A pulley fixed to the ceiling of an elevator car carries a thread whose ends are attached to the loads $m_1$ and $m_2$ ($m_1$ $>$ $m_2$). The car starts going up wih an acceleration $a$ . Assuming the masses of the pulley and the thread, as well as the friction, to be negligible find the acceleration of $m_1$ relative to the elevator shaft and relative to car.

Suppose the acceleration of load $m_1$ with respect to car is a downward, then acceleration of $m_2$ will be $a$ upward.

If their accelerations are $a_1$ and $a_2$ respectively w.r.t. an observer on the ground, then

              $a_1 =(a-a_0)$ downward

  and             $a_2=(a+a_0)$ upward

By Newton’s second law, we have

                       $m_{1}g-T = m_1(a-a_0)$                 …(i)

and                $T-m_{2}g = m_2(a-a_0)$                  …(ii)

Solving equations (i) and (ii), we get

                                $a=\frac{(m_1-m_2)(g+a_0)}{(m_1+m_2)}$

                                $T=\frac{2m_1{m}_2(g+a_0)}{m_1+m_2}$

$\therefore a=a-a_0 =\frac{(m_1-m_2)g -2m_2{a}_0}{(m_1-m_2)}$