**Atwood Machine**

An Atwood machine is a laboratory set-up of two objects connected by a massless string that is passing over an ideal pulley.

Ideal pulley is massless and frictionless. It affects nothing on the object’s motion. The tension is also constant in the whole string.

To solve problems of objects in pulley, the following guides must be followed:

- Draw a free-body diagram of each mass on the string.
- Indicate the direction of the motion of the object.
- Set-up the formula or equation to be used to solve the problem using Newton’s second law of motion and the equation for the different types of forces.

Consider two objects \(m_1\) and \(m_2\) hanging on an Atwood machine.

Since the two objects are connected by the same string over, the tension is the same. Object \(m_1\) is heavier than \(m_2\), thus \(m_1\) accelerates downward pulling \(m_2\) upward, indicating a negative and positive direction, respectively. Thus, the equation for the net force acting on each mass will be expressed as:

For \(m_1\):

\(F_{net}=T-m_1 g=-m_1 a \implies T=m_1 g – m_1 a\)

For \(m_2\):

\(F_{net}=T-m_2 g=m_2 a \implies T=m_2 g + m_2 a\)

Since the tension is the same in all the objects, the above equations can be combined as

\(m_1 g – m_1 a= m_2 g + m_2 a\)

Expressing in terms of \(g\) and \(a\)

\(m_1 g – m_2 g = m_1 a + m_2 a \implies g(m_1 – m_2)=a(m_1 + m_2)\)

It follows that the acceleration of the entire system is the same. Therefore, the acceleration can be solved as

\(a=g \begin{pmatrix} \frac {m_1 – m_2}{m_1 + m_2} \end{pmatrix}\)

**Example 1.**

What is the acceleration of the objects in the pulley as illustrated below.

Given:

\(m_1=10\;kg\\ m_2=5\;kg\)

Solution 1.

Draw a separate free-body diagram for each mass in the pulley. Indicate the direction of the acceleration of the masses.

Mass 1 accelerates downward. This is considered so based on the mass of the objects. Since \(m_1 > m_2\), it indicates that \(m_1\) will pull the string downward causing \(m_2\) to accelerate upward, giving a positive direction to \(m_2\).

Solution 2.

Calculate for the acceleration of the masses in the entire system using \(a=g \begin{pmatrix} \frac {m_1 – m_2}{m_1 + m_2} \end{pmatrix} \).

\(a=9.8\;m/s^2 \begin{pmatrix} \frac {10\;kg – 5\;kg}{10\;kg + 5\;kg} \end{pmatrix}\\\)

\(=9.8\;m/s^2 \begin{pmatrix} \frac {5\;kg}{15\;kg} \end{pmatrix}\\ \)

\(=9.8\;m/s^2 \begin{pmatrix} \frac {1}{3} \end{pmatrix}\\ \)

\(=3.27\;m/s^2\)

\(m_1\) accelerates at a rate of \(3.27\;m/s^2\;\text{downward}\) and \(m_2\) at \(3.27\;m/s^2 \;\text{upward}\).

**Example 2.**

Consider the system of masses below. Let \(a\) be the mass of the first object and \(b\) the mass of the second object.

Given:

\(a= 58\;kg\\ b= 36\;kg\)

Solution 1.

Create a free-body diagram for each object in the system.

Solution 2.

The acceleration of the objects is calculated by the equation \(a=g\begin{pmatrix}\frac{a-b}{a+b} \end{pmatrix}\).

\(a=9.8\;m/s^2 \begin{pmatrix} \frac {58\;kg – 36\;kg}{58\;kg + 36\;kg } \end{pmatrix}\\ \)

\(=9.8\;m/s^2 \begin{pmatrix} \frac {22\;kg}{94\;kg } \end{pmatrix}\\ \)

\(=9.8\;m/s^2 (0.23)\\ \)

\(=2.25\;m/s^2\)

\(a\) accelerates \(2.25\;m/s^2 \;\text{downward}\) and \(b\) at \(2.25\;m/s^2 \;\text{upward}\).