Why can't I choose blocks attached with pulley B as a system?

Question

Three blocks of masses m1, m2 and m3 are connected as shown in the figure. All the surfaces are frictionless and the string and pulleys are light. Find the acceleration of the block of mass m1.

In this problem I know acceleration of the pulley B is same as acceleration of the block of mass $\ {m_{1}}$. But acceleration of bodies with mass $\ {m_{2}}$ and $\ {m_{3}}$ will be different since $\ {m_{2}}$ not equals $\ {m_{3}}$. So, I know that I cannot consider this(Pulley B, Blocks of mass $\ {m_{2}}$ and $\ {m_{3}}$) as a single system. But If the imagine putting those three objects in a box such that what's happening inside will not be visible to me then why I cannot consider this as a single system? The mass will thus be $\ {m_{2}}$ + $\ {m_{3}}$ and acceleration will be that of the block of mass $\ {m_{1}}$. So, I tried to find out the acceleration and I got $a=\dfrac {\left( m_{1}+m_{2}\right) g}{m_{1}+m_{2}+m_{3}}$ but, the answer is given as $\ a =\ \dfrac {g}{1+\dfrac {m_{1}}{4}\left( \dfrac {1}{m_{2}}+\dfrac {1}{m_{3}}\right) }$. So, why I am wrong here? P.S. How could I solve it in a few steps because the original solution I have is quite long.

Answer 1

The key is to realize two things.

First, since pulley B is massless it must be that $$T_A=2T_B$$ where $T_A$ and $T_B$ is the tension of the string around pulley A and pulley B respectively.

Second, since pulley B accelerates downward with the same acceleration of mass 1, and because the string around pulley B has a constant length, it must be that $a_2=-a_1+a_r$ and $a_3=-a_1-a_r$, where $a_r$ is the relative acceleration between the pulley and mass 2. Adding these relations we get $$2a_1+a_2+a_3=0$$

The above key points along with equations from N2L $$m_1a_1=T_A$$ $$m_2a_2=T_B-m_2g$$ $$m_3a_3=T_B-m_3g$$

allow us to determine an "effective mass" pulling on mass 1 by comparing to the case where the pulley B system is replaced with a single hanging mass (work left to you): $$m_{\text {eff}}=\frac{4m_2m_3}{m_2+m_3}$$

Notice how when $m_2=m_3$ we have $m_{\text {eff}}=2m_2$, which is what we would expect. Also notice if, say, $m_2\to0$ that $m_\text{eff}\to0$, which is also what we expected.

This effective mass comes from the key thing you are missing. The acceleration of masses 2 and 3 affect the total force applied to mass 1. You cannot treat the pulley B system as a "black box" whose mass is just the mass of its parts.

A simpler case to understand would be if I was in a box on a scale and you were outside the box. Let's say I was swinging on a swing hanging from the top of the box. If you were looking at the scale, you would be perplexed since the reading would oscillate up and down. Of course, I am not rapidly gaining and losing weight. The forces present "inside of the box" affected the force needed to support the box.

Answer 2

Newton's equations: Upper mass T = m1 A Hanging masses m2 g – t = m2 ( A – a ) and m3 g – t = m3 ( A + a ) where a is the acceleration of the lower cord relative to the lower pulley (assuming m3 > m2 ). Also T = 2t . Divide each of the lower equations by the corresponding mass and add to eliminate “a”: 2g - t / ((1/m2) + (1/m3)) = 2 A replacing t by T/2 and dividing by 2 gives g =[(m1/4)((1/m2) + (1/m3)) + 1] A .