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As the drawing shows, two blocks are connected by a rope that passes over a pulley. The first body, $m_1=600g$, is lying on a flat surface, while the second one, $m_2=200g$, hangs freely off the surface. If the friction is ignored, we have to find the tension force in the rope.enter image description here

My teacher started solving this problem by drawing the two green forces ($m_1a$ and $m_2a$) and writing this equation: $$m_2g=m_1a+m_2a$$

This is where I get confused. What does that acceleration $a$ represent? The way I see this picture is that the first body is moving left and pulling the second one up.

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    $\begingroup$ The diagram is wrong, unless there is some pulling force acting on block 1 there is no way that the block 1 will move towards left. $\endgroup$ – Mitchell May 17 '17 at 12:33
  • $\begingroup$ @BhavyaSharma What about the equation? $\endgroup$ – A6EE May 17 '17 at 12:36
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The diagram doesn't match the situation presented in the question.

There is no way that the block $(1)$ will move towards the left, unless there is some pulling force acting on the block towards left.

$'a'$ represents the acceleration of the system. I am assuming only positive values of $'a'$ in my answer.

Case I, When a pulling force is present :

If there was some pulling force (towards the left) then the equations for getting the acceleration of the system would be as follows :

$P-T=m_1a$ $\tag1$

$T-m_2g=m_2a$ $\tag2$

From equations $(1)$ and $(2)$, you can get the following value value of acceleration and then you can substitute the value of the acceleration either in equation $(1)$ or $(2)$ to get the tension, but you need to know the value of the pulling force for this.

$a=\frac{P-m_2g}{m_1 + m_2}$

Case II , when pulling force is absent :

If the system wasn't acted upon by any external pulling force we would get the following equations,

$T=m_1a$ $\tag3$

$m_2g-T=m_2a$ $\tag4$

From equation $(3)$ and $(4)$, we can conclude,

$m_2g=m_1a+m_2a$

$a=\frac{m_2g}{m_1 + m_2}$

This equation matches with the one in your question. By putting the value of $a$ in either equation $(3)$ or $(4)$, you can get the value of tension. In this case you only require the values of $g$, $m_1$ and $m_2$ in order to get the value of tension. These are provided in the question.

Conclusion :

In the absence of pulling force, the equation presented by your teacher is correct but the diagram associated with it is wrong.

If the blocks were actually moving left then the value of the pulling force have to be provided, if the tension is to be calculated.

Summing it up, the direction of the acceleration given by your teacher is inconsistent with the question.

Although the diagram can be correct if the value of acceleration of the blocks comes out to be -ve. Therefore, by reversing the direction of $'a'$ given in the diagram, you will get the actual direction of acceleration.

Putting into other words, we can either say that block $(1)$ moves with $-a$ acceleration towards left (considering $'a'$ to be positive) or the block moves with $a$ acceleration towards right. Its the same thing.

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    $\begingroup$ The diagram is perfectly fine. It showed both accelerations in the same direction, which is all that matters. The answer will result in negative values for acceleration due to the physical system, but drawing accelerations in that way is perfectly fine. Imagine they were both hanging and you didn't know m1 or m2, you can form the equations for motion knowing acceleration will act along the rope. You don't need to know which sign the accelerations have, just the right orientations (which they have when showing acceleration acting along the rope in consistent direction). $\endgroup$ – JMac May 17 '17 at 14:01

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