Subtopic: Steering and hydraulic control systems used in tractors
Question Statement:
A two-wheel drive tractor with a total weight of 24 kN has a static weight distribution of 30% and 70% at the front and rear axles, respectively. When the tractor is operated on a level ground of pure sand, the maximum tractive force developed is 13 kN. If external weight of 1.5 kN is added to the rear axle, neglecting weight transfer, the change in maximum tractive force in kN is __________ (rounded off to 2 decimal places).
Official Answer Key:
1.15 to 1.20
Step-by-Step Solution & Explanation
For a 2WD tractor, tractive force is developed only at the driven (rear) wheels, so maximum tractive force ∝ weight on the rear axle (F_max = μ×W_rear, μ = traction coefficient of the sand surface).
Initial rear-axle weight, W_rear = 0.70×24 = 16.8 kN. Traction coefficient μ = F_max/W_rear = 13/16.8 = 0.7738.
After adding 1.5 kN to the rear axle (weight transfer neglected, so it adds directly): new W_rear = 16.8+1.5 = 18.3 kN.
New F_max = 0.7738×18.3 = 14.16 kN.
Change in tractive force = 14.16−13.00 = 1.16 kN. Within official range 1.15–1.20.
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