A 3-bottom tractor-mounted mouldboard plough operates at a working depth of $18\text{ cm}$. Each bottom has a furrow cut width of $35\text{ cm}$. A load cell mounted on the tractor drawbar records a steady horizontal draft force of $13.23\text{ kN}$ at a forward speed of $4.5\text{ km/h}$. The unit draft (specific soil resistance) of the plough in $\text{N/cm}^2$ is ________ (round off to 2 decimal places).
Answer Key:
7.00
Step-by-Step Solution & Explanation
```
3-Bottom Furrow Cross Section:
|<--- w = 35 cm --->|<--- 35 cm --->|<--- 35 cm --->|
+-------------------+---------------+---------------+
| | d = 18 cm
+---------------------------------------------------+
Total Cut Area = 3 * w * d
```
**Method 1: First Principles Area-Force Balance**
1. Total cross-sectional area of furrow slice ($A$):
$A = n \times w \times d = 3 \times 35\text{ cm} \times 18\text{ cm} = 1890\text{ cm}^2$
2. Specific draft (soil resistance per unit furrow cross-section):
$R_s = \frac{\text{Draft Force } (D)}{A} = \frac{13230\text{ N}}{1890\text{ cm}^2} = 7.00\text{ N/cm}^2$
**Method 2: SI Unit Conversion ($kPa$)**
$A = 3 \times 0.35\text{ m} \times 0.18\text{ m} = 0.189\text{ m}^2$
$D_s = \frac{13.23\text{ kN}}{0.189\text{ m}^2} = 70.0\text{ kPa} = 7.00\text{ N/cm}^2$