A farm tractor has a rear wheel tread width (wheel track) of $b = 1.60\text{ m}$ and its center of gravity is located at a height of $h_{cg} = 0.80\text{ m}$ above the ground midway between the wheels. Assuming no tire deflection or side slip, the static side slope angle ($\beta$) in degrees at which the tractor is on the verge of lateral rollover is ________ (answer in integer).
Answer Key:
45
Step-by-Step Solution & Explanation
```
Lateral Rollover Equilibrium:
Track width = b = 1.60 m
Half-track = b/2 = 0.80 m
CG height = h_cg = 0.80 m
tan(beta) = (b/2) / h_cg
```
**Step-by-step Calculation:**
1. At the verge of lateral overturning about the downhill wheel contact line:
$\sum M_{\text{downhill}} = 0 \implies W \cos\beta \times \left(\frac{b}{2}\right) - W \sin\beta \times h_{cg} = 0$
$\tan\beta = \frac{b/2}{h_{cg}}$
2. Substituting given values:
$\frac{b}{2} = \frac{1.60}{2} = 0.80\text{ m}$
$h_{cg} = 0.80\text{ m}$
$\tan\beta = \frac{0.80}{0.80} = 1.00$
$\beta = \tan^{-1}(1.00) = 45^\circ$