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I have a slight confusion regarding the application of Newton's 2nd law in a situation where there are two blocks, of mass $m_1$ and $m_2$, which are stacked on top of each other and placed on a frictionless surface. The interface between the two blocks is not frictionless. A force $F$ is applied to the bottom block, of mass $m_1$, causing the blocks to accelerate, as shown in the image below (where I have identified some forces):

enter image description here

There are frictional forces exerted upon the two blocks (due to Newton's Third law), which I have drawn in the image. To find the acceleration of the block at the bottom, I tried to applied Newton's 2nd law in the following way:

$$m_2: \sum F = F - F_f = (m_1+m_2)a$$ $$\iff a = \frac{F-F_f}{m_1+m_2}$$

However, the solution states that

$$a = \frac{F-F_f}{m_1}$$

I am confused as to why the net force acting on the block of mass $m_1$ should be divided by $m_1$ rather than $m_1+m_2$, since the two blocks are stacked together, so could be considered as a single object with mass $m_1+m_2$?

This is probably a silly misconception, but I would greatly appreciate it if anyone would be able to help explain this to me. Thank you in advance!

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  • $\begingroup$ As you said, The blocks are frictionless, No force will act between The two block's contact surfaces, Because there is no friction, So only Bottom block will accelerate $\endgroup$
    – Dheeraj Gujrathi
    Commented Apr 14 at 11:32
  • $\begingroup$ @DheerajGujrathi I believe the OP is saying only the supporting surface is frictionless, not the surface between the blocks. $\endgroup$
    – Bob D
    Commented Apr 14 at 11:36
  • $\begingroup$ @BobD, Got it.. $\endgroup$
    – Dheeraj Gujrathi
    Commented Apr 14 at 11:37
  • $\begingroup$ @BobD Yes, that is what I meant - I will edit the question to clarify that. $\endgroup$
    – ethanolex
    Commented Apr 14 at 11:42
  • $\begingroup$ Is the top block sliding on the bottom block? $\endgroup$
    – Bob D
    Commented Apr 14 at 11:44

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Your application of Newton's Law is incorrect, If you take $m_1+m_2$ as system(Assuming friction is sufficient, Internal forces Cancel Out as vector, In that case, The correct application would be

$$\sum F = \vec F - \vec F_f+\vec F_f = (m_1\vec a_1+m_2\vec a_2)$$ where $a_2$ and $a_1$ are acceleration vectors of Upper and lower block respectively

where $m_2$ and $m_1$ are masses of Upper and lower Blocks respectively Which Gives

$$\vec F=(m_1+m_2) \frac{(m_2\vec a_2+m_1\vec a_1)}{(m_1+m_2)}$$

While now, To find acceleration $a_1$ of Lower block only, We need to consider lower block only into the system

Now for this system whose mass is only $m_1$ and apply newton's second law

$$\sum F = \vec F - \vec F_f =m_1\vec a_1$$

Which gives $$\vec a_1=\frac{\vec F - \vec F_f}{m_1}$$

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  • $\begingroup$ Ahh I see, thank you very much for clearing that up! $\endgroup$
    – ethanolex
    Commented Apr 14 at 11:52

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