1. **Stating the problem:** A rod (beam) of mass 16 kg is placed on a horizontal surface at an angle of $60^\circ$ and is hinged to a wall. The rod length is 10 m, and the center of gravity acts at $\frac{2}{5}$ of the length from the hinge. We need to find the coefficient of friction between the rod and the surface. 2. **Known values:** - Mass, $m = 16$ kg - Length, $L = 10$ m - Angle with horizontal, $\theta = 60^\circ$ - Distance of center of gravity from hinge, $d = \frac{2}{5}L = 4$ m - Gravitational acceleration, $g = 9.8$ m/s$^2$ 3. **Forces and torques:** - Weight force $W = mg = 16 \times 9.8 = 156.8$ N acting downward at 4 m from hinge. - Normal force $N$ acts vertically upward at the base. - Friction force $f$ acts horizontally at the base opposing slipping. 4. **Equilibrium conditions:** - Sum of vertical forces: $N = W$ (since no vertical acceleration) - Sum of horizontal forces: $f$ balances horizontal component of force at hinge. - Sum of torques about hinge must be zero. 5. **Torque calculation:** Taking moments about the hinge: - Torque due to weight: $\tau_W = W \times d \times \cos(\theta)$ (component perpendicular to rod) - Torque due to friction: $\tau_f = f \times L$ (friction acts at base) Since the rod is in equilibrium: $$\tau_f = \tau_W$$ $$f \times L = W \times d \times \cos(\theta)$$ 6. **Calculate friction force $f$:** $$f = \frac{W \times d \times \cos(\theta)}{L} = \frac{156.8 \times 4 \times \cos 60^\circ}{10}$$ $$= \frac{156.8 \times 4 \times 0.5}{10} = \frac{313.6}{10} = 31.36 \text{ N}$$ 7. **Calculate normal force $N$:** $$N = W = 156.8 \text{ N}$$ 8. **Coefficient of friction $\mu$:** $$\mu = \frac{f}{N} = \frac{31.36}{156.8} = 0.2$$ **Final answer:** The coefficient of friction is $\boxed{0.2}$.