1. **Problem Statement:** Determine the velocity and acceleration of the piston at point C in the three-link mechanism given: - Angular velocity of AB: $\omega_{AB} = 10$ rad/s - Angular acceleration of AB: $\alpha_{AB} = 20$ rad/s$^2$ - Angular velocity of BC: $\omega_{BC} = 2.43$ rad/s - Length AB = 0.25 ft, angle AB = 45° - Length BC = 0.75 ft, angle BC = 13.6° 2. **Formulas and Rules:** - Velocity of point B relative to A: $$\vec{v}_B = \vec{\omega}_{AB} \times \vec{r}_{AB}$$ - Velocity of point C relative to B: $$\vec{v}_C = \vec{\omega}_{BC} \times \vec{r}_{BC}$$ - Acceleration of point B relative to A: $$\vec{a}_B = \vec{\alpha}_{AB} \times \vec{r}_{AB} - \omega_{AB}^2 \vec{r}_{AB}$$ - Acceleration of point C relative to B: $$\vec{a}_C = \vec{\alpha}_{BC} \times \vec{r}_{BC} - \omega_{BC}^2 \vec{r}_{BC}$$ - The piston velocity and acceleration are the velocity and acceleration of point C. 3. **Calculate position vectors:** - $\vec{r}_{AB} = 0.25(\cos 45^\circ \hat{i} + \sin 45^\circ \hat{j}) = 0.25(0.7071 \hat{i} + 0.7071 \hat{j}) = 0.1768 \hat{i} + 0.1768 \hat{j}$ ft - $\vec{r}_{BC} = 0.75(\cos 13.6^\circ \hat{i} + \sin 13.6^\circ \hat{j}) = 0.75(0.9724 \hat{i} + 0.2354 \hat{j}) = 0.7293 \hat{i} + 0.1766 \hat{j}$ ft 4. **Calculate velocity of B:** - Angular velocity vector $\vec{\omega}_{AB} = -10 \hat{k}$ (clockwise rotation) - Velocity $\vec{v}_B = \vec{\omega}_{AB} \times \vec{r}_{AB} = -10 \hat{k} \times (0.1768 \hat{i} + 0.1768 \hat{j})$ - Using cross product: $\hat{k} \times \hat{i} = \hat{j}$, $\hat{k} \times \hat{j} = -\hat{i}$ - So, $\vec{v}_B = -10 (0.1768 \hat{j} - 0.1768 \hat{i}) = -1.768 \hat{j} + 1.768 \hat{i}$ ft/s 5. **Calculate velocity of C:** - Angular velocity vector $\vec{\omega}_{BC} = 2.43 \hat{k}$ (counterclockwise assumed positive) - Velocity $\vec{v}_C = \vec{v}_B + \vec{\omega}_{BC} \times \vec{r}_{BC} = (1.768 \hat{i} - 1.768 \hat{j}) + 2.43 \hat{k} \times (0.7293 \hat{i} + 0.1766 \hat{j})$ - Cross product: $2.43 (0.1766 \hat{i} - 0.7293 \hat{j})$ - $= 0.429 \hat{i} - 1.772 \hat{j}$ ft/s - Total velocity $\vec{v}_C = (1.768 + 0.429) \hat{i} + (-1.768 - 1.772) \hat{j} = 2.197 \hat{i} - 3.54 \hat{j}$ ft/s 6. **Calculate acceleration of B:** - Angular acceleration vector $\vec{\alpha}_{AB} = -20 \hat{k}$ - Tangential acceleration: $\vec{\alpha}_{AB} \times \vec{r}_{AB} = -20 \hat{k} \times (0.1768 \hat{i} + 0.1768 \hat{j}) = -20 (0.1768 \hat{j} - 0.1768 \hat{i}) = -3.536 \hat{j} + 3.536 \hat{i}$ ft/s$^2$ - Centripetal acceleration: $-\omega_{AB}^2 \vec{r}_{AB} = -100 (0.1768 \hat{i} + 0.1768 \hat{j}) = -17.68 \hat{i} - 17.68 \hat{j}$ ft/s$^2$ - Total acceleration $\vec{a}_B = (3.536 - 17.68) \hat{i} + (-3.536 - 17.68) \hat{j} = -14.144 \hat{i} - 21.216 \hat{j}$ ft/s$^2$ 7. **Calculate acceleration of C:** - Angular acceleration of BC $\alpha_{BC}$ unknown, assume zero for this instant (or negligible) - Centripetal acceleration of C relative to B: $-\omega_{BC}^2 \vec{r}_{BC} = -(2.43)^2 (0.7293 \hat{i} + 0.1766 \hat{j}) = -5.9049 (0.7293 \hat{i} + 0.1766 \hat{j}) = -4.303 \hat{i} - 1.042 \hat{j}$ ft/s$^2$ - Total acceleration $\vec{a}_C = \vec{a}_B + \vec{a}_{C/B} = (-14.144 - 4.303) \hat{i} + (-21.216 - 1.042) \hat{j} = -18.447 \hat{i} - 22.258 \hat{j}$ ft/s$^2$ **Final answers:** - Velocity of piston (point C): $$\boxed{\vec{v}_C = 2.20 \hat{i} - 3.54 \hat{j} \text{ ft/s}}$$ - Acceleration of piston (point C): $$\boxed{\vec{a}_C = -18.45 \hat{i} - 22.26 \hat{j} \text{ ft/s}^2}$$