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I have a cart with another cart on top which gets pulled down by another cart on the side. There is no friction.

The question is:

How strongly do I have to push with $F$ to keep the cart $m_1$ stable?

http://wstaw.org/m/2012/02/04/m62.png

I work in the system of the cart $m_3$. This system accelerates with some $a$. The force of inertia pulls back on $m_1$ with $m_1 a$. That is opposed by the gravitational force of $m_2$, which is simply $m_2 g$.

The acceleration that I need for this to be stable is:

$$ a = \frac{m_2}{m_1} g $$

Here comes the point I am not too sure about:

The force is on $m_3$ and on $m_2$, so the driving force would be this:

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

On second thought, I think also need to accelerate $m_1$ some way or another, so that my total force would be a little higher:

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

I prefer the latter, but what is the right solution?

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  • $\begingroup$ By the way, inertia is not a force. You could consider it a fictitious force that appears in an accelerating reference frame, which I think may be what you've done here, but using fictitious forces can get you really confused if you don't understand what you're doing very well, so I generally recommend against using fictitious forces if you don't have to. And this problem can definitely be done without them. $\endgroup$
    – David Z
    Commented Feb 5, 2012 at 0:59
  • $\begingroup$ True. I think I translated all the coordinates, velocities and accelerations correctly, so this should work out. $\endgroup$
    – Martin Ueding
    Commented Feb 5, 2012 at 16:21

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The force acts not just on $m_3$ and $m_2$, it acts on $m_1$ as well (via the pulley), so I would prefer the second answer.

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  • $\begingroup$ The “via the pulley“ is the whole point indeed. I guess second answer it is. (After I fully understand Pluckerpluck's answer.) $\endgroup$
    – Martin Ueding
    Commented Feb 4, 2012 at 20:00

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