Powertrain calculation: A car engine produces 125 N·m at 2700 r.p.m. in second gear (gear ratio 1.75) with a final drive ratio of 4.11. If the overall transmission efficiency is 90%, what torque is available at the driving wheels?

Introduction / Context:
Wheel torque determines tractive effort and acceleration. Gear ratios multiply engine torque, while drivetrain losses reduce it. This question practices the fundamental multiplication of torque by gear and final drive ratios with efficiency applied.

Given Data / Assumptions:

• Engine torque T_engine = 125 N·m at 2700 r.p.m.
• Gear ratio (second gear) = 1.75.
• Final drive ratio = 4.11.
• Overall efficiency η = 0.90.

Concept / Approach:
Total torque multiplication equals gear ratio * final drive ratio. The effective wheel torque also includes efficiency losses. Use: T_wheel = T_engine * gear_ratio * final_drive * η

Step-by-Step Solution:

Verification / Alternative check:
Sanity check: Multiplication factor is 1.75 * 4.11 ≈ 7.1925. Then 125 * 7.1925 ≈ 899.1; after 10% loss, ≈ 809.2. The result is consistent.

Why Other Options Are Wrong:

• 8.091 N·m and 80.91 N·m: off by factors of 100 and 10; ignore the large ratio multiplication.
• 8091 N·m: off by factor of 10; ignores efficiency and decimal placement.

Common Pitfalls:

• Forgetting to include final drive ratio.
• Applying efficiency before multiplication or using percentage wrongly.

Final Answer:
809.1 N-m