During operation of a two-wheel drive tractor with a total weight of 20 kN in pure sandy soil (angle of internal friction is 26.5°), the weight distribution at the front and rear axles are found to be 35% and 65%, respectively. If extra weight of 2.5 kN is added to each of the rear wheels, the change in maximum thrust developed by the tractor in per cent will be ______. [round off to 2 decimal places]
Official Answer Key:
38.20 to 38.60
Step-by-Step Solution & Explanation
For a 2WD tractor, driving thrust is developed only at the (driven) rear wheels; on frictional (sandy) soil, maximum thrust ≈ (rear axle weight) × tan(φ). Original rear weight = 0.65×20 = 13 kN → thrust₁ = 13×tan(26.5°) = 13×0.4986 = 6.482 kN. Adding 2.5 kN to EACH rear wheel adds 2×2.5 = 5 kN to the rear axle: new rear weight = 13+5 = 18 kN → thrust₂ = 18×0.4986 = 8.975 kN. % change = (8.975−6.482)/6.482×100 = 38.46%.
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