1. **Problem statement:** A body weighing 4 kg is placed on a rough inclined plane at an angle of 45° to the horizontal. The coefficient of static friction is $\mu_s = \frac{1}{2}$. We need to find the greatest force that can act along the slope to keep the body in equilibrium. 2. **Relevant formulas and concepts:** - Weight $W = mg$, but here weight is given as 4 kg.wt (kilogram-weight), so $W = 4$ kg.wt. - Components of weight along the plane: $W_{\parallel} = W \sin \theta$, perpendicular to the plane: $W_{\perp} = W \cos \theta$. - Maximum static friction force $F_f = \mu_s W_{\perp}$. - For equilibrium, the applied force $F$ along the slope must balance the component of weight and friction. 3. **Calculate components:** - $W_{\parallel} = 4 \sin 45^\circ = 4 \times \frac{\sqrt{2}}{2} = 2\sqrt{2}$ - $W_{\perp} = 4 \cos 45^\circ = 4 \times \frac{\sqrt{2}}{2} = 2\sqrt{2}$ 4. **Calculate maximum friction:** - $F_f = \mu_s W_{\perp} = \frac{1}{2} \times 2\sqrt{2} = \sqrt{2}$ 5. **Determine greatest force for equilibrium:** - The greatest force $F_{max}$ that can act along the slope to preserve equilibrium is the sum of the downslope component of weight and the maximum friction force: $$F_{max} = W_{\parallel} + F_f = 2\sqrt{2} + \sqrt{2} = 3\sqrt{2}$$ **Final answer:** $3\sqrt{2}$ kg.wt. This corresponds to option (a).