# Physics Tutorial – Atwood’s Machine

Atwood’s machine is commonly used in the classroom to demonstrate the mechanical laws of motion with constant acceleration. This device consists of two masses connected by string over a frictionless pulley. This device is commonly used in a variety of grade 11 physics problems dealing with classical mechanics.

Example:

Two unequal masses are hung vertically over a frictionless pulley. If \$m_1\$ is 1.00 Kg and \$m_2\$ is 2.00 Kg, calculate the acceleration of the system and the tension in the string \$(\$Gravity, \$g = 9.81m/s^2)\$.

Solution:

Since \$m_1 > m_2\$ the system will accelerate in favour of \$m_2\$ and therefore assigned a positive value. For mass 1 we find that:

\$(1)\$ \$ \Sigma F_y = T – m_1g = m_1a\$

For mass 2 we find that:

\$(2)\$ \$ \Sigma F_y = T – m_1g = m_1a\$

When equation (2) is subtracted from equation (1), the result is:

\$(3)\$ \$ -m_1g + m_2g = m_1a + m_2a\$

Substitute all known variables and solve:

\$(4)\$ \$ (-1.00 x 9.81) + (2 x 9.81) = 1a + 2a\$

\$(5)\$ \$ 9.81 = 3a\$

\$(6)\$ \$ a = 3.27m/s^2\$

Now that the acceleration of the system is known we can substitute the value of “a” into equation (1) to solve for T:

\$(7)\$ \$ T – m_1g = m_1a\$

\$(8)\$ \$ T – 1.00(9.81) = 1.00(3.27)\$

\$(9)\$ \$ T – 9.81 = 3.27\$

\$(10)\$ \$ T = 13.08\$ Newtons