The goal is to find $a0$ I already have the solution, however, I have a few questions.
1)In the solution they have taken $m2$'s acceleration relative to the ground to be $a0-a$ downwards. However, if $a>a0$, then wouldn't the acceleration relative to the ground end up being being upwards? In that case, how is assuming the acceleration downwards and being equal to $a0-a$ correct?
- If viewed from the accelerating frame of the movable pulley,
pseudo force upwards = $m2a0$ therefore, the eq of motion for m2: $T-m2g+m2a0 = m2a$ => $T-m2g = m2(a-a0)$
Why does the pseudo force method lead to an answer that is "biased" towards $m2$ accelerating upwards relative to the ground? i.e assuming $m2$ will be accelerating upwards from the ground frame?
The final answer is $a0 = g/(a+m1/4(1/m2 + 1/m3))$
