A 20 kN two-wheel-drive tractor works in pure sandy soil, internal friction26.5°. Front/rear loads are35%/65%. Add2.5 kN to each rear wheel. Find percent increase in maximum thrust, to two decimals.
For cohesionless soil, the limiting traction is rear normal load×tanφ. Initial rear load=.65(20)=13 kN;added load=5 kN. Since tanφ cancels, the relative increase is 5/13×100=**38.46%**. The question gives initial axle distribution and directly added rear ballast;it does not redistribute the original20 kN.
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