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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:

  • $m_1\ddot{x}_1 = T_1-m_1g\sin\alpha_1$
  • $m_2\ddot{x}_2 = -2T_2+m_2g$
  • $m_3\ddot{x}_3 = T_3$
  • ${1\over 2}MR^2\ddot{\varphi}=-T_3R+T_2R$
  • $\ddot{x}_3 = \ddot{\varphi}R$
  • $-\ddot{x}_1+2\ddot{x}_2-\ddot{x}_3 = 0$
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    $\begingroup$ 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. $\endgroup$ – garyp Jul 28 '16 at 16:43
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    $\begingroup$ 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. $\endgroup$ – garyp Jul 28 '16 at 16:45
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    $\begingroup$ 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. $\endgroup$ – peterh - Reinstate Monica Jul 28 '16 at 18:36
  • $\begingroup$ @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. $\endgroup$ – 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 $T$.

| cite | improve this answer | |
  • $\begingroup$ 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 $\endgroup$ – hello Jul 28 '16 at 17:56
  • $\begingroup$ I don't see in your revision that you have put $T_1=T_2$. $\endgroup$ – Emil Jul 29 '16 at 13:47
  • $\begingroup$ 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 $\ddot{x}_3 = {m_3\ddot{x}_3\over Mg}$ $\endgroup$ – hello Jul 29 '16 at 14:58
  • $\begingroup$ 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. $\endgroup$ – Emil Jul 29 '16 at 15:05
  • $\begingroup$ @hello : As explained, this is not a homework help site. We do not check your answers. $\endgroup$ – sammy gerbil Jul 30 '16 at 17:11

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