[Physics] Acceleration of a pulley system

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The figure above shows a pulley system consisting of 3 masses ($m_1$, $m_2$ and $m_3$), a homogeneous wheel (radius R, mass M) and 2 massless pulleyes which are connected by a massless rope. Mass $m_1$ is sliding on an inclined surface (inclination angle $\alpha$) and mass $m_3$ is sliding on a horizontal surface. Assume that there are no slipping and considered frictionless, and pulley 1 and pulley 2 are massless.

Assuming that $m_1 = m_3 = m$ and $m_2 = M = 2m$, determine the accelerations of the masses $m_1, m_2$ and $m_3$.

I've tried to answer this question with the results:

• acceleration of $m_1 : \ddot{x}_1 = g \sin \alpha_1 – {T_1 \over m}$
• acceleration of $m_2 : \ddot{x}_2 = {T_2 \over m} – g$
• acceleration of $m_3 : \ddot{x}_3 = {T_3 \over m}$

But since $T_1, T_2$ and $T_3$ are not given, my answers are wrong. Anyone can help me? I would highly appreciate it.

Sorry for the bad drawing.

Additional calculations that I've made:

• $m_1\ddot{x}_1 = T_1-m_1g\sin\alpha_1$
• $m_2\ddot{x}_2 = -2T_2+m_2g$
• $m_3\ddot{x}_3 = T_3$
• ${1\over 2}MR^2\ddot{\varphi}=-T_3R+T_2R$
• $\ddot{x}_3 = \ddot{\varphi}R$
• $-\ddot{x}_1+2\ddot{x}_2-\ddot{x}_3 = 0$

Answer 1

You don't have all equations, and one is not correct. The usual assumption in these problems are:

1. There is no friction.

2. Ropes are glued to pulleys.

   • From 1. it follows that $T_1=T_2$
   • You forgot, that $m_2$ is acted on by $T_2$ twice: ${\ddot{x}_2} = {\frac{2T_2}{m_2} -g}$.
   • $T_3=T_2+N$, where N is force which rotates the big wheel.
   • ${\ddot{\beta}} = {\frac{NR}{I}}$, where $I=MR^2/2$.
   • ${\ddot{\beta}} = {\ddot{x}_3}/R$.

With all these additional equations, you should be able to find all the accelerations. However, pay attention to directions - they depend on your initial choice of signs of $g$ and $T$.