Subtopic: Functional requirements, principles of working, construction and operation of manual, animal, tractor and renewable energy operated equipment for tillage, sowing, planting, fertilizer application, inter-cultivation, spraying, mowing, chaff cutting, harvesting and threshing
Question Statement:
A reciprocating mower has a stroke length of $S = 76.2\text{ mm}$ and operates at a crank speed of $N = 950\text{ rpm}$. The tractor travels forward at $v_f = 5.4\text{ km/h}$. The ratio of average knife speed ($v_{\text{avg}}$) to forward travel speed ($v_f$) is ________ (round off to 2 decimal places).
Answer Key:
1.61
Step-by-Step Solution & Explanation
```
Cutter Bar Kinematics:
Stroke S = 76.2 mm = 0.0762 m
Speed N = 950 rpm
Forward velocity v_f = 5.4 km/h = 1.5 m/s
```
**Step-by-step Calculation:**
1. Distance traveled by knife in one revolution (two strokes: forward and backward):
$d_{\text{rev}} = 2 S = 2 \times 0.0762\text{ m} = 0.1524\text{ m}$
2. Average linear speed of knife ($v_{\text{avg}}$):
$v_{\text{avg}} = \frac{2 S N}{60} = \frac{0.1524 \times 950}{60} = \frac{144.78}{60} = 2.413\text{ m/s}$
3. Forward travel speed ($v_f$):
$v_f = \frac{5.4 \times 1000}{3600} = 1.50\text{ m/s}$
4. Velocity ratio ($K$):
$K = \frac{v_{\text{avg}}}{v_f} = \frac{2.413\text{ m/s}}{1.50\text{ m/s}} = 1.6087 \approx 1.61$