Contact force betweeen 2 objects 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?  
 A: 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.
A: 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.
