1. **State the problem:** A particle of mass 0.4 kg rests on a rough inclined plane at angle $\alpha$ where $\tan \alpha = \frac{3}{4}$. A horizontal force $P$ acts to hold it at rest, and the particle is on the point of sliding up the plane. The coefficient of friction is $\mu = \frac{1}{3}$. 2. **Find $\alpha$:** Given $\tan \alpha = \frac{3}{4}$, use the Pythagorean identity: $$\sin \alpha = \frac{3}{5}, \quad \cos \alpha = \frac{4}{5}$$ 3. **Identify forces:** - Weight $W = mg = 0.4 \times 9.8 = 3.92\,\text{N}$ - Components of weight along plane: - Down the plane: $W \sin \alpha = 3.92 \times \frac{3}{5} = 2.352\,\text{N}$ - Perpendicular to plane: $W \cos \alpha = 3.92 \times \frac{4}{5} = 3.136\,\text{N}$ 4. **Forces on the particle:** - Normal reaction $R$ perpendicular to plane - Friction force $f = \mu R = \frac{1}{3} R$, acting down the plane since particle tends to slide up - Horizontal force $P$ acts horizontally 5. **Resolve forces perpendicular to plane:** The horizontal force $P$ has a component perpendicular to the plane: $$P \cos \alpha$$ Balance perpendicular forces: $$R = W \cos \alpha + P \cos \alpha = (3.92 + P) \cos \alpha$$ 6. **Resolve forces parallel to plane:** Since particle is on the point of sliding up, friction acts down the plane. Balance forces along the plane: $$P \sin \alpha + f = W \sin \alpha$$ Substitute $f = \frac{1}{3} R$: $$P \sin \alpha + \frac{1}{3} R = W \sin \alpha$$ 7. **Substitute $R$ from step 5:** $$P \sin \alpha + \frac{1}{3} (3.92 \cos \alpha + P \cos \alpha) = 3.92 \sin \alpha$$ 8. **Plug in $\sin \alpha = \frac{3}{5}$, $\cos \alpha = \frac{4}{5}$:** $$P \times \frac{3}{5} + \frac{1}{3} \left(3.92 \times \frac{4}{5} + P \times \frac{4}{5} \right) = 3.92 \times \frac{3}{5}$$ 9. **Simplify:** $$\frac{3}{5} P + \frac{1}{3} \left(3.136 + \frac{4}{5} P \right) = 2.352$$ $$\frac{3}{5} P + \frac{1}{3} \times 3.136 + \frac{1}{3} \times \frac{4}{5} P = 2.352$$ $$\frac{3}{5} P + 1.045 + \frac{4}{15} P = 2.352$$ 10. **Combine $P$ terms:** $$\frac{3}{5} P + \frac{4}{15} P = \frac{9}{15} P + \frac{4}{15} P = \frac{13}{15} P$$ 11. **Solve for $P$:** $$\frac{13}{15} P = 2.352 - 1.045 = 1.307$$ $$P = \frac{1.307 \times 15}{13} = 1.509\,\text{N}$$ 12. **Find normal reaction $R$:** $$R = 3.136 + P \times \frac{4}{5} = 3.136 + 1.509 \times 0.8 = 3.136 + 1.207 = 4.343\,\text{N}$$ **Final answer:** $$\boxed{R = 4.34\,\text{N}}$$