1. **Problem statement:** A body weighing 88 newtons is on a rough inclined plane inclined at $60^\circ$ to the horizontal. The body is about to slide, so we need to find the magnitude of the limiting static friction force. 2. **Relevant formulas and concepts:** - Weight $W = 88$ N acts vertically downward. - The component of weight parallel to the incline: $W_\parallel = W \sin 60^\circ$. - The component of weight perpendicular to the incline: $W_\perp = W \cos 60^\circ$. - Limiting friction force $f_{lim}$ equals the component of weight parallel to the incline when the body is just about to slide. 3. **Calculate components:** - $\sin 60^\circ = \frac{\sqrt{3}}{2}$ - $W_\parallel = 88 \times \frac{\sqrt{3}}{2} = 44\sqrt{3}$ N 4. **Interpretation:** - Since the body is about to slide, the limiting friction force balances the parallel component of weight. - Therefore, limiting static friction $= 44\sqrt{3}$ N. 5. **Answer:** The magnitude of the limiting static friction is $44\sqrt{3}$ newtons. **Correct option:** (b) 44√3