1. **Problem statement:** Two particles A and B with masses 0.8 kg and 1.2 kg lie on inclined planes at angles 30° and 60° respectively, connected by a light inextensible string over a smooth pulley. Both planes have the same coefficient of friction $\mu$. Find $\mu$ for which the system is in limiting equilibrium. 2. **Forces on each particle:** - Weight components along the plane: - Particle A: $0.8g \sin 30^\circ$ - Particle B: $1.2g \sin 60^\circ$ - Friction force opposes motion and equals $\mu \times$ normal reaction. - Normal reactions: - Particle A: $0.8g \cos 30^\circ$ - Particle B: $1.2g \cos 60^\circ$ 3. **Limiting equilibrium condition:** Since the system is on the verge of moving, friction forces act to prevent motion. Assume particle B tends to move down, so friction on B acts up the plane, and friction on A acts down the plane. 4. **Equations of motion:** Let tension in the string be $T$. For particle A (up the plane positive): $$ T + \mu \times 0.8g \cos 30^\circ = 0.8g \sin 30^\circ $$ For particle B (down the plane positive): $$ 1.2g \sin 60^\circ = T + \mu \times 1.2g \cos 60^\circ $$ 5. **Eliminate $T$ by equating:** $$ 0.8g \sin 30^\circ - \mu \times 0.8g \cos 30^\circ = 1.2g \sin 60^\circ - \mu \times 1.2g \cos 60^\circ $$ 6. **Simplify and solve for $\mu$:** Divide both sides by $g$: $$ 0.8 \sin 30^\circ - 0.8 \mu \cos 30^\circ = 1.2 \sin 60^\circ - 1.2 \mu \cos 60^\circ $$ Rearranged: $$ 0.8 \sin 30^\circ - 1.2 \sin 60^\circ = 0.8 \mu \cos 30^\circ - 1.2 \mu \cos 60^\circ $$ Factor $\mu$: $$ 0.8 \sin 30^\circ - 1.2 \sin 60^\circ = \mu (0.8 \cos 30^\circ - 1.2 \cos 60^\circ) $$ Calculate values: - $\sin 30^\circ = 0.5$ - $\sin 60^\circ = \frac{\sqrt{3}}{2} \approx 0.866$ - $\cos 30^\circ = \frac{\sqrt{3}}{2} \approx 0.866$ - $\cos 60^\circ = 0.5$ Substitute: $$ 0.8 \times 0.5 - 1.2 \times 0.866 = \mu (0.8 \times 0.866 - 1.2 \times 0.5) $$ $$ 0.4 - 1.0392 = \mu (0.6928 - 0.6) $$ $$ -0.6392 = \mu (0.0928) $$ 7. **Solve for $\mu$:** $$ \mu = \frac{-0.6392}{0.0928} = -6.89 $$ Since $\mu$ cannot be negative, the assumption about friction directions must be reversed. 8. **Reverse friction directions:** Assuming friction on A acts up the plane and on B acts down the plane, the equations become: For A: $$ T - \mu \times 0.8g \cos 30^\circ = 0.8g \sin 30^\circ $$ For B: $$ 1.2g \sin 60^\circ + \mu \times 1.2g \cos 60^\circ = T $$ Equate $T$: $$ 0.8g \sin 30^\circ + \mu \times 0.8g \cos 30^\circ = 1.2g \sin 60^\circ + \mu \times 1.2g \cos 60^\circ $$ Divide by $g$ and rearrange: $$ 0.8 \sin 30^\circ - 1.2 \sin 60^\circ = \mu (1.2 \cos 60^\circ - 0.8 \cos 30^\circ) $$ Substitute values: $$ 0.4 - 1.0392 = \mu (1.2 \times 0.5 - 0.8 \times 0.866) $$ $$ -0.6392 = \mu (0.6 - 0.6928) $$ $$ -0.6392 = \mu (-0.0928) $$ Solve for $\mu$: $$ \mu = \frac{-0.6392}{-0.0928} = 6.89 $$ 9. **Interpretation:** A coefficient of friction $\mu = 6.89$ is unrealistically high, indicating the system cannot be in limiting equilibrium with friction preventing motion under these conditions. **Final answer:** $$ \boxed{\mu = 6.89} $$ This is the coefficient of friction for limiting equilibrium under the reversed friction assumption.