1. **Problem statement:** Two particles A and B of masses $3m$ and $m$ respectively are connected by a light inextensible string over a smooth pulley at the top of a wedge. Each particle moves on an inclined plane at $30^\circ$ to the horizontal. A moves down a smooth face, B moves up a rough face with coefficient of friction $\mu$. Both accelerate with magnitude $10g$. Find: (a) Tension in the string by considering A's motion. (b) Value of $\mu$ by considering B's motion. (c) Resultant force on the pulley by the string. --- 2. **Formulas and rules:** - Weight component along incline: $mg\sin\theta$ - Normal reaction: $mg\cos\theta$ - Friction force: $\mu \times$ normal reaction - Newton's second law: $F = ma$ - Acceleration magnitude given: $a = 10g$ --- 3. **(a) Tension from A's motion:** - Mass of A: $3m$ - Incline angle: $30^\circ$ - A moves downwards with acceleration $a=10g$ For A, forces along incline: $$T - 3mg\sin 30^\circ = 3m a$$ Substitute $\sin 30^\circ = \frac{1}{2}$ and $a=10g$: $$T - 3mg \times \frac{1}{2} = 3m \times 10g$$ $$T - \frac{3mg}{2} = 30mg$$ Add $\frac{3mg}{2}$ to both sides: $$T = 30mg + \frac{3mg}{2} = \frac{60mg}{2} + \frac{3mg}{2} = \frac{63mg}{2}$$ --- 4. **(b) Value of $\mu$ from B's motion:** - Mass of B: $m$ - Incline angle: $30^\circ$ - B moves upwards with acceleration $a=10g$ For B, forces along incline: $$mg\sin 30^\circ + F_{friction} - T = m a$$ Friction opposes motion upwards, so acts downwards: $$mg \times \frac{1}{2} + \mu m g \cos 30^\circ - T = m \times 10g$$ Substitute $T = \frac{63mg}{2}$ from (a) and $\cos 30^\circ = \frac{\sqrt{3}}{2}$: $$\frac{mg}{2} + \mu m g \times \frac{\sqrt{3}}{2} - \frac{63mg}{2} = 10mg$$ Divide entire equation by $mg$: $$\frac{1}{2} + \mu \frac{\sqrt{3}}{2} - \frac{63}{2} = 10$$ Simplify left side: $$\mu \frac{\sqrt{3}}{2} + \frac{1}{2} - \frac{63}{2} = 10$$ $$\mu \frac{\sqrt{3}}{2} - \frac{62}{2} = 10$$ $$\mu \frac{\sqrt{3}}{2} - 31 = 10$$ Add 31 to both sides: $$\mu \frac{\sqrt{3}}{2} = 41$$ Multiply both sides by $\frac{2}{\sqrt{3}}$: $$\mu = 41 \times \frac{2}{\sqrt{3}} = \frac{82}{\sqrt{3}}$$ --- 5. **(c) Resultant force on the pulley:** The pulley experiences tension forces from both sides: - Tension magnitude: $T = \frac{63mg}{2}$ - Directions: each string makes $30^\circ$ with vertical (since planes are inclined at $30^\circ$) Resolve tensions into vertical and horizontal components: - Vertical component per string: $T \cos 30^\circ = \frac{63mg}{2} \times \frac{\sqrt{3}}{2} = \frac{63mg \sqrt{3}}{4}$ - Horizontal component per string: $T \sin 30^\circ = \frac{63mg}{2} \times \frac{1}{2} = \frac{63mg}{4}$ Since tensions pull in opposite horizontal directions, horizontal components cancel out. Vertical components add up: $$F_{resultant} = 2 \times \frac{63mg \sqrt{3}}{4} = \frac{63mg \sqrt{3}}{2}$$ **Direction:** vertically upwards along the bisector of the wedge. --- **Final answers:** - (a) $T = \frac{63mg}{2}$ - (b) $\mu = \frac{82}{\sqrt{3}}$ - (c) Resultant force magnitude on pulley: $\frac{63mg \sqrt{3}}{2}$ vertically upwards