1. **Problem 1: Axial deformation and lateral contraction of a red brass bar** Given: - Axial force $P = 200000$ N (converted from 200 kN) - Length $L = 750$ mm = 0.75 m - Width $w = 50$ mm = 0.05 m - Height $h = 25$ mm = 0.025 m - Young's modulus $E = 101 \times 10^9$ Pa - Poisson's ratio $\nu = 0.35$ **a. Change in length $\Delta L$** 1. The axial stress $\sigma$ is given by: $$\sigma = \frac{P}{A}$$ where $A = w \times h$ is the cross-sectional area. 2. Calculate $A$: $$A = 0.05 \times 0.025 = 0.00125\ \text{m}^2$$ 3. Calculate $\sigma$: $$\sigma = \frac{200000}{0.00125} = 160000000\ \text{Pa} = 160\ \text{MPa}$$ 4. The axial strain $\epsilon$ is: $$\epsilon = \frac{\Delta L}{L} = \frac{\sigma}{E}$$ 5. Calculate $\epsilon$: $$\epsilon = \frac{160 \times 10^6}{101 \times 10^9} = 0.001584$$ 6. Calculate $\Delta L$: $$\Delta L = \epsilon \times L = 0.001584 \times 0.75 = 0.001188\ \text{m} = 1.188\ \text{mm}$$ **b. Change in cross-sectional dimensions (width and height)** 1. Lateral strain $\epsilon_{lat}$ is related to axial strain by Poisson's ratio: $$\epsilon_{lat} = -\nu \epsilon$$ 2. Calculate lateral strain: $$\epsilon_{lat} = -0.35 \times 0.001584 = -0.000554$$ 3. Calculate change in width $\Delta w$: $$\Delta w = \epsilon_{lat} \times w = -0.000554 \times 0.05 = -0.0000277\ \text{m} = -0.0277\ \text{mm}$$ 4. Calculate change in height $\Delta h$: $$\Delta h = \epsilon_{lat} \times h = -0.000554 \times 0.025 = -0.0000139\ \text{m} = -0.0139\ \text{mm}$$ --- 2. **Problem 2: Maximum force on wooden strut under shear stress** Given: - Thickness $t = 25$ mm = 0.025 m - Shear stress limit $\tau_{max} = 600000$ Pa - Angle $\theta = 60^\circ$ 1. The shear area $A_s$ is the area of the glued surface AB: $$A_s = t \times L$$ 2. Length $L = 50$ mm = 0.05 m 3. Calculate $A_s$: $$A_s = 0.025 \times 0.05 = 0.00125\ \text{m}^2$$ 4. Maximum shear force $F_{max}$: $$F_{max} = \tau_{max} \times A_s = 600000 \times 0.00125 = 750\ \text{N}$$ 5. Since the force $P$ acts vertically and the strut is at $60^\circ$, the force on the glued surface is related by: $$P = \frac{F_{max}}{\sin 60^\circ}$$ 6. Calculate $P$: $$P = \frac{750}{\sin 60^\circ} = \frac{750}{0.866} = 866.03\ \text{N}$$ --- 3. **Problem 3: Cable rope pulling a car up an incline** Given: - Mass $m = 20000$ kg - Angle $\alpha = 30^\circ$ - Diameter $d = 50$ mm = 0.05 m - Gravity $g = 9.81$ m/s$^2$ - Young's modulus $E = 220 \times 10^9$ Pa **a. Free body diagram** - Forces: weight $W = mg$ acting vertically down - Tension $T$ in cable acting along incline - Normal and friction forces from track (not asked to calculate here) **b. Stress in cable** 1. Calculate weight: $$W = mg = 20000 \times 9.81 = 196200\ \text{N}$$ 2. Component of weight along incline: $$F = W \sin \alpha = 196200 \times \sin 30^\circ = 196200 \times 0.5 = 98100\ \text{N}$$ 3. Cross-sectional area of cable: $$A = \frac{\pi d^2}{4} = \frac{3.1416 \times 0.05^2}{4} = 0.0019635\ \text{m}^2$$ 4. Stress $\sigma$ in cable: $$\sigma = \frac{F}{A} = \frac{98100}{0.0019635} = 49950000\ \text{Pa} = 49.95\ \text{MPa}$$ **c. Strain in cable** 1. Strain $\epsilon$: $$\epsilon = \frac{\sigma}{E} = \frac{49.95 \times 10^6}{220 \times 10^9} = 0.000227$$ --- 4. **Problem 4: Equilibrium of particle P** Given: - Force $400$ N at $30^\circ$ to vertical - Forces $F_1$ and $F_2$ at angles $0^\circ$ and $60^\circ$ respectively - Ratios $F_1 : F_2 = 5/4 : 3/4$ 1. Resolve forces into components and set sum of forces in $x$ and $y$ directions to zero. 2. Let $F_1 = 5k/4$, $F_2 = 3k/4$ for some $k$. 3. Sum of forces in $x$: $$400 \sin 30^\circ + F_1 - F_2 \cos 60^\circ = 0$$ 4. Sum of forces in $y$: $$400 \cos 30^\circ - F_2 \sin 60^\circ = 0$$ 5. Calculate $k$ from $y$-component: $$400 \times 0.866 = F_2 \times 0.866 \Rightarrow F_2 = 400\ \text{N}$$ 6. Calculate $F_1$: $$F_1 = \frac{5}{3} F_2 = \frac{5}{3} \times 400 = 666.67\ \text{N}$$ --- **Final answers:** 1a. $\Delta L = 1.188$ mm 1b. $\Delta w = -0.0277$ mm, $\Delta h = -0.0139$ mm 2. Maximum force $P = 866.03$ N 3b. Stress in cable $= 49.95$ MPa 3c. Strain in cable $= 0.000227$ 4. $F_1 = 666.67$ N, $F_2 = 400$ N