# Acceleration of a pulley system [closed]

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.

• $$m_1\ddot{x}_1 = T_1-m_1g\sin\alpha_1$$
• $$m_2\ddot{x}_2 = -2T_2+m_2g$$
• $$m_3\ddot{x}_3 = T_3$$
• $${1\over 2}MR^2\ddot{\varphi}=-T_3R+T_2R$$
• $$\ddot{x}_3 = \ddot{\varphi}R$$
• $$-\ddot{x}_1+2\ddot{x}_2-\ddot{x}_3 = 0$$
• Note that you haven't yet taken into account the mass of the wheel. $T_1$, $T_2$ and $T_3$ are not independent of each other. See if these comments help move you forward a bit. Also be aware that this is not a homework help site, so you are likely to get hints and questions directed back at you rather than answers. – garyp Jul 28 '16 at 16:43
• Welcome to Physics Stack Exchange! Please note that Physics.StackExchange is not a homework help site. Please see this Meta post on asking homework questions and this Meta post for “check my work” problems. – garyp Jul 28 '16 at 16:45
• To me, this homework-like question 1) asls about a specific physics concept, and 2) shows some effort to work through the problem. Thus, I vote it to leave open. – peterh - Reinstate Monica Jul 28 '16 at 18:36
• @peterh : Yes the Qn shows some effort to work through the problem, but it is not asking a conceptual question. The Qns asked are (1) what to do next and (2) check my answers. – sammy gerbil Jul 30 '16 at 17:15

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 $T$.

• please check my revision, but I still don't know how to replace the $Ts$. And $m_2= 2m$ that's why the $2s$ canceled – hello Jul 28 '16 at 17:56
• I don't see in your revision that you have put $T_1=T_2$. – Emil Jul 29 '16 at 13:47
• Okay, can you check if my calculations are right... $\ddot{x}_1 = gsin\alpha_1+{\ddot{x}_2\over 2gm_1}$, $\ddot{x}_2 = g-{2\over m_2}(m_3\ddot{x}_3-Mg)$, and $\ddot{x}_3 = {m_3\ddot{x}_3\over Mg}$ – hello Jul 29 '16 at 14:58
• I cannot, this is not homework help site. I can help you with understanding of the problem, but I cannot solve the problem for you. Come on, you have not so much to finish. In your last revision, you have wrote 6 equations in the end. When you add $T_1=T_2$ there will be seven. And so it happens that you have seven unknowns - $T_1, T_2, T_3, \ddot{x}_1, \ddot{x}_2, \ddot{x}_3$ and $\ddot{\phi}$. So the solution is near :) UPD: in your last comment there is something strange - $\ddot{x}_3$ equation doesn't look good. – Emil Jul 29 '16 at 15:05
• @hello : As explained, this is not a homework help site. We do not check your answers. – sammy gerbil Jul 30 '16 at 17:11