If the pulley is massless and moves with an upward acceleration $a_{0}$ , find the acceleration of $m_{1}$ and $m_{2}$ w.r.t. to elevator.

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Solution

Method 1: (Ground frame):
Let acceleration of block $m_{1}$ with respect to pulley to a upward and the acceleration of $m_{2}$ w.r.t. pulley is a (downward)
Equation of motion for $^{'} m_{1}^{'}$
$T - m_{1} g = m_{1} a_{1}$ ..........(i)
$T - m_{2} g = m_{2} a_{2}$ ..........(ii)
${\to a}_{1} = {\to a}_{1. p} + {\to a}_{p} = a + a_{0}$ ..........(iii)
${\to a}_{2} = {\to a}_{2. p} + {\to a}_{p} = - a + a_{0}$ .........(iv)
Substituting $a_{1}$ from (iii) in (i),
$T - m_{1} g = m_{1} (a + a_{0})$ ........(v)
Substituting $a_{2}$ from (iv) in (ii)
$T - m_{2} g = m_{2} (- a + a_{0})$ .......(vi)
Solving (v) and (iv),
$T = \frac{2 m_{1} m_{2}}{m_{1} + m_{2}} (g +)$
and $a = \frac{m_{2} - m_{1}}{m_{1} + m_{2}} (g + a_{0})$
Method 2 : Solving problem from non-inertial frame of reference
Let us build the equations by using Newton's second law sitting on the accelerating pulley. Hence, we impose pseudo force $m_{1} a_{0} ↓$ and $m_{2} a_{0} ↓$ on both $m_{1}$ and $m_{2}$ , respectively, in addition to the upward tension and their weights $m_{1} g ↓ m_{2} g ↓$ , 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 $m_{2} : T m_{2} g + m_{2} a_{0} - T = m_{2} a$ ........(ii)
Solving (i) and(ii), we have
$a = \frac{m_{2} - m_{1}}{m_{1} + m_{2}} (g + a_{0})$

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If the pulley is massless and moves with an upward acceleration a0, find the acceleration of m1 and m2 w.r.t. to elevator.

If the pulley is massless and moves with an upward acceleration $a_{0}$ , find the acceleration of $m_{1}$ and $m_{2}$ w.r.t. to elevator.