Question

The height of the CG of a 4WD tractor with equal tread (1.25 m) in front and rear is 0.85 m. The tractor overturns during turning at a speed of 18 km.h\(^{-1}\). Neglecting the frictional forces on the tractor wheels, the turning radius is _________ m. (Rounded off to 2 decimal places)

Hint:

The turning radius for a tractor can be calculated by considering the speed, the height of the center of gravity, and the tread. Make sure the units are consistent when performing the calculations.

Answer 1

Given: Tread width, $b = 1.25$ m
Height of CG, $h = 0.85$ m
Speed, $V = 18$ km/h = $5$ m/s
Acceleration due to gravity, $g = 9.81$ m/s²
Overturning Condition:

\[ \frac{V^2}{gR} = \frac{b}{2h} \quad \Rightarrow \quad R = \frac{V^2 \cdot 2h}{g \cdot b} \]

Substitute the values:

\[ R = \frac{(5)^2 \cdot 2 \cdot 0.85}{9.81 \cdot 1.25} = \frac{42.5}{12.2625} \approx \boxed{3.47 \, {m}} \]