GATE 2026 AG – Question 52
A 2WD tractor develops 30 kN gross traction with rear rolling radius .73 m. Each rear wheel has a planetary final drive:sun input with 26 teeth, fixed ring with 78 teeth, and carrier driving the hub. Sun–planet external-mesh efficiency is 98.5%,planet–ring internal-mesh efficiency 99%;ignore upstream losses. Find required input sun torque per wheel, in kN·m, to two decimals.
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Correct answer: 2.77 to 2.90
Explanation
Total wheel torque is $Fr=30(.73)=21.9$ kN·m, so each wheel needs 10.95 kN·m. With the ring fixed, the reduction is $i=1+Z_r/Z_s=1+78/26=4$. Using the stated mesh efficiencies as the lumped transmission efficiency gives $\eta=.985(.99)=.97515$. Therefore $T_s=T_c/(i\eta)=10.95/[4(.97515)]=2.80726$ kN·m, rounded **2.81**,within the official range. This is the exam’s lumped two-mesh efficiency model.