1. **Problem statement:** We have a system with a 5 kg particle on a rough horizontal table connected by strings over pulleys to 4 kg and 6 kg hanging masses. The 6 kg mass is 0.5 m above the ground. The system is in limiting equilibrium. (a) Show the coefficient of friction between the 5 kg particle and the table is 0.4. 2. **Step 1: Define forces and variables.** Let the tension in the string on the 4 kg side be $T_1$ and on the 6 kg side be $T_2$. The friction force $F$ acts on the 5 kg particle opposing motion. 3. **Step 2: Write equilibrium equations for the 5 kg particle.** Since the system is in limiting equilibrium, the 5 kg particle is on the verge of moving but stationary. Horizontal forces on 5 kg particle: $$T_2 - T_1 = F$$ Friction force $F = \mu R$, where $R$ is the normal reaction equal to weight of 5 kg particle: $$R = 5 \times 9.8 = 49 \text{ N}$$ 4. **Step 3: Write equilibrium for hanging masses.** For 4 kg mass: $$T_1 = 4 \times 9.8 = 39.2 \text{ N}$$ For 6 kg mass: $$T_2 = 6 \times 9.8 = 58.8 \text{ N}$$ 5. **Step 4: Calculate friction force and coefficient.** $$F = T_2 - T_1 = 58.8 - 39.2 = 19.6 \text{ N}$$ Coefficient of friction: $$\mu = \frac{F}{R} = \frac{19.6}{49} = 0.4$$ **Answer (a):** Coefficient of friction is 0.4. --- (b) Now the 6 kg mass is replaced by 8 kg and the system is released from rest. 6. **Step 1: Define acceleration and tensions.** Let acceleration of 4 kg mass be $a$ upwards, 8 kg mass be $a$ downwards, and 5 kg particle move horizontally with acceleration $a$. Tensions in strings are $T_1$ (left) and $T_2$ (right). 7. **Step 2: Write equations of motion for each mass.** For 4 kg mass: $$T_1 - 4 \times 9.8 = 4a$$ For 8 kg mass: $$8 \times 9.8 - T_2 = 8a$$ For 5 kg particle (horizontal): $$T_2 - T_1 - F = 5a$$ Friction force $F = \mu R = 0.4 \times 49 = 19.6$ N opposing motion. 8. **Step 3: Substitute and solve.** From 4 kg mass: $$T_1 = 39.2 + 4a$$ From 8 kg mass: $$T_2 = 78.4 - 8a$$ Substitute into 5 kg equation: $$78.4 - 8a - (39.2 + 4a) - 19.6 = 5a$$ Simplify: $$78.4 - 8a - 39.2 - 4a - 19.6 = 5a$$ $$19.6 - 12a = 5a$$ $$19.6 = 17a$$ $$a = \frac{19.6}{17} \approx 1.15 \text{ m/s}^2$$ 9. **Step 4: Find tensions.** $$T_1 = 39.2 + 4 \times 1.15 = 39.2 + 4.6 = 43.8 \text{ N}$$ $$T_2 = 78.4 - 8 \times 1.15 = 78.4 - 9.2 = 69.2 \text{ N}$$ **Answer (b):** Acceleration $a \approx 1.15$ m/s$^2$, $T_1 \approx 43.8$ N, $T_2 \approx 69.2$ N. --- (c) After the 8 kg particle hits the ground and stops, find the time until the other two particles come to rest. 10. **Step 1: New system after 8 kg hits ground.** Now only 4 kg mass and 5 kg particle connected by one string remain. The 4 kg mass moves upward with initial velocity $u = a t_1$ where $t_1$ is time taken for 8 kg mass to fall 0.5 m. 11. **Step 2: Find $t_1$ for 8 kg mass to fall 0.5 m with acceleration $a$.** Using: $$s = ut + \frac{1}{2} a t^2$$ Initial velocity $u=0$, displacement $s=0.5$ m, acceleration $a=1.15$ m/s$^2$ downward. $$0.5 = 0 + \frac{1}{2} \times 1.15 \times t_1^2$$ $$t_1^2 = \frac{2 \times 0.5}{1.15} = \frac{1}{1.15} \approx 0.8696$$ $$t_1 = \sqrt{0.8696} \approx 0.93 \text{ s}$$ 12. **Step 3: Velocity of 4 kg mass at $t_1$.** Velocity $v = u + at = 0 + 1.15 \times 0.93 = 1.07$ m/s upward. 13. **Step 4: After 8 kg hits ground, 4 kg and 5 kg connected by string with friction acting on 5 kg particle.** For 5 kg particle: $$T - F = 5a'$$ For 4 kg mass: $$4g - T = 4a'$$ Friction force $F = 19.6$ N opposing motion. 14. **Step 5: Combine equations to find deceleration $a'$.** Add equations: $$4g - F = 9a'$$ $$9.8 \times 4 - 19.6 = 9a'$$ $$39.2 - 19.6 = 9a'$$ $$19.6 = 9a'$$ $$a' = \frac{19.6}{9} \approx 2.18 \text{ m/s}^2$$ This is deceleration (opposite to velocity). 15. **Step 6: Time to stop from velocity 1.07 m/s with deceleration 2.18 m/s$^2$.** $$t = \frac{v}{a'} = \frac{1.07}{2.18} \approx 0.49 \text{ s}$$ **Answer (c):** Time elapsed after 8 kg hits ground before other particles stop is approximately 0.49 seconds.