A two-wheel drive tractor weighing 20 kN has a wheel base of 2.1 m with a static weight distribution of 35% and 65% at the front and rear axles respectively. On a level ground, the tractor moves at a speed of 4 km h⁻¹. Considering small steer angle, the cornering force acting on each of the front tyre for a turning radius of 1.8 m is
Options:
(A) 0.244 kN
(B) 0.454 kN
(C) 0.489 kN
(D) 0.907 kN
Official Answer Key:
A
Step-by-Step Solution & Explanation
Cornering force is generated by the mass on the steered (front) axle undergoing circular motion: F_total = (m_front)v²/R. W_front = 0.35×20 = 7.0 kN ⟹ m_front = 7000/9.81 = 713.6 kg. v = 4 km/h = 1.111 m/s. F_total = 713.6×(1.111)²/1.8 = 489.4 N = 0.489 kN (both front tyres together). Per tyre: 0.489/2 = 0.244 kN. (C) is the total front-axle force, not per-tyre.
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