Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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?

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?

share|improve this question
    
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. –  David Z Feb 5 '12 at 0:59
    
True. I think I translated all the coordinates, velocities and accelerations correctly, so this should work out. –  queueoverflow Feb 5 '12 at 16:21
add comment

1 Answer

up vote 1 down vote accepted

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.

share|improve this answer
    
The “via the pulley“ is the whole point indeed. I guess second answer it is. (After I fully understand Pluckerpluck's answer.) –  queueoverflow Feb 4 '12 at 20:00
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.