly equal masses barely move, which is why Atwood used this machine to measure $g$ slowly, and that the acceleration can approach $g$ but never exceed it.

Watch out. The tension is not either weight. The heavy mass falls, so $T$ is less than its weight; the light one rises, so $T$ is more than its weight.

A block on a table and a hanging block

A block of mass $M$ on a smooth table, tied over the edge to a hanging mass $m$. Only the hanging weight drives the system:

A worked example

Masses of $2.5\ \text{kg}$ and $1.5\ \text{kg}$ hang over a smooth pulley.

1. System. $a = \dfrac{(2.5 - 1.5)\ \text{kg} \times 9.8\ \text{m/s}^2}{4.0\ \text{kg}} = 2.45\ \text{m/s}^2$.
2. The lighter mass. $T = 1.5\ \text{kg} \times (9.8\ \text{m/s}^2 + 2.45\ \text{m/s}^2) = 18.4\ \text{N}$.
3. Check, heavier mass. $2.5\ \text{kg} \times 9.8\ \text{m/s}^2 - 18.4\ \text{N} = 6.1\ \text{N} = 2.5\ \text{kg} \times 2.45\ \text{m/s}^2$. It agrees.

Think first. The tension, $18.4\ \text{N}$, lies between the weights $14.7\ \text{N}$ and $24.5\ \text{N}$, as it must.

Practise with new numbers

Try it in Python

The program works out the Atwood machine for heavier and heavier left masses.

Remember these

• A light string: the same acceleration for both bodies, the same tension at both ends.
• System first for $a$, one body alone for $T$.
• Atwood: $a = \dfrac{(m_1 - m_2)g}{m_1 + m_2}$, $T = \dfrac{2m_1m_2g}{m_1 + m_2}$.
• Table and hanging block: $a = \dfrac{mg}{M + m}$, $T = Ma$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Two equal masses hang over a smooth pulley and are given a small push. What happens?

2. A $3.0\ \text{kg}$ block on a smooth table is pulled by a hanging $1.0\ \text{kg}$ block. Find the acceleration. ($g = 9.8\ \text{m/s}^2$.)

3. In question 2, what is the tension?

4. Why can the acceleration of an Atwood machine never be more than $g$?