system of blocks

Like this:

T is tension in string; N is normal reaction force; F is force of frictional resistance.

.

The frictional force must exactly balance the horizontal force supplied by the string.  The tension in the string angled at 40° to the vertical is the same as the weight of C (because the pulley is frictionless).  The horizontal component of this is just 1kg*9.8m/s²*sin(40°)

$${\mathtt{1}}{\mathtt{\,\times\,}}{\mathtt{9.8}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{40}}^\circ\right)} = {\mathtt{6.299\: \!318\: \!574\: \!932\: \!6}}$$

or approximately 6.3N

.

Alan 

can you make free body digram for each object  ? 

for the free body digram for B 

why Tsin40 directid to the left ? 

On B, mass A is applying a force that is trying to pull B to the left.  This is balanced by the pull from the sloping string over the pulley, which has a horizontal component of Tsin(40°).  Because the system is stationary, the pull to the left must have the same magnitude as the pull to the right.  (Note: the vertical component of the pull from the sloping string exactly balances the weight of B).

.

thank you