Pseudo force/negative acceleration

Question

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

Answer 1

$a_0-a$ , here $a$ means the acceleration of two blocks with respect to the pulley. Hence, if $a$ > $a_0$ then also this relation perfectly works. They are assuming $m_2$ to be accelerating upwards because they have preasumed that $m_3$>$m_2$. A very simple way to assume this is that suppose the whole system was at rest. Then the mass $m_2$ would move upwards and mass $m_3$ would move downwards with respect to the ground frame. Now in the present state pully B and both the masses are accelerating downwards with respect to the ground. But we still want to find the only downwards acceleration of both the masses with respect to the ground then for heavier mass acceleration is $a_0$ + $a$ and for the smaller mass is $a_0$ - $a$

Answer 2

Just the longer version of the answer @xasthor

Suppose the acceleration of $m_1$ is $a_0$ towards right. That will also be the downward acceleration of the pulley B because the string connecting $m_1$ and $B$ is constant in length. Also the string connecting $m_2$ and $m_3$ has a constant length. This implies that the decrease in the seperation between $m_2$ and $B$ equals the increase in seperation between $m_3$ and B. So, the upward acceleration of $m_2$ with respect to $B$ equals the downward acceleration of $m_3$ with respect to $B$. Let this acceleration be $a$. The acceleration of $m_2$ with respect to the ground downwards is $a_0$-$a$, and the acceleration of $m_3$ with respect to the ground ism$a_0$+$a$. These accelerations will be used in Newton's laws.

Let the tension be T in the upper string and 2T' in lower string.

Consider the motion of lower pulley. Now the forces acting on the lower pulley are T upwards by the upper string, and 2T' downwards by the lower string. As the mass of the pulley is negligible, $$2T'-T = 0$$

giving

$$T' = T/2$$ ......(I)

Now consider the motion of $m_1$:

The acceleration is $a_0$ in the horizontal direction, the forces on $m_1$ are:

T by the upper string, though also normal reaction but it won't matter since the surface is frictionless.

In the horizontal direction the equation is $$T = m_1a_0$$ ......(II)

Now take motion of $m_2$: acceleration is $a_0-a$ downwards. The forces on $m_2$ are $m_2$g forward by the earth and T' = T/2 upward by the lower string. $$m_2g - T/2 = m_2(a_0-a)$$ .....(III) (AND HERE YOU WERE MAKING THE MISTAKE)

Similarly by motion of $m_3$ we obtain $$m_3g-T/2 = m_3(a_0+a)$$ ....(IV)

Using (II), ( III) and (IV) we obtain, $$a_0-a = g-(m_1a_0/2m_2)$$ $$a_0+a = g-(m_1a_0/2m_3)$$ By Adding, $$2a_0 = 2g-(m_1a_0/2)(1/m_2 + 1/m_3)$$

which yields $$a_0 = g/1+(m_1/4)(1/m_2+1/m_3)$$