Question

Two masses \[8{\text{ }}kg{\text{ }}and{\text{ }}12{\text{ }}k\]g are connected at 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

Hint:
The relation between tension and the mass is to be utilized.
All the forces acting on the bodies should be balanced accordingly.

Complete step by step solution:
 We are given with two masses, \[{m_2} = 12kg\] Smaller mass, \[{m_1} = 8kg\]

Larger mass \[{m_1}\]

Tension in the string \[ = {\text{ }}T\]

Mass owing to its weight moves downward with acceleration, and mass moves upwards.

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:
\[T - {m_1}g = {m_1}a\](i)

For mass \[{m_2}\]

The equation of motion can be written as:

 \[{m_2}g - T = {m_1}a\]… (ii)

Adding equations (i) and (ii), we get:

\[({m_2} - {m_1})g = ({m_1} + {m_2})a\]

 \[a = \dfrac{{({m_2} - {m_1})g}}{{({m_1} + {m_2})}}\]​ ....(iii)

\[a = \dfrac{{(12 - 8)10}}{{(12 + 8)}}\]
\[a = 2m{s^{ - 1}}\]

Therefore, the acceleration of the masses is \[a = 2m{s^{ - 1}}\].

Substituting the value of a in equation (ii), we get:

\[{m_2}g - T = {m_2}\dfrac{{({m_2} - {m_1})g}}{{({m_1} + {m_2})}}\]​

\[T = \left( {{m_2} - {m_2}\dfrac{{({m_2} - {m_1})}}{{({m_1} + {m_2})}}} \right)g\]

\[T = \left( {12 - 12\dfrac{{(12 - 8)}}{{(8 + 12)}}} \right)10\]
\[T = 96N\]

Therefore, the tension in the string is \[96{\text{ }}N.\]

Note:
1) Note that calculating the acceleration is important to get the tension, or else, it’ll be difficult to get the acceleration from tension.
2) The acceleration for 8kg mass is directed upwards \[\left( {hence{\text{ }} - ve} \right)\], while the acceleration for 12 kg mass is directed downwards \[\left( {hence{\text{ }} + ve} \right)\] with respect to gravity.