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Three blocks of masses m1, m2 and m3 are connected as shown in the figure. All the surfaces are frictionless and the string and pulleys are light. Find the acceleration of the block of mass m1. Figure

In this problem I know acceleration of the pulley B is same as acceleration of the block of mass $\ {m_{1}}$. But acceleration of bodies with mass $\ {m_{2}}$ and $\ {m_{3}}$ will be different since $\ {m_{2}}$ not equals $\ {m_{3}}$. So, I know that I cannot consider this(Pulley B, Blocks of mass $\ {m_{2}}$ and $\ {m_{3}}$) as a single system. But If the imagine putting those three objects in a box such that what's happening inside will not be visible to me then why I cannot consider this as a single system? The mass will thus be $\ {m_{2}}$ + $\ {m_{3}}$ and acceleration will be that of the block of mass $\ {m_{1}}$. So, I tried to find out the acceleration and I got $a=\dfrac {\left( m_{1}+m_{2}\right) g}{m_{1}+m_{2}+m_{3}}$ but, the answer is given as $\ a =\ \dfrac {g}{1+\dfrac {m_{1}}{4}\left( \dfrac {1}{m_{2}}+\dfrac {1}{m_{3}}\right) }$. So, why I am wrong here? P.S. How could I solve it in a few steps because the original solution I have is quite long.

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    $\begingroup$ By saying acceleration of the block (mass m1) and pulley B is same I mean magnitude only. $\endgroup$ – Aditya Kshitiz Aug 23 at 11:15
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    $\begingroup$ Hint: think about what happens as the mass of 2 gets really small and the mass of 3 gets really big. $\endgroup$ – Aaron Stevens Aug 23 at 12:00
  • $\begingroup$ @Aaron Stewens then the block of mass 3 will be in almost free fall and will not cause the pulley to accelerate and therefore the block of mass 1 will not accelerate with the same acceleration of block of mass 3? But why does pulley is accelerating in the first place( in both cases m1 + m2 will be same) $\endgroup$ – Aditya Kshitiz Aug 23 at 12:06
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    $\begingroup$ The tension around pulley B depends on the accelerations of blocks 2 and 3. Therefore, how blocks 2 and 3 accelerate determines the acceleration of block 1, right? $\endgroup$ – Aaron Stevens Aug 23 at 12:17
  • $\begingroup$ Have in mind that the important thing to accelerate mass m1 is the tension on the string that, as it is massless, will be the same all along. Furthermore, it will be twice the tension in the string between m2 and m3. But tension around pulley B must be different from (m2+m3)g for the masses to accelerate. $\endgroup$ – MaxWell Aug 23 at 12:33
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The key is to realize two things.

First, since pulley B is massless it must be that $$T_A=2T_B$$ where $T_A$ and $T_B$ is the tension of the string around pulley A and pulley B respectively.

Second, since pulley B accelerates downward with the same acceleration of mass 1, and because the string around pulley B has a constant length, it must be that $a_2=-a_1+a_r$ and $a_3=-a_1-a_r$, where $a_r$ is the relative acceleration between the pulley and mass 2. Adding these relations we get $$2a_1+a_2+a_3=0$$

The above key points along with equations from N2L $$m_1a_1=T_A$$ $$m_2a_2=T_B-m_2g$$ $$m_3a_3=T_B-m_3g$$

allow us to determine an "effective mass" pulling on mass 1 by comparing to the case where the pulley B system is replaced with a single hanging mass (work left to you): $$m_{\text {eff}}=\frac{4m_2m_3}{m_2+m_3}$$

Notice how when $m_2=m_3$ we have $m_{\text {eff}}=2m_2$, which is what we would expect. Also notice if, say, $m_2\to0$ that $m_\text{eff}\to0$, which is also what we expected.

This effective mass comes from the key thing you are missing. The acceleration of masses 2 and 3 affect the total force applied to mass 1. You cannot treat the pulley B system as a "black box" whose mass is just the mass of its parts.

A simpler case to understand would be if I was in a box on a scale and you were outside the box. Let's say I was swinging on a swing hanging from the top of the box. If you were looking at the scale, you would be perplexed since the reading would oscillate up and down. Of course, I am not rapidly gaining and losing weight. The forces present "inside of the box" affected the force needed to support the box.

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  • $\begingroup$ Thanks a lot. I finally figured it out! $\endgroup$ – Aditya Kshitiz Aug 23 at 14:45
  • $\begingroup$ @AdityaKshitiz Always remember to upvote useful answers, and to accept answers for future readers. If you want to wait and see if more answers come in you can. $\endgroup$ – Aaron Stevens Aug 23 at 14:49
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Newton's equations: Upper mass T = m1 A Hanging masses m2 g – t = m2 ( A – a ) and m3 g – t = m3 ( A + a ) where a is the acceleration of the lower cord relative to the lower pulley (assuming m3 > m2 ). Also T = 2t . Divide each of the lower equations by the corresponding mass and add to eliminate “a”: 2g - t / ((1/m2) + (1/m3)) = 2 A replacing t by T/2 and dividing by 2 gives g =[(m1/4)((1/m2) + (1/m3)) + 1] A .

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    $\begingroup$ This terse recipe doesn't have the type of conceptual explanation that we hope for in our answers to homework-like questions. $\endgroup$ – rob Aug 23 at 19:21

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