Question

Two masses 8 kg and 12 kg are connected at the two ends of a light inextensible string that goes over a frictionless pulley. Find the acceleration of the masses, and the tension in the string when the masses are released.

Answer 1

The given system of two masses and a pulley can be represented as shown in the following figure:

Smaller mass, \(m_1\) = \(8\; kg\) 
Larger mass, \(m_2\) = \(12 \;kg\) 
Tension in the string = \(T\) 
Mass \(m_2\), owing to its weight, moves downward with acceleration \(a\), and mass \(m_1\) moves upward.
Applying Newton’s second law of motion to the system of each mass:
For mass \(m_1\): The equation of motion can be written as:
\(T \,–\, m_1\,g\) = \(ma\) …………… (i)
For mass \(m_2\): The equation of motion can be written as:
\(m_2\,g \,– \,T\)= \(m_2a\) ………….… (ii)
Adding equations (i) and (ii), we get:
\((m_2\,-m_1)\)\(g\) = \((m_1+m_2)\)\(a\)

\(\therefore a\) = \(\bigg(\frac{m_2-m_1}{m1+m2}\bigg)g\) .................(iii)

= \(\bigg(\frac{12-8}{12+8}\bigg) \times\ 10\) 

= \(\frac{4}{20} \times 10\) = \(2 \; m/s^2\)
Therefore, the acceleration of the masses is \(2 \;m/s^2.\)
Substituting the value of \(a\) in equation (ii), we get:
\(m^2g - T\) = \(m_2\bigg(\frac{m_2-m_1}{m_1+m_2}\bigg)g\)

\(T\) = \(\bigg(\frac{m_2 - m_2^2-m_1m_2}{m_1+m_2}\bigg)g\)

= \(\bigg(\frac{2m_1m_2}{m_1+m_2}\bigg)g\)

= \(\bigg(\frac{2 \times 12 \times 8}{12+8}\bigg) \times 10\)

= \(\bigg(\frac{2 \times 12 \times 8}{20}\bigg) \times 10\)
= \(96 \,\text N\)

Therefore, the tension in the string is \(96 \,\text N\).