A tractor rear drive wheel with an effective rolling radius of $r = 0.70\text{ m}$ under zero-slip conditions makes $N = 150$ revolutions while the tractor travels forward through an actual measured distance of $d_a = 593.76\text{ m}$ under drawbar load. The wheel travel reduction (slip $S$) expressed as a percentage is ________ $\%$ (round off to integer).
Answer Key:
10
Step-by-Step Solution & Explanation
```
Wheel Slip Calculation:
Rolling radius: r = 0.70 m
Revolutions: N = 150
Actual distance: d_a = 593.76 m
```
**Step-by-step Calculation:**
1. Theoretical distance traveled ($d_t$) without slip:
$d_t = 2 \pi r N = 2 \times \pi \times 0.70\text{ m} \times 150 = 210\pi = 659.734\text{ m}$
2. Percentage wheel slip ($S$):
$S = \frac{d_t - d_a}{d_t} \times 100 = \frac{659.734 - 593.76}{659.734} \times 100 = \frac{65.974}{659.734} \times 100 = 10.00\%$
Slip is exactly $10\%$.