1. **Problem Statement:** Given a linkage with two guided blocks at points A and B moving in fixed slots, point A moves downward with velocity $v_A = 2$ m/s. At the instant when $\theta = 45^\circ$, determine the velocity of point B, $v_B$. 2. **Relevant Formula:** The velocity of point B relative to A is given by the vector equation: $$v_B = v_A + \omega \times r_{B/A}$$ where $\omega$ is the angular velocity of the link and $r_{B/A}$ is the position vector from A to B. 3. **Vector Components:** - Velocity of A: $v_A = -2\mathbf{j}$ (downward direction) - Position vector from A to B: $$r_{B/A} = 0.2\sin 45^\circ \mathbf{i} - 0.2\cos 45^\circ \mathbf{j}$$ - Angular velocity vector: $\omega = \omega \mathbf{k}$ (perpendicular to the plane) 4. **Cross Product Calculation:** Calculate $\omega \times r_{B/A}$: $$\omega \mathbf{k} \times (0.2\sin 45^\circ \mathbf{i} - 0.2\cos 45^\circ \mathbf{j}) = 0.2\omega \sin 45^\circ \mathbf{j} + 0.2\omega \cos 45^\circ \mathbf{i}$$ 5. **Velocity of B Components:** $$v_B = v_A + \omega \times r_{B/A} = -2\mathbf{j} + 0.2\omega \sin 45^\circ \mathbf{j} + 0.2\omega \cos 45^\circ \mathbf{i}$$ 6. **Equate Components:** - In the $\mathbf{i}$ direction (horizontal): $$v_B = 0.2\omega \cos 45^\circ$$ - In the $\mathbf{j}$ direction (vertical): $$0 = -2 + 0.2\omega \sin 45^\circ$$ 7. **Solve for $\omega$:** $$0.2\omega \sin 45^\circ = 2 \implies \omega = \frac{2}{0.2 \times \sin 45^\circ} = \frac{2}{0.2 \times \frac{\sqrt{2}}{2}} = \frac{2}{0.1414} \approx 14.1 \text{ rad/s}$$ 8. **Calculate $v_B$:** $$v_B = 0.2 \times 14.1 \times \cos 45^\circ = 0.2 \times 14.1 \times \frac{\sqrt{2}}{2} = 2 \text{ m/s}$$ 9. **Interpretation:** The velocity of point B is 2 m/s directed horizontally to the right at the instant $\theta = 45^\circ$. The angular velocity of the link is approximately 14.1 rad/s. **Final Answer:** $$v_B = 2 \text{ m/s}$$