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A large box (1) is on a flat surface and is towed by a constant force in the horizontal direction, left to right, but it is not moving due to the presence of a little box (2) in front of the larger one.

The force that box (2) exerts on the larger box (1) is the same force that the large box (1) exerts on the little box (2), $F_{12} = F_{21}$, right? Is this why the larger box is not moving?

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How many boxes are there? "A box..", "..a little box.." and then "...the largest box...". A picture is worth a lot of words. – Vijay Murthy May 8 '12 at 16:58
What is the geometry of the configuration? Also, what is your question? The second sentence is basically an incorrect statement of Newton III (forces are opposite, not the same). – tmac May 8 '12 at 17:06
Thanks @Vijay Murthy, I fixed it now – Zignd May 8 '12 at 17:07
There's a problem with my question? Nobody knows the answer? – Zignd May 8 '12 at 18:01
What you are thinking is pretty close to correct. A picture would really help clarify a few issues with the math. Be carefully not to confuse the force applies to the box(1) from towing with the force applied from box(1) on box(2). – Argus Aug 6 '12 at 19:10
up vote 0 down vote accepted

The problem with your question is that you are working with an incomplete picture of the physical system. You can see this as follows:

If the boxes are not moving then the velocity and acceleration must both be zero. Writing out Newton II for each gives something like this:

$$\Sigma F_1 = m_1 a_1 = 0 = T + F_{21} + (F_{other-1})$$ $$\Sigma F_2 = m_2 a_2 = 0 = F_{12} + (F_{other-2})$$

Where I have included the forces $(F_{other})$ as placeholders for stuff you may need to account for.

There are a few details, such as $F_{21} = -F_{12}$ (from Newton III), but right away you should see that in order for the second equation to be satisfied, $(F_{other-2})$ will need to be different from zero.

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F-other includes a coefficient of static friction. – Timbo Aug 21 '14 at 1:12

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