Question Bank • QB_FP_133 NAT • 2 Marks

Section 3: Farm Power

Farm Power

Subtopic: Thermodynamic principles of I.C. engines

Question Statement:

A tractor is parked heading directly uphill on a slope of angle $\theta = 20^\circ$. The tractor weight is $20\text{ kN}$, wheelbase is $1.8\text{ m}$, and its CG is located midway along the wheelbase ($0.9\text{ m}$ from both axles) and $0.7\text{ m}$ above ground. Calculate the normal reaction on the front wheels in $\text{kN}$.

Answer Key:

6.74

Step-by-Step Solution & Explanation

Taking moments about the rear wheel contact point on the inclined slope: $R_f \times L = W \cos\theta \times x_{cg} - W \sin\theta \times h_{cg}$ Where: - $W = 20\text{ kN}$ - $\theta = 20^\circ$ ($\cos 20^\circ = 0.93969$, $\sin 20^\circ = 0.34202$) - $L = 1.8\text{ m}$ - $x_{cg} = 0.9\text{ m}$ - $h_{cg} = 0.7\text{ m}$ $R_f \times 1.8 = [20 \times 0.93969 \times 0.9] - [20 \times 0.34202 \times 0.7]$ $R_f \times 1.8 = 16.9145 - 4.7883 = 12.1262$ $R_f = \frac{12.1262}{1.8} \approx 6.74\text{ kN}$
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