1. **State the problem:** A 5 kg block rests on a rough inclined plane at 30° connected by a light string over a smooth pulley to a 6 kg hanging block. The system is in equilibrium. Find the magnitude and direction of the friction force on the 5 kg block. 2. **Identify forces on the 5 kg block:** - Weight component down the incline: $W_1 = mg \sin 30^\circ = 5 \times 9.8 \times 0.5 = 24.5$ N - Weight component perpendicular to incline: $N = mg \cos 30^\circ = 5 \times 9.8 \times \frac{\sqrt{3}}{2} \approx 42.44$ N - Tension $T$ in the string acts up the incline. - Friction force $f$ acts to oppose motion; direction unknown yet. 3. **For the 6 kg hanging block:** - Weight $W_2 = 6 \times 9.8 = 58.8$ N - Tension $T$ acts upward. 4. **Equilibrium conditions:** - For 6 kg block: $T = W_2 = 58.8$ N - For 5 kg block along incline: $T + f = W_1$ 5. **Calculate friction force:** $$f = W_1 - T = 24.5 - 58.8 = -34.3 \text{ N}$$ Negative means friction acts opposite to assumed direction. 6. **Interpretation:** Since friction is negative, it acts down the incline (same direction as weight component). 7. **Convert friction force to kg wt:** $$f_{kg} = \frac{34.3}{9.8} \approx 3.5 \text{ kg wt}$$ **Final answer:** The friction force is 3.5 kg wt acting downward along the incline. **Answer choice:** (b) 3.5 kg wt downward.