A 2WD tractor’s driving axle experiences a total dynamic normal load of 28 kN. The driving wheels have 0.60 m rolling radius. The total driving axle torque measured is 10.7 kN·m. At 18% wheel slip, the coefficient of net traction is 0.35. Tractive efficiency of the driving wheels, in %, is nearest to
Options:
(A) 45
(B) 35
(C) 55
(D) 58
Official Answer Key:
A
Step-by-Step Solution & Explanation
Tractive efficiency is the ratio of useful drawbar output power to the axle input power:
TE = (Net traction × actual travel speed)/(Axle torque × angular speed)
Actual travel speed with slip s: v = r·ω(1 − s). Substituting and cancelling ω:
TE = [F_net × r × (1 − s)]/T
Step 1 — net traction force:
F_net = coefficient of net traction × dynamic load = 0.35 × 28 = 9.8 kN
Step 2 — substitute:
TE = (9.8 × 0.60 × (1 − 0.18))/10.7 = (9.8 × 0.60 × 0.82)/10.7 = 4.822/10.7 = 0.4506
TE ≈ 45 %
Interpretation: TE combines the two loss mechanisms — motion resistance (captured in the net-traction coefficient) and slip (the 1 − s factor).
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