# Newton's second law: direction of forces

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.

• 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 $a$? Oct 8, 2016 at 9:55
• Yes, exactly. The "table" could actually be a lift. Oct 8, 2016 at 10:02
• 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.) Oct 8, 2016 at 12:00