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enter image description here

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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closed as off-topic by sammy gerbil, CuriousOne, John Rennie, heather, honeste_vivere Aug 6 '16 at 15:08

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – sammy gerbil, CuriousOne, John Rennie, heather, honeste_vivere
If this question can be reworded to fit the rules in the help center, please edit the question.

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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 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
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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 $T$.

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  • $\begingroup$ 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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