A pulley fixed to the ceiling of an elevator car carries a thread whose ends are attached to the loads of masses `m_1` and `m_2`. The car starts going up with an acceleration `w_o`. Assuming the masses of the pulley and the thread, as well as the friction, to be negligible find:
(a) the acceleration of the load `m_1` relative to the elevator shaft and relative to the car,
(b) the force exerted by the pulley on the ceiling of the car.

Explanation

Let us write Newton's second law in vector from `vecF=mvecw`, for both the blocks (in the frame of ground).
`vecT+m_1vecg=m_1vecw_1` (1)
`vecT+m_2vecg=m_2vecw_2` (2)
These two equations contain three unknown quantities `vecw_1` and `vecw_2` and `vecT`. The third equation is provided by the kinematic relationship between the accelerations:
`vecw_1=vecw_0+vecw`, `vecw_2=vecw_0-vecw` (3)
where `vecw` is the acceleration of the mass `m_1` with respect to the pulley or elevator car.
Summing up termwise the left hand and the right-hand sides of these kinematical equations, we get
`vecw_1+vecw_2=2vecw_0` (4)
The simultaneous solution of Eqs. (1), (2) and (4) yields
`vecw_1=((m_1-m_2)vecg+2m_2vecw_0)/(m_(1+)m_2)`
Using this result in Eq. (3), we get,
`vecw=(m_1-m_2)/(m_1+m_2)(vecg-vecw_0)` and `vecT=(2m_1m_2)/(m_1+m_2)(vecw_0-vecg)`
Using the results in Eq. (3) we get `vecw=(m_1-m_2)/(m_1+m_2)(vecg-vecw_0)`
(b) obviously the force exerted by the pulley on the celing of the car
`vecF=-2vecT=(4m_1m_2)/(m_1+m_2)(vecg-vecw_0)`
Note: one could also solve this problem in the frame of elevator car.