A farm tractor has a rear wheel tread width of $b = 1.60\text{ m}$ and its center of gravity is located on the vehicle centerline at a height of $h = 0.80\text{ m}$ above ground. The tractor stands statically across a side hill slope of angle $\alpha = 30^\circ$. The proportion of total tractor weight carried by the downhill wheels ($R_{\text{down}}/W$) is ________ (round off to 2 decimal places).
Answer Key:
0.68
Step-by-Step Solution & Explanation
1. Taking moments about the downhill wheel ground contact line:
$R_{\text{up}} \cdot b = W \left( \frac{b}{2} \cos\alpha - h \sin\alpha \right)$
$\frac{R_{\text{up}}}{W} = 0.5 \cos(30^\circ) - \frac{h}{b} \sin(30^\circ) = (0.5 \times 0.8660) - \left(\frac{0.80}{1.60} \times 0.50\right) = 0.4330 - 0.2500 = 0.1830$
2. Normal force on downhill wheels:
$\frac{R_{\text{down}}}{W} = \cos(30^\circ) - \frac{R_{\text{up}}}{W} = 0.8660 - 0.1830 = 0.6830 \approx 0.68$