Consider this arrangement of pulley:


The problem is to find the acceleration of $m_1$.

This is my solution:


The problem explicitly wants $\ddot y_1$.

So starting by applying Newton's second law we get:

$$T_1-m_1g=m_1\ddot y_1$$ $$T_2-m_2g=m_2\ddot y_2$$

Where $T$'s are tension forces acting on the masses with the same subscript.

We need two more equations. First is obtained by geometry of the system:

$$l_2=y_3+(y_3-y_2)+\pi R=2y_3-y_2+\pi R\implies 2\ddot y_3=\ddot y_2$$ $$l_1=h-y_1+h-y_3+\pi R=2h-y_1-y_3+\pi R\implies \ddot y_1=-\ddot y_3$$

$$\ddot y_2=-2\ddot y_3$$

But I still need another constraint to relate $T_1$ to $T_2$.

How to write that?


1 Answer 1


Since the pulley is weightless, according to Newton's second law the net force acting on it should be zero. so: $$T_1=2T_2$$


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