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Suppose box A is connected to a larger box B, and both are on top of a frictionless surface. My hand applies a force to box A in the direction of box B ($F_{ha}$), causing boxes A and B to move. So box B moves. By Newton's third law, the force exerted by box B on to box A ($F_{ba}$) = -the force exerted by box A on to box B ($F_{ab}$). Now, the only force (that we care about for now) external to the system is $F_{ha}$, then $F_{ha} = F_{ab} = -F_{ba}$. So mathematically, Box A should not move.

Technically, though, we can apply a force such that Box A and Box B both move. So the net force acting on box A > $F_{ba}$. How is this possible?

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  • $\begingroup$ Why do you think Fha=Fab? $\endgroup$ – nasu May 30 at 18:28
  • $\begingroup$ @nasu: If it's not, I cannot explain why. $\endgroup$ – moonman239 May 30 at 18:37
  • $\begingroup$ How do you explain your claim that it is true? $\endgroup$ – Dvij Mankad May 30 at 18:41
  • $\begingroup$ @FeynmansOutforGrumpyCat: Because the only force acting on box A in the direction of desired motion is Fha. Box A does not create forces, it only transfers existing forces. (objects undergoing uniform motion tend to stay in uniform motion) $\endgroup$ – moonman239 May 30 at 18:49
  • $\begingroup$ Box A does produce forces because it is composed of molecules which interact electromagnetically with other objects like the hand, Box B, and the floor. It also exerts a force on the earth if you consider gravity. You cannot assume that the force of the hand is transmitted to Box B. Box A exerts a force on box B, but it is NOT NECESSARILY the same magnitude as the force of the hand on A. $\endgroup$ – Bill N May 30 at 21:14
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The issue with your reasoning lies in the assumption (or, mistaken conclusion) that $F_{ha}=F_{ab}$. I can imagine why one might be led to think this is true. One might think that the force that the hand applies to box $A$ is somehow transferred to box $B$ due to its contact with box $A$ on which we are applying the force $F_{ha}$. And since this force $F_{ha}$ is simply transferred to box $B$, the force on box $B$ via box $A$ must be the same as $F_{ha}$. However, this is not true for the following reason.

It is true that the force acting on $B$ due to $A$ is essentially due to the force being applied on $A$ by our hand. So, in that sense, the box $A$ is transferring the force applied on it onto box $B$. But, the point is that box $A$ doesn't transfer all of the force onto box $B$. It transfers some of it and the difference is precisely equal to the force that would be needed to produce that acceleration with which the box $A$ would end up moving. Let's make this story more precise via figuring out an expression for the force applied on $B$ by $A$ in terms of the force applied by the hand on $A$.

Let's say the system $A+B$ accelerates with an acceleration $a$. Then, since the only external force on the system $A+B$ is $F_{ha}$, we can easily deduce that $$a=\frac{F_{ha}}{m_a+m_b}$$ Now, the only force acting on $B$ is $F_{ab}$. Thus, this acceleration must be produced by $F_ab$ in $B$. Thus, we obtain $$F_{ab}=m_ba=\frac{F_{ha}m_b}{m_a+m_b}=\frac{F_{ha}}{1+\frac{m_a}{m_b}}$$Or, in other words, the fraction of the force that is transferred by the box $A$ onto box $B$ is given by $$\frac{F_{ab}}{F_{ha}}=\frac{m_b}{m_a+m_b}=\frac{1}{1+\frac{m_a}{m_b}}$$We can verify that the net force acting on $A$ would thus be equal to $F_{ha}-F_{ab}=\frac{F_{ha}m_a}{m_a+m_b}=am_a$. This means that the net force on box $A$ is exactly what is needed for it to accelerate with the acceleration $a=\frac{F_{ha}}{m_a+m_b}$. In some rough sense, box $A$ "absorbs" this much force and transfers the rest to box $B$.

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  • $\begingroup$ I just thought of a slightly more intuitive explanation - Realize that every particle (quantum stuff aside) is itself a system of particles. So, all one body problems are technically multi-body problems. Ergo, the laws of physics, if they are to be consistent, must apply to any system - to say system A+B obeys different laws than system A or system B is inconsistent. $\endgroup$ – moonman239 May 30 at 20:56
  • $\begingroup$ @moonman239 Well, as long as forces between the constituent particles follow the third law of Newton, you are correct. And we are assuming that to be the case. But your question was precisely about explicitly showing that this general expectation is indeed satisfied in this particular example. $\endgroup$ – Dvij Mankad May 30 at 22:41
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Consider box A and B together as comprising a system. See the diagrams below. You apply a force to box A in contact with box B, call it $F_{ext}$. At the interface between the boxes Box A applies a force to Box B, call it $F_{AB}$ and per Newton's third law box B applies an equal and opposite force on Box A, call it $F_{BA}$. The equal and opposite action/reaction forces $F_{AB}$ and $F_{BA}$ are internal to the system and cancel each other

But the force you apply, $F_{ext}$ is external to the system of the two boxes. Since the boxes are on a frictionless surface there is no external friction force acting on the boxes opposing your applied force.

But you also need to consider that when you push on the two boxes they together push back on you with an equal and opposite force again per Newton's third law. In order for that reaction not to cancel your externally applied force, there needs to be another external force, the static friction force between you and the surface, to oppose the reaction force of the two boxes. Imagine trying to move the boxes if you were standing on ice. As long as your external force does not exceed the maximum static friction force between your feet and the surface, you will not slip.

If you look at the bottom diagram showing all the forces you can see that they all cancel each other, except for your externally applied force to the two boxes, and thus

$$F_{ext}= (M_{A}+ M_{B})a$$

As a matter of fact, it is the static friction force between your feet and the surface that is responsible for you being able to accelerate the boxes.

Hope this helps.

enter image description here

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Force balance on box A: $$m_Aa=F-F_{ba}$$ Force balance on box B: $$m_Ba=F_{ab}$$ Adding the force balances together:$$(m_A+m_B)a=F-F_{ba}+F_{ab}=F$$

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There is no reason why $F_{ha} = F_{ab}$. If we write down the forces on the system:

$$ F_{ha} + F_{ab} + F_{ba} = (M_A + M_B)a $$

And as $F_{ba} = - F_{ab}$ then $F_{ha} = (M_A + M_B)a$

So, internal forces cancell each other and won't produce any acceleration.

If you think about it, you don't need two separate boxes. When you push any object, everý slice of it is in contact with the next one. If reaction forces cancelled external forces, nothing would ever move.

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  • $\begingroup$ More explicitly, $F_{ha}$ cannot be equal to $F_{ab}$ unless the box $A$ is massless. $\endgroup$ – Dvij Mankad May 30 at 18:43
  • $\begingroup$ This is a correct and reasonable response to the OP. Can the downvoter explain the reasoning behind the down-vote? $\endgroup$ – Dvij Mankad May 30 at 18:44

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