Let's solve the **quick return mechanism** problem step by step: --- ## **Given Data** - Crank radius, \( r_2 = 10 \) cm - Slotted lever (link), \( r_3 = 50 \) cm - Offset (distance from O to slider), \( r_1 = 20 \) cm - Crank angular velocity, \( \omega = \) rad/sec (clockwise) --- ## **Mechanism Type** This is a **slotted lever (quick return) mechanism**. --- ## **(a) Angle of Sweep for Forward Stroke (\(\theta_w\))** The angle swept by the crank during the forward stroke (\(\theta_w\)) is given by: \[ \theta_w = 2 \cos^{-1} \left( \frac{r_1}{r_2} \right) \] But here, since this is a slotted lever (Whitworth/quick return), the **angle for the forward stroke** is: \[ \theta_w = 2 \cos^{-1} \left( \frac{r_1}{r_2} \right) \] But in this offset crank-slider, the formula is: \[ \theta_{w} = 2 \cos^{-1} \left( \frac{r_1}{r_2} \right) \] Plug in the values: \[ \theta_{w} = 2 \cos^{-1} \left( \frac{20}{10} \right) \] But here, \(\frac{r_1}{r_2} = 2\), and \(\cos^{-1}(2)\) is not defined (since cosine values are between -1 and 1). So, let's correct: For the **crank and slotted lever mechanism** (Whitworth), the sweep angle is: \[ \theta_{w} = 180^\circ + 2\phi \] where \[ \sin \phi = \frac{r_1}{r_3} \] So, \[ \phi = \sin^{-1} \left( \frac{20}{50} \right) = \sin^{-1}(.4) = 23.58^\circ \] So, \[ \theta_{w} = 180^\circ + 2 \times 23.58^\circ = 180^\circ + 47.16^\circ = 227.16^\circ \] --- ## **(b) Angle of Return Stroke (\(\theta_r\))** \[ \theta_r = 360^\circ - \theta_w = 360^\circ - 227.16^\circ = 132.84^\circ \] --- ## **(c) Time Ratio (tr)** Time ratio: \[ \text{Time ratio (tr)} = \frac{\text{Time for forward stroke}}{\text{Time for return stroke}} = \frac{\theta_w}{\theta_r} \] \[ tr = \frac{227.16}{132.84} = 1.71 \] --- ## **(d) Stroke Length (s)** Stroke length is: \[ s = 2 \times r_3 \times \sin \phi \] \[ s = 2 \times 50 \times \sin(23.58^\circ) = 100 \times .4 = 40~\text{cm} \] --- ## **(e) Average Speed in Forward Stroke** First, find time for forward stroke: - Crank angular velocity, \( \omega = 10 \) rad/s - Sweep angle in radians: \( \theta_w = 227.16^\circ \times \frac{\pi}{180} = 3.966 \) rad Time for forward stroke: \[ t_f = \frac{\theta_w}{\omega} = \frac{3.966}{10} = .397~\text{sec} \] Average speed, \( V_{avg} \): \[ V_{avg} = \frac{\text{stroke length}}{\text{time}} \] \[ V_{avg} = \frac{40}{.397} = 100.8~\text{cm/s} \] --- ## **Summary Table** | Quantity | Value | |---------------------------|------------------------| | Angle of Forward Stroke | \( 227.16^\circ \) | | Angle of Return Stroke | \( 132.84^\circ \) | | Time Ratio (tr) | \( 1.71 \) | | Stroke Length (s) | \( 40 \) cm | | Avg. Speed (Forward) | \( 100.8 \) cm/s | --- ### **Let me know if you need step-by-step explanations for any part!**