Question 4888Deck General70% to pass
A jack-up 210 feet in length is level during transit. The LCF is 140 feet aft of the bow. How much weight should be applied at the bow to level the jack-up if 150 kips are loaded at the transom?
AI Explanation
The correct answer is B) 75 kips.
To level the jack-up during transit, the moment created by the 150 kips load at the transom needs to be balanced by the moment created by the weight applied at the bow. Since the LCF (Longitudinal Center of Flotation) is 140 feet aft of the bow, applying 75 kips at the bow will create a moment of 10,500 kip-ft (75 kips x 140 feet), which exactly balances the 150 kips load at the transom (150 kips x 70 feet = 10,500 kip-ft).
The other options are incorrect because:
A) 50 kips at the bow would create a moment of 7,000 kip-ft, which is insufficient to balance the moment created by the 150 kips load at the transom.
C) 100 kips at the bow would create a moment of 14,000 kip-ft, which is excessive and would overcorrect the imbalance.
D) 200 kips at the bow would create a moment of 28,000 kip-ft, which is far too much and would significantly over-correct the imbalance.
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