1. **Problem statement:** A particle of mass 12 kg is stationary on a rough plane inclined at an angle of 25° to the horizontal. A pulling force $P$ acts at an angle of 8° above the line of greatest slope of the plane. The coefficient of friction between the particle and the plane is 0.3. Find the greatest possible value of $P$ that keeps the particle in equilibrium. 2. **Forces involved:** - Weight $W = mg = 12 \times 9.8 = 117.6$ N acting vertically downward. - Pulling force $P$ at 8° above the slope. - Friction force $F$ opposing motion, with maximum $F_{max} = \mu R$ where $R$ is the normal reaction. 3. **Resolve weight into components along and perpendicular to the plane:** Along slope: $W_{\parallel} = W \sin 25^\circ$ Perpendicular to slope: $W_{\perp} = W \cos 25^\circ$ 4. **Resolve pulling force $P$ into components along and perpendicular to the plane:** Along slope: $P_{\parallel} = P \cos 8^\circ$ Perpendicular to slope: $P_{\perp} = P \sin 8^\circ$ 5. **Equilibrium conditions:** - Along slope: $P_{\parallel} + F = W_{\parallel}$ - Perpendicular to slope: $R = W_{\perp} - P_{\perp}$ 6. **Friction force at maximum:** $F = \mu R = 0.3 (W_{\perp} - P_{\perp})$ 7. **Substitute and solve for $P$:** Along slope: $$P \cos 8^\circ + 0.3 (W \cos 25^\circ - P \sin 8^\circ) = W \sin 25^\circ$$ Rearranged: $$P \cos 8^\circ + 0.3 W \cos 25^\circ - 0.3 P \sin 8^\circ = W \sin 25^\circ$$ Group $P$ terms: $$P (\cos 8^\circ - 0.3 \sin 8^\circ) = W \sin 25^\circ - 0.3 W \cos 25^\circ$$ Calculate numeric values: $\cos 8^\circ \approx 0.9903$, $\sin 8^\circ \approx 0.1392$, $\sin 25^\circ \approx 0.4226$, $\cos 25^\circ \approx 0.9063$ $$P (0.9903 - 0.3 \times 0.1392) = 117.6 \times 0.4226 - 0.3 \times 117.6 \times 0.9063$$ $$P (0.9903 - 0.0418) = 49.66 - 31.96$$ $$P \times 0.9485 = 17.7$$ Divide both sides by 0.9485: $$P = \frac{17.7}{0.9485}$$ $$P \approx 18.65$$ **Final answer:** $$\boxed{18.7 \text{ N}}$$