Question

1. the masses $m_1 = 1\text{ kg}$ and $m_2 = 6\text{ kg}$, in the system shown in the figure, are attached to each other with weightless pulleys and strings. if the pulleys and the surface are smooth a) what are the accelerations of $m_1$ and $m_2$? b) what are the tensions in the strings?

Explanation:

Step1: Analyze the relationship between accelerations

Let the acceleration of \(m_1\) be \(a_1\) and the acceleration of \(m_2\) be \(a_2\). From the string - pulley system, we know that \(a_2=\frac{a_1}{2}\) (by the property of the pulley - string motion, the length of the string is related to the displacements of the masses).
Apply Newton's second law to \(m_1\): \(T_1 = m_1a_1\) (since the only horizontal force on \(m_1\) is the tension \(T_1\) as the surface is smooth).
Apply Newton's second law to \(m_2\): \(m_2g-2T_2 = m_2a_2\). Also, \(T_1 = T_2\) (because of the weightless string).

Step2: Substitute and solve for accelerations

Substitute \(a_2=\frac{a_1}{2}\) and \(T_1 = T_2\) into the equations.
We have \(T_1 = m_1a_1\) and \(m_2g-2T_1 = m_2\frac{a_1}{2}\).
Substitute \(m_1 = 1\space kg\) and \(m_2 = 6\space kg\) into \(m_2g-2T_1 = m_2\frac{a_1}{2}\), and \(T_1 = m_1a_1\).
\(6g-2\times(1\times a_1)=6\times\frac{a_1}{2}\).
\(6g-2a_1 = 3a_1\).
\(6g=3a_1 + 2a_1\).
\(5a_1=6g\). Taking \(g = 10\space m/s^2\), we get \(a_1=\frac{6\times10}{5}=12\space m/s^2\).
\(a_2=\frac{a_1}{2}=6\space m/s^2\).

Step3: Calculate the tensions

Since \(T_1 = m_1a_1\) and \(m_1 = 1\space kg\), \(a_1 = 12\space m/s^2\), then \(T_1=1\times12 = 12\space N\).
\(T_2=T_1 = 12\space N\).

Answer:

a) The acceleration of \(m_1\) is \(a_1 = 12\space m/s^2\) and the acceleration of \(m_2\) is \(a_2 = 6\space m/s^2\).
b) The tensions \(T_1=T_2 = 12\space N\).