Masses $\mathrm{M_1}$ and $\mathrm{M_2}$ are connected to a system of strings and pulleys as shown. The strings are massless and inextensible, and the pulleys are massless and frictionless. Find the acceleration of $\mathrm{M_1}$.
Clue: If $\mathrm{M_2} = \mathrm{M_1}$, acceleration ($A$) will be $A = \dfrac{g}{5}$
Source: An Introduction to Mechanics – Kleppner & Kolenkow
My attempt:
Let $T$ be the tension of the rope connected to $\mathrm{M_2}$. So the tension of the rope connected to $\mathrm{M_1}$ will be $2T$.
The acceleration of both the masses is $A$.
Now,
$\mathrm{M_2}g – T = \mathrm{M_2}A \label{1}\tag{1}$
$2T – \mathrm{M_1}g = \mathrm{M_1}A \label{2}\tag{2}$
From $\ref{1}$ and \ref{2},
$$A = \dfrac{g(2\mathrm{M_2} – \mathrm{M_1})}{(2\mathrm{M_2} + \mathrm{M_1})}$$
Now if $\mathrm{M_2} = \mathrm{M_1}$,
I get,
$A = \dfrac{g}{3}$.
But the answer is $A = \dfrac{g}{5}$
Where am I wrong?
Will the accelerations of $\mathrm{M_1}$ and $\mathrm{M_2}$ not be the same? or, are there anything about the tensions ?
Best Answer
Start from the beginning. Why constraint relations? Why are they there? Let me emphasize :
Let's take origin at top pulley which is at rest.
Note that length of top rope is constant : $a+b=k\implies a''+b''=0 \implies a''=-b''$
Also Length of second rope is constant : $(c-b)+(d-b)=k\implies c''+d''=2b''$
Note that $d$ is a constant as the top pulley and ground is rest : $c''=2b''$
Hence, $c''=-2a''$ as stated in comments.
Also, everything we have done is futile and the block $M_2$ will hit the ground very quickly.