I have a block $A$ of mass $m_1$ resting on a block $B$ of mass $m_2$.

Both are resting on a table. My problem is how to write Newton's second law applied on the system; it is not known if the system is accelerating or not.

Supposing that the positive orientation of the (monodimensional) vectors is in the downward direction, i think I see four forces, or two action-reaction couples:

1) the force acted on the block $B$ by the block $A$, i.e. $F_1 = m_1g$;

2) the reaction (normal) force $R_1$, acted on $A$ by $B$;

3) the force acted on the table by both blocks, $F_2 = (m_1+m_2)g$;

4) the reaction force acted by the table, $R_2$;

I'd like to write the following equation: \begin{equation} \sum F = ma\quad\Longrightarrow\quad F_1+F_2-R_1-R_2=ma \end{equation} which yelds \begin{equation} (2m_1+m_2)g-R_1-R_2=(m_1+m_2)a \end{equation} My question is: is this setup right? I am sure it is not, since by Newton's third law the reactions should be equal and opposite to $F_1$ and $F_2$, so the LHS should be zero. This leads me to think that I'm mixing internal and external forces, but I'm lost.

  • $\begingroup$ Do you mean that the system (block $A$+block $B$ + table) might be accelerating upwards or downwards due to some external force that causes the acceleration $a$? $\endgroup$
    – DelCrosB
    Commented Oct 8, 2016 at 9:55
  • $\begingroup$ Yes, exactly. The "table" could actually be a lift. $\endgroup$ Commented Oct 8, 2016 at 10:02
  • $\begingroup$ You have to clearly define your system. Newton's second law involves only the external forces on the system. Internal forces, like action-reaction pairs within the system, are ignored. (They cancel.) $\endgroup$
    – garyp
    Commented Oct 8, 2016 at 12:00

2 Answers 2


My problem is how to write Newton's second law applied on the system

Then all you need are the forces on the system (two blocks) due to the surroundings, external forces, not any internal forces which you can identify as Newton third law action-reaction pairs.

As the question is written there only seem to be two of these external forces.


Let's take an example,Suppose a block of mass m is dropped from a height and you can neglect air resistance,now the force acting on block is only mg,no other force is acting on it(because normal reaction occur when two bodies are in contact and there is no contact).The direction of force(mg) due to vertically downward force,ball reaches to earth. Now,lets come to your example: The two blocks are in contact and the larger block is in contact with the table. since each body is in contact with a other body,every force(action) will have a reaction and thus system will not move.

  • $\begingroup$ This is not true. Every force always has a reaction force; the problem here is (I think) how to define the external forces. In ultimate analysis I think it's a problem of system definition. $\endgroup$ Commented Oct 8, 2016 at 10:44
  • $\begingroup$ you are correct,the block is being pulled by mg downward force by earth ,in turn block attract the earth by a force mg(reaction) but this reaction force have nothing to do with the motion of block,that's why i didn't mentioned it in my annswer. $\endgroup$ Commented Oct 8, 2016 at 10:54
  • $\begingroup$ Thank you. (I was addressing your sentence " since each body is in contact with a other body,every force(action) will have a reaction". This a false implication) $\endgroup$ Commented Oct 8, 2016 at 11:21

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