# Acceleration of masses hanging from a system of two pulleys

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 ?

• The tensions are correct. The problem is that you assume $a_1 = a_2$. If $M_2$ descends 1cm, how far does pulley 2 descend? May 18 '14 at 13:47
• I think, if $M_2$ descends 1cm, then pulley 2(i.e the movable one) also descends 1cm and $M_1$ ascends 1cm as well. Isn't it? But how are the accelerations different? May 18 '14 at 13:58