1. **Problem Statement:** We have a uniform rod of length 3 m and mass 2 kg hinged at one end. Two loads of 10 kg and 5 kg are attached at 1 m and 2 m from the hinge respectively. A cable attached at the rod's far end (3 m) runs to the ceiling 2 m horizontally from the hinge. We need to: (a) Draw the free body diagram (FBD). (b) Calculate the tension in the cable. (c) Calculate the reaction force vector at the hinge. (d) Find the horizontal distance $d$ at the ceiling to minimize cable tension if the cable connection can slide. 2. **Key formulas and concepts:** - Weight force: $W = mg$ where $g = 9.81$ m/s$^2$. - Torque (moment) about hinge: $\tau = r \times F = rF \sin(\theta)$. - Equilibrium conditions: - Sum of forces in $x$ and $y$ directions = 0. - Sum of moments about hinge = 0. 3. **Step (a): Free Body Diagram** - Forces acting on the rod: - Weight of rod $W_r = 2 \times 9.81 = 19.62$ N acting at rod center (1.5 m from hinge). - Load 1: $W_1 = 10 \times 9.81 = 98.1$ N at 1 m. - Load 2: $W_2 = 5 \times 9.81 = 49.05$ N at 2 m. - Tension $T$ in cable at 3 m, acting at angle $\theta$ where $\tan \theta = \frac{2}{3}$. - Reaction force at hinge with components $H_x$ and $H_y$. 4. **Step (b): Calculate tension $T$** - Calculate angle $\theta$: $$\theta = \arctan\left(\frac{2}{3}\right) \approx 33.69^\circ$$ - Sum moments about hinge (counterclockwise positive): $$\sum \tau = 0 = T \times 3 \times \sin(\theta) - W_1 \times 1 - W_2 \times 2 - W_r \times 1.5$$ - Substitute values: $$T \times 3 \times \sin(33.69^\circ) = 98.1 \times 1 + 49.05 \times 2 + 19.62 \times 1.5$$ $$T \times 3 \times 0.5547 = 98.1 + 98.1 + 29.43 = 225.63$$ $$T = \frac{225.63}{3 \times 0.5547} = \frac{225.63}{1.664} \approx 135.6 \text{ N}$$ 5. **Step (c): Calculate reaction force at hinge $(H_x, H_y)$** - Sum forces in $x$ direction: $$\sum F_x = 0 = H_x - T \cos(\theta)$$ $$H_x = T \cos(33.69^\circ) = 135.6 \times 0.8321 = 112.9 \text{ N}$$ - Sum forces in $y$ direction: $$\sum F_y = 0 = H_y + T \sin(\theta) - W_1 - W_2 - W_r$$ $$H_y = W_1 + W_2 + W_r - T \sin(\theta) = 98.1 + 49.05 + 19.62 - 135.6 \times 0.5547$$ $$H_y = 166.77 - 75.25 = 91.52 \text{ N}$$ - Reaction force vector: $$\vec{H} = (112.9, 91.52) \text{ N}$$ 6. **Step (d): Minimize tension by varying horizontal distance $d$** - Let $d$ be the horizontal distance from hinge to cable attachment on ceiling. - Cable length vector: horizontal $d$, vertical $h$ (height unknown but constant). - Angle $\theta = \arctan(\frac{h}{d})$. - Moment equilibrium: $$T \times 3 \times \sin(\theta) = 225.63$$ $$T = \frac{225.63}{3 \sin(\theta)} = \frac{225.63}{3 \times \frac{h}{\sqrt{d^2 + h^2}}} = \frac{225.63 \sqrt{d^2 + h^2}}{3h}$$ - To minimize $T$, minimize $\sqrt{d^2 + h^2}$ with respect to $d$. - Since $h$ is constant, $T$ is minimized when $d=0$. - **Interpretation:** The cable should be attached directly above the hinge (no horizontal offset) to minimize tension. **Final answers:** - (b) Tension $T \approx 135.6$ N - (c) Reaction force at hinge $\vec{H} = (112.9, 91.52)$ N - (d) Minimum tension at $d=0$ m