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Suppose that we have 2 objects next to each other.The object at the left is A and the object at right is B.now we have these situations: 1) we push B to the left. 2) we pull A to the left.(in this case A and B are stuck to each other.) Do the forces A exert on B have the same magnitude and direction in both situations?

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It depends; are the forces with which you are pushing and pulling equal in magnitude? Are the objects rigid? –  joshphysics Jan 30 '13 at 8:46
    
yes, both pushing and pulling forces have the same magnitude and direction (to the left) and objects are rigid. –  Sogol Jan 30 '13 at 11:53
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2 Answers

Let us consider the two scenarios separately. Let $B$ have a mass of $m_b$ and $A$ have a mass of $m_a$. Since both bodies are connected they will have the same acceleration, which I will call $a$.

Scenario 1)

Let us first consider object $B$. There are two forces acting on this object. The external force, and the contact force due to $A$, opposing the external force.

Let $F$ be the external force applied to object $B$ and let $F_a$ be the force on $B$ due to $A$, opposing the externally applied force. Therefore, net force acting on object B is:

$F_{netB}= m_Ba = F - F_a$.

Therefore,

$a = \frac{F - F_a}{m_B} (1)$

Now let us consider object $A$. There is one force acting on this object, that is the contact force between $B$ and $A$. Because of Newton's third law, this is the same as the contact force $A$ applies to $B$, that is, $F_a$. Therefore the net force acting on $A$ is:

$F_{netA} = m_aa = F_a$.

Therefore,

$a = \frac{F_a}{m_a} (2)$

Equating (1) and (2) gives: $\frac{F_a}{m_a} = \frac{F - F_a}{m_B}$

which can be re-arranged to arrive at:

$F_a = \frac{Fm_a}{m_a + m_b}$

Scneario 2) Let us consider the forces acting on object A. There is the externally applied force $F$, and a force pulling against this, due to object $B$, which I will label $F_a$. The net force acting on $A$ is thus:

$F_{netA} = m_aa = F - F_a$

and therefore,

$a = \frac{F - F_a}{m_a}$ (1)

Let us now consider object $B$. There is one force acting on object $B$, and this is $F_b$, by Newton's third law. The net force on object B is thus:

$F_{netB} = m_ba = F_a$

and therefore,

$a = \frac{F_a}{m_b}$ (2)

Equating (1) and (2) gives:

$\frac{F_a}{m_b} = \frac{F - F_a}{m_a}$

and solving gives:

$F_a = \frac{Fm_b}{m_a + m_b}$

Therefore, the force acting on B due to A will differ depending on the scenario.

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The answer is no! In both direction and magnitude.

I will present a formal way and a heuristic argument.

Formal: write Newton's laws. I assume we exert the same force $F$ in both cases. take left moving as positive.

Case 1) The applied force is on B. Let the force of A on B be $-R$. (so $R$ is the force of B on A). So, \begin{align} &m_B a = F - R \\ & m_A a = R \\ \implies & a = \frac{F}{m_A + m_B} \\ \implies & -R = -\frac{F m_A }{m_A + m_B} \end{align} where $a$ is the common acceleration of both blocks.

Case 2) The applied force is on A. $N$ is the force of A on B. \begin{align} & m_A a = F - N \\ & m_B a = N \\ \implies & a = \frac{F}{m_A + m_B} \\ \implies & N = \frac{F m_B}{m_A + m_B}. \end{align} Clearly, $-R \neq N$.

Heuristic argument: since the force applied is assumed to be the same, the accelerations in both cases are equal $ = a$. now imagine block B is much more massive than A. so to accelerate A , we don't need much force on it. Correspondingly, for case 1), the force that B applies on A is small, i.e. $m_A a$, and by Newton's 3rd law the force on A on B is also small, i.e. $-m_A a$. For case 2), something has to accelerate B, and the only agent is A. Since B's mass is so large, it needs a lot of force, i.e. $m_B a$. So A must pull B with a lot of force, $m_B a$. We see that $-m_A a \neq m_B a$.

There you go.

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thanks, and something else.. does it change the answer for the second situation if there's friction between A and the horizontal surface? –  Sogol Jan 30 '13 at 11:58
    
@ Sogol: change what answer? You mean the expression for $N$? Yes. From the heuristic argument we see that the force A exerts on B is $m_B a$. But because of the friction, $a$ this time is smaller than without friction. So the force that A exerts on B is less. Think of the limiting case: if the friction coefficient is infinite, then the blocks won't budge no matter how much force you apply. Since the system doesn't accelerate, the net force on B must be 0. And since the only force on B is from A, the force of A on B must be 0. –  nervxxx Jan 30 '13 at 19:11
    
anndd one more thing, the direction of N is to the left, am I right? –  Sogol Jan 30 '13 at 19:43
    
@ Sogol: yes, left is positive. $N > 0$ so it is to the left. –  nervxxx Jan 30 '13 at 19:57
    
then why do we have mAa=F−N and not mAa=F+N? –  Sogol Jan 30 '13 at 20:22
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