Question

Front wheel pivot points of a 2WD tractor are 1.2 m apart. When making turn on a flat concrete surface, the inner front wheel makes 50$^\circ$ and the outer front wheel makes 35$^\circ$ steering angles. To ensure turning without front wheel skid, the wheel base of the tractor should be _________ m. (Rounded off to 2 decimal places)

Hint:

To ensure turning without skid, the ratio of the tangent of the steering angles should be equal to the ratio of the wheelbase to the distance between the pivot points. Use this relationship to calculate the wheelbase.

Answer 1

Given: Track width between front wheels: $t = 1.2$ m Inner wheel steering angle: $\theta_{{in}} = 50^\circ$ Outer wheel steering angle: $\theta_{{out}} = 35^\circ$
Ackermann relation: \[ \tan(\theta_{{in}}) - \tan(\theta_{{out}}) = \frac{t}{L} \cdot \tan(\theta_{{in}}) \cdot \tan(\theta_{{out}}) \] 
Substitute: \[ \tan(50^\circ) \approx 1.1918, \quad \tan(35^\circ) \approx 0.7002 \] \[ 1.1918 - 0.7002 = \frac{1.2}{L} \cdot (1.1918 \cdot 0.7002) \Rightarrow 0.4916 = \frac{1.2}{L} \cdot 0.8349 \] \[ L = \frac{1.2 \cdot 0.8349}{0.4916} = \frac{1.00188}{0.4916} \approx \boxed{2.04\ {m}} \]