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The goal is to find $\boldsymbol {a_0}$ I already have the solution, however, I have a few questions.

1. In the solution they have taken $\boldsymbol {m_2}$'s acceleration relative to the ground to be $\boldsymbol {a_0-a}$ downwards. However, if $\boldsymbol {a>a_0}$, 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 $\boldsymbol {a_0-a}$ correct?

2. If viewed from the accelerating frame of the movable pulley, pseudo force upwards = $\boldsymbol {m_2a_0}$ therefore, the eq of motion for $\boldsymbol {m_2}$: $$T-m_2g+m_2a_0 = m_2a \Longrightarrow T-m_2g = m_2(a-a_0)$$

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 $\boldsymbol {a_0} = \dfrac {g}{a+ \dfrac{m_1}{4(\dfrac{1}{m_2} + \dfrac {1}{m_3})}}$

The goal is to find $\boldsymbol {a_0}$ I already have the solution, however, I have a few questions.

1. In the solution they have taken $\boldsymbol {m_2}$'s acceleration relative to the ground to be $\boldsymbol {a_0-a}$ downwards. However, if $\boldsymbol {a>a_0}$, 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 $\boldsymbol {a_0-a}$ correct?

2. If viewed from the accelerating frame of the movable pulley, pseudo force upwards = $\boldsymbol {m_2a_0}$ therefore, the eq of motion for $\boldsymbol {m_2}$: $$T-m_2g+m_2a_0 = m_2a \Longrightarrow T-m_2g = m_2(a-a_0)$$

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 $\boldsymbol {a_0} = \dfrac {g}{1+ m_1/4(\dfrac{1}{m_2} + \dfrac {1}{m_3})}$

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?

2. 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 $a0 = g/(1+(m1/4)(1/m2 + 1/m3))$

The goal is to find $\boldsymbol {a_0}$ I already have the solution, however, I have a few questions.

1. In the solution they have taken $\boldsymbol {m_2}$'s acceleration relative to the ground to be $\boldsymbol {a_0-a}$ downwards. However, if $\boldsymbol {a>a_0}$, 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 $\boldsymbol {a_0-a}$ correct?

2. If viewed from the accelerating frame of the movable pulley, pseudo force upwards = $\boldsymbol {m_2a_0}$ therefore, the eq of motion for $\boldsymbol {m_2}$: $$T-m_2g+m_2a_0 = m_2a \Longrightarrow T-m_2g = m_2(a-a_0)$$

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 $\boldsymbol {a_0} = \dfrac {g}{a+ \dfrac{m_1}{4(\dfrac{1}{m_2} + \dfrac {1}{m_3})}}$

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?

2. 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 $a0 = g/(a+m1/4(1/m2 + 1/m3))$

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?

2. 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 $a0 = g/(1+(m1/4)(1/m2 + 1/m3))$