- #1

- 103

- 5

## Homework Statement

## Homework Equations

##F_{net} = ma##

## The Attempt at a Solution

If ##m_2## accelerates, then the force on this box is ##m_1g## and its acceleration is ##\frac{m_1g}{m_2}##. If I disregard the presence of ##m_1## for a minute, I'll have to push on ##M## with a force such that its acceleration is the same to that of ##m_2##. And I'll need to push two masses: ##M## and ##m_1##, so ##F = (M+m_1)\frac{m_1g}{m_2}##.

I'm not sure whether this conclusion is correct. ##m_1## has a downward acceleration and a horizontal force ##F## cannot prevent it from falling; unless there is friction between ##M## and ##m_1##. On the other hand, if a push on ##M##, by Newton's First Law ##m_2## will attempt to remain at rest, which is the same as exerting a force ##F## on ##m_2## in the opposite direction and ##M## remaining at rest. Since I want the smaller boxes to remain stationary, my problem is reduced to one where I need to counteract the force making them move, hence ##F## would have to be ##m_1g##.

What is the correct approach?