Friction and Newton's Laws of motion I came across the following problem.

Ok, I thought that assuming there is no friction between blocks B and C, when the blocks B and A will move, the block C will go to the right due to the force by the mass less pulley. I thought that the acceleration of B to the left will be numerically same as the vertical acceleration of block A, because there is no friction to oppose it. Just free motion. But it turned out that the acceleration of B gets affected by the acceleration of C.
And it is acceleration of B if C was fixed - acceleration of C itself.
My question is why is this so? What does B have to do with C and if they are not interacting by any means, then why would B get the acceleration of C?
 A: B and C do interact via the pulley.  It appears you're aware that the string is pushing on C (so that C accelerates to the right).  As forces arise in couples, this implies that the behavior of the rest of the system must depend on C.
In this case the motion of C also moves A horizontally.  Moving C to the right effectively moves A and B closer together.
If C were fixed, the descent of A requires B to move left by the same amount.  But if C can move, then less motion of B is needed.  In the limit, if we force C to move to the right with acceleration equal to $g$, then no motion (and no force) is required from B.  B will remain motionless.
A: In the question it is given that the pulley is attached to block $C$ , so any force on pulley can affect the motion of block $C$
Now note that there's a tension in string (because of block $A$) and this tension in string pushes pulley as shown

This causes horizontal motion of block $C$ and increase the normal force on it.
Let vertical acceleration of $A$ be $v$, horizontal be $h$
acceleration of $B$ be $b$ and of $C$ be $c$
First apply string constraint
Let $l_1$ be horizontal length of string, $l_2$ be vertical length of string
$l_1+l_2=\text{constant}$
Differentiating we get
$v_1+v_2=0$ or $v_1=-v_2$ where $v_1$ is the rate at which $l_1$ decreases that is seperation between pulley and block $B$ decreases and $v_2$ the rate at which $l_2$ increases
differentiate both sides again
$a_1=-a_2$ where $a_1$ is the rate at which rate of decrease of $l_1$ changes and $a_2$=rate at which rate of increase of l1 changes
$a_1=b+c$ and $a_2=v$
Thus you get $v=-(b+c)$
Therefore it is clear that motion of $B$, $C$ and $A$ are interrelated
