# Force exerted on three blocks

Ok so I am stuck on one of the problems about three blocks being pushed by a force:

Three blocks on a frictionless horizontal surface are in contact with each other, as shown in Fig. 4–51. A force is applied to block 1 (mass m1). (a) Draw a free-body diagram for each block. Determine (b) the acceleration of the system (in terms of m1, m2 and m3), (c) the net force on each block, and (d) the force of contact that each block exerts on its neighbor. (e) If m1 = m2 = m3 = 12.0 Kg and F = 96.0 N , give numerical answers to (b), (c), and (d).

Now I am having problems with part c

So I first solved for the net force as a whole:

$$F = a(m_1+m_2+m_3)$$ $$a = \frac{F}{m_1+m_2+m_3}$$

Now I have found acceleration, which will be the same throughout this example.

So I move on to finding the net force on each block

$$net force = F - (Force of m_2 + Force of m_3)$$ $$netforce = F - \frac{(m_2 + m_3)F}{m_1+m_2+m_3}$$

So that means that what is left is the force of

$$netforce = \frac{m_1F}{m_1+m_2+m_3}$$

But now to get the net force of the second box is where I get confused:

$$netforce_2 = Force_1 + Force_2 - Force_3$$

I was thinking I would have to add the forces of box 1 + box 3 and subtract the force of box 3.

So I did and I got:

$$netforce = \frac{(m_1F + m_2F) - m_3F}{m_1+m_2+m_3}$$

However the book tells me the answer is:

$$netforce = \frac{m_2F}{m_1+m_2+m_3}$$

Does anyone know why?

The net force on $m_2$ must be just the force needed to accelerate that block with $a$ . Since you know $a=\frac{F}{m_1+m_2+m_3}$, and the net force needed is just $m_2 a$, the result from the book follows.
This problem is all about Newton's Second Law, applying it first to the group of blocks together and then to each one individually. The book's answer satisfies Newton's Second Law $$net force_2=m_2a$$
The problem with your approach, $$net force_2=Force_1+Force_2-Force_3$$ is that $Force_1,$ $Force_2,$ and $Force_3$ have no clear meaning.
It would be true to say $$netforce_2=F-netforce_1-netforce_3,$$ or $$m_2a=F-m_1a-m_3a,$$ which is what you did in finding $netforce_1$