1. **Problem Statement:** Given a mechanism with a linear actuator attached at a fixed point O, and a sliding block A moving with velocity $\vec{v}_A = 5$ mm/s horizontally, we need to find expressions for the actuator's rotational speed, sliding speed, and acceleration for any angle $\theta$, and then calculate these values at $\theta = 20^\circ$. 2. **Setup and Known Data:** - Velocity of block A: $v_A = 5$ mm/s (horizontal direction) - Angle of actuator with vertical: $\theta$ - Length of vertical segment: 12.5 cm = 125 mm - Angle of line from bottom segment to block A: 30° 3. **Step (a): Expressions for rotational speed $\omega$ and sliding speed $v$ of actuator** - Let the actuator length be $r$. - The actuator rotates about O with angular velocity $\omega = \frac{d\theta}{dt}$. - The sliding speed of the actuator is $v = \frac{dr}{dt}$. - Position of block A depends on $r$ and $\theta$: The horizontal position of A is: $$x_A = r \sin\theta + L \cos 30^\circ$$ where $L$ is the length along the sloped surface to block A. - Since block A moves horizontally with velocity $v_A$, differentiate $x_A$ w.r.t time: $$v_A = \frac{dx_A}{dt} = \frac{d}{dt}(r \sin\theta) + 0 = \dot{r} \sin\theta + r \cos\theta \dot{\theta}$$ - Rearranged: $$v_A = v \sin\theta + r \cos\theta \omega$$ - This is the general expression relating sliding speed $v$, rotational speed $\omega$, and angle $\theta$. 4. **Step (b): Expression for acceleration of actuator** - Differentiate velocity expression w.r.t time: $$0 = \frac{d}{dt}(v_A) = \frac{d}{dt}(v \sin\theta + r \cos\theta \omega)$$ - Since $v_A$ is constant (5 mm/s), its derivative is zero: $$0 = \dot{v} \sin\theta + v \cos\theta \omega + \dot{r} \cos\theta \omega - r \sin\theta \omega^2 + r \cos\theta \dot{\omega}$$ - Simplify and solve for $\dot{v}$ (sliding acceleration) or $\dot{\omega}$ (angular acceleration) as needed. 5. **Step (c): Calculate $v$ and $\omega$ at $\theta = 20^\circ$** - Given $v_A = 5$ mm/s, $\theta = 20^\circ$, and $r = 125$ mm (length of vertical segment): - From step (a): $$5 = v \sin 20^\circ + 125 \cos 20^\circ \omega$$ - We need a second equation to solve for $v$ and $\omega$. Assuming the actuator length $r$ is fixed or given, or if $v$ and $\omega$ are related by the mechanism constraints, but since not provided, we can express one variable in terms of the other: - For example, solve for $v$: $$v = \frac{5 - 125 \cos 20^\circ \omega}{\sin 20^\circ}$$ - Without additional constraints, this is the relation. 6. **Step (d): Calculate sliding acceleration $\dot{v}$ at $\theta = 20^\circ$** - Using the acceleration expression from step (b), substitute known values and solve for $\dot{v}$. - Since $v_A$ is constant, acceleration of block A is zero, so the expression simplifies accordingly. **Final answers:** - General velocity relation: $$v_A = v \sin\theta + r \cos\theta \omega$$ - General acceleration relation: $$0 = \dot{v} \sin\theta + v \cos\theta \omega + r \cos\theta \dot{\omega} - r \sin\theta \omega^2$$ - At $\theta = 20^\circ$, $r=125$ mm, $v_A=5$ mm/s: $$5 = v \sin 20^\circ + 125 \cos 20^\circ \omega$$ - Sliding acceleration $\dot{v}$ can be found from the acceleration equation once $v$, $\omega$, and $\dot{\omega}$ are known.