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There's no friction and the pulleys and strings are massless. The aim is to solve the system and determine all the tensions and accelerations Okay, so for the record, I've solved this system considering each object as an individual system but what I can't comprehend is why I can't consider the hanging pulley and the 2 hanging masses as a system with a combined mass of (m1+m2) and total force acting as [(m1+m2)g-T] and then solve for the tension by equating accelerations. This always yields an incorrect answer and I've serious troubles justifying that to myself. Some help would really do me good and help me gain a better perspective. The picture encloses what I wanna consider as a system just so it's clear. Thanks. (Friction is absent, strings and pulleys are ideal)

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    $\begingroup$ Do you include in your analysis how the centre of mass of your defined system moves and accelerates? $\endgroup$
    – DJohnM
    Jul 30, 2017 at 15:23
  • $\begingroup$ @DJohnM Nope. I'm a beginner in my physics course and am not acquainted with that. $\endgroup$
    – ditsuke
    Jul 30, 2017 at 15:29
  • $\begingroup$ @garyp here goes: the systems total mass is (m1+m2) and the total force acting on the system is [(m1+m2)g-T]. This system's (or what I think, wrongly) acceleration equals that of the mass m0's, equating the accelerations yields the tension equaling (m1m0+m2m0)g/(m1+m2+m3). DJohnM has kind of made me realize what my analysis lacks. $\endgroup$
    – ditsuke
    Jul 30, 2017 at 15:37
  • $\begingroup$ My advice: You should forget about $m_1$ and $m_2$ and consider what happens if you replace it by a single mass $m_3$ fixed to the robe. Then you use d'Alembert's principle to solve for the acceleration of $m_0$. Have you done that? $\endgroup$
    – Semoi
    Jul 30, 2017 at 15:49
  • $\begingroup$ @Semoi , my consideration did replace them by a single mass m3. You see, I'm not far-off in my physics course and hence, happen not to know of the principal which I'm gonna look up now. Also, I'm not aware of centre of mass considerations but my Intuition does seem to be simulated now and I get the thing better and can, qualitatively, analyse the system. $\endgroup$
    – ditsuke
    Jul 30, 2017 at 15:54

3 Answers 3

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You can consider the hanging pulley and the masses hanging from it as a system of mass $m_1+m_2$. However, analysing the forces on this sytem does not tell you about the motion of individual parts of the system. It only tells you about the acceleration of its centre of mass. Often this is not useful.

If the system is rigid, then the acceleration of the centre of mass is the same as the acceleration of each part of the system. (This is also the case if $m_1= m_2$, even if the two masses are in relative motion.) If the system is not rigid, these accelerations are not the same. To find out how the separate parts of the system move, you have to consider internal forces. Ultimately this is the same as analysing the forces on each part of the system separately.

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  • $\begingroup$ Thank you for the answer Sammy. I now do understand what the combined mass consideration yields and why it doesn't correspond to the components' individual motions: the mass isn't uniform and nor are the forces. My reputation doesn't yet allow me to upvote your answer but it surely deserves one :)). $\endgroup$
    – ditsuke
    Jul 31, 2017 at 14:46
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The forces acting on the lower pulley is twice the tension of the rope connecting $m_1$ and $m_2$, which we denote as $T'$. In this situation, we have the equations $T'-m_2g=-m_2 a$ and $T'-m_1g=m_1a$. The reason $2T'$ is not the sum of $m_1$ and $m_2$ is because of the acceleration of the system. Imagine standing in a lift that is accelerating upwards. You would feel "pressed" into the floor since an additional force is exerted on you to accelerate you upwards in addition to your weight, i.e. the reaction force exerted on you by the floor is greater than your weight. For a lift accelerating downwards, you feel "lifted" off the ground because the reaction force acting between you and the floor is LESS than your weight. A similar thing is happening here as the weight of the mass accelerating downwards (let us assume it's $m_2$ for simplicity) is not equal to the tension. Rather, we see that $T'=m_2(g-a)$ is less than the weight of the mass $m_2$. For the bob accelerating upwards, the tension is greater than the mass $m_1$, as we see from $T'=m_1(g+a)$. After a bit of algebra, we can see then that the force acting on the lower pulley is less than the sum of the weights , with the force being equal to the sum only when the two masses are of the same weight.

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Basically, we have six unknown in this problem:

  • the accelerations of the three masses $a_0$, $a_1$, and $a_2$.
  • the accelerations of the lower pulley $a_x$, which I give the subscript $x$
  • the tension $T$ in the ropes of the upper pulley.
  • the tension $t$ in the ropes of the lower pulley.

Hence, in order to solve this problem, we need to get six linear independent equations. I have never understood, why people bother to define different directions for each mass. It just complicates the calculations, because you can mess up the signs. Therefore, I always assume that all masses accelerate in the positive $z$ direction (=anti-parallel to the gravitation). If this is not true, the solution will contain a minus sign.

Using this convention, here are the six equations:

  • d'Alembert's principle of mass $m_0$ $$m_0 a_0 = T$$ This of course is not in the $z$-direction. That's why the $g$-term is missing.
  • d'Alembert's principle of the lower pulley $m_x a_x = T - m_x g - 2t$. Note that the tension $t$ is acting twice on the pulley, because each mass ($m_1$ and $m_2$) is pulling on it. Since the pulley is assumed to be massless $m_x=0$, this equation simplifies to $$T = 2t$$
  • d'Alembert's principle of mass $m_1$ $$m_1 a_1 = t - m_1 g$$
  • d'Alembert's principle of mass $m_2$ $$m_2 a_2 = t - m_2 g$$
  • If we assume that the upper rope is non-elastic, its length is constant. Expressing the length in a formula and taking the second derivative yields $$a_0 = - a_x$$ which is rather obvious: The acceleration of the mass $m_0$ and the lower pulley are equal in magnitude, but in opposite directions.
  • Same for the lower rope: Constant length yields $$a_1^{(x)} = -a_2^{(x)}$$ where the superscript indicates, that these accelerations are taken with respect to the center point of the lower pulley. The transformation into absolute coord. is given by $$a_j = a_j^{(x)} + a_x$$ for $j=1, 2$. The absolute acceleration $a_j$ is just the sum of the acceleration of the pulley $a_x$ and the relative acceleration $a_j^{(x)}$.

Now, all you have to do is to manipulate these equation. If you don't know how, please look up the Gauss algorithm.

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  • $\begingroup$ Thank you, Semoi. I did indeed solve the system using this system of equations before I posted this problem here but what I couldn't comprehend was WHY replacing the hanging system by a single uniform mass didn't yield the correct results. As others have pointed out, that consideration will yield the centre of mass' motion rather than the individual components which will not do us any good in this analysis which is to do with the individual components. I really appreciate you sparing the time to answer my problem elaborately. Thanks. $\endgroup$
    – ditsuke
    Jul 31, 2017 at 14:43

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