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## GR0877 Problem 57

PROBLEM STATEMENT: This problem is still being typed.

SOLUTION: (B) For both figures accelerations of masses $2m$ and $m$ are the same and can be calculated via Newton’s second law for the whole system of two bodies: $3ma = F$, $a = F/(3m)$. For the first figure $ma = F_{12}$ and $F_{12} = F/3$. For the second figure $2ma = F_{12}$ and $F_{12} = 2F/3$. Thus, choice (B).

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1. 21.10.2012 at 04:08

Can you explain why in the first figure, ma = F_12 and in the second, it is 2ma = F_12 instead of the reverse (where 2ma = F_12 in the first figure, and ma = F_12 in the second)?

• 29.08.2016 at 09:39

This reply is 4 years late. Woops.

In figure 1 we have that F_12 = ma because, simply put, the small mass (of mass ‘m’) is accelerating (with acceleration ‘a’), so there must be a force causing it to accelerate. That force is precisely F_12. If you are confused with why, for figure 1, I only considered the small mass rather than the larger one, this is because small mass is being acted on ONLY by the larger mass, whereas the larger mass is being acted on by BOTH the external force and the smaller box. In other words, the smaller box is simpler to deal with. Of course, the force that one exerts on the other is equal in magnitude and opposite in direction.