A 35 kW two-wheel drive tractor weighing 20 kN is fitted with 6-16 8PR tyre at the front axle and 13.6-28 12PR tyre at the rear axle. The ratio of section height and section width for all tyres is 0.75. The tractor has a wheel base of 2.1 m and the center of gravity is located 0.7 m ahead of the rear axle center on a horizontal plane. The tractor is to be towed on a level ground having sandy clay loam soil at 10% moisture content with a cone index of 1200 kPa. If the same tractor is to be towed on a level ground with compacted dry clay soil, the force required for towing is
Options:
(A) 0.27 kN
(B) 0.40 kN
(C) 0.53 kN
(D) 0.80 kN
Official Answer Key:
B
Step-by-Step Solution & Explanation
Compacted (hard, dry) clay behaves as a firm/rigid surface rather than a deformable soil, so the Wismer–Luth cone-index-based rolling-resistance model (used for soft/tilled soils) is not applicable; a typical firm-surface rolling-resistance coefficient of ≈0.02 (from standard rolling-resistance tables for hard/compacted ground) is used instead. Total towing force = C_rr × total weight = 0.02×20 kN = ≈0.40 kN.
Verification note: Because the surface changes from soft sandy-clay-loam (Q71/72, CI-based) to compacted dry clay, the appropriate rolling-resistance approach also changes to a fixed empirical coefficient for firm ground, rather than the cone-index formula.
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