If the pulley is massles and moves with an upward acceleration `a_(0)`. Find the acceleration of `(m_1)and (m_2)` w.r.t to elevator.

Explanation

Method 1: (Ground frame) :

Let acceleration of block `m_(1)` with respect to pulley be a (upward) and the acceleration of `m_(2)` w.r.t pulley is a (down ward)
Equation of motion for '`m_(1)`'
`T-m_(1)g=m_(1)a_(1)` ...(i)
`T-m_(2)g=m_(2)a_(2)` ...(i)
`vec(a)_(1)=vec(a)_(1,p)+vec(a)_(p)=a+a_(0)` ...(iii)
`vec(a)_(2)=vec(a)_(2,p)+vec(a)_(p)= -a+a_(0)` ...(iv)
subsituting `a_(1)` from (iii) in (i),
`T-m_(1)g=m_(1)(a+a_(0))` ...(v)
Subsituting `a_(2)` from (iv) in (ii)
`T-m_(2)g=m_(2)(-a+a_(0))` ...(vi)
Solving (v) and (iv),
`T=(2m_(1)m_(2))/(m_(1)+m_(2))(g+a_(0))`
and `a=(m_(2)-m_(1))/(m_(1)+m_(2))(g+a_(0))` .
Method 2: Solving problem from non-inertial frame of reference
Let us build the equations by using Newton's second law sittion on the accelerationg pulley. Hence, we impose pseudo force `m_(1)a_(0)` darr and `m_(2)a_(0)darr` on both `m_(1)` and `m_(2)`, respectively, in addition to the upward tension and their weight 1 `m_(1)gdarr m_(2)g darr`, respectively. If `m_(1)` accelerates up relative to the pulley, `m_(2)` must accelerate down relative to the pulley with acceleration a.

Force equation:
for `m_(1):T-m_(1)g-m_(1)a_(0)=m_(1)a` ..(i)
for `m_(2):m_(2)a_(0)-T=m_(2)a`...(ii)
Solving (i) and (ii) we have
`a=(m_(2)-m_(1))/(m_(1)+m_(2))(g+a_(0))`.