FAA Aviation Mechanic General (AMG)Materials, Hardware and ProcessesMedium
A mechanic must torque a fitting to 200 in-lb but cannot fit a standard wrench due to clearance, so an extension is added that increases the effective length from the pivot to the fastener by 2 inches, giving a total length of 12 inches (original wrench length was 10 inches). What torque wrench reading should the mechanic use to achieve the true 200 in-lb at the fastener?
- A150 in-lb
- B166.7 in-lb
- C200 in-lb
- D240 in-lb
Show answer & explanationAnswer & explanation
Correct answer: B. 166.7 in-lb
Using TW = TE × L/(L+A): TW = 200 × 10/(10+2) = 2000/12 = 166.7 in-lb. Because the extension lengthens the lever arm, the wrench will under-read the true torque, so a lower dial reading is required to actually apply 200 in-lb at the fastener.
Why the other options are wrong
- A. Not derived from the correct formula or given lengths.
- C. Ignores the effect of the extension entirely.
- D. Uses the formula inverted (multiplying by (L+A)/L).
Torque Wrench Extension Correction
When an extension changes the effective lever arm of a torque wrench, the dial reading must be adjusted using TW = TE × L/(L+A) so the actual torque applied to the fastener matches the specification.
- L = wrench length, A = extension length
- Extensions in-line with the wrench increase the lever arm
- Wrench reading must be LOWER than actual desired torque when adding length
Memory trick: 'Longer arm, lower reading' — the extension steals leverage