1. **State the problem:** A box of mass 2 kg is held in equilibrium on a rough inclined plane at 30° to the horizontal by a rope inclined at angle $\alpha$ to the plane, where $\tan \alpha = \frac{3}{4}$. The coefficient of friction $\mu = \frac{1}{3}$. The box is on the point of slipping up the plane. Find the tension $T$ in the rope. 2. **Known values:** - Mass $m = 2$ kg - Incline angle $\theta = 30^\circ$ - $\tan \alpha = \frac{3}{4} \Rightarrow \sin \alpha = \frac{3}{5}, \cos \alpha = \frac{4}{5}$ (from the 3-4-5 triangle) - Coefficient of friction $\mu = \frac{1}{3}$ - Gravitational acceleration $g = 9.8$ m/s$^2$ 3. **Forces acting on the box:** - Weight $W = mg = 2g$ vertically downwards - Normal reaction $R$ perpendicular to the plane - Friction force $F = \mu R$ acting down the plane (since slipping up) - Tension $T$ in the rope, inclined at angle $\alpha$ to the plane 4. **Resolve weight into components:** - Parallel to plane: $W_{\parallel} = mg \sin \theta = 2g \sin 30^\circ = 2g \times \frac{1}{2} = g$ - Perpendicular to plane: $W_{\perp} = mg \cos \theta = 2g \cos 30^\circ = 2g \times \frac{\sqrt{3}}{2} = g\sqrt{3}$ 5. **Normal reaction $R$:** Since the box is in equilibrium perpendicular to the plane, $$ R = W_{\perp} + T \sin \alpha = g\sqrt{3} + T \sin \alpha $$ 6. **Friction force:** $$ F = \mu R = \frac{1}{3} \left(g\sqrt{3} + T \sin \alpha \right) $$ 7. **Resolve tension $T$ into components along and perpendicular to the plane:** - Along plane: $T \cos \alpha$ - Perpendicular to plane: $T \sin \alpha$ 8. **Equilibrium along the plane (taking up the plane as positive):** Since the box is on the point of slipping up, friction acts down the plane. $$ T \cos \alpha = W_{\parallel} + F $$ Substitute values: $$ T \cos \alpha = g + \frac{1}{3} \left(g\sqrt{3} + T \sin \alpha \right) $$ 9. **Substitute $\sin \alpha = \frac{3}{5}$ and $\cos \alpha = \frac{4}{5}$:** $$ T \times \frac{4}{5} = g + \frac{1}{3} \left(g\sqrt{3} + T \times \frac{3}{5} \right) $$ 10. **Multiply both sides by 15 to clear denominators:** $$ 15 \times T \times \frac{4}{5} = 15g + 5 \left(g\sqrt{3} + T \times \frac{3}{5} \right) $$ $$ 12T = 15g + 5g\sqrt{3} + 3T $$ 11. **Rearrange to isolate $T$:** $$ 12T - 3T = 15g + 5g\sqrt{3} $$ $$ 9T = g(15 + 5\sqrt{3}) $$ 12. **Solve for $T$:** $$ T = \frac{g(15 + 5\sqrt{3})}{9} = \frac{5g(3 + \sqrt{3})}{9} = \frac{5g}{9}(3 + \sqrt{3}) $$ 13. **Numerical value:** Using $g = 9.8$ m/s$^2$, $$ T = \frac{5 \times 9.8}{9} (3 + 1.732) = \frac{49}{9} \times 4.732 = 5.444 \times 4.732 \approx 25.75 \text{ N} $$ **Final answer:** $$ \boxed{T \approx 25.75 \text{ N}} $$