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I am confused about a method used in the following problem. There is an arrangement as shown below. The surface is smooth, and the pulleys are light. We have to find the acceleration $a_0$ of $m_1$.

Problem

The method I used to solve it was to consider the pulley B and masses $m_2$ and $m_3$ as a single system that goes down with the same acceleration as that of $m_1$. If this acceleration be $a_0$, then the equations of motion give $$a_0=\frac {m_2+m_3}{m_1+m_2+m_3}g$$

However, the textbook solution treats motions of all objects individually, where $m_1$ has an acceleration $a_0$, $m_2$ has an acceleration $a_0-a$ and $m_3$ has an acceleration $a_0+a$, all from the lab frame(inertial). The answer calculated thus does not match with mine. The texbook gives $$a_0=\frac {g}{1+ \frac {m_1(m_2+m_3)}{4m_2m_3}}$$

The question is, what is the problem with considering the pulley B and the masses $m_2$ and $m_3$ as a single system of mass $(m_2+m_3)$? Or do we have to take some precautions, when the system is accelerated? (The textbook solution is perfectly alright and I understood it too, but what is the problem with mine?)

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    $\begingroup$ Imagine what happens when $m_2\rightarrow 0$: $m_3$ will fall freely and $m_1$ will not move, right? So your reasoning does not work very well. You need a way to find the force that $m_2$ and $m_3$ generates on the pulley, which in general is different from the sum of their weights ;) $\endgroup$ – DarioP Apr 15 '14 at 11:16
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The problem in yours is that you are taking the net force acting downward to be $(m_2+m_3)g$ is incorrect and that led you to take the total mass to be $m_1+m_2+m_3$ which is again incorrect because $m_2\neq m_3$. If $m_2=m_3$ then the center of mass of $m_2$ and $m_3$ will lie on the straight vertical line through the center of the pulley B and the force would act exactly at the center of the pulley B but $m_2\neq m_3$ so the center of mass will shift so at the center of the pulley B the effective mass, $m$, due to which the net force is acting downward is to be found out.

The net force acting is the two tension in the string where the masses $m_2$ and $m_3$ are suspended. From free body diagram of $m_2$ and $m_3$ tension $T$ can be found out and net force acting on pulley B will be $2T$.$$F_{net}=2T=4m_2m_3g/(m_2+m_3)$$ $F_{net}/g=m$ where $m=4m_2m_3/(m_2+m_3)$ is the effective mass of $m_2$ and $m_3$ with pulley B

So the new problem consists of two masses $m_1$ and $m$ with pulley A, $m$ replacing $m_2$ and $m_3$.

$F_{net}=mg$ and the total mass now is $M=m_1+m$ and $$a_0=F_{net}/M$$

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The vertically moving object is an Atwood machine and the two masses have their own accelerations that are in different directions. The acceleration of $m_2$ and $m_3$ (separate from the total system) is given by $$ a=\frac{m_3-m_2}{m_2+m_3}g\tag{1} $$

Mass $m_2$ is accelerating upwards, hence the acceleration in your case of $a_0-a$; likewise mass $m_3$ is accelerating downwards with an acceleration of $a_0+a$.

Newton's 2nd law says that the sum of the forces is equal to $ma$, so you should be using all the forces in the set up.

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    $\begingroup$ When you say, "separate from the total system", what does it mean? Assuming it talks about their individual motions, is it w.r.t. the lab frame or the frame of the lower Atwood machine. $\endgroup$ – Shubham Apr 14 '14 at 14:03
  • $\begingroup$ No, I'm considering only the solution to the stationary Atwood machine as given in the Wikipedia link. $\endgroup$ – Kyle Kanos Apr 14 '14 at 14:04
  • $\begingroup$ But that would be like...'easy'. But what is the problem with my assumption? $\endgroup$ – Shubham Apr 14 '14 at 19:57
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    $\begingroup$ @Shubham: if $m_{2}\neq m_{3}$, the center of mass of the system B moves relative to the pulley, so its net acceleration is $a_{0} + a_{cm}$. It is almost certainly easier to just solve this by doing free-body diagrams for all three masses, rather than having to worry about treating the internal system properly. $\endgroup$ – Jerry Schirmer Apr 14 '14 at 22:37

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