1. **Problem Statement:** We have two blocks A and B, each weighing 10 lb, on double inclined planes with angles 60° and 30° respectively. The coefficient of kinetic friction is $\mu_k = 0.1$. We need to find the acceleration of each block. 2. **Formulas and Concepts:** - Weight component along the incline: $W \sin \theta$ - Friction force: $f_k = \mu_k N = \mu_k W \cos \theta$ - Net force on each block: $F = W \sin \theta - f_k$ - Using Newton's second law: $F = ma$, where $m = \frac{W}{g}$ and $g = 32.2$ ft/s² (standard gravity) - Since blocks are connected by a rope over a pulley, their accelerations have the same magnitude but opposite directions. 3. **Calculate forces for block A (60° incline):** - Weight $W_A = 10$ lb - Component down the slope: $W_A \sin 60^\circ = 10 \times \frac{\sqrt{3}}{2} = 8.66$ lb - Normal force: $N_A = W_A \cos 60^\circ = 10 \times 0.5 = 5$ lb - Friction force: $f_{kA} = \mu_k N_A = 0.1 \times 5 = 0.5$ lb - Net force on A: $F_A = 8.66 - 0.5 = 8.16$ lb 4. **Calculate forces for block B (30° incline):** - Weight $W_B = 10$ lb - Component down the slope: $W_B \sin 30^\circ = 10 \times 0.5 = 5$ lb - Normal force: $N_B = W_B \cos 30^\circ = 10 \times \frac{\sqrt{3}}{2} = 8.66$ lb - Friction force: $f_{kB} = \mu_k N_B = 0.1 \times 8.66 = 0.866$ lb - Net force on B: $F_B = 5 - 0.866 = 4.134$ lb 5. **Set up equations for acceleration:** - Let acceleration of A be $a$ down the slope, and B be $a$ up the slope (same magnitude). - Mass of each block: $m = \frac{W}{g} = \frac{10}{32.2} = 0.3106$ slugs - Net force on system: $F_{net} = F_A - F_B = 8.16 - 4.134 = 4.026$ lb 6. **Calculate acceleration:** $$ a = \frac{F_{net}}{m} = \frac{4.026}{0.3106} = 12.97 \text{ ft/s}^2 $$ 7. **Answer:** The acceleration of each block is approximately $12.97$ ft/s², with block A accelerating down its incline and block B accelerating up its incline.