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!

  • $\begingroup$ 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 $\endgroup$ 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.. $\endgroup$ 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

1 Answer 1


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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.