Question Bank • QB_FP_096 NAT • 2 Marks

Section 3: Farm Power

Farm Power

Subtopic: Thermodynamic principles of I.C. engines

Question Statement:

A tractor drive tire rotates at $30\text{ rpm}$ with a rolling radius of $0.75\text{ m}$. The actual forward travel speed of the tractor measured over a $50\text{ m}$ run is $2.0\text{ m/s}$. Calculate the travel reduction (wheel slip) in percentage.

Answer Key:

15.12

Step-by-Step Solution & Explanation

Theoretical forward velocity without slip: $v_t = \omega \times r = \frac{2 \pi N}{60} \times r = \frac{2 \pi \times 30}{60} \times 0.75 = \pi \times 0.75 = 2.3562\text{ m/s}$ Actual velocity: $v_a = 2.0\text{ m/s}$ Wheel slip ($s$): $s = \frac{v_t - v_a}{v_t} \times 100 = \frac{2.3562 - 2.0}{2.3562} \times 100 = \frac{0.3562}{2.3562} \times 100 = 15.12\%$
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