# How do I tackle this problem WITHOUT focusing the whole system as a “single mass”?

Problem/Solution

! !

In the third FBD, they made the whole system as one object. What happens if I don't want to do that? What if I jsut apply Newton's second Law to the mass M? I tried it out but it didn't work. Could someone point out my mistake? Note that my other force diagram are like the ones in the solutions. Note that $n_1'$ and $n_2'$ are the reaction forces for mass $m_1$ and $m_2$ respectively.

!

Writing out Newton's Second law for all the other massses

For M

$\begin{cases} \sum F_x = F - n'_2 = Ma \\ \sum F_y = n = n_1' + Mg + m_2g \end{cases}$

For $m_1$

$\begin{cases} \sum F_x = T = m_1 a \\ \sum F_y = n = m_1g \end{cases}$

For $m_2$

$\begin{cases} \sum F_x = n_2 = m_2 a \\ \sum F_y = T = m_2g \end{cases}$

If I eliminate the system of equations, I would get something absurd like $F = (M + m_2)a$

You're not following the definition of Newton's law correctly... the FBD is of the single mass M, so there should be no gravitational force $m_2g$!! Anyway, you can only apply this law to a body, and there are are a finite amount of bodies here (including packaging multiple bodies into one)... you will see that you won't get very far in some of these cases. –  Chris Gerig Dec 27 '11 at 4:15
So it's impossible? I had $m_2g$ in there because doesn't M support both $m_1$ and $m_2$? –  Hawk Dec 27 '11 at 4:27
Could I also ask again on the $m_2$'s motion? Remember when I asked about its vertical motion being 0 and you convinced me that it has horizontal motion with respect to the cart? Something is still off. When the mass on top accelerates forward, doesn't the string "shorten" on the top and therefore the string "below" attached to mass on pulley "extends"? And doesn't that mean there is in fact motion in the vertical direction? –  Hawk Dec 27 '11 at 4:29