Block $A$ of weight $100 \, N$ rests on a frictionless inclined plane of slope angle $30^{\circ}$. $A$ flexible cord attached to $A$ passes over a frictionless pulley and is connected to block $B$ of weight $w$. Find the weight $w$ for which the system is in equilibrium.

(C) For the system to be in equilibrium,the net force on each block must be zero.
For block $A$ on the inclined plane,the component of weight acting down the plane is $W_A \sin \theta$,where $W_A = 100 \, N$ and $\theta = 30^{\circ}$.
The tension $T$ in the cord must balance this component:
$T = W_A \sin 30^{\circ} \quad \dots (i)$
For block $B$ hanging vertically,the tension $T$ must balance its weight $w$:
$T = w \quad \dots (ii)$
Equating $(i)$ and $(ii)$:
$w = W_A \sin 30^{\circ}$
$w = 100 \times \sin 30^{\circ}$
$w = 100 \times \frac{1}{2}$
$w = 50 \, N$
Thus,the weight $w$ required for equilibrium is $50 \, N$.