CDL General Knowledge TestTransporting Cargo SafelyMedium
According to federal regulations, what is the minimum aggregate working load limit (WLL) required for all tiedowns securing a piece of cargo weighing 50,000 pounds?
- A10,000 pounds
- B50,000 pounds
- C25,000 pounds
- D12,500 pounds
Show answer & explanationAnswer & explanation
Correct answer: C. 25,000 pounds
Federal Motor Carrier Safety Regulations (FMCSA 393.108(d)) state that the aggregate Working Load Limit (WLL) of any securement system used to secure an article of cargo against movement in any direction must be at least one-half (50%) of the weight of the article. For a 50,000-pound piece of cargo, 50% of its weight is 25,000 pounds (50,000 lbs * 0.50 = 25,000 lbs).
Why the other options are wrong
- A. Incorrect. This is 20% of the cargo weight, which is too low.
- B. Incorrect. While more WLL is safer, the minimum required is 50% of the cargo weight, not 100%.
- D. Incorrect. This is 25% of the cargo weight, which is too low.
Aggregate WLL Rule
The total Working Load Limit (WLL) of all tiedowns used to secure a piece of cargo must be at least half the weight of the cargo itself.
- WLL is the maximum weight a tiedown can safely handle.
- Aggregate WLL is the sum of the WLLs of all tiedowns.
- Formula: Aggregate WLL ≥ 0.50 * Cargo Weight.
- Ensures adequate strength to prevent cargo movement.
Memory trick: Half the weight, or it's not alright!