1. **Problem statement:** A box of mass 2 kg rests on a rough inclined plane inclined at 30° to the horizontal. The box is held in equilibrium by a rope inclined at angle $\alpha$ to the plane, where $\tan \alpha = \frac{3}{5}$. The coefficient of friction between the box and the plane is $\mu = \frac{1}{3}$. The box is on the point of slipping up the plane. Find the tension in the rope. 2. **Known values:** - Mass $m = 2$ kg - Incline angle $\theta = 30^\circ$ - $\tan \alpha = \frac{3}{5}$ so $\sin \alpha = \frac{3}{5}$ and $\cos \alpha = \frac{4}{5}$ (from the right triangle with sides 3,4,5) - Coefficient of friction $\mu = \frac{1}{3}$ - Gravitational acceleration $g$ (leave as $g$ for generality) 3. **Forces acting on the box:** - Weight $W = mg = 2g$ - Normal reaction $R$ from the plane - Friction force $F = \mu R$ - Tension $T$ in the rope, inclined at angle $\alpha$ to the plane 4. **Resolve weight into components:** - Perpendicular to plane: $W_\perp = mg \cos \theta = 2g \cos 30^\circ$ - Parallel to plane (down the slope): $W_\parallel = mg \sin \theta = 2g \sin 30^\circ$ 5. **Equilibrium conditions:** Since the box is on the point of slipping up the plane, friction acts down the plane (opposing upward slip). - Perpendicular direction: $$ R = W_\perp = 2g \cos 30^\circ $$ - Parallel direction (along the plane): Sum of forces = 0 $$ T \cos \alpha - F - W_\parallel = 0 $$ Substitute $F = \mu R$: $$ T \cos \alpha - \mu R - W_\parallel = 0 $$ 6. **Substitute known values:** - $\cos 30^\circ = \frac{\sqrt{3}}{2}$ - $\sin 30^\circ = \frac{1}{2}$ - $\cos \alpha = \frac{4}{5}$ Calculate: $$ R = 2g \times \frac{\sqrt{3}}{2} = g \sqrt{3} $$ $$ W_\parallel = 2g \times \frac{1}{2} = g $$ 7. **Plug into parallel force equation:** $$ T \times \frac{4}{5} - \frac{1}{3} \times g \sqrt{3} - g = 0 $$ Rearranged: $$ T \times \frac{4}{5} = g + \frac{g \sqrt{3}}{3} = g \left(1 + \frac{\sqrt{3}}{3}\right) $$ 8. **Solve for $T$:** $$ T = \frac{5}{4} g \left(1 + \frac{\sqrt{3}}{3}\right) = \frac{5g}{4} \left(1 + \frac{\sqrt{3}}{3}\right) $$ 9. **Final answer:** The tension in the rope is $$\boxed{T = \frac{5g}{4} \left(1 + \frac{\sqrt{3}}{3}\right)}$$ This expression gives the tension in terms of $g$. Numerically, $\sqrt{3} \approx 1.732$, so $$ 1 + \frac{1.732}{3} \approx 1 + 0.577 = 1.577 $$ Thus, $$ T \approx \frac{5g}{4} \times 1.577 = 1.971 g $$ If $g = 9.8$, then $$ T \approx 1.971 \times 9.8 \approx 19.3 $$