Question

A 2WD tractor's driving axle experiences 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

• A. 45
• B. 35
• C. 55
• D. 58

Hint:

Combine the rim-pull to axle-force ratio with the slip factor to get tractive efficiency.

Answer 1

The Correct Option is A

Step 1: Get the wheel rim force from torque.
$F_r = T / r = 10.7\ \text{kN.m} / 0.60\ \text{m} = 17.83\ \text{kN}$ is the force delivered at the tyre rim.

Step 2: Get the actual drawbar pull.
Net traction coefficient times dynamic load gives the pull available at the drawbar: $P_{db} = 0.35 \times 28\ \text{kN} = 9.8\ \text{kN}$.

Step 3: Find the pull ratio (how much of the rim force reaches the drawbar).
$\text{Pull ratio} = P_{db}/F_r = 9.8/17.83 = 0.550$.

Step 4: Apply the slip loss.
Slip wastes part of the travel distance without doing drawbar work, so multiply by $(1-s) = (1-0.18) = 0.82$.
$\eta_t = 0.550 \times 0.82 = 0.451$, so tractive efficiency is about 45%.

Final Answer:
The driving wheels convert axle power to drawbar power at close to 45% efficiency. \[ \boxed{\eta_t \approx 45\%} \]