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Q5.16 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 masses' acceleration and the tension in the string when the masses are released.

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Since both the masses are connected with string they will have the same acceleration, let's say 'a'.

                                                    

We will apply Newton's law for each block individually.

For smaller the block (8 Kg)  :

                                      T\ -\ m_sg\ =\ m_sa                                 ...................................................(i)

For larger the block (12 Kg) : 

          The equation of motion is given by : 

                                     m_lg\ -\ T\ =\ m_la                                 ......................................................(ii)

Adding both equations we get : 

 The acceleration is given by :

                                      a\ =\ \left ( \frac{m_l\ -\ m_s}{m_l\ +\ m_s} \right )g

or                                         =\ \left ( \frac{12\ -\ 8}{12\ +\ 8} \right )10

or                                         =\ 2\ m/s^2

Now put the value of acceleration in any of the equations to get the value of T.

                                T\ =\ \left ( m_l\ -\ \frac{m_l^2\ -\ m_s m_l}{m_s\ +\ m_l} \right )g

or                                    =\ \left ( \frac{2m_s m_l}{m_s\ +\ m_l} \right )g

or                                    =\ \left ( \frac{2\times 12\times 8}{12\ +\ 8} \right )10

                                       =\ 96\ N

Thus the tension in the string is 96 N.

Posted by

Devendra Khairwa

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