1. **Problem Statement:** a. Draw the free body diagram of a cable car of mass 20,000 kg pulled up an inclined track at angle $\alpha = 30^\circ$. b. Calculate the stress in the cable rope with diameter 50 mm. c. Calculate the strain in the cable rope given elastic deformation with Young's modulus $E = 220$ GPa. 2. **Free Body Diagram Explanation:** The forces acting on the car are: - Weight $W = mg$ acting vertically downward. - Tension $T$ in the cable pulling up the incline. - Normal reaction $N$ perpendicular to the incline. 3. **Calculations:** **Step 1: Calculate the weight** $$W = mg = 20000 \times 9.81 = 196200\ \text{N}$$ **Step 2: Resolve weight into components parallel and perpendicular to incline** Parallel component pulling car down the incline: $$W_{\parallel} = W \sin \alpha = 196200 \times \sin 30^\circ = 196200 \times 0.5 = 98100\ \text{N}$$ Perpendicular component: $$W_{\perp} = W \cos \alpha = 196200 \times \cos 30^\circ = 196200 \times 0.866 = 169800\ \text{N}$$ **Step 3: Tension in cable** Since the car is pulled slowly (assumed equilibrium), tension $T$ balances the parallel component: $$T = W_{\parallel} = 98100\ \text{N}$$ **Step 4: Calculate cross-sectional area of cable** Diameter $d = 50$ mm = 0.05 m $$A = \pi \left(\frac{d}{2}\right)^2 = \pi \times (0.025)^2 = \pi \times 0.000625 = 0.0019635\ \text{m}^2$$ **Step 5: Calculate stress $\sigma$ in cable** $$\sigma = \frac{T}{A} = \frac{98100}{0.0019635} = 49950000\ \text{Pa} = 49.95\ \text{MPa}$$ **Step 6: Calculate strain $\varepsilon$ using Young's modulus** Given $E = 220$ GPa = $220 \times 10^9$ Pa $$\varepsilon = \frac{\sigma}{E} = \frac{4.995 \times 10^7}{2.2 \times 10^{11}} = 2.27 \times 10^{-4}$$ --- 4. **Problem Statement:** Determine magnitudes of forces $F_1$ and $F_2$ so that particle $P$ is in equilibrium with a 400 N force acting at $30^\circ$, and $F_1$, $F_2$ acting at $60^\circ$ angles. 5. **Equilibrium conditions:** Sum of forces in x and y directions must be zero: $$\sum F_x = 0, \quad \sum F_y = 0$$ 6. **Resolve forces:** Let the 400 N force act at $30^\circ$ from horizontal. Components: $$F_{400,x} = 400 \cos 30^\circ = 400 \times 0.866 = 346.4\ \text{N}$$ $$F_{400,y} = 400 \sin 30^\circ = 400 \times 0.5 = 200\ \text{N}$$ For $F_1$ at $60^\circ$ to the left: $$F_{1,x} = -F_1 \cos 60^\circ = -0.5 F_1$$ $$F_{1,y} = F_1 \sin 60^\circ = 0.866 F_1$$ For $F_2$ at $60^\circ$ to the right: $$F_{2,x} = F_2 \cos 60^\circ = 0.5 F_2$$ $$F_{2,y} = F_2 \sin 60^\circ = 0.866 F_2$$ 7. **Set equilibrium equations:** $$\sum F_x = 346.4 - 0.5 F_1 + 0.5 F_2 = 0$$ $$\sum F_y = 200 + 0.866 F_1 + 0.866 F_2 = 0$$ 8. **Solve system:** From $\sum F_x = 0$: $$0.5 F_1 - 0.5 F_2 = 346.4$$ $$F_1 - F_2 = 692.8 \quad (1)$$ From $\sum F_y = 0$: $$0.866 F_1 + 0.866 F_2 = -200$$ $$F_1 + F_2 = -231.0 \quad (2)$$ Add (1) and (2): $$2 F_1 = 692.8 - 231.0 = 461.8$$ $$F_1 = 230.9\ \text{N}$$ Substitute into (1): $$230.9 - F_2 = 692.8 \Rightarrow F_2 = 230.9 - 692.8 = -461.9\ \text{N}$$ Negative $F_2$ means direction opposite assumed. --- **Final answers:** - Stress in cable: $49.95$ MPa - Strain in cable: $2.27 \times 10^{-4}$ - Forces for equilibrium: $F_1 = 230.9$ N, $F_2 = 461.9$ N (opposite assumed direction)